Brownian Motion Construction and Continuity — Examples

For each $j$, let ${\sigma }_j=\sqrt{h_j}>0$ and $g_j(y)=\frac{1}{{\sigma }_j}\varphi (y\mathrm{/}{\sigma }_j)=\frac{1}{\sqrt{2\pi h_j}}{e}^{-y^2\mathrm{/}(2h_j)}\mathrm{.}$ For a Borel set $A\subseteq \mathbb{R}$, apply [F6] to the dilation $T_j(z)={\sigma }_{j}z$ and the nonnegative Borel function ${\mathbf{1}}_A{g}_j$. Since $T_j^{\mathrm{\prime }}={\sigma }_j$, this gives ${\int }_A{g}_j(y)dy={\int }_{\mathbb{R}}{\mathbf{1}}_A({\sigma }_{j}z)g_j({\sigma }_{j}z){\sigma }_{j}dz={\int }_{\mathbb{R}}{\mathbf{1}}_A({\sigma }_{j}z)\varphi (z)dz\mathrm{.}$ By the pushforward definition in [F2], $g_j$ is therefore a density of $N(0,h_j)$.

Define $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ and $S:{\mathbb{R}}^n\to {\mathbb{R}}^n$ by $(Ty{)}_j={\sum }_{k=1}^j{y}_k,(Sx{)}_j=x_j-x_{j-1},x_0=0.$ Direct telescoping gives $ST=TS=\mathrm{id}$. Their coordinate partial derivatives are constant, so [F7] makes them $C^1$ with their displayed matrices. The matrix of $S$ is lower triangular with every diagonal entry one; hence $\mathrm{\mid }\det DS\mathrm{\mid }=1$. Thus $S$ is a $C^1$ diffeomorphism with inverse $T$. Moreover, $T\mathrm{\Delta }=(B_{t_1},\dots ,B_{t_n})$ almost surely by telescoping and $B_0=0$ almost surely.

Put $q(y_1,\dots ,y_n)={\prod }_{j=1}^n{g}_j(y_j)$. Induction on $n$, using Tonelli and the Borel equality ${\lambda }_{n-1}\times {\lambda }_1={\lambda }_n$, shows that the measure $Q(E)={\int }_{E}qd{\lambda }_n$ has on every Borel rectangle the value ${\prod }_j{\int }_{A_j}{g}_{j}d{\lambda }_1$. The base $n=1$ is step 1.1; the induction also gives $Q({\mathbb{R}}^n)=1$. Thus [F4] and uniqueness of the product measure identify $Q$ with the product of the $N(0,h_j)$ laws. By [F1] and [F3], $Q$ is exactly the joint law of $\mathrm{\Delta }=({\mathrm{\Delta }}_1,\dots ,{\mathrm{\Delta }}_n)$.

For a Borel set $A\subseteq {\mathbb{R}}^n$, step 2.1 and the almost-sure identity in step 1.2 give $P((B_{t_1},\dots ,B_{t_n})\in A)={\int }_{T^{-1}A}q(y)dy\mathrm{.}$ Apply [F6] to $S$ and $f(y)={\mathbf{1}}_{T^{-1}A}(y)q(y)$. Because $T(Sx)=x$ and $\mathrm{\mid }\det DS\mathrm{\mid }=1$, the right side becomes ${\int }_{{\mathbb{R}}^n}{\mathbf{1}}_A(x)q(Sx)dx={\int }_A{\prod }_{j=1}^n{g}_j(x_j-x_{j-1})dx\mathrm{.}$ Substituting the formula from step 1.1 is exactly the stated density.

The strict inequalities make every $h_j$ positive, so no division by zero or singular normal density occurs. For $n=1$, the formula is the $N(0,t_1)$ density with $x_0=t_0=0$; the empty case $n=0$ is excluded. The triangular determinant is one even when $n=1$. AC is used through the Brownian and normal-law suppliers, and it supplies the countable choice required by [F5] and [F6]; the finite triangular transformation makes no additional choice.

Bilinearity and the Brownian covariance kernel give $\mathrm{Cov}(B_t-B_s,B_v-B_u)=\min (t,v)-\min (t,u)-\min (s,v)+\min (s,u)\mathrm{.}$ Every term is finite because the Brownian values are square-integrable.

Both sides of the claimed formula are unchanged when the ordered pairs $(s,t)$ and $(u,v)$ are exchanged, the left side by symmetry of covariance. It therefore suffices to assume $s\le u$. If $t\le u$, the four minima in step 1.1 are respectively $t,t,s,s$, so the covariance is zero; also $\min (t,v)\le u=\max (s,u)$, so the stated overlap length is zero. This includes $t=u$ and all zero-length first intervals.

Still assuming $s\le u$, suppose $u<t\le v$. The four minima in step 1.1 are $t,u,s,s$, so the covariance is $t-u$. Here $\min (t,v)=t$ and $\max (s,u)=u$, giving the same positive overlap length. This case includes $s=u$ and $t=v$, but excludes $t=u$, already handled in step 2.1.

Finally, if $s\le u\le v<t$, the four minima in step 1.1 are $v,u,s,s$, so the covariance is $v-u$. Here $\min (t,v)=v$ and $\max (s,u)=u$. The value is zero exactly when $u=v$, so zero-length second intervals are included. These three cases exhaust $s\le u$; pair symmetry handles $u<s$.

Steps 2.1, 3.1, and 4.1 prove the formula for every allowed endpoint order, including coincident endpoints, $s=t$, $u=v$, and $s=u=t=v=0$. There is no empty family or biconditional. AC is used only through [F1]; expanding four covariances and comparing endpoints uses no further choice.

First suppose $n\ge 1$ and write $X=(B_{t_1},\dots ,B_{t_n})$ and $a=(a_1,\dots ,a_n)$. By [F1]--[F2], $X$ is multivariate normal with mean vector zero and covariance matrix $K_{ij}=\min (t_i,t_j)$, even if some coordinates repeat or are deterministic. Its projection $a\cdot X$ therefore has law $N\text{\hspace{0.17em}⁣}(0,a^{T}Ka)=N\text{\hspace{0.17em}⁣}(0,{\sum }_{i,j=1}^n{a}_i{a}_j\min (t_i,t_j))\mathrm{.}$

Independently, covariance bilinearity computes $\mathrm{Var}\text{\hspace{0.17em}⁣}\left({\sum }_{j=1}^n{a}_j{B}_{t_j}\right)={\sum }_{i,j=1}^n{a}_i{a}_j\mathrm{Cov}(B_{t_i},B_{t_j})={\sum }_{i,j=1}^n{a}_i{a}_j\min (t_i,t_j),$ confirming that the second parameter in step 1.1 is the actual variance. It is nonnegative by [F4], including when cancellations make it zero; in that case [F2] interprets the law as the point mass $N(0,0)$.

If $n=0$, the sum and the double sum are both empty and equal zero, so the random variable is the constant zero and has law $N(0,0)$ by [F2]. Zero coefficients, $t_j=0$, and repeated times require no deletion and are already covered by the possibly singular matrix in steps 1.1--2.1. AC is used only through [F1]--[F2]; the finite algebra and the positive-semidefinite calculation make no further choice.

Fix $n\ge 1$, times $t_1,\dots ,t_n\in [0,1]$, and coefficients $a_1,\dots ,a_n$. Then ${\sum }_{j=1}^n{a}_j{\beta }_{t_j}={\sum }_{j=1}^n{a}_j{B}_{t_j}-\left({\sum }_{j=1}^n{a}_j{t}_j\right)B_1\mathrm{.}$ This is a finite linear combination of Brownian values, with time $1$ appended if necessary, so [F1] makes it normal; repeated occurrences of time $1$, repeated $t_j$, and zero coefficients are allowed. Its mean is zero by finite linearity because all Brownian values are centered. Hence $\beta$ is a centered Gaussian process.

For $s,t\in [0,1]$, covariance bilinearity gives $\mathrm{Cov}({\beta }_s,{\beta }_t)=\min (s,t)-t\min (s,1)-s\min (1,t)+st\mathrm{Var}(B_1)\mathrm{.}$ Since $s,t\le 1$ and $\mathrm{Var}(B_1)=1$, this is $\min (s,t)-ts-st+st=\min (s,t)-st$.

Let $A$ be the probability-one event on which $t\mapsto B_t(\omega )$ is continuous on $[0,\mathrm{\infty })$, and let $A_0=\{B_0=0\}$. Their intersection has probability one by [F5]. For $\omega \in A\cap A_0$, the map $t\mapsto t{B}_1(\omega )$ is continuous and [F4] makes $t\mapsto {\beta }_t(\omega )$ continuous on $[0,1]$. On this event ${\beta }_0=B_0=0$, while for every $\omega$ one has ${\beta }_1=B_1-B_1=0$.

Steps 1.1--1.3 establish Gaussianity, centering, the covariance, path continuity, and both endpoints. The cases $s=0$, $t=0$, $s=t$, and $s=t=1$ follow directly from the same covariance formula, including its zero endpoint variances. The empty finite-dimensional list, if admitted, has the unique empty-tuple law. AC is used only through [F1]; the deterministic linear transformation and continuity argument make no new choice.

For $m\ge 1$ and $t\ge 0$, put $R_m(t)=\frac{t}{m}{\sum }_{k=0}^{m-1}{B}_{kt\mathrm{/}m}\mathrm{.}$ This is a measurable random variable by [F2]. On $A$, [F3] makes $R_m(t)(\omega )$ converge to the displayed Riemann integral as $m\to \mathrm{\infty }$; for $t=0$, every sum and the integral are zero. Define $X_t$ by this limit on $A$ and by zero on $A^c$. It is measurable: the convergence set and limsup of the measurable sequence are measurable by [F2], and pasting that finite limit on the measurable set $A$ with zero on its complement preserves every Borel inverse image. Thus the statement defines a real stochastic process rather than an integral that might be undefined on exceptional paths.

For $s,t>0$, covariance bilinearity and [F1] give $C_m(s,t):=\mathrm{Cov}(R_m(s),R_m(t))=\frac{st}{m^2}{\sum }_{k,\mathrm{\ell }=0}^{m-1}\min (ks\mathrm{/}m,\mathrm{\ell }t\mathrm{/}m)\mathrm{.}$ This is the lower-corner product-grid Riemann sum for the continuous function $(u,v)\mapsto \min (u,v)$ on $[0,s]\times [0,t]$. Its mesh tends to zero, so [F4] gives $C_m(s,t)⟶C(s,t):={\int }_0^s{\int }_0^t\min (u,v)dvdu\mathrm{.}$ If $s=0$ or $t=0$, both $C_m(s,t)$ and the declared degenerate-rectangle integral are zero, so the same conclusion holds.

Fix $n\ge 1$, times $t_1,\dots ,t_n\ge 0$, coefficients $a_1,\dots ,a_n$, and set $Y_m={\sum }_i{a}_i{R}_m(t_i)$ and $Y={\sum }_i{a}_i{X}_{t_i}$. By [F1], $Y_m$ is centered normal. Step 1.2 and covariance bilinearity show that its variance $V_m={\sum }_{i,j=1}^n{a}_i{a}_j{C}_m(t_i,t_j)$ converges to $V={\sum }_{i,j=1}^n{a}_i{a}_{j}C(t_i,t_j)\mathrm{.}$ On $A$, step 1.1 gives $Y_m\to Y$, hence the convergence is almost sure. In particular, $V={\lim }_m{V}_m\ge 0$.

Now let $0<s\le t$. Every section of $\min (u,v)$ is continuous, so [F4] writes $C(s,t)={\int }_0^s({\int }_0^{u}vdv+{\int }_u^{t}udv)du={\int }_0^s(ut-\frac{u^2}{2})du\mathrm{.}$ The polynomial antiderivatives justified by [F9] give $C(s,t)=\frac{t{s}^2}{2}-\frac{s^3}{6}=\frac{s^2(3t-s)}{6}\mathrm{.}$ For $s=0$ both the double integral and polynomial are zero by their endpoint conventions.

For each real $z$, [F6] gives $E[e^{iz{Y}_m}]=e^{-z^2{V}_m\mathrm{/}2}$. The left side converges to $E[e^{izY}]$ by [F7], because $e^{iz{Y}_m}\to e^{izY}$ almost surely and every modulus is one; the right side converges to $e^{-z^{2}V\mathrm{/}2}$. By [F8], $Y\sim N(0,V)$. Since the finite list and coefficients were arbitrary, $X$ is a centered Gaussian process. This argument proves Gaussian closure from the actual almost-sure Riemann-sum limit; it does not assume that arbitrary pointwise limits of Gaussian variables remain Gaussian.

Apply step 3.1 to the singleton coefficients and to $(X_s+X_t)$. It gives $\mathrm{Var}(X_r)=C(r,r)$ and $\mathrm{Var}(X_s+X_t)=C(s,s)+2C(s,t)+C(t,t)$. Covariance bilinearity also gives $\mathrm{Var}(X_s+X_t)=\mathrm{Var}(X_s)+2\mathrm{Cov}(X_s,X_t)+\mathrm{Var}(X_t)$. Comparing and cancelling proves $\mathrm{Cov}(X_s,X_t)=C(s,t)$, including $s=0$ or $t=0$.

Steps 1.1--4.1 prove every claim. Repeated times, zero coefficients, and singular linear combinations are retained in steps 2.1 and 3.1; $t=0$ is handled without a nondegenerate rectangle, and the empty finite list has the unique empty-tuple law. Changing the chosen probability-one continuity event changes $X_t$ only on a null set for each $t$, so the asserted finite laws and covariance are unaffected. AC is used exactly through [F1], [F3], [F6], and [F8], including the countable-choice input inherited by the continuous-integrability interface in [F3]; the fixed uniform left-endpoint sums, pasting, and finite algebra require no further choices.

Write $B_t=(B_t^1,\dots ,B_t^d)$. By [F1], each $B_t^{\alpha }$ has law $N(0,t)$, so [F6] gives $E[(B_t^{\alpha }{)}^2]=t$. The coordinate-square sum in [F5] and finite $L^1$ linearity in [F7] therefore give $E\parallel B_t{\parallel }_2^2={\sum }_{\alpha =1}^{d}E[(B_t^{\alpha }{)}^2]=dt\mathrm{.}$ Thus $\parallel B_t{\parallel }_2^2$ and $M_t$ are integrable. The map $x\mapsto \parallel x{\parallel }_2^2$ is continuous, hence Borel by [F5]; since $B_t$ is an observation generating ${\mathcal{F}}_t$, both $\parallel B_t{\parallel }_2^2$ and $M_t$ are ${\mathcal{F}}_t$-measurable. Consequently $M$ is adapted. At $t=0$, the same calculation gives expectation zero although $B_0=0$ is only an almost-sure identity.

Fix $0\le s<t$ and put $\mathrm{\Delta }=B_t-B_s$. Let ${\mathrm{\Pi }}_s$ consist of $\mathrm{\Omega }$ and all finite intersections ${\bigcap }_{j=1}^m\{B_{u_j}\in A_j\},0\le u_j\le s,A_j\in \mathcal{B}({\mathbb{R}}^d)\mathrm{.}$ It is a pi-system and its generated sigma-algebra is ${\mathcal{F}}_s$ by [F2]. Given one such cylinder, sort its distinct positive observation times and adjoin $0,s,t$. On the probability-one event $\{B_0=0\}$, every displayed past observation is a finite cumulative sum of the vector increments ending no later than $s$. Those increments and $\mathrm{\Delta }$ are disjoint groups of the mutually independent increment family in [F1]; [F3] makes their grouped vectors, and then the past observation tuple and $\mathrm{\Delta }$, independent. Replacing the tuple by its almost-surely equal cumulative-sum expression changes the cylinder event by a null set, so for every Borel $C\subseteq {\mathbb{R}}^d$ and every $A\in {\mathrm{\Pi }}_s$, $P(A\cap \{\mathrm{\Delta }\in C\})=P(A)P(\mathrm{\Delta }\in C)\mathrm{.}$ Empty cylinders give $\mathrm{\Omega }$. If $s=0$, every finite past tuple is almost surely the constant zero tuple, and the same null-set argument gives the identity.

Fix a Borel $C\subseteq {\mathbb{R}}^d$ and let ${\mathrm{\Lambda }}_C$ be the events $A\in {\mathcal{F}}_s$ satisfying the factorization in step 1.2. The class contains $\mathrm{\Omega }$; finite differences of nested members follow by subtracting the two finite probability identities, and increasing countable unions follow from probability continuity in [F4]. Thus ${\mathrm{\Lambda }}_C$ is a lambda-system containing ${\mathrm{\Pi }}_s$. By [F4], it contains $\sigma ({\mathrm{\Pi }}_s)={\mathcal{F}}_s$. Since $C$ was arbitrary, $\mathrm{\Delta }$ is independent of ${\mathcal{F}}_s$. For $s=t$, $\mathrm{\Delta }=0$ identically and the same independence conclusion is immediate.

Let $h=t-s\ge 0$. By [F1], the coordinates of $\mathrm{\Delta }$ have laws $N(0,h)$; [F6] and [F7] give $E[{\mathrm{\Delta }}^{\alpha }\mid {\mathcal{F}}_s]=0,E[\parallel \mathrm{\Delta }{\parallel }_2^2\mid {\mathcal{F}}_s]=E\parallel \mathrm{\Delta }{\parallel }_2^2=dh\mathrm{.}$ Indeed each scalar coordinate and the squared norm are Borel functions of the independent vector from step 2.1, and the squared norm is integrable by [F5]--[F7]. Also $B_s^{\alpha }$ and ${\mathrm{\Delta }}^{\alpha }$ are square-integrable by [F1] and [F6], so [F8] makes their product integrable. Because $B_s^{\alpha }$ is ${\mathcal{F}}_s$-measurable, taking out what is known yields $E[B_s^{\alpha }{\mathrm{\Delta }}^{\alpha }\mid {\mathcal{F}}_s]=B_s^{\alpha }E[{\mathrm{\Delta }}^{\alpha }\mid {\mathcal{F}}_s]=0.$ The formulas also hold when $h=0$, where $\mathrm{\Delta }=0$ pointwise.

The pathwise Euclidean identity $\parallel B_t{\parallel }_2^2=\parallel B_s{\parallel }_2^2+2{\sum }_{\alpha =1}^d{B}_s^{\alpha }{\mathrm{\Delta }}^{\alpha }+\parallel \mathrm{\Delta }{\parallel }_2^2$ and conditional linearity now give $E[\parallel B_t{\parallel }_2^2\mid {\mathcal{F}}_s]=\parallel B_s{\parallel }_2^2+d(t-s)\mathrm{.}$ Subtracting the deterministic $dt$ proves $E[M_t\mid {\mathcal{F}}_s]=\parallel B_s{\parallel }_2^2+d(t-s)-dt=M_s\mathrm{.}$ With the adaptation and integrability from step 1.1, this is exactly the all-pairs martingale definition in [F2].

Steps 1.1--4.1 prove both displayed claims. The case $d=1$ is Durrett's scalar square martingale; $d=0$ is excluded. Time zero, $s=0$, $s=t$, and $t=0$ are all covered, and no completed or right-continuously augmented filtration has been substituted for the stated natural filtration. There is no biconditional. AC is used only through [F1], [F6], and [F7]; the finite grid, pi-lambda promotion, and coordinate sum make no additional choices.

Take $I=[0,1]$. For each finite $F\subseteq I$, let ${\mu }_F$ be the uniform probability on $\{0,1{\}}^F$. Its mass is $2^{-\mathrm{\mid }F\mathrm{\mid }}$ at every point; for $F=\mathrm{\varnothing }$ this is the unique probability on the singleton $\{0,1{\}}^{\mathrm{\varnothing }}$. Marginalizing from $G$ to $F\subseteq G$ sums over $2^{\mathrm{\mid }G\setminus F\mathrm{\mid }}$ extensions and gives $2^{\mathrm{\mid }G\setminus F\mathrm{\mid }}{2}^{-\mathrm{\mid }G\mathrm{\mid }}=2^{-\mathrm{\mid }F\mathrm{\mid }}$, so the family is consistent. By [F1], under AC it has a probability extension on the cylinder sigma-algebra of $\mathrm{\Omega }=\{0,1{\}}^{[0,1]}$, and $X_t(\omega )=\omega (t)$ is a measurable coordinate process with these finite laws. For a finite $J\subseteq I$ and sets $A_j\subseteq \{0,1\}$, uniform counting gives $P\text{\hspace{0.17em}⁣}({\bigcap }_{j\in J}\{X_j\in A_j\})=\frac{{\prod }_{j\in J}\mathrm{\mid }{A}_j\mathrm{\mid }}{2^{\mathrm{\mid }J\mathrm{\mid }}}={\prod }_{j\in J}P(X_j\in A_j)\mathrm{.}$ This also gives $1$ for $J=\mathrm{\varnothing }$, by the empty-product convention. Since every subset of $\{0,1\}$ is measurable, [F1] makes the whole coordinate family independent, and each coordinate is a fair bit.

For $n\in \mathbb{N}$, put $q_n=1\mathrm{/}(n+1)$. These are distinct points of $(0,1]$. Given $\epsilon >0$, [F6] gives a positive integer $N$ with $1\mathrm{/}N<\epsilon$; whenever $n+1\ge N$, positivity and order reversal under inversion give $0<q_n\le 1\mathrm{/}N<\epsilon$. Hence $q_n\to 0$.

Put $A_n=\{X_{q_n}=1\}$. By the independence and fair laws in step 1.1, the events $(A_n)$ are pairwise independent and $P(A_n)=1\mathrm{/}2$. For each pair, [F2] also makes their complements $A_n^c=\{X_{q_n}=0\}$ independent, and $P(A_n^c)=1\mathrm{/}2$. Both probability series diverge because their first $m$ terms sum to $m\mathrm{/}2$. Applying [F2] twice gives probability-one events $E_1=\{A_n\text{ occurs infinitely often}\},E_0=\{A_n^c\text{ occurs infinitely often}\}\mathrm{.}$ Thus on $E_0\cap E_1$ the bit sequence $(X_{q_n})$ has infinitely many zeros and infinitely many ones.

Suppose for contradiction that $Y=(Y_t{)}_{t\in [0,1]}$ is a continuous modification of $X$: for some measurable event $C$ with $P(C)=1$, every path $t\mapsto Y_t(\omega )$ with $\omega \in C$ is continuous on $[0,1]$. By [F4], each $H_n=\{Y_{q_n}=X_{q_n}\}$ is measurable and has probability one. The complement of $H={\bigcap }_n{H}_n$ is the countable union of the null events $H_n^c$, so [F3] gives $P(H)=1$. A finite union bound likewise gives $P(C\cap H\cap E_0\cap E_1)=1$, so this intersection is nonempty. Fix $\omega$ in it.

The path $f(t)=Y_t(\omega )$ is continuous at the endpoint $0$. Since $q_n\to 0$, the choice-free forward implication in [F5] gives $Y_{q_n}(\omega )\to Y_0(\omega )$. But $\omega \in H\cap E_0\cap E_1$, so $Y_{q_n}(\omega )=X_{q_n}(\omega )$ for every $n$, with both values $0$ and $1$ occurring infinitely often. This sequence cannot converge: if it converged to $a$, its tail would eventually lie within $1\mathrm{/}3$ of $a$; a tail containing both $0$ and $1$ would then give $1\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }1-a\mathrm{\mid }<2\mathrm{/}3$, a contradiction. Therefore no such continuous modification $Y$ exists.

The witness has consistent finite laws and a genuine cylinder-space Kolmogorov extension, yet lacks a continuous modification, which refutes the statement. The empty finite support was checked in step 1.1; $t=0$ is the continuity endpoint in step 4.1; $t=1=q_0$ and the values $0,1$ occur in the construction; repeated coordinates are handled by the coordinate process rather than treated as independent copies. There is no biconditional. AC is used exactly through the arbitrary-index extension invoked in step 1.1, while Borel--Cantelli, the fixed sequence, and the countable intersection add no choice.

Let $\mathcal{F}=\{A\cap \mathrm{\Omega }:A\in \mathcal{L}(\mathbb{R})\}$ and define $P(E)={\lambda }_1(E)$ for $E\in \mathcal{F}$. Because $\mathrm{\Omega }$ is Lebesgue measurable, every $E\in \mathcal{F}$ is Lebesgue measurable. The trace family $\mathcal{F}$ contains $\mathrm{\Omega }$, is closed under complements relative to $\mathrm{\Omega }$ and under countable unions, so it is a sigma-algebra. Countable additivity of $P$ is inherited from ${\lambda }_1$, and $P(\mathrm{\Omega })={\lambda }_1([0,1])=1$ by [F1]. Thus $(\mathrm{\Omega },\mathcal{F},P)$ is a probability space by [F3]; this is normalized Lebesgue measure on the interval.