RV with finite mean implies RV in $L^1$

Question

Is it true that

$E[X]<\infty \Rightarrow E[|X|] < \infty $, in other words if a RV X has finite mean than $X \in L^1$?

Answer 1

$EX <\infty$ does not imply $E|X| <\infty$ ($EX$ could be $-\infty$). But if $EX$ exists and is a real number then $E|X| <\infty$. This is a consequence of the basic definition of integration in measure theory. Indeed, if $EX$ exists and is finite then $EX^{+} <\infty$ and $EX^{-} <\infty$, so $E|X|=EX^{+}+EX^{-} <\infty$.

Answer 2

Recall that a measurable function $u$ can be decomposed into $u=u^+-u^-$ with $u^\pm$ measurable positive functions and that $|u|=u^+ +u^-$. $u$ is Lebesgue inegrable, if $u^\pm$ are both measurable and $\int u^\pm d\mu<\infty$. So

$$\int |u|d\mu=\int u^+ d\mu+\int u^-d\mu <\infty.$$

Also $$\mathbb{E}X=\int_{\Omega} X~d\mathbb{P}.$$