Assumption of almost supermartingagles convergence theorem

Question

My question is about the assumption of the following result, which is due to Robbins and Siegmund.

Suppose that $(\varOmega,\mathscr{F},P)$ is a probability space. Let $(\mathscr{F}_n)_{n\in\mathbb{N}}$ be a filtration in $\mathscr{F}$. For every $n\in\mathbb{N}$, let $X_n\colon\varOmega\to\left[0,{+}\infty\right[$ be a random variable which is $\mathscr{F}_n$-measurable, and let $Y_n\colon\varOmega\to\left[0,{+}\infty\right[$ be a random variable. Suppose that $\sum_{n\in\mathbb{N}}Y_n<{+}\infty$ $P$-a.s. and that $$(\forall n\in\mathbb{N})\quad\mathbb{E}\big(X_{n+1}~|~\mathscr{F}_n\big)\leq X_n+Y_n\quad \text{$P$-a.s.} $$ Then $(X_n)_{n\in\mathbb{N}}$ converges almost surely.

The main idea is to define $$(\forall n\in\mathbb{N})\quad Z_n=X_n-\sum_{k=0}^{n-1}Y_k $$ and show that $(Z_n,\mathscr{F}_n)_{n\in\mathbb{N}}$ is a supermartingale. This leads to the following questions:

1. Do we have to assume that $(X_n)_{n\in\mathbb{N}}$ are integrable? Otherwise the conditional expectation $\mathbb{E}\big(X_{n+1}~|~\mathscr{F}_n\big)$ does not make sense.
2. Similarly, do we have to assume that $(Y_n)_{n\in\mathbb{N}}$ are integrable so that $\mathbb{E}\big(Z_{n}~|~\mathscr{F}_n\big)$ is well defined? I could not see how one can infer from the almost sure summability of $\sum_{n\in\mathbb{N}}Y_n$ that $(Y_n)_{n\in\mathbb{N}}$ are integrable.

Robbins and Siegmund did not mention these assumptions in their original work, but it seems to me that these assumptions are important.

Any help/hint is highly appreciated.

Answer 1

If $X$ is a non-negative random variable on $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G}$ is a sub $\sigma$-algebra of $\mathcal{F}$, it is still possible to define the conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ when $X$ has infinite expectation:

Recall that the existence of $\mathbb{E}[X\mid \mathcal{G}]$ in the usual situation where $\mathbb{E}[|X|]<\infty$ is guaranteed by the Radon-Nikodym theorem. The standard version of this theorem assumes that both of the measures in question are $\sigma$-finite, but there is actually an extension to the case where $\nu$ is a positive measure (but not necessarily $\sigma$-finite), $\mu$ is $\sigma$-finite, and $\nu\ll\mu$, provided that one is willing to allow the Radon-Nikodym derivative to take the value $\infty$. This extension is generally relegated to the exercises of measure theory textbooks (see this question for some references), but is precisely what is needed to define the conditional expectation in the situation where $X$ is non-negative but has infinite expectation.

Answer 2

There's no need to assume that $X_n$ or $Y_n$ is integrable. It's not hard to check that if the basic inequality is satisfied, then so is the same inequality with $X_n$ and $Y_n$ replace by $\min(X_n,c)$, $\min(Y_n,c)$ (respectively) for each constant $c>0$. The supermartingale argument then shows that $\lim_n\min(X_n,c)$ exists a.s. for each $c>0$, which implies that $\lim_nX_n$ exists, a.s., as well.