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Pointwise, an integer-valued nonnegative random variable has the representationwhen $T=k$, exactly the terms $n=0,\dots ,k-1$ equal one. Applying the Tonelli theorem, or the monotone convergence theorem to the finite sums of indicator random variables, gives the tail-sum formula for expectationBoth sides may be infinite; no integrability assumption is necessary.

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 $|X_{n\wedge T}|\le Y$.

If $T$ is future time to dementia from age 50, its survival function isThe tail-sum formula for expectation gives the expected future time alive without dementia:

With administrative censoring at $c$, one patient is observed to experience $A$ with probabilitysoThe observed time is $\min (T_A,T_B,c)$, whose mean follows from the tail-sum formula for expectation:Hence

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.