I got the following task:
Let $(\Omega,F,P)$ be a probability space and $X:\Omega\rightarrow R$ a random variable with $X\geq0$ P-a.s. Prove that $E[X]=0$ implies that $X=0$ P-a.s.
I tried to prove it using the monotone convergence theorem and simple functions. Is it a valid proof? If no, I would be glad for corrections or just hints.
Thank you
Note: i'm sure it is pretty sloppy, but this is my first ever attempt of a proof ever:)
Proof: Since $X$ is a discrete random variable, taking values $a_i$ = ($a_1, a_2, ... , a_n$), we have
\begin{equation} X(\omega)=\sum^{\infty}_{i=1}a_iI_{A_i}(\omega) \end{equation}
with $A_i\subseteq F$, $a_i\geq0$. Now let \begin{equation} X_n(\omega)=\sum^{n}_{i=1}a_iI_{A_i}(\omega) \end{equation}
be a sequence of measurable functions that converges to $X(\omega)$ P-a.s $\forall\omega\in\Omega$.
Thus, by the monotone convergence theorem:
\begin{align} E(X) &= \lim_{n\rightarrow\infty} E(X_n)\\ &= \lim_{n\rightarrow\infty} \sum_{i=1}^{n}a_i P(A_i)\\ &= \sum_{i=1}^{\infty}a_iP(X=a_i). \end{align}
Thus $E(X)=0$ implies $a_i=0$ P-a.s. and therefore $P(X=0)=1$ or $X=0$ P-a.s.
For each $n>0$, note that $$ P(X>1/n)\leq nE(X)=0 $$ by Chebyshev's inequality. In particular $P(X>0)=0$. Thus $X=0$ a.s.