Question: For a super-martingale $X$ with stopping time $T$, when can it be that $E[X_T] = \infty$?
Here is a more precise formal statement of the question. We follow (almost) the standard setting for the optional stopping theorem. Let $X = (X_t)_{t∈\mathbb{N}_0}$ be a discrete-time super-martingale with $X_0=0$, and $T$ a stopping time with values in $\mathbb{N}_0 \cup \{\infty\}$, both with respect to a filtration $(F_t)_{t∈\mathbb{N}_0}$.
Assume that $X_{t+1} - X_t \ge -1$ for all $t\le T$. Assume also that $T$ is always finite ($\Pr[T = \infty] = 0$).
Can it be the case that $E[X_T] = \infty$?
I'm looking for an example, or a proof that it cannot happen.
Remarks. Note that, in any example, it must be that $E[T] = \infty$. Otherwise the optional stopping theorem in its usual form implies that $E[X_T] \le X_0 = 0$. For the record, here is a proof of that for the case when $E[T]$ is finite.
Define $Y_t = X_t - X_{t-1}$ (for $t\ge 1$, and define $Y_0 = X_0 = 0$).
Then $X_t = \sum_{s=1}^t Y_t$, and $E[Y_t \,|\, T \ge t] \le 0$ (because $X$ is a super-martingale). Then the expectation in question is
$\begin{align} E\Big[\sum_{t=1}^T Y_t\Big] &{} =~ \sum_{s=1}^\infty \Pr[ T=s ]\, E\Big[\sum_{t=1}^T Y_t ~|~ T=s\Big] \\ &{} =~ \sum_{s=1}^\infty \sum_{t=1}^s \Pr[T=s] \, E[Y_t \,|\, T=s] && \textit{(linearity of expectation)}~~~~~~~~~~(1)\\ &{} \mathbin{\stackrel{*}{=}}~ \sum_{t=1}^\infty \sum_{s=t}^\infty \Pr[T=s] \, E[Y_t\, |\, T=s] && \textit{(holds given that $E[T]$ is finite; see below)} \\ &{} =~ \sum_{t=1}^\infty \Pr[T\ge t]\,E[Y_t\,|\, T \ge t] \\ &{} =~ \sum_{t=1}^\infty 0 &&\textit{(as $E[Y_t\,|\, T \ge t] \le 0$)} \\ &{} =~ 0. \end{align}$
All equalities above except the third (marked $\stackrel{*}{=}$) follow from basic properties of probability. Given that $E[T]$ is finite, the third must also hold, because (by the assumption that $Y_{t+1}-Y_t \ge -1$) the total of the negative terms in the sum (1) is at least $$\sum_{s=1}^\infty \sum_{t=1}^s -1 \Pr[T=s] = -1 E[T],$$ so the total of the negative terms is finite, so changing the order of summation in (1) does not change the value.
Note that that, in any suitable example for my question, the sum (1) in question must be infinite, but changing the order of summation (as above) must make the value finite.
Note also that (by a standard application of the optional stopping theorem) for every $n$ we have that $E[X_{\min\{n, T\}}] \le 0$, so $\lim_{n\rightarrow \infty} E[X_{\min\{n, T\}}] \le 0$ (if the limit exists). Also, the quantity in question is $E[\lim_{n\rightarrow\infty} X_{\min\{n, T\}}]$. So the question is whether the expectation of the limit is the limit of the expectations.
Here's an example (following this question) that violates the assumptions that $\Pr[T=\infty] = 0$ and $Y_t \ge -1$ (although it does satisfy $Y_t \le 1$), and for which $E[\sum_{t=1}^T Y_t] = \infty$.
Take $\Pr[T=\infty]=1$, and define $Y_t$ to be $1-t^2$ with probability $1/t^2$, or $1$ with probability $1-1/t^2$. Then, for any $m$, the probability of the event ``$\exists t\ge m$ such that $Y_t \ne 1$'' is less than $\sum_{t=m}^\infty 1/t^2$, which tends to zero as $m\rightarrow\infty$. So with probability one the sum equals $\infty$. (Unless I'm missing something?)
Note: This question is an update of this one, in which the technical assumptions weren't quite right.