# Implications between $\mathbb P [\tau < \infty] =1$ and $\tau \in L_1 (\mathbb P)$

We've got the usual filtered stochastic basis $(\Omega, \mathcal F, (\mathcal F_n). \mathbb P), \space \tau : \Omega \to \mathbb{N}\cup \{\infty\}, [\tau \le n] \in \mathcal F_n$

($\tau$ is an $\mathcal F_n$ stopping time).

I'm wondering whether the implication below holds:

$$\tau \in L^1 ( \mathbb P) \Rightarrow \mathbb P [\tau < \infty]=1$$

I think it should hold, because

$$\tau \in L^1 ( \mathbb P) \Rightarrow \mathbb E [\tau] =\int_{\Omega}\tau d \mathbb P < \infty \\ \mathbb P [\tau < \infty] = \int_{[\tau < \infty]} d \mathbb P \le \int_{\Omega} d \mathbb P \le \int_{\Omega}\tau d \mathbb P < \infty$$

The only problem I can see is that $$\int_{\Omega} d \mathbb P \le \int_{\Omega}\tau d \mathbb P$$ might not hold in general. For instance if $X \sim N(0,\sigma^2)$ we've got $\mathbb P (\Omega) =1$ and $\mathbb{E} X =0$. So is this the reason why my proof does not work?

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You've identified one problem with your reasoning, but even if that inequality were true I don't see how it would prove that $\tau$ is finite with probability one. – Gibarian Jun 1 '13 at 10:08

The implication holds for any random variable, not just stopping times. It easily seen by showing the contrapositive: $$P(X<\infty)<1\;\Rightarrow\;X\notin L^1(P).$$ So suppose $P(X<\infty)<1$, then $P(X=\infty)>0$ and hence $${\rm E}[|X|]\geq{\rm E}[|X|\mathbf{1}_{\{X=\infty\}}]=\infty\cdot P(X=\infty)=\infty,$$ and hence $X\notin L^1(P)$. The other implication, however, does not hold in general.
In general, instead of showing that A implies B you can show that not-A implies not-B. See this. What I mean by "the other implication does not hold" is that you can't say $P(X<\infty)=1\;\Rightarrow\; X\in L^1(P)$. I don't get the first line of your proof. – Stefan Hansen Jun 1 '13 at 9:41