Strong Markov fails at a nonstopping random time

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.