Theorem
\[E(\vert X \vert)<\infty \quad \Rightarrow \quad \Pr(\vert X \vert<\infty)=1 \quad\text{i.e.}\quad \vert X \vert<\infty \;\text{a.s.}\]Proof
If not $\vert X \vert<∞$ a.s., then there is a set of positive measure where $\vert X \vert=∞$. The integral
\[∫ΩX\,\mathrm{d}P=∫Ω\max\{0,X\}\,\mathrm{d}P−∫Ω\max\{0,−X\}\,\mathrm{d}P\]is defined only when at most one of the summands is infinite.
If $X=+∞$ on a set of positive measure, then $X=−∞$ only on a zero-set, and then $E(X)=+∞$.
Similarly, if $X=−∞$ on a set of positive measure, then $E(X)=−∞$.
Thus we have the following possibilities:
- $\vert E(X) \vert<∞$ and $\vert X \vert<∞$ a.s.
- $E(X)=+∞$ and $X>−∞$ a.s.
- $E(X)=−∞$ and $X<∞$ a.s.
- $X$ is not integrable ($E(X)$ does not exixts)
Note: If $E(X)<∞$ without absolute value, it is possible to have $\vert X \vert=∞$ with positive probability.
Refenrence
probability - Does finite expectation imply bounded random variable?