Why does $\mathrm{E}[|X|] < \infty$ imply $\mathrm{P}[X < \infty] = 1$?

Question

I found this one

Finite almost surely implies integrable?

which shows that $X$ finite a.s. does not imply $X$ integrable...

But how can I show that the converse is true?

Many thanks in advance.

Answer 1

Let $S_{n}=\{\omega\in \Omega: |X(\omega)|\geq n\}$ and $S=\displaystyle\bigcap_{n=1}S_{n}$, then $P(S)\leq P(S_{n})\leq n^{-1}E[|X|]\rightarrow 0$, so $P(S)=0$. Then $P[|X|<\infty]=1-P(S)=1$. Since $P[|X|<\infty] \leq P[X<\infty]$, now we have $P[X<\infty] = 1$.