Confusion on integrability of stopping times

Question

I know the definition of a $\mathbb F-$stopping time $\tau$ is that for all $n \in\mathbb N $ that $\{ \tau \leq n\} \in \mathcal{F}_{n}$

How do the ideas of integrability and well-definedness of $\tau$ actually fit in to the concept of a stopping. I realize the question is not concise.

For example, if $P(\tau < \infty)=1$ does this mean that $\tau$ is integrable? I know this is the case if $\tau$ is $-$a.s. bounded, i.e. there exists $c\in \mathbb N$ so that $\tau \leq c-$a.s. and hence $E[\tau] \leq E[c]=c<\infty$

Answer 1

The answer is NO. The sigma fields $\mathcal F_n$ and the fact that $\tau$ is a stopping time are not relevant to the question. You can take $\mathcal F_n$ to be the Borel sigma field of $[0,1]$ for each $n$ and take any non-negative random variable with infinite mean, say $\tau =\sum n I_{(\frac 1 {n+1}, \frac 1 n)}$.

Answer 2

You should just think of $\tau$ as a random variable which takes values in $\mathbb N$ with the additional measurability property that $\{\tau \le n\} \in\mathcal F_n$ for each $n\in\mathbb N$.

Hence, $\tau$ being integrable and well-defined means the same thing as what it does for any other random variable to be integrable or well-defined. For example, integrability means that $\mathbb E[\vert\tau\vert] = \mathbb E[\tau] < \infty$.

That $\mathbb P(\tau < \infty) = 1$ just means that $\tau$ is a.s. finite. As with any other random variable, this is weaker than integrability. That is, if $\tau$ is integrable, then it is almost surely finite, but not vice-versa. For an example of a random variable that is a.s. finite but not integrable, consider the one given in the St. Petersburg Paradox.