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The required condition is uniform integrability, written entirely in terms of the given probability laws:Together with the given almost sure convergence, this implies L1 convergence to an integrable $M_{\infty }$. Passing to the limit in $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$, using the contraction property of conditional expectation in L1 space, gives $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For every bounded stopping time $S$, the same identity holds with ${\mathcal{F}}_S$: split an event of its stopping-time sigma-algebra according to $\{S=k\}$ and use the deterministic-time identity. In particular $M_{T\wedge n}=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_{T\wedge n}]$. These conditional expectations are uniformly integrable: for an integrable $Y$ and $Z=\mathbb{E}[Y\mid \mathcal{G}]$, the event $A=\{|Z|>K\}$ satisfies $\mathbb{P}(A)\le \mathbb{E}|Y|/K$, andFirst choose $R$, then $K$, uniformly in $\mathcal{G}$. Define $M_T=M_{\infty }$ on $\{T=\infty \}$. Now $M_{T\wedge n}\to M_T$ almost surely and in L1 space, so the bounded-time mean identities imply $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ even when $T$ can be infinite.

A standard Brownian motion is a real stochastic process $(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous sample paths almost surely, and independent increments such that $B_t-B_s$ has normal distribution $N(0,t-s)$ whenever $0\le s<t$. With its completed, right-continuous natural filtration $({\mathcal{F}}_t)$, its Strong Markov property is the following: for every almost-surely finite stopping time $T$, the processis a standard Brownian motion independent of ${\mathcal{F}}_T$. Here the stopping-time sigma-algebra isIn particular, for every bounded ${\mathcal{F}}_T$-measurable random variable $H$ and bounded Borel functional $G$ on the continuous path space,Thus the independence is from the entire stopped past, not only from $B_T$, and is a statement about the whole future path, not just a single increment.

To prove the Brownian reflection principle without assuming that a level is hit in finite time, fix $t>0$, $m>0$, and set $T_m=\inf \{s\ge 0:B_s\ge m\}$ and $R=T_m\wedge t$. Continuity of the sample paths makes $T_m$ a stopping time and gives $B_{T_m}=m$ whenever $T_m<\infty$. The stopping time $R$ is bounded. Reflect the path after $R$:By the Strong Markov property, the future increments after $R$ are an independent Brownian motion. Negating them preserves their probability distribution, by symmetry of their joint normal distributions. Hence the reflected process $\hat{B}$ has the same path law as $B$.

If $T_m>t$, reflection changes nothing on $[0,t]$ and ${\hat{B}}_t=B_t<m$. If $T_m\le t$, both paths first hit $m$ at the same time and ${\hat{B}}_t=2m-B_t$. Therefore, for $x\le m$, the following events agree path by path:Indeed, the right-hand endpoint is at least $m$, so the reflected path must have hit $m$ before or at $t$; this also forces the original path to have done so. The reflection relation then makes the endpoint inequality equivalent to $B_t\le x$. Equality cases at $T_m=t$ satisfy the same relation. Since ${\hat{B}}_t$ and $B_t$ have the same probability distribution,For $t=0$, both sides are zero, because $B_0=M_0=0$ and $2m-x\ge m>0$.

For $t>0$, set $x=m$ in this identity. Continuity implies $\{B_t>m\}\subseteq \{M_t\ge m\}$, and the normal distribution of $B_t$ has no atom at $m$. Splitting according to whether $B_t\le m$ givesBoth random variables are nonnegative. Their tails agree for all $m>0$, which determines their probability distributions, including the mass at zero. Thus the Brownian running maximum satisfies

For $x>0$ and $t>0$, continuity again gives $\{T_x\le t\}=\{M_t\ge x\}$. Since $B_t/\sqrt{t}$ is a standard normal random variable, the distribution function of the Brownian first-passage time isIt tends to one as $t\to \infty$ and to zero as $t\downarrow 0$, proving almost-sure finiteness and absence of an atom at zero. For the standard normal random variable $B_1$,The value of this ratio on the zero-probability event $\{B_1=0\}$ may be defined arbitrarily. Matching the distribution functions provesFinally, differentiate the distribution function on $t>0$, using ${\Phi }^{\prime }=\varphi$ and $\frac{d}{dt}(x/\sqrt{t})=-x/(2t^{3/2})$. The first-passage-time density isand is zero for $t\le 0$. Its integral is one by the endpoint limits of $F_{T_x}$, so no missing mass at an infinite hitting time is present.

A discrete-time martingale relative to a filtration $({\mathcal{F}}_n)$ is an adapted process $(M_n)$ with $\mathbb{E}|M_n|<\infty$ andIterating conditional expectation also gives $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ for $m\ge n$. It is bounded in $L^2$ when each $M_n$ is square integrable andThis is a uniform bound on the second moments, not a pathwise bound on all sample values.

The optional stopping theorem for bounded stopping times says that, if $\sigma \le \tau \le K$ for a deterministic finite $K$, thenHere ${\mathcal{F}}_{\sigma }$ is the stopping-time sigma-algebra. The second assertion follows by taking $\sigma =0$ and then expectations. An $L^2$ bound is not needed: integrability at the finitely many deterministic times suffices.

For a proof, both stopped values are integrable since $|M_{\tau }|,|M_{\sigma }|\le {\sum }_{j=0}^K|M_j|$. ExpandIf $A\in {\mathcal{F}}_{\sigma }$, then $A\cap \{\sigma \le k\}\in {\mathcal{F}}_k$ by the definition of the stopping-time sigma-algebra, and $\{\tau >k\}\in {\mathcal{F}}_k$ because $\tau$ is a stopping time. Thus the indicator of $A\cap \{\sigma \le k<\tau \}$ is ${\mathcal{F}}_k$-measurable. Multiplying the sum by $1_A$, taking expectations, and conditioning each summand on ${\mathcal{F}}_k$ gives zero, by the martingale increment property. Therefore $\mathbb{E}[1_A{M}_{\tau }]=\mathbb{E}[1_A{M}_{\sigma }]$ for every $A\in {\mathcal{F}}_{\sigma }$. Since $M_{\sigma }$ is ${\mathcal{F}}_{\sigma }$-measurable, this is exactly the claimed conditional expectation identity.

Put $Z_n=X_{n\wedge T}$. The stopped martingale in discrete time identityshows that $Z$ is a martingale: the indicator function is ${\mathcal{F}}_n$-measurable and the increment has zero conditional expectation. Integrability follows because $Z_n$ is selected from the finitely many integrable values $X_0,\dots ,X_n$.

For the bounded stopping time $\tau =n\wedge T$, the optional stopping theorem in its conditional form givesHere ${\mathcal{F}}_{\tau }$ is the stopping-time sigma-algebra. Indeed, for $A\in {\mathcal{F}}_{\tau }$, partition $A$ into $A\cap \{\tau =k\}\in {\mathcal{F}}_k$ and apply the martingale identity $\mathbb{E}[X_n\mid {\mathcal{F}}_k]=X_k$ on each piece. This proves the displayed conditional expectation identity directly.

Let $C={\sup }_n\mathbb{E}|X_n|<\infty$ and $A_n=\{|Z_n|>K\}\in {\mathcal{F}}_{n\wedge T}$. Conditional absolute-value domination and the Markov inequality give $\mathbb{P}(A_n)\le C/K$ and, for any $R>0$,First choose $R$ using the uniform integrability of $X$, and then choose $K$. The estimate is uniform in $n$, proving uniform integrability of a stopped uniformly integrable martingale. No finiteness assumption on $T$ is needed.

Recall the stopping-time sigma-algebra:For the proposed random time,The first set belongs to ${\mathcal{F}}_t$. Since $S\le T$, the second can be writtenwhich also belongs to ${\mathcal{F}}_t$. Therefore $U$ is a stopping time, as in pasting ordered stopping times. Moreover $S\le U\le T$, so $U$ is a bounded stopping time.