Finite almost surely implies integrable?

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space.

If a random variable $X$ satisfies $\mathbb{P}[X<\infty]=1$ (which means $X$ is finite almost surely, doesn't it?) then $\mathbb{E}[X]<\infty$?

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No. Take $P(X=2^n)=2^{-n}$. Then $P(X<\infty)=1$ and $E(X)=\infty$.
@seek The best condition I can think of is .... that $\Bbb E[X]<\infty$. – Hagen von Eitzen Feb 14 at 14:08