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When the drift bound is a deterministic summable sequence ${\delta }_n\ge 0$, the displayed correction is a nonnegative supermartingale. Its added tail tends to zero, so the almost sure supermartingale convergence theorem directly implies finite almost sure convergence of $X_n$. This simpler tail correction requires deterministic bounds; an arbitrary future random drift tail is not generally measurable at the current time.

Suppose $X_n\ge 0$ is an integrable adapted process, $Y_n\ge 0$ is finite and adapted, and $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]\le (1+Y_n)X_n$. The displayed correction is a nonnegative supermartingale: its next denominator is measurable at time $n$, and its reciprocal is at most one. If ${\sum }_n{Y}_n<\infty$ almost surely, the denominators converge to a finite positive limit since $\log (1+y)\le y$. The almost sure supermartingale convergence theorem then gives finite almost sure convergence of $X_n$. No integrability of the infinite product or of ${\sum }_n{Y}_n$ is needed.

A convenient precise version is the almost sure submartingale convergence theorem: if $(S_n)$ is a real submartingale and ${\sup }_n\mathbb{E}{S}_n^{+}<\infty$, then $S_n$ converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|S_n|<\infty$. It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.

Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed upcrossings by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\le a$, exit when $S_k\ge b$. Witheach completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a{)}^{-}$. Hence $G_n\ge (b-a)U_n[a,b]-(S_n-a{)}^{-}$. The submartingale property and $0\le H_k\le 1$ give $\mathbb{E}{G}_n\le \mathbb{E}(S_n-S_0)$: the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since $(S_n-a{)}^{+}\le S_n^{+}+|a|$. By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso ${\sup }_n\mathbb{E}|S_n|<\infty$. The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to $-Z_n$ gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.

For the given adapted multiplicative drift, putThe next denominator $P_{n+1}$ is ${\mathcal{F}}_n$-measurable, and $P_n\ge 1$ makes $Z_n$ integrable. ThusThe nonnegative supermartingale $Z_n$ has a finite almost sure convergence limit $Z_{\infty }$. On the probability-one event where the specified series is finite,Consequently $P_n\uparrow P_{\infty }\in [1,\infty )$ and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.

Use the natural filtration ${\mathcal{G}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$. Define the deterministic tail $r_n={\sum }_{j=n}^{\infty }{\delta }_j$ and put $Z_n={\xi }_n+r_n$. This is nonnegative and integrable. Since $r_n={\delta }_n+r_{n+1}$,Thus the deterministic summable drift correction produces a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives a finite limit $Z_{\infty }$ almost surely. Its finiteness is also reflected in the bound $\mathbb{E}{Z}_n\le \mathbb{E}{Z}_1$ and Fatou's lemma, which gives $\mathbb{E}{Z}_{\infty }\le \mathbb{E}{Z}_1<\infty$. The convergence theorem follows from the upcrossing argument: bounded expected negative parts of $-Z$ rule out infinitely many crossings of rational intervals, and the expectation bound excludes an infinite limit.

Since $r_n\to 0$, the required conclusion is $\boxed{{\xi }_n⟶Z_{\infty }\text{ almost surely, with }{Z}_{\infty }<\infty }$. The original sequence need not itself be a supermartingale; subtracting the vanishing deterministic tail recovers it from one that is.

Work from time one; a finite initial prefix does not affect convergence. Fix a positive integer $a$, and use the stopping time ${\tau }_a$ and budget $A_n$ from part (b). DefineIt is nonnegative. For each finite $n$, the stopped variable $X_{n\wedge {\tau }_a}$ is integrable: it is bounded by the finite sum $X_1+\cdots +X_n$, and the budget is bounded by $a$.

The stopped increment of $X$ isTaking conditional expectations, the assumed drift bound givesThe last term is bounded by $a$; no integrability assumption on the unlocalized $Y_n$ has been used. Its contribution is exactly cancelled by the budget decrement from part (b), givingThus $Z^{(a)}$ is an integrable nonnegative supermartingale. By the almost sure supermartingale convergence theorem it has a finite almost-sure limit. The bounded decreasing budget $A_n$ has a finite pathwise limit as well, so $X_{n\wedge {\tau }_a}=Z_n^{(a)}-A_n$ converges to a finite limit almost surely.

Take the intersection of these probability-one convergence events over all positive integers $a$. On the further probability-one event ${\sum }_{j\ge 1}{Y}_j<\infty$, choose an integer $a$ strictly larger than that total. Then ${\tau }_a=\infty$, so the stopped sequence is the original one. ConsequentlyThis proves supermartingale convergence with summable adapted drift under almost-sure summability alone; it does not require $\mathbb{E}{\sum }_j{Y}_j<\infty$.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.

For every rational $0\le a<b$, part a implies $U[a,b]<\infty$ almost surely. The intersection of these probability-one events over the countable collection of rational pairs still has probability one. On this event, ifsome rational interval $[a,b]$ lies strictly between them, forcing infinitely many upcrossings, a contradiction. Thus $X_n$ has an extended limit almost surely.

The limit cannot be $+\infty$ on a set of positive probability: Fatou lemma and the supermartingale property giveNonnegativity excludes $-\infty$. Therefore $X_n$ converges almost surely to a finite random variable, proving the almost sure supermartingale convergence theorem in this case.

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

Let $A_n=\{X_m=0\text{ for some }m\ge n\}$. By the Markov property,The events $A_n$ decrease to the event that the walk visits zero infinitely often, which has probability zero by the stated transience assumption. HencePart a and the almost sure supermartingale convergence theorem give an almost-sure limit $V\ge 0$. By Fatou lemma, $\mathbb{E}V\le {\mathrm{liminf}}_n\mathbb{E}[v(X_n)]=0$, so $V=0$ almost surely.

The almost sure supermartingale convergence theorem says that a supermartingale whose negative parts have uniformly bounded expectations converges almost surely to a finite integrable limit. In particular, every nonnegative supermartingale converges almost surely.

Here the process is uniformly bounded, say $0\le X_n\le C$. Let $X_{\infty }={\lim }_n{X}_n$, whose existence follows from the theorem. The dominated convergence theorem then givesThe limit on the right also exists directly because the expectations of a supermartingale form a decreasing sequence.