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Brownian Path Properties

1 · Prerequisites

2 · Summary

This page studies the sample paths of the Brownian motion of brownian-motion-construction-and-continuity and brownian-motion-markov-properties-and-hitting-times, as paths rather than as a Markov process or a martingale.

The roughness of the paths is quantified first: no nondegenerate interval admits a finite one-half Hölder constant Brownian paths are nowhere locally one-half Hölder, the paths are nowhere differentiable Brownian paths are nowhere differentiable, and their total variation is infinite on every nondegenerate interval Brownian paths have infinite total variation. Against this, the quadratic sums along a named partition sequence behave regularly: the definition Quadratic variation along a partition sequence fixes the two conventions and the partition dependence, the dyadic sums converge to elapsed time Brownian quadratic variation on dyadic partitions and even uniformly in time Uniform dyadic Brownian quadratic variation process, yielding the one- versus two-variation dichotomy Brownian one- and quadratic variation. The quantifier boundary of that construction is recorded separately Quadratic variation needs a partition convention.

The all-path continuous jointly measurable version Brownian motion has a jointly measurable continuous version makes the zero set a well-behaved random closed set The Brownian zero set, which is Lebesgue-null The Brownian zero set has Lebesgue measure zero yet has no isolated points The Brownian zero set has no isolated points and is therefore uncountable The Brownian zero set is uncountable.

Growth at infinity and at zero is governed by the two-sided Mills bounds Two-sided Mills bounds for the standard normal tail, the law of the iterated logarithm at infinity Brownian law of the iterated logarithm at infinity, its time-inverted form at zero Brownian law of the iterated logarithm at zero, and the resulting critical Hölder boundary at the origin The critical Hölder boundary at zero.

Finally the page proves the first arcsine law for the last zero before a fixed time The last Brownian zero has the arcsine law and the second arcsine law for the occupation time of the positive half-line Brownian positive occupation time has the arcsine law, the latter through the step-potential resolvent Brownian step-potential resolvent at zero.

Choice is declared wherever the Brownian, conditional-expectation, Borel-Cantelli or integration-by-parts interfaces require it, and the countable-choice use in the monotone-differentiability step is declared at the p-variation computation. The companion page brownian-path-properties-examples carries the dyadic moment computations, the p-variation threshold, the LIL consequence for square-root bounds, and the three counterexamples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian paths are nowhere locally one-half Hölder

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion Brownian motion. Almost surely there is no nondegenerate interval I[0,) and no finite constant C such that BtBsCts1/2for every s,tI. The assertion concerns intervals only: no claim is made here about the exceptional times at which a deterministic pointwise one-half Hölder bound might hold, and no uniform modulus theorem is asserted.

Facts & Assumptions

Given: AC and a standard Brownian motion B on nonnegative times.

[F1]

For every finite list 0=t0<t1<<tn the increments BtjBtj1 are mutually independent with laws N(0,tjtj1). Brownian motion

[F2]

N(0,h) is by definition the law of hZ for ZN(0,1), and Z has the strictly positive density φ(x)=ex2/2/2π of total mass one; hence pC:=P(ZC)<1 for every finite real C. Standard normal and normal laws The standard normal density has total mass one

[F3]

The rationals are dense in R: every nondegenerate interval contains a nondegenerate interval with rational endpoints. The rationals embed densely in the reals

[F4]

AC is the ambient assumption of the Brownian and normal-law interfaces. The Axiom of Choice

[F5]

Countable unions of measurable null events are null by countable subadditivity. Basic identities for a probability measure

Proof

technique · direct
1.1

Fix rationals 0a<b and an integer C1, and put Ea,b,C:=r,qQ[a,b]{BrBqCrq1/2}. This is measurable because it is a countable intersection of coordinate events. For every integer n1, with h=(ba)/n and grid points tk=a+kh (all rational), the event Ea,b,C is contained in An:={ω: Btk(ω)Btk1(ω)Ch for k=1,,n}, because consecutive grid points are rational pairs in [a,b] at distance h.

givenF1
2.1

Insert the endpoint 0 before a when a>0; [F1] then applies to the increasing grid starting at zero, and its subfamily of increments on [a,b] is independent. By [F1] and [F2] the increments BtkBtk1, k=1,,n, are independent with the law of hZ, so each satisfies P(BtkBtk1Ch)=P(ZC)=pC<1, and independence gives P(An)=pCn; hence P(Ea,b,C)pCn for every n1 and therefore P(Ea,b,C)=0. For completeness, the standard normal probability of [C+1,C+2] is at least e(C+2)2/2/2π>0, proving p_C<1 for the positive integers C used here.

F1F2step 1.1
3.1

The family of ordered pairs of rationals and of integers is countable, so [step 2.1], countable subadditivity [F5] and [F4] give P(0a<b, a,bQC1Ea,b,C)=0.

step 2.1F4F5
4.1

On the complement of that null event there is no nondegenerate interval I with a finite one-half Hölder constant: if I were such an interval with any finite real constant C, then by [F3] we could choose rationals 0a<b with [a,b]I, and with C:=max(C,1)N the bound would in particular hold for all rational s,t[a,b], that is, Ea,b,C would occur.

step 3.1F3
5.1

The intended cases are covered: the interval is required to be nondegenerate, so the empty and singleton interval cases are excluded; the value n=1 in [step 2.1] is the degenerate single-increment case of the estimate and already gives P(A1)=pC<1; the union over integers C1 covers every finite real constant up to rounding up; the estimates in [step 2.1] hold for every positive integer n and imply nullness without requiring the mesh events to be nested; and AC is used only through [F4] via [F1] and [F2].

step 2.1step 3.1step 4.1F4given

Source notes

Durrett's remark after Theorem 7.1.6 records that one-half is the critical exponent for uniform interval bounds and that the exceptional set of times at which a pointwise one-half Hölder bound holds is not ruled out by this theorem. Yoshida proves the subcritical uniform statement in Section 6.3 and the nowhere alpha-Hölder statement for alpha > 1/2 in Section 6.4 of the same notes (Proposition 6.4.1 there); the argument above is instead the direct mesh computation: on a fixed rational interval a one-half Hölder bound forces all n increments of the uniform n-mesh to be of size at most Ch, an event of probability pCn whose intersection over n is null.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian paths are nowhere differentiable

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion Brownian motion. Almost surely, the path tBt(ω) has no finite two-sided derivative at any t>0, and no finite right derivative B+(0) at t=0. The assertion is uniform over the possible times: it is not the statement that the path fails to be differentiable at any single prescribed time.

Facts & Assumptions

Given: AC, a standard Brownian motion B, and an integer C1 together with rationals 0a<b.

[F1]

The increments of B over disjoint time intervals are independent with laws N(0,h) for interval length h, and one probability-one event carries all continuous paths. Brownian motion

[F2]

If a real function f has a finite two-sided derivative f(s) at an interior point s, or a finite right derivative f+(s) at a left endpoint, or a finite left derivative f(s) at a right endpoint, then with ε=1 in The derivative f(c)=limxcf(x)f(c)xc of f:AR at a point cA that is a limit point of A, and differentiability on a set and The left and right derivatives of a real function as one-sided limits of its difference quotient there is δ>0 such that f(t)f(s)L(ts)ts for the relevant t with 0<ts<δ, where L is the corresponding derivative; in particular f(t)f(s)(L+1)ts there.

[F3]

ZN(0,1) has the strictly positive density φ(x)=ex2/2/2π; consequently P(Zy)y for every 0<y1. Standard normal and normal laws The standard normal density has total mass one

[F4]

If events Gn satisfy nP(Gn)<, then almost surely only finitely many Gn occur, that is, P(lim supnGn)=0. First Borel-Cantelli lemma for events

[F5]

The rationals are dense in R: every point of [0,) lies in a nondegenerate interval with rational endpoints. The rationals embed densely in the reals

[F6]

AC is the ambient assumption of the Brownian and normal-law interfaces. The Axiom of Choice

[F7]

Countable unions of measurable null events are null. Basic identities for a probability measure

Proof

technique · direct
1.1

Suppose the continuous path f:=B(ω) has a finite two-sided derivative at some s(a,b), or a finite right derivative at s=a, or a finite left derivative at s=b, with absolute value at most C; use both sides for an interior point, the right side at a and the left side at b, applying [F2] to obtain δ>0 such that f(t)f(s)(C+1)ts for every t[a,b] on the permitted side or sides of s with 0<ts<δ.

givenF2
2.1

Fix n6 with (ba)/n<δ/6, write h:=(ba)/n and tk:=a+kh, and put k0:=max{k{0,,n1}:tks}. Then tk0stk0+1, including k0=n1 when s=b. If k0+5n take the block of five increments beginning at tk0, and otherwise take the block of five increments ending at tn. In either case all endpoints lie in [a,b], on the side of s allowed in step 1.1 for the endpoint cases, and within distance 6h<δ of s, so each increment has absolute value at most 2(C+1)6h=12(C+1)h=:D/n with D:=12(C+1)(ba).

step 1.1given
3.1

For n=0,1,2,3,4,5 set G_n to be the empty event, without defining h or a mesh for those indices. For integers n>=6 define Gn to be the event that some block of five consecutive increments Btk+iBtk+i1 (k=0,,n5, i=1,,5) has all five absolute values at most D/n, where h=(ba)/n. Every G_n is a finite union of finite intersections of measurable coordinate events. Insert 0 before a if a>0 and use the subfamily of grid increments in [a,b]; the increments of one block are independent with laws N(0,h) by [F1], so by [F3] and independence the probability for a fixed block is at most yn5, where yn:=D/(ba)n=12(C+1)(ba)/n1 for large n. Hence P(Gn)nyn5=125(C+1)5(ba)5/2n3/2, and nP(Gn)<: the finitely many remaining initial terms are at most one each, and n=2j2j+11n3/22j/2 bounds the tail by a geometric series.

givenF1F3step 2.1
4.1

By [F4] and step 3.1, almost surely Gn fails for all sufficiently large n; by step 2.1 this means that almost surely the path has no finite derivative with absolute value at most C at any point of [a,b] (two-sided on (a,b), right at a, left at b).

step 2.1step 3.1F4
5.1

Intersect the common continuity event from [F1] with the complements of all the measurable limsup events of [F4]. Taking the union of those null events over the countably many rational pairs 0a<b and over integers C1, and using [F5] to place every s>0 in the interior of such an interval (while s=0 is the left endpoint of one), we obtain: almost surely no time s0 has a finite two-sided derivative (for s>0) or finite right derivative (for s=0).

step 4.1F1F4F5F7
6.1

The boundary cases are covered by the block choices of step 2.1: s=0 uses the right-handed block beginning at a, s=b the left-handed block ending at b, and interior times either the forward or the backward block, all of which stay inside [a,b]; the cases n<6 are defined to be empty events in step 3.1, so the sequence is indexed by all natural numbers and no division by zero is performed; rounding the derivative bound up to an integer C loses nothing, and the finite-difference ratio of [F2] is the definition-level form of The derivative f(c)=limxcf(x)f(c)xc of f:AR at a point cA that is a limit point of A, and differentiability on a set; AC is inherited through [F6] from the Brownian and normal-law interfaces.

step 2.1step 4.1F2F6given

Source notes

This is the Dvoretsky-Erdős-Kakutani mesh argument as in Durrett, Theorem 7.1.6 and its proof: differentiability at a single time forces five consecutive increments of every sufficiently fine uniform mesh to be small. The calculation above bounds the union over the O(n) possible blocks by the summable quantity 125(C+1)5(ba)5/2n3/2. A fixed-time argument would only produce an uncountable intersection of null events; the mesh argument converts this into one countable Borel-Cantelli statement.

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian paths have infinite total variation

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion Brownian motion. Almost surely, on every nondegenerate compact interval [a,b][0,) the variation sums of the path are unbounded above: supPi<nBti+1Bti=+ over all partitions P=(n,t) of [a,b] in the sense of Bounded variation and total variation on an interval. Equivalently, almost surely the path is not of bounded variation on any nondegenerate compact interval.

Facts & Assumptions

Given: AC, a standard Brownian motion B, and nonnegative rationals a<b.

[F1]

For disjoint time intervals the increments of B are independent with laws N(0,h) for interval length h. Brownian motion

[F2]

N(0,h) is the law of hZ for ZN(0,1), and Z has the strictly positive density φ(x)=ex2/2/2π with Rφ=1. Standard normal and normal laws The standard normal density has total mass one

[F3]

Improper integrals of nonnegative measurable functions are the limits of their integrals over [0,R], and substitution by u=x2/2 computes 0Rxex2/2dx=1eR2/2. Monotone convergence for the integral Substitution: if φ is differentiable on [c,d] with φ integrable and f is continuous on an interval containing φ([c,d]), then φ(c)φ(d)f=cd(fφ)φ. Under Countable Choice, continuous compact-interval integrands have equal Riemann and Lebesgue integrals; expectation is integration against the law, and a density can be moved into the integrand. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral Change of variables for expectation Integrating against a density agrees with integrating the product

[F4]

For a real function on [a,b] with a<b, the variation of a partition P is V(f,P)=i<nf(ti+1)f(ti), and the sums over all partitions are nonempty; a partition of a subinterval refines to a partition of the larger interval, so the variation sums are monotone under passing to subintervals. Bounded variation and total variation on an interval

[F5]

Chebyshev: P(XEXλ)Var(X)/λ2 for a square-integrable real X and λ>0. Chebyshev's inequality for random variables

[F6]

First Borel-Cantelli: if nP(Gn)< then almost surely only finitely many Gn occur. First Borel-Cantelli lemma for events

[F7]

The rationals are dense and countable; enumerating ordered pairs by diagonals gives a countable list of rational intervals. The rationals embed densely in the reals Q is countably infinite

[F8]

AC is the ambient assumption of the Brownian and normal-law interfaces. The Axiom of Choice

[F9]

For ZN(0,1) the even-moment formula gives EZ2m=(2m1)!! for m1; in particular EZ2=1 (Gaussian even moments for Brownian increments).

[F10]

Products of integrable Borel functions of independent real random variables have factored expectations. For square-integrable real random variables, covariance is bilinear and Cov(X,Y)=E[XY]EXEY, with Var(X)=EX2(EX)2 (Expectations factor over finite products of independent random variables, Variance and covariance identities for random variables).

Proof

technique · direct
1.1

For ZN(0,1) one has EZ=2/π and Var(Z)=12/π: by [F2] and [F3], EZ=20xφ(x)dx=2(2π)1/2limR(1eR2/2)=(2/π)1/2, while EZ2=EZ2=1 by [F9] with m=1, so [F10] gives variance 1(2/π). The compact substitution integrals in [F3] are transferred to Lebesgue integrals before the nonnegative monotone limit; expectation and density identities in [F3] justify the displayed Gaussian integral.

F2F3F9F10
2.1

Fix n1, put h=(ba)/2n and Sn:=k=12nBa+khBa+(k1)h; by [F1] and [F2] the 2n increments are independent with the law of hZ, so ESn=2n2h/π=2(ba)/π2n/2 and Var(Sn)=2nh(12/π)=(ba)(12/π) is constant in n. Indeed every absolute increment is square-integrable by step 1.1; [F10] applied to the absolute-value Borel functions on any two distinct increments gives zero covariance, and finite bilinearity gives the variance sum. If a>0, the independent increment list in [F1] is obtained by including 0,a in the time grid and discarding its first increment; for a=0 no extra interval is needed.

step 1.1F1F2F10
3.1

By [F5] applied to Sn with λ=12ESn, P(Sn12ESn)4Var(Sn)/(ESn)2=4(ba)(12/π)2(ba)π2n=2π(12/π)2n, which is summable in n.

step 2.1F5
4.1

By [F6] and [step 2.1], almost surely Sn>12ESn for all sufficiently large n, and hence Sn; since each Sn=V(B,Pn) is the variation of the path over the dyadic partition Pn of [a,b], the variation sums over partitions of [a,b] are almost surely unbounded above.

step 2.1step 3.1F4F6
5.1

The argument of steps 1.1-4.1 depends on 0a<b only through the single number ba, so it applies verbatim to every ordered pair of nonnegative rationals a<b; by [F7] the pairs form a countable family. For each pair use the measurable event NnN{Sn>12ESn} of probability one from step 4.1; intersecting these explicit events gives a measurable probability-one event on which the variation sums of B are unbounded above on every compact interval in [0,) with rational endpoints.

step 1.1step 4.1F7F8
6.1

On that event every nondegenerate compact interval [c,d][0,) has unbounded variation sums as well: choose, by [F7], nonnegative rationals a<b with c<a<b<d, note that the dyadic partitions of [a,b] extend to partitions of [c,d] by adding the points c and d, and that adding points can only increase a variation sum by the triangle inequality, so the sums over partitions of [c,d] dominate the unbounded family for [a,b].

step 5.1F4F7
7.1

The boundary cases are covered: the interval is required to be nondegenerate, so a=b and the singleton convention are excluded; 2n2 increments are used, so the n-sums are genuine variation sums over partitions in the sense of [F4]; the variance bound is uniform in the mesh index for each fixed interval, while the full-measure intersection in step 5.1 is legitimate by countability, not by an interval-independent variance constant. AC includes the Countable Choice required for the compact Riemann/Lebesgue bridge and is inherited through the Brownian and normal-law interfaces in [F8].

step 6.1F4F8given

Source notes

The proof uses the published Gaussian even-moment calculation at order two, a compact substitution calculation for the absolute first moment, the general independent-product and covariance interfaces, Chebyshev and first Borel–Cantelli. It requires neither bounded-variation differentiability nor a quadratic-variation theorem.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-22Open item page →

Quadratic variation along a partition sequence

Definition

Fix T>0 and a continuous function x:[0,T]R (Continuity of f:AR at a point of A and on A: the ε-δ condition, its agreement with limxcf(x)=f(c) at a limit point, and continuity at an isolated point). A partition sequence of [0,T] is a sequence (πn)n0 of partitions of [0,T] in the sense of Partition of [a,b] as a finite strictly increasing list a=t0<t1<<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, written πn=(mn,s(n)),0=s0(n)<s1(n)<<smn(n)=T, whose mesh mesh(πn):=max1kmn(sk(n)sk1(n)) tends to zero as n. The partitions need not refine one another and no regularity of the points beyond mesh convergence is assumed. A family originally indexed by positive integers is read as the zero-indexed family qn=πn+1; this changes none of its limiting assertions.

For t[0,T] two partial sums are attached to (πn). If t is a partition point, both are defined by the same formula; the cases t=0 and t=T are included, and an empty sum is 0.

  1. Step convention. With k(t) the largest index in {0,,mn} with sk(t)(n)t, [x]tπn,step:=k=1k(t)(xsk(n)xsk1(n))2.
  2. Partial-increment convention. On the interval [sk(t)(n),sk(t)+1(n)] containing t one also adds the terminal increment, [x]tπn,part:=[x]tπn,step+(xtxsk(t)(n))2(k(t)<mn), and [x]Tπn,part:=[x]Tπn,step.

The quadratic variation of x along (πn) is the limit of either family of functions, in a mode that is part of every later statement: the pointwise claim is that [x]tπn converges as n for each fixed t[0,T], and the uniform claim is that it converges uniformly on [0,T]. No other mode and no other partition family is included.

Three conventions are built into the definition and are used later in this form.

  1. The two conventions differ by at most the squared maximal oscillation. For every n and t, [x]tπn,part[x]tπn,step=(xtxsk(t)(n))2  (maxk supu,v[sk1(n),sk(n)]xuxv)2, and the right-hand side tends to 0 as n because x is uniformly continuous on the compact interval [0,T] and the mesh tends to 0. For completeness this uniform-continuity assertion is choice-free: for each c[0,T] and ε>0, continuity supplies a least integer j(c)0 such that xyxc<ε/2 whenever y[0,T] and yc<21j(c). The intervals of radius rc=2j(c) centered at c cover [0,T]. Its compactness Heine-Borel by bisection: every closed bounded interval [a,b] is compact gives finitely many such intervals covering it. Put δ equal to the minimum of their positive radii. If uv<δ and u belongs to the interval centered at c, then both u,v are within 2rc of c, so xuxv<ε. This proves uniform continuity without selecting arbitrary radii. In particular the two conventions have the same limit whenever either limit exists.
  2. No partition-independent object is defined. The symbol [x]πn names the nth sum along the named sequence (πn), and any quadratic-variation limit is attached to that sequence; it is not a claim that the sums converge along every refining sequence, nor that a path-dependent choice of partitions leaves the limit unchanged. When a statement below says "quadratic variation", the partition sequence is part of the data.
  3. Dependence on t. Each [x]tπn is a genuine real number, being a finite sum of nonnegative terms; the family t[x]tπn is nondecreasing in t for the step convention, and for the partial-increment convention it agrees with the step value at partition points.

No choice principle is used: the partitions are given as a sequence of finite lists, the sums are finite sums of real numbers, and the limits are the usual uniqueness-of-limit limits.

Source notes

Lawler, Section 2.8, defines the quadratic sums along a partition sequence, proves the convergence results for meshes tending to zero (Theorems 2.8.1-2.8.2) and warns explicitly that the mesh condition is load-bearing: without a prescribed partition family the sums may depend on the partitions chosen. The definition above separates the two partial-sum conventions used in that section and records only a difference bound, so that later items can state their limits for either convention without redefining the symbol.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian quadratic variation on dyadic partitions

Statement

Let B be a standard Brownian motion Brownian motion and fix T>0. For n1 let πn be the dyadic partition of [0,T] with points kT/2n, k=0,,2n, and let Qn:=k=12n(BkT/2nB(k1)T/2n)2=[B]Tπn in the notation of Quadratic variation along a partition sequence. Then QnT in L2 and almost surely as n.

Facts & Assumptions

Given: AC, a standard Brownian motion B, T>0, and the dyadic partitions πn above with h=T/2n.

[F1]

The increments of B over disjoint intervals are independent with laws N(0,h) for interval length h. Brownian motion

[F2]

If XtXs has law N(0,ts) then EXtXs2m=cmtsm with cm=(2m1)!!; in particular E(ΔB)2=h and E(ΔB)4=3h2 for an increment of length h. Gaussian even moments for Brownian increments

[F3]

Chebyshev: P(XEXλ)Var(X)/λ2 for a square-integrable real X and λ>0. Chebyshev's inequality for random variables

[F4]

First Borel-Cantelli: if nP(Gn)< then almost surely only finitely many Gn occur. First Borel-Cantelli lemma for events

[F5]

[B]Tπn denotes the terminal quadratic sum along the named partition sequence πn, whose mesh T/2n tends to zero. Quadratic variation along a partition sequence

[F6]

AC is the ambient assumption of the Brownian and normal-law interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Writing Δk:=BkhB(k1)h for k=1,,2n, [F1] and [F2] give EΔk2=h and EΔk4=3h2, so the centered variables Yk:=Δk2h satisfy EYk=0 and Var(Yk)=EΔk4(EΔk2)2=3h2h2=2h2.

F1F2
2.1

QnT=k=12nYk has mean 0, and [F1] makes the Yk independent, so Var(QnT)=kVar(Yk)=2n2(T/2n)2=2T2/2n; hence E(QnT)2=2T2/2n0 and QnT in L2.

step 1.1F1
3.1

For every ε>0, [F3] gives P(QnTε)Var(QnT)/ε2=2T2/(2nε2), which is summable in n; applying [F4] to Gn={QnT1m} for each m1 and intersecting the resulting probability-one events over m shows that almost surely QnT.

step 2.1F3F4
4.1

The degeneracies are covered: T>0 is required, so h>0 and the sums are nonempty with 2n2 terms; the partition sequence is the one named in [F5], with mesh T/2n0 and with consecutive refinements πnπn+1, so no ambiguity of convention arises at t=T, where the step and partial-increment conventions coincide by Quadratic variation along a partition sequence; and AC enters only through [F6].

step 2.1step 3.1F5F6given

Source notes

Lawler, Theorems 2.8.1 and 2.8.2, proves the mean-square convergence of the dyadic quadratic sums and their almost-sure convergence along meshes whose sizes are summable (here T/2n). The computation above is the direct one: the second and fourth Gaussian moments of the increments give Var(QnT)=2T2/2n, Chebyshev gives summable error probabilities, and Borel-Cantelli upgrades to almost-sure convergence.

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Uniform dyadic Brownian quadratic variation process

Statement

Let B be a standard Brownian motion Brownian motion and fix T>0. For n1 let πn be the dyadic partition of [0,T] with points kT/2n, k=0,,2n. Then almost surely the partial quadratic-variation processes t[B]tπn converge to t uniformly on [0,T], for the step convention and for the partial-increment convention of Quadratic variation along a partition sequence alike: sup0tT[B]tπnt0.

Facts & Assumptions

Given: AC, a standard Brownian motion B, T>0, the dyadic partitions πn with mesh h=T/2n, and grid points tj=jT/2n.

[F1]

The increments of B over disjoint intervals are independent with laws N(0,h) for interval length h, and one probability-one event carries all continuous paths. Brownian motion

[F2]

For an increment ΔB of length h, E(ΔB)2=h and E(ΔB)4=3h2. Gaussian even moments for Brownian increments

[F3]

Kolmogorov's maximal inequality: for independent centered square-integrable X1,,Xn with partial sums Sk, P(max1knSkλ)Var(Sn)/λ2. Kolmogorov maximal inequality

[F4]

First Borel-Cantelli: if nP(Gn)< then almost surely only finitely many Gn occur. First Borel-Cantelli lemma for events

[F5]

The two conventions of [B]tπn agree at partition points and differ by the squared terminal increment (BtBsk(t)(n))2; the mesh of πn is T/2n0. Quadratic variation along a partition sequence

[F6]

A subset of R is compact if and only if it is closed and bounded; hence [0,T] is compact. A subset of R is compact if and only if it is closed and bounded

[F7]

AC is the ambient assumption of the Brownian and normal-law interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Put Δj:=BtjBtj1, Xj:=Δj2h and Sj:=i=1jXi for j=1,,2n; at partition points the step convention reads [B]tjπn,step=Sj+tj, and the two conventions agree there by [F5].

givenF5
2.1

By [F1] and [F2], EXj=hh=0 and Var(Xj)=3h2h2=2h2; the Xj are independent because they are functions of disjoint increments, so Var(S2n)=2n2h2=2T2/2n, and [F3] gives P(max1j2nSjε)2T2/(2nε2) for every ε>0.

step 1.1F1F2F3
3.1

The bound of [step 2.1] is summable in n for each fixed ε>0; applying [F4] to the events {maxjSj1/m} for m1 and intersecting the resulting probability-one events over m yields: almost surely, for every rational ε>0 one has max1j2nSj<ε for all sufficiently large n, hence maxj[B]tjπn,steptj0.

step 2.1F4
4.1

For t[tj,tj+1) the step convention satisfies [B]tπn,steptSj+(ttj)maxjSj+h, and the same bound with j=2n holds at t=T; since h=T/2n0, [step 3.1] gives almost-sure uniform convergence to t for the step convention on [0,T].

step 3.1F5
5.1

For the partial-increment convention, [F5] gives [B]tπn,part[B]tπn,step=(BtBtj)2ω(T/2n)2, where ω(δ):=sup{BuBv:u,v[0,T], uvδ}; by [F6] and the finite-subcover argument applied to the continuous path on the compact interval [0,T], ω(δ)0 as δ0, so the two conventions have the same uniform limit t.

step 4.1F5F6
6.1

The boundary cases are covered: T>0 so h>0 and the sums have at least two terms; t=0 is a partition point with [B]0πn=0 for both conventions; t=T is a partition point where the conventions coincide by [F5]; the mesh tends to zero and the partitions refine, so the named sequence is a partition sequence in the sense of [F5]; and AC enters only through [F7].

step 4.1step 5.1F5F7given

Source notes

Lawler, Section 2.8, obtains the uniform statement by controlling the maximal partial sum at the grid points and observing that the path increments are small between them. The proof above uses Kolmogorov's maximal inequality directly on the centered squared increments Xj=Δj2h, whose variance is 2h2 by the fourth Gaussian moment, and then handles the two partial-sum conventions with the difference bound recorded in the definition.

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian one- and quadratic variation

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion Brownian motion. Then:

  1. Almost surely, the path has unbounded total-variation sums on every nondegenerate compact interval [a,b][0,): it is not of bounded variation there.
  2. For every fixed T>0, almost surely the dyadic partial quadratic-variation processes on [0,T] converge uniformly to t. These assertions hold simultaneously for every horizon in any prescribed countable subset of (0,), in particular for all positive integer horizons.

The two conclusions are not in conflict: the first is an assertion about sums of first powers Bti+1Bti over partitions, the second about sums of squares along the named dyadic sequence.

Facts & Assumptions

Given: AC and a standard Brownian motion B.

[F1]

Almost surely the variation sums of the path are unbounded above on every nondegenerate compact interval in [0,), so the path is not of bounded variation there. Brownian paths have infinite total variation

[F2]

For each fixed T>0, almost surely the dyadic partial quadratic-variation processes of [0,T] converge to t uniformly on [0,T], for both conventions of Quadratic variation along a partition sequence. Uniform dyadic Brownian quadratic variation process

[F3]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Proof

technique · direct
1.1

By [F1] there is one probability-one event on which the variation sums are unbounded on every nondegenerate compact interval in [0,); this is already a single almost-sure statement and needs no further intersection.

F1
1.2

For each fixed T>0, [F2] gives a probability-one event on which the dyadic partial quadratic-variation processes formed from the dyadic partitions of [0,T] converge uniformly to t on [0,T]. For a prescribed countable C(0,), select such a measurable full-measure event ET for each TC using the stated AC, and intersect with the single full-measure continuity event supplied by Brownian motion. The event EC=TCET (also intersected with that continuity event) has probability one, giving simultaneous convergence for C. This includes finite and empty C and all positive integer horizons. The partition used for each T has points kT/2n; convergence is not transferred between differently scaled grids. [F3]

F2F3
2.1

For a fixed horizon, or the prescribed countable set in step 1.2, intersecting the probability-one event of [step 1.1] with the corresponding event of [step 1.2] gives both assertions simultaneously; the dyadic partition sequence is named, so the second conclusion is a statement about that sequence and not about arbitrary partitions.

step 1.1step 1.2
3.1

The degenerate cases are covered: the interval in the first assertion and the horizon T in the second are required to be nondegenerate and positive respectively; the value t=0 is a partition point at which both quadratic sums vanish; only prescribed countable families of horizons are intersected, because the dyadic partitions supplied by [F2] depend on the horizon; and AC enters through the suppliers and the countable selection of their full-measure events in [F3].

step 1.2step 2.1F3given

Source notes

Lawler, Section 2.8, records both faces of the dichotomy: the absolute-increment sums diverge while the squared-increment sums converge to elapsed time. The corollary collects the two independently proved statements on the page and makes explicit that the quadratic variation is asserted along the named dyadic sequence.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Brownian motion has a jointly measurable continuous version

Statement

Let B be a standard Brownian motion Brownian motion on a probability space (Ω,F,P). Then there is a process B^=(B^t)t0 with the following properties.

  1. B^ is indistinguishable from B, and B^0=0 on all of Ω.
  2. Every path tB^t(ω) is continuous on [0,), so ωB^(ω) is a map into the path space C([0,),R) of Uniform-on-compacts metric on continuous path space.
  3. That map is Borel measurable for the uniform-on-compacts Borel σ-algebra, and (t,ω)B^t(ω) is measurable for B([0,))F.

The version is obtained by keeping the original process on one probability-one event and replacing the whole path by the zero path outside it; no distribution of B is altered and no path is selected by any choice principle.

Facts & Assumptions

Given: AC and a standard Brownian motion B on (Ω,F,P).

[F1]

There is a measurable event A with P(A)=1 on which every path tBt(ω) is continuous, and B0=0 almost surely. Brownian motion

[F2]

On C([0,),R) the uniform-on-compacts metric induces the topology of uniform convergence on compact subsets of [0,). Uniform-on-compacts metric on continuous path space

[F3]

The Borel σ-algebra of C([0,),R) is generated by the coordinate maps πt(f)=f(t), t0. Borel sigma-algebra of continuous path space is generated by coordinates

[F4]

The evaluation map e(f,t):=f(t) on C([0,),R)×[0,) is defined for every pair, and it is continuous when the path space carries the uniform-on-compacts topology. The evaluation map e:C(X,Y)×XY, e(f,x)=f(x)

[F5]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Put A:=A{B0=0}, a measurable event with P(A)=1 by [F1], and define B^t(ω):=Bt(ω) for ωA and B^t(ω):=0 for ωA; then B^ is indistinguishable from B because it differs from B only on the null set (A)c, and B^0(ω)=0 for every ω.

givenF1
1.2

The evaluation map of [F4] is continuous: if fmf in the metric of [F2] and tmt in [0,), choose N with tmN for all m and f(s)f(t)<ε/2 for st<η, which is possible because the continuous f is continuous at t; convergence in the metric of [F2] gives sup0sNfm(s)f(s)<ε/2 for all large m, and then fm(tm)f(t)sup0sNfm(s)f(s)+f(tm)f(t)<ε for all large m.

F2F4
2.1

Every path of B^ is continuous: for ωA it is the Brownian path, continuous by [F1], and for ωA it is the zero path.

step 1.1F1
2.2

Each time coordinate of B^ is measurable: B^t=1ABt is the product of a measurable indicator with the measurable function Bt.

step 1.1
3.1

The map ωB^(ω) is a random element of C([0,),R) with its Borel σ-algebra: by [F3] that σ-algebra is σ(πt:t0), and for every t and every Borel ER one has {ω:πt(B^(ω))E}={B^tE}, measurable by [step 2.2]; the family of Borel sets with measurable preimage is a σ-algebra containing the generating sets πt1(E), hence all of B(C).

step 2.1step 2.2F3
4.1

The pair map (t,ω)(B^(ω),t) is measurable from B([0,))F into the product of the Borel σ-algebras, because its first component is measurable by [step 3.1] and its second is a coordinate projection; composing with the continuous, hence Borel measurable, evaluation map of [step 1.2] shows that (t,ω)B^t(ω) is B([0,))F-measurable.

step 3.1step 1.2
5.1

The boundary cases are covered: the normalization changes the process only on the null set (A)c, so indistinguishability and every almost-sure statement are preserved; the initial value is 0 on all outcomes as required; the case t=0 enters the measurability statements through the coordinate B^0, which is identically zero; and AC enters only through [F5] as the ambient assumption of the Brownian definition.

step 1.1step 2.1step 4.1F5given

Source notes

Durrett, Section 7.1, fixes a continuous version of Brownian motion as part of the definition of the process; Sousi's Chapter 6 treats Brownian motion as a random continuous path. The lemma records the two measurability consequences used later on the page: the path map is a random element of the continuous path space, and the evaluation map makes the process jointly measurable, which is what the Tonelli and Fubini arguments for the zero set and the occupation time need.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

The Brownian zero set

Definition

Assume the Axiom of Choice and let B be a standard Brownian motion. Replace B, as permitted by Brownian motion has a jointly measurable continuous version, by the indistinguishable version B^ whose every path is continuous and whose evaluation (t,ω)B^t(ω) is jointly measurable. The Brownian zero set is Z:=Z(B^):={t0:B^t=0}, and for a horizon T>0 one writes ZT:=Z[0,T].

The following are part of the definition and are used later in this form.

  1. Pathwise closedness. For every outcome the set Z is closed in [0,) and nonempty: it is the preimage of the closed set {0} under the continuous path, and B^0=0 for every outcome. Hence ZT is compact for every T<.
  2. Version independence. Replacing B^ by the original B changes Z only on a P-null set: B and B^ are indistinguishable. Every almost-sure assertion about Z proved below is therefore an almost-sure assertion about the zero set of B, and no statement below quantifies over versions.
  3. Measurability of the section integrals. Joint measurability makes (t,ω)1{B^t=0} measurable for B([0,))F, so for each T the section integral ωλ(ZT(ω))=0T1{B^t(ω)=0}dt is a measurable function of ω, and the Tonelli identity for the product measure dtP applies to it.
  4. Endpoint conventions. The point t=0 belongs to Z for every outcome, the singleton {0} has Lebesgue measure zero, and a horizon T may be replaced by any larger horizon since ZTZT for TT.

No further structure is assigned: in particular Z is not asserted to be perfect, uncountable or of measure zero by this definition; those are statements proved separately on this page.

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

The Brownian zero set has Lebesgue measure zero

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion and let Z be its zero set as in The Brownian zero set, understood through the all-path continuous jointly measurable version. Then for every 0T< the Lebesgue measure of ZT is zero almost surely: P(λ(ZT)=0)=1. Here set Z0=Z{0}, extending the positive-horizon notation. There is one measurable full event on which λ(Z)=0 and all finite horizons have zero measure. On its intersection with the supplied event of all-time agreement, the same pathwise assertion holds for the original B.

Facts & Assumptions

Given: AC, a standard Brownian motion B, its normalized version B^ and zero set Z, and a horizon T(0,).

[F1]

The zero set is the closed, nonempty set Z={t0:B^t=0} of the all-path continuous version B^, and ZT=Z[0,T]. The Brownian zero set

[F2]

The map (t,ω)B^t(ω) is B([0,))F-measurable, so (t,ω)1{B^t=0} is product measurable. Brownian motion has a jointly measurable continuous version

[F3]

For t>0 the Brownian definition gives BtB0 law N(0,t), and B_0=0 almost surely, hence B_t has that law. By the definition of the normal law, it is the law of sqrt(t) times a standard normal variable. Brownian motion Standard normal and normal laws

[F4]

Tonelli: for a product-measurable f0 on a product of sigma-finite spaces, the section integrals are measurable and the two iterated integrals agree. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

[F5]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

[F6]

A nonnegative measurable function has integral zero exactly when it vanishes almost everywhere. Countable unions of probability-zero events are null. A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere Basic identities for a probability measure

Proof

technique · direct
1.1

For t>0, [F3] gives P(Bt=0)=γ({0})=0, because sqrt(t)>0 and the standard normal density integrates to zero over the Lebesgue-null singleton {0}. The latter singleton convention is in [F1]. The normalized version agrees with B on one measurable full event by [F2], so P(B^t=0)=0 as well. At t=0 the probability is one, but that time singleton has Lebesgue measure zero. The zero indicator is product measurable by [F2].

F1F2F3
2.1

Applying [F4] to that indicator over the sigma-finite product [0,T]×Ω with the product measure dtP gives Eλ(ZT)=0TP(B^t=0)dt=0, and it also shows that ωλ(ZT(ω)) is measurable.

step 1.1F4
3.1

Since 0λ(ZT)T, its measurability and zero expectation in step 2.1 permit [F6], giving λ(ZT)=0 almost surely.

F6step 2.1
4.1

Intersect the measurable probability-one events from step 3.1 over the explicitly listed horizons N>=1. By [F6] their intersection A is measurable with probability one. On A, every finite nonnegative T has Z[0,T]Z[0,N] for some integer N>=T, so its measure is zero by monotonicity. Also Z=N1(Z[0,N]), so countable subadditivity of Lebesgue measure gives λ(Z)=0 on A. Conversely a zero-measure whole zero set has zero-measure intersections, which explains the global formulation.

F1F6step 3.1
5.1

At T=0 the zero set is the singleton {0}, so its measure is zero pathwise; the time endpoint t=0 does not affect step 2.1. Let A_* be the measurable full event on which the supplied normalized process agrees with B at all times. On A intersect A_* from step 4.1 the original B path has exactly the same zero set, proving its pathwise nullness there. No claim is needed that arbitrary exceptional paths of B have measurable zero sets or that the entire all-time equality event is measurable. The countable horizon list is fixed; full AC is inherited from [F5], with no further selection of paths or exceptional events.

F1F2F5step 2.1step 4.1

Source notes

Durrett, Section 7.4.1, computes EZ[0,T]=0TP(Bt=0)dt=0 and concludes that the zero set has measure zero; the argument above makes the Tonelli step explicit through the all-path continuous jointly measurable version, so the section integrals are measurable without any auxiliary regularity assumption.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

The Brownian zero set has no isolated points

Statement

Assume the Axiom of Choice. Let B be standard Brownian motion and let Z be the zero set of the everywhere-continuous, zero-start representative X fixed in The Brownian zero set. On one measurable event of probability one, every t in Z is a limit point of Z{t}. Thus Z is closed and has no isolated points in [0,infinity). On the supplied common full event, this assertion also holds for the original B path.

Facts & Assumptions

Given: AC, B, its normalized representative X, and its zero set Z.

[F1]

X has measurable coordinates, every path is continuous and starts at zero, and X agrees with B on a measurable full event. Thus its finite-dimensional distributions, including the independent Gaussian increments, agree with those of B and X itself is a standard Brownian motion. Z is closed and contains zero. The Brownian zero set Brownian motion

[F2]

Form the raw natural filtration and usual augmentation of X itself. This augmentation contains the raw filtration of X. The proof applies the strong Markov theorem directly to X and requires no identification with the filtration of the original B. Natural and usual augmented Brownian filtrations

[F3]

Stopping times use the non-strict sublevel test. For a bounded stopping time S of the usual augmentation of a Brownian motion X and a bounded product-measurable functional G, the conditional future identity is E[G((XS+r)r0)GS]=Ψ(XS), where Ψ(x)=G(x+w)μ(dw) and mu is Wiener measure. Since X is everywhere continuous and S finite, the theorem's measurable random-time version equals the literal evaluation everywhere. Continuous-time stopping times and stopped sigma-algebras Strong Markov property of Brownian motion Wiener measure on continuous path space

[F4]

The maximum of a continuous zero-start Brownian motion on [0,h], h>0, has distribution P(Mhx)=2Φ(x/h)1 for x>=0; in particular P(M_h=0)=0. The process -X is also Brownian by negating its independent centered Gaussian increments. Law of the Brownian maximum Brownian motion

[F5]

Rational density and countable subadditivity for null events. The rationals embed densely in the reals Basic identities for a probability measure

[F6]

Conditional-expectation equalities are almost-sure equalities of versions characterized by event integrals. Full AC is inherited from the Brownian, completion and conditional-expectation interfaces. Conditional expectation as an ae class The Axiom of Choice

Proof

technique · direct
1.1

Apply [F4] to X and -X on each interval [0,1/m], m>=1. Off a countable union of null events [F5], X has both a positive and a negative value on each such interval. Neither value is at zero. The intermediate value theorem between those two times gives a zero at a positive time <=1/m. Therefore positive zeros accumulate at zero. The same argument applies to the coordinate process under Wiener measure, which is Brownian by [F3].

F1F3F4F5
1.2

Fix a nonnegative rational q and define τq=inf{sq:Xs=0}, with infimum of the empty set infinity. For t<q its sublevel event is empty. For t>=q, compactness and closedness give {τqt}={minr[q,t]Xr=0}, which equals the event that the infimum over (Q[q,t]){q,t} is zero. At t=q this is the singleton test X_q=0. Every coordinate used has time <=t, so the event belongs to the raw filtration of X and thus to its usual augmentation [F2]. This proves that tau_q is a stopping time. If tau_q is finite, closedness gives X_{tau_q}=0. No recurrence or finiteness assertion is needed.

F1F2F3F5
2.1

Define G(w)=1m1k>m{infrQ[1/k,1/m]w(r)=0}. This is product measurable, since only countably many coordinates and set operations occur. For continuous w the compact infimum is a minimum and equals the rational infimum, so G(w)=1 precisely when positive zeros accumulate at zero. Step 1.1 shows Ψ(0)=G(w)μ(dw)=1. For any nonzero x, continuity of w at zero and w(0)=0 exclude zeros of x+w sufficiently near zero; hence Ψ(x)=0. These assertions hold mu-almost surely even if a realization of Wiener measure contains nonzero-start null paths.

F1F3F5step 1.1
3.1

For integers N>=1 put S=tau_q wedge N. It is a bounded stopping time: its sublevel event is {tau_q<=t} for t<N and the whole space for t>=N. By [F3] and step 2.1, E[G((XS+r)r0)GS]=1{XS=0}. The event {X_S=0} is in the stopped sigma-algebra by [F3]. Integrating the nonnegative variable 1{XS=0}(1G((XS+r))) over the whole space gives zero by the conditional event-integral identity [F6]; since this variable is an indicator, the event it indicates is null. Thus, almost surely on {X_S=0}, positive zeros of the post-S path accumulate at zero. On {tau_q<=N} step 1.2 gives S=tau_q and X_S=0. We conclude that, outside a measurable null event, whenever tau_q<=N the zero tau_q is approached by other zeros from the right.

F3F6step 2.1step 1.2
4.1

Take the intersection of the full event in step 1.1 with the full events in step 3.1 over the countable set of nonnegative rationals q and positive integers N. It is a measurable full event by [F5]. On it, every finite tau_q is approached by other zeros from the right, since some N is larger than tau_q.

F5step 1.1step 3.1
5.1

On that event zero is not isolated by step 1.1. If t>0 were an isolated zero, some delta with 0<delta<t would satisfy Z(tδ,t+δ)={t}. By rational density choose q in (t-delta,t). There are no zeros in [q,t), and t is a zero, so tau_q=t. Step 4.1 supplies other zeros in (t,t+delta), a contradiction. Hence every zero is a limit point of the other zeros. Z is already closed by [F1].

F1F5step 1.1step 4.1
6.1

The empty hitting set and tau_q=infinity cause no problem because only bounded S are used, and the conclusion about tau_q is conditional on its finiteness. The time-zero endpoint has its own full event. The only intersections of full events are countable; no rational time is claimed itself to be a positive Brownian zero. On the common agreement event in [F1] the original B and X have identical zero sets, so the original path assertion follows there. AC is inherited as in [F6]. In particular, the proof never places an arbitrary normalization null event in the original Brownian filtration.

F1F6step 1.2step 4.1step 5.1

Source notes

The standard rational-next-zero and strong-Markov architecture is described in Durrett, Section 7.4.1, and Sousi, Theorem 6.39. Here immediate zero accumulation follows from the maximum law applied to both signs. Using tau_q wedge N removes the need for a recurrence argument. The auxiliary filtration is the usual augmentation of the normalized Brownian motion itself.

CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

The Brownian zero set is uncountable

Statement

Let B be a standard Brownian motion and let Z be its zero set The Brownian zero set. Almost surely, for every T>0 the set ZT=Z[0,T] is uncountable. In particular the zero set is almost surely uncountable in every nondegenerate interval [0,T], although by The Brownian zero set has Lebesgue measure zero it has Lebesgue measure zero there.

Facts & Assumptions

Given: AC, a standard Brownian motion B, its zero set Z, and T>0.

[F1]

Z is a closed subset of [0,) containing 0, and ZT=Z[0,T] is compact. The Brownian zero set

[F2]

Almost surely every point of Z is a limit point of Z, so Z has no isolated points. The Brownian zero set has no isolated points

[F3]

Almost surely λ(ZN)=0 for every integer horizon N1. The Brownian zero set has Lebesgue measure zero

[F4]

A set PR is perfect when it is closed and has no isolated points; every nonempty perfect subset of R is uncountable. Perfect subset of R: closed with no isolated points Every nonempty perfect subset of R is uncountable

[F5]

The rationals are dense in R. The rationals embed densely in the reals

[F6]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Fix a rational q>0 with qZ; then Z[0,q] is nonempty because 0Z, it is closed in R as the intersection of the closed set Z with the closed interval [0,q], and it has no isolated points: for tZ[0,q] one has t<q, and by [F2] there are zeros of Z different from t in every neighbourhood of t, which for a neighbourhood of radius <min(qt,t+1) lie in [0,q].

F1F2F4given
1.2

Almost surely, for every T>0 there is a rational q(0,T) with qZ: by [F3] and monotonicity of Lebesgue measure, λ(ZT)=0<T=λ([0,T]), so ZT[0,T]; picking s(0,T)Z, openness of the complement of the closed set Z supplies a neighbourhood of s disjoint from Z, and [F5] supplies a rational q in that neighbourhood with 0<q<T.

F3F5given
2.1

On the probability-one event of [step 1.2] and [F2], and for a rational q as there, [step 1.1] exhibits Z[0,q] as a nonempty perfect subset of R; by [F4] it is uncountable, and since Z[0,q]ZT, the set ZT is uncountable.

step 1.1step 1.2F4
3.1

The cases are covered: T>0 is required, so the interval is nondegenerate; the rational q is chosen strictly inside (0,T) so that the potential isolated point q of Z[0,q] is excluded by qZ; the statement is asserted simultaneously for all T>0 on one probability-one event, obtained by intersecting the countably many events of [step 1.2] over rational T and using monotonicity in T; and AC enters only through [F6].

step 1.2step 2.1F1F6given

Source notes

Sousi, Theorem 6.39, states the zero set is almost surely closed with no isolated points, and Durrett, Section 7.4.1, derives uncountability from closedness and the absence of isolated points by the perfect-set theorem. The corollary adds the explicit choice of a rational point outside Z inside every horizon, which is what makes the subset used in the perfect-set theorem nonempty and genuinely free of the terminal-point exception.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Two-sided Mills bounds for the standard normal tail

Statement

Let φ(x)=(2π)1/2ex2/2 and Φ(x)=xφ(y)dy for xR. Then for every x>0 xφ(x)1+x2  Φ(x)  φ(x)x, and consequently, for every x>1, (1x1x3)φ(x)  Φ(x)  φ(x)x. Both bounds are sharp as x in the sense that the ratio of each side to Φ(x) tends to 1.

Facts & Assumptions

Given: AC, ACω, DC, and a real x>0.

[F1]

φ is the strictly positive, Borel measurable standard normal density, with Rφ=1, and N(0,1) is the probability measure φdy. Standard normal and normal laws The standard normal density has total mass one

[F2]

On a compact interval every C1 function is Lipschitz, hence absolutely continuous and of bounded variation; in particular φ and y1/y are absolutely continuous on [x,R] for 0<x<R. C1 implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

[F3]

Integration by parts for absolutely continuous functions: abFG+abFG=F(b)G(b)F(a)G(a), under ACω and DC. Integration by parts for absolutely continuous functions The Axiom of Countable Choice (ACω) The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain

[F4]

Substitution computes xRyey2/2dy=ex2/2eR2/2, and monotone convergence justifies passing to the limit R in the integrals of the nonnegative functions φ and φ(y)/y2 over [x,R]. Substitution: if φ is differentiable on [c,d] with φ integrable and f is continuous on an interval containing φ([c,d]), then φ(c)φ(d)f=cd(fφ)φ Monotone convergence for the integral

[F5]

AC is the ambient assumption; ACω and DC are the hypotheses of the integration-by-parts interface used in [F3]. The Axiom of Choice The Axiom of Countable Choice (ACω) The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain

Proof

technique · direct
1.1

For yx>0 one has φ(y)(y/x)φ(y); integrating over [x,R] and letting R with [F4] gives Φ(x)1xxyφ(y)dy, and [F4] computes xRyφ(y)dy=(2π)1/2(ex2/2eR2/2)φ(x), so Φ(x)φ(x)/x.

givenF1F4
1.2

For 0<x<R, [F2] and [F3] applied to F(y)=1/y and G=φ on [x,R], together with φ(y)=yφ(y), give xRφ(y)dy=φ(x)/xφ(R)/RxRφ(y)y2dy.

F2F3
2.1

Letting R in [step 1.2] with [F4] gives Φ(x)=φ(x)/xxφ(y)y2dy, and since y2x2 on [x,) one gets Φ(x)φ(x)/xΦ(x)/x2, that is, Φ(x)(1+x2)φ(x)/x and hence Φ(x)xφ(x)/(1+x2).

step 1.2F4
3.1

The algebraic comparison x/(1+x2)(1/x1/x3) for x>0 is equivalent to x4(x21)(1+x2)=x41, which holds; combining it with [step 2.1] gives the displayed form for x>1, and the ratio claim follows because (1/x1/x3)/(1/x)=1x21 and (x/(1+x2))/(1/x)=1/(1+x2)1 while the upper bound already is φ(x)/x.

step 1.1step 2.1
4.1

The endpoint and degenerate cases are covered: x>0 is required so that 1/x and the integration interval [x,R] are meaningful and F=1/y is C1 on it; x=0 is excluded by the statement because the upper bound would divide by zero, while Φ(0)=12 is finite; the limit R is handled by monotone convergence over the increasing family [x,R]; the constants ACω and DC are those declared for [F3] and are used nowhere else; and AC enters only through [F5].

step 1.2step 2.1F3F5given

Source notes

Durrett, Lemma 1.2.6 and the estimates (8.5.2) in the proof of Theorem 8.5.1, states the two-sided bound (x1x3)ex2/2xey2/2dyx1ex2/2 for x>0 (up to the normalization constant), together with the asymptotic ratio 1 used in the law of the iterated logarithm. The proof above derives the stronger lower bound xφ(x)/(1+x2) from the identity obtained by integrating by parts, which is the form consumed by the LIL item.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian law of the iterated logarithm at infinity

Statement

Let B be a standard Brownian motion Brownian motion. Then on one measurable event of probability one lim suptBt2tloglogt=1,lim inftBt2tloglogt=1 the normalizing function being taken for t>e so that loglogt>0.

Facts & Assumptions

Given: AC, a standard Brownian motion B, rationals α>1, β>1 and the geometric sequence tn=αn.

[F1]

The increments of B over disjoint intervals are independent with laws N(0,h) for interval length h; the path is continuous on a probability-one event. Brownian motion

[F2]

Use the everywhere-continuous zero-start representative fixed in the maximum-law theorem (zero the path outside a measurable full event of continuity and zero start). It equals the original Brownian motion at all times on that event, so the final path conclusion transfers back. Law of the maximum: with MT=sup0sTBs one has P(MT>x)=2Φ(x/T) for x>0, where Φ(z)=zφ(y)dy. Law of the Brownian maximum

[F3]

Mills bounds: for z>0, Φ(z)φ(z)/z; for z>1, (z1z3)φ(z)Φ(z); and φ(z)=(2π)1/2ez2/2 is decreasing in z. Two-sided Mills bounds for the standard normal tail Standard normal and normal laws The standard normal density has total mass one

[F4]

First Borel-Cantelli: summable probabilities give almost surely finitely many occurrences; second Borel-Cantelli: independent events with divergent probability sum occur infinitely often almost surely. First Borel-Cantelli lemma for events Second Borel-Cantelli lemma under pairwise independence

[F5]

If X is a standard Brownian motion then so is X: the covariance characterisation exhibits the increments of X as independent stationary Gaussian increments, and continuity is preserved. Brownian covariance is equivalent to independent stationary normal increments

[F6]

The rationals are dense in R. The rationals embed densely in the reals

[F7]

AC is inherited from the Brownian, normal-law and maximum-law interfaces. Both cited Borel–Cantelli statements are choice-free; no choice assumption is added to them. The Axiom of Choice

Proof

technique · direct
1.1

For n with tn>e put Un:={Mtn>2βtnloglogtn} and zn:=2βloglogtn; by [F2] and [F3], P(Un)=2Φ(zn)2φ(zn)/znCα,βnβ(logn)1/2 for a constant Cα,β, because φ(zn)=(2π)1/2(logtn)β=(2π)1/2(nlogα)β; since β>1 the probabilities are summable. For example, grouping n in [2j,2j+1) bounds the upper series by a constant times j2j(1β), which is geometric. Set the finitely many early events with tne to the empty event.

givenF1F2F3
1.2

For the lower bound fix α>1, put β:=α/(α1)>1 and Dn:=Btn+1Btn, so that by [F1] the Dn are independent with law N(0,tn+1tn)=N(0,tn+1/β); let En:={Dn>γ2tn+1loglogtn+1} with γ:=1/β, and note γ2β=1.

givenF1
2.1

By [F4] and [step 1.1] there is a probability-one event on which Un fails for all sufficiently large n; on that event, for every ttN with N large, choosing n with tnt<tn+1 gives MtMtn+12βtn+1loglogtn+1 and hence Mt2tloglogtβαloglog(αt)loglogtαβ; therefore lim suptBt2tloglogtαβ almost surely.

step 1.1F1F4
2.2

For large n the probability of En satisfies P(En)=Φ(γ2βloglogtn+1)12zn1φ(zn)=122π12loglogtn+1logtn+1, with zn=2loglogtn+1, by the lower Mills bound of [F3]; since logtn+1=(n+1)logα, for sufficiently large n this is at least cα/((n+1)log(n+1)) with cα>0. In each block 2jn+1<2j+1, the sum of these lower bounds is at least a positive constant times 1/j+1; hence the series diverges (grouping that latter series into square blocks already gives a fixed positive contribution per block). Define the finitely many early events with tn+1e to be empty.

step 1.2F3
3.1

Intersecting the events of [step 2.1] over the countably many pairs of rationals α,β(1,2) and using [F6] to choose, for every ε>0, rationals with αβ1+ε, we obtain lim suptBt2tloglogt1 almost surely.

step 2.1F6
4.1

By [F5] the process B is again a standard Brownian motion, so [step 3.1] applies to it and gives lim inftBt2tloglogt1 almost surely.

step 3.1F5
5.1

By [F4] and [step 2.2] the events En occur infinitely often almost surely; on the event of [step 4.1] intersected with this one, for infinitely many n one has Btn+1=Btn+Dn(1+ε)2tnloglogtn+γ2tn+1loglogtn+1=2tn+1loglogtn+1(γ(1+ε)α1/2(1+o(1))).

step 4.1step 1.2step 2.2F1F4
6.1

Since γ=11/α1 and α1/20 as α, for every δ>0 there are rational α>1 and ε>0 with γ(1+ε)α1/21δ; hence [step 5.1] gives lim suptBt2tloglogt1δ almost surely for every rational δ>0, and intersecting over the countably many δ yields lim sup1 almost surely.

step 5.1F7
7.1

Combining [step 3.1] and [step 6.1] gives lim suptBt2tloglogt=1 almost surely; applying this conclusion to the standard Brownian motion B of [F5], whose limit superior is the negative of the limit inferior of B, gives lim inftBt2tloglogt=1 almost surely.

step 3.1step 6.1F5
8.1

The boundary and degeneracy cases are covered: the normalizer is positive precisely for t>e, and the statement is asymptotic as t; the geometric sequences are indexed from n=0, with only finitely many terms below e; the parameters over which probability-one events are intersected may be restricted to rational α,β>1 and rational δ>0, a countable family. The auxiliary value γ=11/α need not be rational and creates no additional event: once α is fixed it is a deterministic threshold in the same block events En. The upper and lower bounds are established using the first Borel-Cantelli lemma for the upper bound and the second for the lower bound; AC enters only through [F7].

step 3.1step 6.1step 7.1F7given

Source notes

This follows the geometric-block architecture of Durrett, Theorem 8.5.1 (printed pp.416–418), with an explicit critical-threshold variant. Durrett's lower bound uses threshold coefficient 1/β and a subcritical exponent 1/β; here γ=1/β gives exponent one, whose remaining 1/logn factor still makes the probability series divergent, as proved in step 2.2. The upper bound, interpolation, independent-block lower bound and sign symmetry follow the same route. The threshold γ need not be rational: it is determined by the rational parameter α.

CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian law of the iterated logarithm at zero

Statement

Let B be a standard Brownian motion Brownian motion. Then almost surely lim supt0Bt2tloglog(1/t)=1,lim inft0Bt2tloglog(1/t)=1, the normalizer being taken for 0<t<e1 so that loglog(1/t)>0.

Facts & Assumptions

Given: AC and a standard Brownian motion B.

[F1]

Almost surely lim supsYs/2sloglogs=1 and the corresponding limit inferior is 1 for every standard Brownian motion Y. Brownian law of the iterated logarithm at infinity

[F2]

Time inversion: the process Y0=0, Ys:=sB1/s for s>0, is again a standard Brownian motion, with continuity at s=0 part of the conclusion. Brownian time inversion

[F3]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Let Y be the time-inverted process of [F2]; by [F1] applied to Y there is a probability-one event on which lim supsYs/2sloglogs=1 and lim infsYs/2sloglogs=1.

givenF1F2
2.1

On that event, substituting s=1/t with t0 and using Y1/t=Bt/t gives Y1/t2t1loglog(1/t)=Bt/t2t1loglog(1/t)=Bt2tloglog(1/t), so lim supt0Bt2tloglog(1/t)=1 and lim inft0Bt2tloglog(1/t)=1 almost surely.

step 1.1F2
3.1

The cases are covered: the substitution t1/t is a bijection of (0,) onto itself, so t0 corresponds to s; the normalizer is positive exactly for 0<t<e1; the endpoint t=0 is not evaluated, continuity at zero being part of [F2]; and AC enters only through [F3].

step 2.1F2F3given

Source notes

Time inversion converts the law of the iterated logarithm at infinity, Theorem 8.5.1 of Durrett, into the corresponding statement at zero, with the normalizing factor transforming exactly as displayed.

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

The critical Hölder boundary at zero

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion Brownian motion. Almost surely both of the following hold.

  1. For every exponent 0<α<1/2 and every T>0 the path is α-Hölder on [0,T], that is, locally below the critical exponent.
  2. The path is not one-half Hölder at zero: there is no finite constant C and no δ>0 with BtCt for all 0<t<δ. In fact Bt/t is unbounded as t0.

The second assertion concerns the critical exponent at the single point 0; the first concerns uniform subcritical Hölder bounds on each compact interval.

Facts & Assumptions

Given: AC and a standard Brownian motion B.

[F1]

There is a probability-one event on which, for every T>0 and every 0<γ<1/2, a finite K=K(ω,T,γ) satisfies BtBsKtsγ for all 0s,tT. Brownian paths are locally Holder below one half

[F2]

Almost surely lim supt0Bt2tloglog(1/t)=1 and lim inft0Bt2tloglog(1/t)=1. Brownian law of the iterated logarithm at zero

[F3]

The rationals are dense in R. The rationals embed densely in the reals

[F4]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Proof

technique · direct
1.1

On the probability-one event of [F1], for every T>0 and every 0<γ<1/2 there is a finite constant K with BtBsKtsγ on [0,T]; since [0,T] contains 0 and the exponents are ordered, this is precisely assertion 1.

F1given
1.2

On the probability-one event of [F2], put r(t)=Bt/2tloglog(1/t) for 0<t<e1. For every ε>0 the limsup and liminf bounds imply 1ε<r(t)<1+ε for all sufficiently small t, and r(t)>1ε occurs at arbitrarily small positive times. Hence lim supt0r(t)=1; in particular r(t)>1/2 at arbitrarily small positive times.

F2
2.1

Fix any finite C0 and δ>0. Since 2loglog(1/t), choose η>0 smaller than δ and e1 such that this factor exceeds 2C whenever 0<t<η. Step 1.2 supplies such a t with r(t)>1/2. Then Bt/t=2loglog(1/t)r(t)>C. As this works for every C and δ, the ratio is unbounded in every right neighborhood of zero and assertion 2 follows. A negative C cannot bound the nonnegative ratio either.

step 1.2
3.1

Intersect the events in steps 1.1 and 1.2 with {B0=0}, also of probability one by the Brownian definition. Both assertions then hold simultaneously; since B0=0, the critical bound written with Bt is precisely the pointwise Hölder bound at zero. The correct exponent comparison is downward: for any 0<α<1/2, choose a rational q with α<q<1/2 using [F3] and an integer Nmax(1,T). A bound with exponent q on [0,N] implies BtBsKNqαtsα on [0,T], since tsqαNqα; the diagonal is immediate. Thus rational exponents above each desired exponent suffice, not exponents below it. In this proof [F1] already supplies the single event for every exponent and horizon, so no further uncountable intersection is made. The normalizer in step 1.2 is used only at positive t<e1. AC is inherited through [F4] and the two Brownian suppliers; the arbitrarily-small-time argument requires no selected sequence of times.

step 1.1step 1.2step 2.1F3F4given

Source notes

The local Hölder supplier gives one full-measure event for all subcritical positive exponents and compact horizons. The zero-time LIL supplier gives arbitrarily small times at which its normalized absolute value exceeds one half. The proof combines these interfaces and gives the explicit downward power comparison.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-22Open item page →

Quadratic variation needs a partition convention

Remark

The identity [B]t=t established on this page is an almost-sure assertion along the fixed dyadic partition sequence of Quadratic variation along a partition sequence; the uniform form of that assertion is Uniform dyadic Brownian quadratic variation process. It is not a simultaneous assertion over all refining sequences, and the path-dependent or arbitrary refinements of a realized path are not covered.

What the quantifiers allow. The theorem cited above supplies almost-sure uniform convergence for the fixed dyadic sequence. For a general prescribed deterministic sequence whose mesh tends to zero, the standard conclusion without an additional summability or regularity hypothesis is convergence in probability, not almost-sure convergence along the whole sequence. In particular there is no single event on which every refining sequence simultaneously has the same limit, and the definition deliberately builds in no partition-independent object.

What fails without regularity. The convergence proofs use the independence of increments over a preselected mesh together with a summable mesh estimate. A refinement adapted to the oscillations of one realization destroys that independence and can change the sums; the deterministic partition dependence of quadratic sums is illustrated on the companion examples page, and the boundary is recorded here so that later semimartingale statements do not silently inherit a claim about arbitrary partitions.

No proof is attached to this remark: it records the quantifier boundary of the preceding definition and theorem rather than a new mathematical assertion.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

The last Brownian zero has the arcsine law

Statement

Assume the Axiom of Choice. Let B be a standard Brownian motion and use the everywhere-continuous, zero-start representative B^ fixed by The Brownian zero set. For t>0 put Lt=max{s[0,t]:B^s=0}. This is a random variable and, for 0ut, P(Ltu)=2πarcsinu/t. Consequently Lt/t has density 1/(πx(1x)) on (0,1), with no mass at either endpoint. On the supplied measurable full event of all-time agreement, this maximum is also the last zero of the original B. The distribution is independent of the normalized representative.

Facts & Assumptions

Given: AC, B and its normalized representative, and t>0.

[F1]

The normalized zero set is closed, contains zero, and agrees with the original zero set on a measurable full event; its normalized coordinates are measurable. The Brownian zero set

[F2]

For a bounded product-measurable future functional G and deterministic u, its conditional expectation given the raw Brownian past is the Borel function xG(x+w)μ(dw) evaluated at B_u, where mu is Wiener measure on continuous paths. Future-path Markov property

[F3]

For normalized Brownian motion W and s>0, its maximum has continuous distribution P(Msx)=2Φ(x/s)1 for x>=0. Law of the Brownian maximum

[F4]

B_u has law N(0,u) for u>0: its increment from zero has that law and B_0=0 almost surely. This law is the pushforward of φ(z)dz under z mapped to sqrt(u)z, where φ(z)=ez2/2/2π. Negation preserves all independent centered Gaussian increments and continuity, so -W is Brownian as well. Brownian motion Standard normal and normal laws

[F5]

Tonelli for nonnegative product-measurable functions on sigma-finite spaces. One-dimensional C1 diffeomorphisms transport integrable functions with their absolute derivative. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions

[F6]

Absolutely continuous functions obey the Lebesgue fundamental theorem; monotone convergence exhausts nonnegative integrals. Every C1 function on a compact interval is Lipschitz by its bounded derivative and the mean value theorem, hence absolutely continuous directly from the definition. Fundamental theorem of calculus for absolutely continuous functions Monotone convergence for the integral

[F7]

Probability is continuous along increasing or decreasing sequences of events, and a Borel probability law is uniquely determined by its distribution function. Basic identities for a probability measure Probability laws correspond to distribution functions

[F8]

Full AC is inherited from the Brownian and conditional-expectation interfaces and directly supplies all dependent or countable witness choices used by the integration and distribution interfaces. The Axiom of Choice

Proof

technique · direct
1.1

By [F1], the zero set in [0,t] is nonempty compact, so its maximum exists and lies in [0,t]. For 0<v<=t, the event L_t<v is exactly that the path has no zero in [v,t]. For v<t this is the event that the infimum of B^r over (Q[v,t]){v,t} is positive, since this dense infimum equals the attained compact minimum. For v=t it is simply B^t>0. Both are measurable; the cases v<=0 and v>t are empty and whole. Thus L_t is measurable.

F1
2.1

Fix 0<u<t and s=t-u. Define the bounded product-measurable functional G(w)=1{infr(Q[0,s]){s}w(r)>0}. On continuous paths it is the indicator of no zero in [0,s]. On the common full event in [F1], and outside {B_u=0}, the indicator of L_t<=u equals G applied to the original future (Bu+r)r0. The excluded event has probability zero by [F4], since the normal density gives zero mass to a singleton. Taking expectations in [F2] therefore gives P(Ltu)=EΨ(Bu), where Ψ(x)=G(x+w)μ(dw). No all-time event on the full cylinder space or shifted hitting law is used.

F1F2F4step 1.1
3.1

For x>0 a continuous zero-start W makes x+W zero-free on [0,s] precisely when it stays positive there, or equivalently when the maximum of -W is strictly less than x. By [F3] and [F4], Ψ(x)=2Φ(x/s)1; strict versus weak inequality makes no difference because the maximum law has no atom at x. For x<0 apply the same argument to -x-W. At x=0, G(x+W)=0 since W_0=0, also agreeing with 2Φ(0)1=0 by symmetry of the normal density. Hence Ψ(x)=2Φ(x/s)1 for every x.

F3F4step 2.1
4.1

Using the pushforward law in [F4], not an unproved density transformation, step 3.1 gives P(Ltu)=I(a), where a=u/(tu)>0 and I(a)=R(2Φ(az)1)φ(z)dz. Symmetry of the even density gives I(a)=40φ(z)0azφ(y)dydz. For fixed z>0, apply [F5] to the diffeomorphism v mapped to zv from (0,a) onto (0,az); the normal density is integrable on this bounded interval. Thus the inner integral equals 0azφ(zv)dv. Endpoints have Lebesgue measure zero.

F4F5step 2.1step 3.1
5.1

Tonelli [F5] now gives I(a)=2π0a0ze(1+v2)z2/2dzdv. The explicit primitive e(1+v2)z2/2/(1+v2) on [0,R], followed by monotone convergence R increasing to infinity, makes the inner integral 1/(1+v2). The primitive arctan(v) on [0,a] therefore gives I(a)=2arctan(a)/π by [F6]. Since a>0, the angle arctan(a) is in (0,pi/2) and has sine a/1+a2=u/t; hence it equals arcsin(sqrt(u/t)). This proves the asserted formula for 0<u<t without a polar substitution.

F5F6step 4.1
6.1

Since 0<=L_t<=t, its distribution function equals one at t. Decreasing u to zero and increasing u to t through explicit sequences in (0,t), [F7] and step 5.1 give P(Lt=0)=0 and P(Lt<t)=1. Thus there is no atom at t either, and both endpoint values of the formula follow.

F7step 1.1step 5.1
7.1

Put H(v)=2arcsin(v)/π for 0<v<1. Its derivative is f(v)=1/(πv(1v))>0. On every compact subinterval of (0,1), H is C1, so [F6] gives bcf=H(c)H(b). Let b decrease to zero and c increase to one. Monotone convergence gives total integral one and 0vf=H(v). Extend f by zero off (0,1). The probability measure with this density has the same distribution function as L_t/t by steps 5.1 and 6.1, and uniqueness in [F7] identifies the laws.

F6F7step 5.1step 6.1
8.1

On the supplied measurable full event, the original B and normalized process have identical zero sets and hence identical last zeros. Two permitted normalized representatives agree on the intersection of their supplied full events, so give the same distribution. No measurability of the original last-zero functional on exceptional paths is asserted. The assumption t>0 is essential to the ratio; u=0,t and x=0 were handled separately. AC is used exactly through [F8]; the exhaustion sequences are fixed.

F1F8step 1.1step 3.1step 6.1step 7.1

Source notes

Durrett, Example 7.4.3, printed p.374, equation (7.4.7), proves the last-zero law by conditioning and a nonnegative iterated integral. Here the equivalent Gaussian integral is evaluated by one-dimensional substitution and Tonelli. The normalized zero-set convention and endpoint/density justifications are explicit.

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian step-potential resolvent at zero

Statement

Assume the Axiom of Choice. Let α,β>0, let V(x):=α+β1{x>0} for xR, let B be a standard Brownian motion Brownian motion and let B^ be its all-path continuous jointly measurable version. Define, for xR, u(x):=0E[exp(0tV(x+B^r)dr)]dt. Then 0u1/α, the function u is Borel, C1 on R and C2 off 0, and 12u(x)=αu(x)1(x<0),12u(x)=(α+β)u(x)1(x>0), while u(0)=1α(α+β).

Facts & Assumptions

Given: AC, ACω, DC, reals α,β>0, the potential V=α+β1{x>0}, a standard Brownian motion B and its jointly measurable continuous version B^.

[F1]

Every path of B^ is continuous and (t,ω)B^t(ω) is product measurable; hence (x,t,ω)V(x+B^t(ω)) is product measurable and the version agrees with B at all times on one measurable full event. We use only this full-event conclusion, not measurability of the entire equality set. Brownian motion has a jointly measurable continuous version

[F2]

Markov property: E[Φ((Bs+r)r0)Fs]=ΨΦ(Bs) almost surely for bounded Borel Φ, with ΨΦ(y)=Φ(y+w)μ(dw); and for f bounded Borel, Psf(x)=E[f(x+Bs)]=f(y)ps(x,y)dy with ps the Brownian transition density. Future-path Markov property The Brownian transition semigroup The Brownian kernels form a semigroup The future-path theorem also supplies its continuous-path Borel formulation; below it is applied to the Brownian process B^ with its own natural filtration.

[F3]

FTC package: the indefinite integral of an L1 function is absolutely continuous and has the integrand as its derivative almost everywhere; for absolutely continuous F one has F(x)F(a)=axF(t)dt for all x. This interface carries ACω and DC. The indefinite integral of an L1 function is absolutely continuous The indefinite integral of an L1 function is differentiable almost everywhere Fundamental theorem of calculus for absolutely continuous functions The Axiom of Countable Choice (ACω) The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain

[F4]

Tonelli applies to nonnegative product-measurable integrands, giving measurability of the section integrals and equality of the iterated integrals; φ(x)=(2π)1/2ex2/2 has total mass one and Rev2dv=π. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product Standard normal and normal laws The standard normal density has total mass one The Gaussian integral ex2dx=π

[F5]

The Lebesgue change-of-variables formula holds for a C1 diffeomorphism and an integrable function, using the absolute Jacobian, under Countable Choice. Monotone convergence passes nonnegative exhaustion limits, and dominated convergence applies under an integrable majorant. A differentiable function with zero derivative on an interval is constant. A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions Monotone convergence for the integral Dominated convergence A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant

[F6]

Taking out what is known: an Fs-measurable bounded factor may be moved inside a conditional expectation. Taking out what is known

[F7]

AC is the ambient assumption of the Brownian, Markov and conditional expectation interfaces. The Axiom of Choice

Proof

technique · direct
1.1

Write X for the supplied all-path continuous version. For parameters (x,t,omega,r), the nonnegative function 1{0rt}V(x+Xr(ω)) is product measurable by [F1]. Its section integral in r is measurable by [F4]; for a measurability-only application one may equip the remaining parameter space with the zero measure, which is finite, so no parameter measurability is lost. Thus the exponential and its successive integrals in omega and t are measurable, and u is Borel. Since αt0tV(x+Xr)dr(α+β)t, one has 0u0eαtdt=1/α. The last integral follows from [F3] on finite intervals and then [F5] by monotone convergence. X is itself Brownian, since all its finite-dimensional laws agree with B.

F1F3F4F5given
1.2

Put I(a,b)=0eas2b/s2ds for a>0,b>=0. For b=0, scaling the Gaussian integral gives I(a,0)=π/(2a). For b>0 put c=b/a and g(s)=eas2b/s2. This is integrable, bounded by eas2. Reciprocal substitution s=c/v gives I(a,b)=0(c/s2)g(s)ds. The increasing diffeomorphism v=asb/s maps (0,infinity) onto the real line, with derivative a(1+c/s2). As v2=as2+b/s22ab, [F4] and [F5] yield π=2ae2abI(a,b). All substitutions can first be applied on compact subintervals to integrable continuous functions, then exhausted by monotone convergence; the absolute Jacobian handles the reciprocal map. Now set t=s^2 in the Gaussian time integral, likewise by positive exhaustion. For λ=2α this gives 0eαtpt(z)dt=2/πI(α,z2/2)=λ1eλz, including z=0 by the separate b=0 case.

F4F5
2.1

For bounded Borel f define Gαf(x)=0eαtPtf(x)dt. For nonnegative f, Tonelli and step 1.2 give Gαf(x)=λ1Reλxyf(y)dy. The kernel has integral 2/λ2=1/α, as computed by integrating its exponential on each half-line using [F3] and exhaustion. For signed bounded f apply Tonelli separately to its positive and negative parts; both integrals are bounded by f/α, so subtraction is legitimate and gives the same formula.

F2F3F4F5step 1.2
2.2

Fix x and t>0. On each outcome, A(s)=0s1{x+Xr>0}dr is absolutely continuous with derivative 1{x+Xs>0} almost everywhere by [F3]. Set W(s)=eαsexp(stV(x+Xr)dr). The inner exponent is absolutely continuous and bounded on [0,t]. Composition with the exponential is absolutely continuous because the exponential is Lipschitz on its bounded range; the ordinary chain rule applies at every point where the inner derivative exists. Hence W(s)=β1{x+Xs>0}W(s) almost everywhere. The AC fundamental theorem gives eαtexp(0tV(x+Xr)dr)=β0teαs1{x+Xs>0}exp(stV(x+Xr)dr)ds. No derivative at every point of an open-set boundary is asserted.

F1F3step 1.1
3.1

Let f be bounded Borel and K(z)=eλz/λ. For fixed x and |h|<=1, the difference quotient K(x+hy)K(xy)/h is at most eλeλxy, by the one-sided derivatives of K and integration along the segment. It converges for every y unequal to x to sgn(xy)eλxy. Dominated convergence, with majorant feλeλxy, therefore differentiates the kernel formula. Put L(x)=y<xeλ(xy)f(y)dy and R(x)=y>xeλ(yx)f(y)dy. Then Gαf=(L+R)/λ and (Gαf)=L+R. These L,R are continuous: on a compact range of x write them as exponential factors times indefinite integrals of locally bounded functions, with fixed finite tails. At any continuity point of f, the difference quotient of its weighted indefinite integral tends to the integrand value, since the average error is bounded by the supremum error near that point. Thus L=λL+f(x) and R=λRf(x) there. It follows that (Gαf)=λ(L+R)2f=2αGαf2f at each such point. The first derivative is continuous everywhere.

F3F5step 2.1
3.2

For tau>=0, the function ϕx,τ(w)=exp(0τV(x+w(r))dr) is Borel on continuous-path space: evaluation is jointly measurable as in [F1], and the section-integral argument of step 1.1 applies. Apply the continuous-path formulation in [F2] to X and this bounded functional. At time zero, X_0=0 makes its expectation the Wiener integral. At general s this gives E[exp(ss+τV(x+Xr)dr)FsX]=hτ(x+Xs),hτ(y)=Eϕy,τ(X). The indicator 1{x+Xs>0} is measurable for this filtration. Multiplying by it using [F6] and taking expectations yields an equality for each s,tau; the expectation property here is the defining conditional-expectation event identity with the whole event. Tonelli integrates these nonnegative quantities in s and tau, so no simultaneous choice of conditional-expectation versions over uncountably many times is required. Integrate step 2.2 in t and expectation, then translate t=s+tau using [F5]. Since 0hτ(y)dτ=u(y), the result is u(x)=1αβ0eαsE[1{x+Xs>0}u(x+Xs)]ds.

F1F2F4F5F6step 1.1step 2.2
4.1

Set f(y)=1{y>0}u(y), bounded Borel by step 1.1. The semigroup formula and Gα1=1/α in step 2.1 turn step 3.2 into u=Gα1βGαf. Therefore step 3.1 already proves u is continuously differentiable on the whole real line. In particular f is continuous on each open half-line.

F2step 1.1step 2.1step 3.1step 3.2
5.1

At every x unequal to zero, apply step 3.1 to 1 and to f from step 4.1. The coefficient beta must multiply both terms of its second derivative: 12u=αGα11β(αGαff)=αu1+β1{x>0}u. This is continuous separately on the half-lines, so u is C2 there and satisfies the stated two equations. At zero only the already established C1 regularity is used.

step 3.1step 4.1
6.1

To solve the ODE without an unproved general-solution assertion, on an interval where v=k2v, k>0, set F=vkv. Then (ekxF)=0, so [F5] makes F a constant times ekx. Differentiating ekxv and then subtracting the explicit primitive of that exponential gives, again by [F5], v=Aekx+Dekx. Apply this with v=u1/α, k=lambda, on the negative half-line and with v=u1/(α+β), k=μ=2(α+β), on the positive half-line. Boundedness in step 1.1 excludes the exponentially growing term at the respective infinite endpoint. Hence u(x)=1/α+Aeλx (x<0),u(x)=1/(α+β)+Deμx (x>0).

F5step 1.1step 5.1
7.1

Continuity of u and of u at 0, from [step 5.1], gives 1/α+A=1/(α+β)+D and λA=μD; substituting the second into the first yields u(0)=1/α+A with α(u(0)1/α)=α+β(u(0)1/(α+β)), whose solution is u(0)=1/α(α+β).

step 6.1
8.1

The boundary and degeneracy cases are covered: α,β>0 keep V bounded between α and α+β, so 0u1/α and all the integrals converge absolutely; the potential has its single discontinuity at x=0, so the second-order equation is asserted only off 0, where 1{y>0}u is continuous; the case x=0 is handled by the continuity of u and u rather than by the differential equation; the time integral starts at t=0 where the exponent vanishes; and the choice principles used are exactly those declared: ACω and DC enter through the FTC package of [F3], and AC is the ambient assumption of [F7].

step 2.2step 5.1step 7.1F3F7given

Source notes

Yoshida, Lemmas 6.8.1-6.8.3, computes the step-potential resolvent at the origin by an ODE matching argument after identifying the resolvent kernel of the Gaussian semigroup. The proof above separates the two analytical inputs: the Gaussian resolvent kernel λ1eλxy from the time integral of the heat kernel, and the Duhamel identity for the potential V=α+β1{x>0}, which is proved pathwise from the fundamental theorem for absolutely continuous functions. The conditional expectation step uses the future-path Markov property of the page and takes the bounded Fs-measurable factor out.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-22Open item page →

Brownian positive occupation time has the arcsine law

Statement

Let B be a standard Brownian motion Brownian motion, choose its all-path continuous jointly measurable version B^, and for t>0 let At:=0t1{B^s>0}ds be the occupation time of the positive half-line up to time t. Then for every 0x1, P(Attx)=2πarcsinx, and At/t has the arcsine density f(x)=1πx(1x),0<x<1. Thus the occupation-time proportion of Brownian motion has the same distribution as the last-zero proportion of the theorem The last Brownian zero has the arcsine law, although the two random variables are of a different nature.

Facts & Assumptions

Given: AC, a standard Brownian motion B with its all-path continuous jointly measurable version B^, t>0, and reals α,β>0.

[F1]

With V=α+β1{y>0} the function u(x)=0E[exp(0tV(x+B^r)dr)]dt satisfies u(0)=1/α(α+β), and its defining integral is an E-integral against the occupied time. Brownian step-potential resolvent at zero

[F2]

Every path of B^ is continuous and the evaluation is jointly measurable, so s1{B^s>0} is measurable and At=0t1{B^s>0}ds is a random variable with 0Att. Any two such jointly measurable indistinguishable versions give the same occupation time almost surely by Tonelli. Brownian motion has a jointly measurable continuous version Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

[F3]

Scaling: for c>0 the process rc1/2Bcr is again standard Brownian motion, so the occupation times satisfy At=dtA1. Brownian scaling

[F4]

Tonelli for nonnegative product-measurable integrands. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

[F6]

Stone-Weierstrass: the unital point-separating algebra of polynomials is uniformly dense in C([0,1],R). Real Stone–Weierstrass theorem for compact Hausdorff spaces

[F7]

Dominated convergence justifies interchanging limits with expectations and integrals under an integrable dominating function. Dominated convergence

[F8]

AC is the ambient assumption of the Brownian and resolvent interfaces. The Axiom of Choice

Proof

technique · direct
1.1

By [F2], At is a random variable with values in [0,t]; by [F3] applied with c=t one has At=dtA1, because the time change maps the set of positive times for B to the corresponding set for the scaled motion, and hence At/t has the law of A1.

givenF2F3
1.2

The probability measure μ on [0,1] with density π1x1/2(1x)1/2 satisfies 01(α+βx)1μ(dx)=1/α(α+β) for all α,β>0: substituting x=sin2θ turns the integral into 2π0π/2(α+βsin2θ)1dθ, then s=tanθ turns it into 2π0dt/(α+(α+β)t2), and finally w=t(α+β)/α and [F5] give 2π1α(α+β)π2.

givenF5
2.1

For α,β>0, 0eαtE[eβAt]dt=E[(α+βA1)1]: [step 1.1] gives E[eβAt]=E[eβtA1], so the left side is 0E[e(α+βA1)t]dt, and [F4] equals it to E[0e(α+βA1)tdt]=E[1/(α+βA1)], the integrand being nonnegative and α+βA1α>0.

step 1.1F4
3.1

For V=α+β1{y>0} and t>0 one has 0tV(B^r)dr=αt+βAt up to the single point r=0, which is Lebesgue-null; hence the function u of [F1] satisfies u(0)=0eαtE[eβAt]dt, and [F1] with [step 2.1] yields E[1/(α+βA1)]=1/α(α+β) for all α,β>0.

step 2.1F1F2
4.1

The law of A1 and μ have the same moments: fixing α=1 and expanding 1/(1+βx)=k0(βx)k for x[0,1] and 0<β<1, uniformly on the square, [F7] shows that E[1/(1+βA1)]=k(β)kE[A1k] and (1+βx)1μ(dx)=k(β)kxkμ(dx) for every such β; since [step 3.1] and [step 1.2] make the two sides equal for all β(0,1), subtracting the two power series gives k(β)k(E[A1k]xkμ(dx))=0 on an interval, so every coefficient vanishes and all moments agree.

step 3.1step 1.2F7
5.1

Consequently E[f(A1)]=fdμ for every continuous f:[0,1]R: given ε>0, [F6] supplies a polynomial p with fp<ε, and E[f(A1)]fdμ2ε+E[p(A1)]pdμ=2ε by [step 4.1]; taking continuous fn1[0,x] and applying [F7] to both sides gives P(A1x)=μ([0,x])=2πarcsinx for 0x1.

step 4.1F6F7
6.1

By [step 1.1], P(At/tx)=P(A1x)=2πarcsinx for every t>0 and 0x1, which is the displayed distribution function.

step 1.1step 5.1
7.1

Differentiating the distribution function on (0,1) gives ddx2πarcsinx=2π12x(1x)=1πx(1x); this density is integrable on (0,1) (substitute x=sin2θ), so it is the density of At/t and both endpoints carry zero mass.

step 6.1F5
8.1

The boundary cases are covered: x=0 gives P(A1=0)=0=2πarcsin0 and x=1 gives 1=2πarcsin1=2ππ2; the value θ=π/2 of the substitution is the endpoint of the principal branch of [F5]; the parameters satisfy α,β>0 in [step 3.1] and 0<β<1 in [step 4.1]; the occupation time is taken over the half-line {y>0} so the single instant s=0 is excluded by a null set; and AC enters only through [F8].

step 3.1step 4.1step 7.1F5F8given

Source notes

Yoshida, Proposition 6.8.4, obtains the occupation-time arcsine law from the Laplace transform E[1/(α+βA1)]=1/α(α+β) produced by the step-potential resolvent of Lemmas 6.8.1-6.8.3. The proof above proves the same transform identity directly from the resolvent lemma of this page, identifies the arcsine law as the unique probability measure on [0,1] with that transform by moment matching and Stone-Weierstrass, and transfers the result from A1 to At by scaling.

5 · Examples, counterexamples and false statements

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