Does finite expectation imply bounded random variable?

Solution 1:

If not $|X|<\infty$ a.s., then there is a set of positive measure where $|X|=\infty$. The integral $$ \int_\Omega X\,\mathrm dP= \int_\Omega \max\{0,X\}\,\mathrm dP-\int_\Omega \max\{0,-X\}\,\mathrm dP$$ is defined only when at most one of the summands is infinite. If $X=+\infty$ on a set of positive measure, then $X=-\infty$ only on a zero-set, and then $\mathbb E(X)=+\infty$. Similarly, if $X=-\infty$ on a set of positive measure, then $\mathbb E(X)=-\infty$. Thus we have the following possibilities:

• $|\mathbb E(X)|<\infty$ and $|X|<\infty$ a.s.
• $\mathbb E(X)=+\infty$ and $X>-\infty$ a.s.
• $\mathbb E(X)=-\infty$ and $X<\infty$ a.s.
• $X$ is not integrable ($\mathbb E(X)$ does not exixts)

(So specifically, as you wrote $\mathbb E(X)<\infty$ without absolute value, it is possible to have $|X|=\infty$ with positive probability)