probability - Does finite expectation imply bounded random variable?

In general if I have that
\begin{equation}
\mathbb{E}(X)<\infty
\end{equation}
does this imply that $|X|<\infty$ a.s? An attempted proof:

Let $(\Omega,\mathbb{F},\mathbb{P})$ be a probability space and $K>0$ be a constant, then

\begin{equation}
\mathbb{E}(X)=\int_{\Omega}XdP\,=\int_{\{|X|\leq K \}}XdP\,+\int_{\{|X|>K\}}XdP\, <\infty
\end{equation}
Taking the limit as $K\rightarrow \infty$
\begin{equation}
\int_{\{|X|>K\}}XdP\, \rightarrow 0
\end{equation}
Struggling how to finish from here or posssibly this isn't true? Thanks!

Answer

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)