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For standard three-dimensional Brownian motion started at $x$ with $r<|x|<R$,The harmonic function $x\mapsto 1/|x|$ gives a bounded stopped martingale in the annulus, so the optional stopping theorem gives the formula. Taking $R\to \infty$ yields ${\mathbb{P}}_x({\tau }_r<\infty )=r/|x|$ for $|x|>r$. This limit uses path continuity on the finite interval before a hit, rather than assuming transience to prove transience.

Define the stopped martingaleIt is nonnegative: before the first visit to $1$, the nearest-neighbour random walk is at most $0$, and afterwards the stopped position is $1$. Each $M_n$ is integrable, with $|M_n|\le n+1$. MoreoverThe indicator is ${\mathcal{F}}_n$-measurable and the increment has conditional mean zero, so $M_n$ is a martingale with $\mathbb{E}{M}_n=M_0=1$.

Part (a) shows $T_1<\infty$ almost surely, so every such path eventually has $M_n=0$. Thus $M_n\to 0$ almost surely. But $\mathbb{E}|M_n-0|=1$ for every $n$, so there is no convergence in L1. Any possible L1 convergence limit would agree with the almost sure limit. This is a nonnegative martingale with almost sure convergence but no convergence in L1, and hence it is not uniformly integrable.

The bounded optional stopping theorem gives $\mathbb{E}{X}_{T\wedge n}=\mathbb{E}{X}_0$. Decompose the difference from the terminal value:ConsequentlyThe first term tends to zero by hypothesis. The second tends to zero by the dominated convergence theorem, because $X_T$ is integrable and $T$ is finite with probability one. Thus the stopped martingale converges to $X_T$ with convergence in L1, which permits passage of expectations to the limit:The explicit tail condition supplies exactly the missing control for an unbounded stopping time.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

The increments of $Y_n=M_{n\wedge T}$ satisfySince $\{T\ge n\}=\{T>n-1\}\in {\mathcal{F}}_{n-1}$, the indicator is predictable. Part (a), applied to this bounded integrand, shows that $Y$ is a martingale. Equivalently, this is the stopped martingale theorem.

For an almost surely finite stopping time $T$, if $\mathbb{E}|M_T|<\infty$ and $\mathbb{E}[|M_n|1_{\{T>n\}}]\to 0$, then $M_{n\wedge T}\to M_T$ with convergence in L1. Thus L1 convergence implies uniform integrability makes the stopped martingale uniformly integrable.

For a simple symmetric random walk, the infinite mean first passage of a simple symmetric random walk gives a finite almost sure first passage $T_1$ to $1$. The stopped martingale $M_n=1-S_{n\wedge T_1}$ is nonnegative and eventually zero almost surely. Nevertheless $\mathbb{E}{M}_n=1$ for every $n$, so it has almost sure convergence but no convergence in L1. It is therefore not uniformly integrable. Before absorption the nearest-neighbour walk is at most zero, ensuring nonnegativity.