Brownian Motion, Markov Properties and Hitting Times — Examples

Put $A=\frac{1}{2s}+\frac{1}{2t}=\frac{s+t}{2st}$, $m=\frac{tx+sz}{s+t}$ and $C=\frac{(z-x{)}^2}{2(s+t)}$. Expanding squares gives $\frac{(y-x{)}^2}{2s}+\frac{(z-y{)}^2}{2t}=A(y-m{)}^2+C$: the coefficient of $y^2$ is $A$, the coefficient of $-2y$ is $2Am=\frac{z}{t}+\frac{x}{s}$, and subtracting the square leaves the constant $\frac{x^2}{2s}+\frac{z^2}{2t}-\frac{(z\mathrm{/}t+x\mathrm{/}s{)}^2}{4A}$, which equals $\frac{(z-x{)}^2}{2(s+t)}$.

Consequently $p_s(x,y)p_t(y,z)=\frac{1}{2\pi \sqrt{st}}{e}^{-C}{e}^{-A(y-m{)}^2}$ for every $y$, a nonnegative continuous function of $y$.

For $L>0$, the affine substitution $u=\sqrt{A}(y-m)$ on $[m-L\mathrm{/}\sqrt{A},m+L\mathrm{/}\sqrt{A}]$ and [F2] give ${\int }_{m-L\mathrm{/}\sqrt{A}}^{m+L\mathrm{/}\sqrt{A}}{e}^{-A(y-m{)}^2}dy=\frac{1}{\sqrt{A}}{\int }_{-L}^L{e}^{-u^2}du$; letting $L\to \mathrm{\infty }$ with [F2] gives ${\int }_{\mathbb{R}}{e}^{-A(y-m{)}^2}dy=\frac{\sqrt{\pi }}{\sqrt{A}}=\sqrt{\frac{2\pi st}{s+t}}$.

Multiplying by the constant of step 2.1, ${\int }_{\mathbb{R}}{p}_s(x,y)p_t(y,z)dy=\frac{1}{2\pi \sqrt{st}}\sqrt{\frac{2\pi st}{s+t}}{e}^{-C}=\frac{1}{\sqrt{2\pi (s+t)}}{e}^{-(z-x{)}^2\mathrm{/}(2(s+t))}=p_{s+t}(x,z)$, which is the displayed identity; the semigroup identity for the operators follows from it as in the cited lemma.

The degenerate cases are excluded or harmless as stated: $s,t>0$ keeps $A$ finite and positive and all square roots real, the case $s=t$ and $x=z$ are included, and the cases $s=0$ or $t=0$ belong to the identity operator convention of the semigroup rather than to this convolution. AC is used only through [F3].

By [F1], $P(M_t\ge a)=1-P(M_t<a)=1-P(M_t\le a)=1-(2\mathrm{\Phi }(a\mathrm{/}\sqrt{t})-1)=2(1-\mathrm{\Phi }(a\mathrm{/}\sqrt{t}))$ for every $a>0$ and $t>0$, which is the displayed value.

As $a\downarrow 0$ one has $a\mathrm{/}\sqrt{t}\downarrow 0$, so continuity of $\mathrm{\Phi }$ at $0$ with $\mathrm{\Phi }(0)=1\mathrm{/}2$ gives $2(1-\mathrm{\Phi }(a\mathrm{/}\sqrt{t}))\to 2(1-1\mathrm{/}2)=1$; as $a\to \mathrm{\infty }$ one has $a\mathrm{/}\sqrt{t}\to \mathrm{\infty }$ and $\mathrm{\Phi }\to 1$, so the value tends to $0$. Since $\mathrm{\Phi }$ is nondecreasing, the value is nonincreasing in $a$.

The boundary cases are consistent: at $a=0$ the formula would give $1$, while the stated range is $a>0$; the case $t>0$ is essential because the normalization $a\mathrm{/}\sqrt{t}$ uses a positive square root. AC is used only through [F3].

Since ${\tau }_a\ge 0$ and its law has density $g$ by [F1], [F2] gives $\mathbb{E}[{\tau }_a]={\int }_0^{\mathrm{\infty }}tg(t)dt=a(2\pi {)}^{-1\mathrm{/}2}{\int }_0^{\mathrm{\infty }}{t}^{-1\mathrm{/}2}{e}^{-a^2\mathrm{/}(2t)}dt$, the last integrand being nonnegative.

For $t\ge a^2$ the exponent satisfies $a^2\mathrm{/}(2t)\le 1\mathrm{/}2$, so $e^{-a^2\mathrm{/}(2t)}\ge e^{-1\mathrm{/}2}$; hence the integrand in step 1.1 is bounded below on $[a^2,\mathrm{\infty })$ by $a(2\pi {)}^{-1\mathrm{/}2}{e}^{-1\mathrm{/}2}{t}^{-1\mathrm{/}2}$.

For every integer $j\ge 0$, on $I_j=[2^j{a}^2,2^{j+1}{a}^2]$ one has $t^{-1\mathrm{/}2}\ge (2^{j+1}{a}^2{)}^{-1\mathrm{/}2}$, so ${\int }_{I_j}{t}^{-1\mathrm{/}2}dt\ge a{2}^{(j-1)\mathrm{/}2}$. These lower bounds do not tend to zero and their partial sums diverge. Monotone convergence over the increasing finite unions of the $I_j$, together with step 2.1, therefore gives ${\int }_0^{\mathrm{\infty }}{t}^{-1\mathrm{/}2}{e}^{-a^2\mathrm{/}(2t)}dt=+\mathrm{\infty }$ and hence $\mathbb{E}[{\tau }_a]=+\mathrm{\infty }$.

The comparison with finite almost-sure values is the point of the example: [F1] gives ${\tau }_a<\mathrm{\infty }$ almost surely, so the random variable is finite-valued almost surely while its expectation is infinite; the divergence comes from the polynomial tail $t^{-1\mathrm{/}2}$ of the first-moment integrand and not from any exceptional path. The cases $a=0$ (where ${\tau }_0=0$) and $a<0$ are excluded by the hypothesis $a>0$. AC is used only through [F4].

Apply [F1] with $a=-2$, $x=0$, $b=3$: $P_0(T_3<T_{-2})=\frac{0-(-2)}{3-(-2)}=\frac{2}{5}$.

The events $\{T_3<T_{-2}\}$ and $\{T_{-2}<T_3\}$ are disjoint, and their union has probability one because both hitting times are finite almost surely by [F4] and the process cannot be at both levels at once; hence $P(T_{-2}<T_3)=1-\frac{2}{5}=\frac{3}{5}$.

The values $\frac{2}{5}+\frac{3}{5}=1$ sum to one, the starting point $0$ lies strictly between the endpoints, and the common denominator $b-a=5$ is nonzero. AC is used only through [F3].

The normalized process X has measurable coordinates, all its paths are continuous, and X_0=0 everywhere. It retains every finite-dimensional Brownian law, since its coordinates were changed on one measurable null event. By [F4] it is a C-valued random element, supported on the closed subspace C_0 of paths starting at zero. Work with the usual augmentation of this process, as defined in [F2], when applying strong Markov. This does not transfer a stopping-time claim to the original B filtration.

For w in C_0 define $\sigma (w)=\inf \{t\ge 0:\mathrm{\mid }w(t)\mathrm{\mid }=1\}$. This is measurable: for t>=0, continuity gives $\{\sigma \le t\}=\{{\sup }_{0\le s\le t}\mathrm{\mid }w(s)\mathrm{\mid }\ge 1\}$, and the supremum is over the countable set of rationals in [0,t] together with t. Continuity, the intermediate value property and attainment of the maximum on a compact interval prove this equality. It also shows sigma is strictly positive on C_0. For a continuous Brownian W, [F2] makes sigma(W) a stopping time. If sigma(W)>m, then |W_m|<1; hence [F1] gives $P(\sigma >m)\le P(\mathrm{\mid }{W}_m\mathrm{\mid }<1)\le 2\mathrm{/}\sqrt{2\pi m}$. Continuity from above in [F5] proves sigma finite almost surely. At a finite sigma, continuity gives $W_{\sigma }\in \{-1,1\}$.

Define the segment map $e(w)=(\sigma (w),(w(t\wedge \sigma (w)){)}_{t\ge 0})$ into $[0,\mathrm{\infty }]\times C$, and the remainder $r(w)(t)=w(\sigma (w)+t)-w(\sigma (w))$ if sigma is finite, and the zero path otherwise. Both are Borel maps. Indeed evaluation (w,s) maps to w(s) continuously for finite s: if w_j converges uniformly on compact sets and s_j tends to s, bound by the uniform error on one common compact interval plus continuity of w there. Compose with measurable sigma for each fixed coordinate, using t wedge infinity=t and the zero convention for r; then [F4] proves path-valued measurability. The duration sigma is measurable by step 2.1.

For a continuous Brownian W and its sigma, e(W) is measurable for the stopped sigma-algebra. Sigma itself is measurable there by its defining tests. For a fixed t and u, on {sigma<=u} the value W_{t wedge sigma} equals W_{t wedge sigma wedge u}, which is measurable at time u: approximate t wedge sigma wedge u from below by a finite grid in [0,u], use adaptation and continuity. Thus the Borel inverse images of W_{t wedge sigma}, intersected with {sigma<=u}, lie in F_u. Rational coordinates and [F4] finish this claim. By [F3], r(W) is independent of F_sigma and has Brownian finite-dimensional laws. It is C_0-valued by construction; [F4] and the pi-lambda theorem identify its C_0 law with that of X and promote coordinate independence to independence of every Borel path event. Therefore e(W) and r(W) are independent, and r(W) again has the law of X.

Set $W^0=X$, $W^{n+1}=r(W^n)$, ${\sigma }_n=\sigma (W^n)$, $E_n=e(W^n)$, and $S_n={\sum }_{j<n}{\sigma }_j$. These are measurable by step 3.1. Inductively each W^n has the law of X by step 4.1, so each sigma_n is finite and strictly positive almost surely by step 2.1. Their countable intersection is a measurable full event by [F5]. On it, finite induction gives $W_t^n=X_{S_n+t}-X_{S_n}$ simultaneously for all t, and hence the formulas for S_n and E_n in the statement. Outside this event the just-defined measurable E_n provide the promised convention; S_n are extended sums, so subsequent times after infinity stay infinite.

Prove by induction that W^n is independent of H_n=sigma(E_0,...,E_{n-1}) and that the preceding E_j are iid with law nu=law(e(X)). For n=0 this is vacuous. If it holds at n, the pair (e(W^n),r(W^n)) is independent of H_n, since it is a measurable function of W^n. Its components are independent by step 4.1. Thus for A in H_n and Borel segment and path sets D,L, $P(A\cap \{E_n\in D\}\cap \{W^{n+1}\in L\})=P(A)\nu (D)P(X\in L)\mathrm{.}$ For fixed L, the pi-lambda theorem extends this identity from intersections A intersect {E_n in D} to all of H_{n+1}. It follows that W^{n+1} is independent of H_{n+1}; taking L to be the whole path space also proves E_n independent of H_n with law nu. Induction proves mutual independence of every finite family of segments, hence the asserted iid sequence. It does not assert independence of the nested entire future processes W^n.

Negation preserves the law of X on C_0: centered independent normal increments are unchanged jointly under sign reversal by the even normal density and reflection substitution in [F6], so finite-dimensional laws agree, and [F4] and pi-lambda give equality of path laws. The map d(w)=w(sigma(w)) on finite sigma, zero otherwise, is Borel by the evaluation argument in step 3.1. Also sigma(-w)=sigma(w) and d(-w)=-d(w). Step 2.1 makes its value a sign almost surely, so its two probabilities are equal and sum to one. Each displacement d(W^n) is a function of E_n: for finite duration it is the stopped path evaluated at that duration. Step 6.1 therefore makes these displacements independent fair signs, including the first d(W^0)=X_{S_1}. Telescoping gives X_{S_n}=sum_{j<n}d(W^j) on the full event in step 5.1, proving the embedded-walk claim.

At time zero X starts at zero and each duration is strictly positive; at exit the value is exactly one of the two endpoints. Infinite durations lie in a measurable null event and have explicit segment/remainder conventions. The empty history in the induction is the trivial sigma-algebra. The claims concern exit segments and signs; no inference about moments of a two-sided exit time is made from a one-sided hitting-time density. Full AC is inherited through [F5]; normalization uses one supplied full event, not a choice of paths.

Fix $R>\mathrm{\mid }x-y\mathrm{\mid }$. For every $n$ with $1\mathrm{/}n<\mathrm{\mid }x-y\mathrm{\mid }$, the event $\{T_y<T_R\}$ is contained in $\{S_{1\mathrm{/}n}<T_R\}$: a path that reaches $y$ before leaving the disc of radius $R$ passes through the circle of radius $1\mathrm{/}n$ about $y$ first, by continuity. Hence by [F1], $P_x(T_y<T_R)\le {\lim }_n\frac{\log R-\log \mathrm{\mid }x-y\mathrm{\mid }}{\log R-\log (1\mathrm{/}n)}=0$, the denominator tending to $+\mathrm{\infty }$.

For a given $\delta >0$ choose $0<\epsilon <\min \{\delta ,\mathrm{\mid }x-y\mathrm{\mid }\}$; then $P_x(S_{\epsilon }<T_R)=\frac{\log R-\log \mathrm{\mid }x-y\mathrm{\mid }}{\log R-\log \epsilon }\to 1$ as $R\to \mathrm{\infty }$, so the $\epsilon$-circle about $y$ is hit almost surely, hence the disc of radius $\delta$ about $y$ is hit almost surely. Applying this to the countably many discs with rational centre and rational radius and intersecting the resulting probability-one events via [F3], while every nonempty open disc contains such a rational disc by [F4], gives the almost-sure statement of assertion 3.

By [F2] each coordinate increment $W^{(i)}-x_i$ is a standard one-dimensional Brownian motion, so it visits every neighbourhood of the level $y_i-x_i$ at arbitrarily large times; equivalently, $W^{(i)}$ visits every neighbourhood of $y_i$ at arbitrarily large times almost surely. This is assertion 1, and it does not synchronize the two coordinates.

The event $\{T_y<\mathrm{\infty }\}$ is the union over the countably many integers $R>\mathrm{\mid }x-y\mathrm{\mid }$ of the increasing events $\{T_y<T_R\}$: if $T_y<\mathrm{\infty }$ then the path on $[0,T_y]$ is a compact subset of ${\mathbb{R}}^2$, hence stays in some disc of integer radius about $y$, and conversely $T_y<T_R<\mathrm{\infty }$ implies $T_y<\mathrm{\infty }$. By [F3] and step 1.1, $P_x(T_y<\mathrm{\infty })\le {\sum }_R{P}_x(T_y<T_R)=0$.

By step 2.1 there is a probability-one event on which the planar path never equals $y$; on that event no time can satisfy both coordinate equations simultaneously, so $\{t:W_t^{(1)}=y_1\}\cap \{t:W_t^{(2)}=y_2\}=\mathrm{\varnothing }$, even though each of the two sets is almost surely unbounded by step 1.3. This is assertion 2.

The cases are consistent: the point $x=y$ is excluded, so $T_y=0$ is not possible; the dimension is two, and the polarity statement is not asserted in dimension one, where the same computation fails because the logarithm is replaced by a bounded harmonic function; the radius parameter $\delta >0$ and the inner radius $\epsilon$ are chosen strictly positive. AC is used only through [F5].

Define $A:={\bigcup }_{m\ge 1}{\bigcap }_{s\in [0,1\mathrm{/}m]}\{{\pi }_s=0\}={\bigcup }_{m\ge 1}{\bigcap }_{q\in \mathbb{Q}\cap [0,1\mathrm{/}m]}\{{\pi }_q=0\}$, the second description using continuity of every path; thus $A$ is measurable and nonempty, since the zero path belongs to it, while the path $s\mapsto s$ does not.

For every $t>0$ one has $A\in {\mathcal{F}}_t^0$: choose $m$ with $1\mathrm{/}m<t$; then ${\bigcap }_{q\in \mathbb{Q}\cap [0,1\mathrm{/}m]}\{{\pi }_q=0\}\in {\mathcal{F}}_{1\mathrm{/}m}^0\subseteq {\mathcal{F}}_t^0$, and the union over $m$ lies in ${\mathcal{F}}_t^0$. Hence $A\in {\mathcal{F}}_{0+}^0={\bigcap }_{t>0}{\mathcal{F}}_t^0$ The Brownian germ sigma-algebra at zero.

$A\mathrm{\notin }{\mathcal{F}}_0^0=\sigma ({\pi }_0)$: the zero path is in $A$ and the path $s\mapsto s$ is not, while both have ${\pi }_0=0$, so membership in $A$ is not determined by ${\pi }_0$.

$\mathsf{W}(A)=0$: $A$ is contained in ${\bigcup }_{m\ge 1}\{{\pi }_{1\mathrm{/}(2m)}=0\}$, and each of those events has probability zero because the law of ${\pi }_{1\mathrm{/}(2m)}$ has the strictly positive density $p_{1\mathrm{/}(2m)}(0,\cdot )$ and is therefore atomless The Brownian kernels form a semigroup Standard normal and normal laws; countable subadditivity gives the claim Basic identities for a probability measure.

Combining steps 2.1, 2.2 and 2.3, $A\in {\mathcal{F}}_{0+}^0\setminus {\mathcal{F}}_0^0$, so the raw natural filtration is not right-continuous at time zero; its usual augmentation contains $A$ as a null event and is right-continuous by construction Natural and usual augmented Brownian filtrations. The witness is nonempty and of probability zero, so it is invisible to any probability computation alone. AC is used only through the ambient Brownian and Wiener interfaces.

The zero set $\{t\in [0,1]:B_t=0\}$ is closed and contains $0$, so $L$ is a maximum, not merely a supremum. Moreover $L<1$ almost surely: if $L=1$ then $1$ lies in the zero set and $B_1=0$, an event of probability zero because the law of $B_1$ has the strictly positive density $p_1(0,\cdot )$ and is atomless The Brownian kernels form a semigroup Standard normal and normal laws.

A Brownian path has a zero in every interval $(0,s]$ almost surely. By Brownian scaling Brownian scaling it suffices to treat $s=1$; a path with no zero in $(0,1]$ has constant sign there, so $P(\text{no zero in }(0,1])=2P(B_t>0\mathrm{\forall }t\in (0,1])\le 2P({\min }_{[0,1]}B\ge 0)$ by symmetry of the Brownian law, and since $-B$ is again a standard Brownian motion the maximum law Law of the Brownian maximum gives $P({\min }_{[0,1]}B\ge 0)=P({\max }_{[0,1]}(-B)\le 0)=2\mathrm{\Phi }(0)-1=0$.

The time $L$ is a measurable $[0,1]$-valued random variable and is $G_L$-measurable. First, for $0<a\le 1$, continuity gives $\{L<a\}=\{{\inf }_{a\le t\le 1}\mathrm{\mid }{B}_t\mathrm{\mid }>0\};$ the infimum equals the infimum over the countable dense set $(\mathbb{Q}\cap [a,1])\cup \{a,1\}$, so this event is measurable. Hence $L$ is measurable. For each fixed $s$, approximating $s\wedge L$ by finite-valued dyadic times and using path continuity shows that $B_{s\wedge L}$ is measurable, so $G_L$ is well defined. Next, almost surely $B$ is constant on no nondegenerate interval: otherwise some rational $u<v$ lie in a constant interval and $B_v-B_u=0$, while that increment has the atomless law $N(0,v-u)$; a countable union over rational pairs is still null Standard normal and normal laws. On this no-flat-interval event, for every $a>0$, $\{L<a\}={\bigcup }_{q\in \mathbb{Q},0\le q<a}{\bigcap }_{r\in \mathbb{Q},r\ge q}\{B_{r\wedge L}=B_{q\wedge L}\}\mathrm{.}$ Indeed, if $L<a$, choose $q\in (L,a)$; conversely, constancy of the stopped path after some $q<a$ forces $L\le q$, since $q<L$ would make $B$ constant on $[q,L]$. The displayed right side belongs to $G_L$, and the completed null ideal absorbs the exceptional flat-path set. Thus $\{L<a\}\in G_L$ for every $a$, proving the claimed $G_L$-measurability.

The path has no zero in $(L,1]$, so by continuity it has constant sign there; consequently the shifted path $t\mapsto B_{L+t}$ has no zero in $(0,1-L]$, an event of probability one, and $1-L>0$ almost surely.

The conditional law of the shifted increment process $W_t:=B_{L+t}-B_L$ given $G_L$ is not Wiener measure. Suppose it were. For each positive integer $n$, the event $H_n:=\{1-L\ge 1\mathrm{/}n\}$ belongs to $G_L$ by step 1.3, while the continuous-path event $C_n:=\{W\text{ has no zero in }(0,1\mathrm{/}n]\}$ is Borel: it is the intersection, over sufficiently large integers $k$, of the events that the infimum of $\mathrm{\mid }W\mathrm{\mid }$ on $[1\mathrm{/}k,1\mathrm{/}n]$ is positive, and each such infimum is determined by countably many rational evaluations. Step 1.2 gives Wiener measure zero to $C_n$. The assumed conditional Wiener law would therefore imply $P(C_n\cap H_n)=E[1_{H_n}P(C_n\mid G_L)]=0$. By step 2.1 the event $A:=\{W\text{ has no zero in }(0,1-L]\}$ has probability one, but $A\cap H_n\subseteq C_n\cap H_n$ for every $n$. Since $L<1$ almost surely, $A\subseteq {\bigcup }_{n\ge 1}(A\cap H_n)$ up to a null event, forcing $P(A)=0$, a contradiction. Hence the conditional law is not Wiener measure, and the future is not a Brownian motion independent of the stopped past at $L$.

The time $L$ is not a stopping time for the completed filtration. Suppose it were; then $\{L<t\}\in {\mathcal{F}}_t$ for every $t\in (0,1)$, and $\{L<t\}\cap \{B_t\mathrm{\ne }0\}=\{B_t\mathrm{\ne }0\}\cap \{\text{no zero in }[t,1]\}$. By the future-path theorem Future-path Markov property the conditional probability of the second factor given ${\mathcal{F}}_t$ is $q_t(B_t)$, where $q_t(y)$ is the probability that a Brownian motion started at $y$ has no zero in $[0,1-t]$; the first factor is ${\mathcal{F}}_t$-measurable, so $E[1_{\{L<t\}}{1}_{\{B_t\mathrm{\ne }0\}}\mathrm{\mid }{\mathcal{F}}_t]=1_{\{B_t\mathrm{\ne }0\}}{q}_t(B_t)$ almost surely Taking out what is known, while ${\mathcal{F}}_t$-measurability of $\{L<t\}$ also makes that conditional expectation $1_{\{L<t\}}{1}_{\{B_t\mathrm{\ne }0\}}$ Conditional expectation as an ae class Conditional expectation is unique almost surely.

For $y\mathrm{\ne }0$ the shifted hitting-time law gives $P_y(T_0\le 1-t)=2(1-\mathrm{\Phi }(\mathrm{\mid }y\mathrm{\mid }\mathrm{/}\sqrt{1-t}))\in (0,1)$ Distribution of a one-sided Brownian hitting time Brownian motion started at x, so $0<q_t(y)<1$, while $q_t(0)=0$ by step 1.2. Comparing the two expressions of step 3.2 on $\{B_t\mathrm{\ne }0\}$ forces $P(B_t\mathrm{\ne }0)=0$: where $L<t$ one would need $q_t(B_t)=1$, and where $L\ge t$ one would need $q_t(B_t)=0$, hence $B_t=0$. This contradicts the atomlessness of the law of $B_t$ Standard normal and normal laws, so $L$ is not a stopping time.

The witness therefore has all three claimed properties: $L\le 1$ is finite, it is $G_L$-measurable but not a stopping time, and the strong-Markov conclusion fails at it in the precise sense of step 3.1. The case $t=1$ is excluded in step 3.2 where $1-t>0$ is needed; the case of an interval without a zero is impossible by step 1.2; and the degenerate case $B_1=0$ is the null event excluded in step 1.1. AC is used only through the ambient Brownian, completion and conditional-expectation interfaces.

By One-dimensional Brownian motion hits every point almost surely, ${\tau }_a$ is a measurable $[0,\mathrm{\infty }]$-valued random time and the measurable event $N:=\{{\tau }_a=+\mathrm{\infty }\}$ has probability zero.

Define $T_a(\omega )={\tau }_a(\omega )$ for $\omega \mathrm{\notin }N$ and $T_a(\omega )=0$ for $\omega \in N$. Then $T_a$ takes values in $[0,\mathrm{\infty })\subset \mathbb{R}$ on every outcome. It is measurable: for $c<0$ the set $\{T_a\le c\}$ is empty, while for $c\ge 0$ it is $N\cup \{{\tau }_a\le c\}$. Thus $T_a$ is a real random variable in the sense of Random elements and real random variables, and in particular a real-valued random time.

The variables $T_a$ and ${\tau }_a$ differ only on the null event $N$. Consequently, for every Borel set $C\subseteq [0,\mathrm{\infty })$, the events $\{T_a\in C\}$ and $\{{\tau }_a\in C\}$ have symmetric difference contained in $N$, so they have the same probability. Hence $T_a$ has the density $a(2\pi t^3{)}^{-1\mathrm{/}2}{e}^{-a^2\mathrm{/}(2t)}$ from A Brownian hitting time has infinite mean, and the same supplier's calculation gives $E[T_a]=+\mathrm{\infty }$.

Therefore $T_a$ is finite on every outcome, hence finite almost surely, but is not integrable. This real-valued witness refutes the stated implication. The null-set modification is explicit, and AC is used only through the Brownian construction and the two cited hitting-time suppliers.