TITLE: Assumption of almost supermartingagles convergence theorem
QUESTION [1 upvotes]: 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:

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.
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.

REPLY [2 votes]: 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.
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