Let $X$ be a random variable on a probability space $(\Omega, \mathcal{F},\Bbb P)$. If $\Bbb E |X| \le 0$, show that $X=0$ almost surely. Let $X$ be a random variable on a probability space $(\Omega, \mathcal{F},\Bbb P)$. If $\Bbb E |X| \le 0$, show that $X=0$ almost surely.
Definition.
Let $X$ be a random variable on a probability space $(\Omega,\mathcal{F},\Bbb P)$. The expectation of $X$ is defined to be
\begin{equation*}
\Bbb E X = \int_{\Omega} X(\omega) d\Bbb P(\omega),
\end{equation*}
if $X \ge 0$ almost surely.
Definition.
Let $(\Omega,\mathcal{F},\Bbb P)$ be a probability space. If a set $A \in \mathcal{F}$ satisfies $\Bbb P (A) = 1$, then the event $A$ occurs almost surely.
attempt:
By definition,
\begin{equation*}
\Bbb E |X| = \int_{\Omega} |X(\omega)| d\Bbb P(\omega) \le 0.
\end{equation*}
On the other hand, since $|X|\ge 0$, then $\Bbb E |X| \ge 0$.
Hence, $\Bbb E |X| = 0$. Then, $|X|=0$. Since $X \ge 0$, then $X=0$ almost surely.
I believe that this is not in the correct way, and the definition of "almost surely" doesn't occur, that is, $\Bbb P (X=0) =1$. Any ideas? How to approach it?
Thanks in advanced.
 A: Some comments on your solution:

By definition,
\begin{equation*}
\Bbb E |X| = \int_{\Omega} |X(\omega)| d\Bbb P(\omega) \le 0.
\end{equation*}
On the other hand, since $|X|\ge 0$, then $\Bbb E |X| \ge 0$.
Hence, $\Bbb E |X| = 0$.

Good so far.

Then, $|X|=0$.

I suspect you're supposed to say more here. Why would this follow from the above, and in what sense? It is true that what you've written so far implies that $|X| = 0$ almost surely, but I suspect from context that this is something you should show or argue.

Since $X \ge 0$, then $X=0$ almost surely.

I don't think it's clear (or true) that $X \geq 0$, and I don't see why that would imply that $X = 0$ almost surely.
HINT: To repair this, consider a contrapositive approach. If $X \neq 0$ almost surely, then there exists some $A \subset \Omega$ with $\mathbb P(A) > 0$ so that $\omega \in A \implies |X(\omega)| > \epsilon$ for some positive $\epsilon$. (Why?) What would that imply?
A: Let $X$ be a random variable on a probability space $(\Omega, \mathcal{F}, \Bbb P)$ and $\Bbb EX \le 0$. We want to show that $X=0$ almost surely, that is, $\Bbb P(A)=0$, where $A = \{\omega: X(\omega) \ne 0\}$. Define $A_n := \{\omega: |X(\omega)| \ge \frac1n\}$ for all $n \in \Bbb N$.
Then, $A = \bigcup_{n=1}^\infty A_n$. Notice that $\{\omega:X(\omega) \ne 0 \} = \{\omega: |X(\omega)| \ne 0 \}$. Now, note that
\begin{align*}
0 \le \Bbb P(A) \le \Bbb P(\bigcup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty \Bbb P(A_n) &= \sum_{n=1}^\infty \int_{A_n} d\Bbb P(\omega) \\
&\le \sum_{n=1}^\infty \int_{A_n} n|X(\omega)| d\Bbb P(\omega) \\
&\le \sum_{n=1}^\infty n \int_{\Omega} |X(\omega)| d\Bbb P(\omega) \\
&= \sum_{n=1}^\infty n\Bbb E|X| \\
&\le 0.
\end{align*}
Hence, $0 \le \Bbb P(A) \le 0$, which forces $\Bbb P(A) = 0$. Therefore, $X=0$ almost surely. $\qquad \qquad \Box$
