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Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

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.

A uniformly integrable martingale $M_n$ has almost sure convergence and convergence in L1 to an integrable random variable $M_{\infty }$, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.