Is a non-negative random variable with zero mean almost surely zero? We have proven the following in class:

If $X$ is a finite random variable with $X\geq 0$ then $$E(X)=0  \iff  P(X=0)=1$$ (By finite I meant that the range has finitely many elements).

Does it hold for any non-negative random variable $X:\Omega\to\mathbb R_{\geq0}$?
(I've tried proving it with the indicator function yet it didn't help.)
 A: Note that, for every $x\gt0$,
$$
X\geqslant x\mathbf 1_{X\geqslant x}.
$$
(This is the step where one uses that $X\geqslant0$ almost surely.) It follows that
$$
E(X)\geqslant xP(X\geqslant x).
$$
If $E(X)=0$ this inequality implies that $P(X\geqslant x)=0$. Finally,
$$
[X\gt0]=\bigcup_{n\geqslant1}[X\geqslant1/n],
$$
hence, if every event in the RHS has probability zero the event on the LHS has probability zero.
A: Let $\langle \Omega,\mathcal{F},P\rangle$ denote our probability space.  The expected value of $X$ is $$E[X]=\int_{\Omega}X\, \textrm{d}P.$$
If we let $U\in\mathcal{F}$ be $X^{-1}\{0\}$, then we can decompose the exectation:
$$E[X]=\int_{U}X\, \textrm{d}P+\int_{\Omega\setminus U}X\, \textrm{d}P=\int_{\Omega\setminus U}X\, \textrm{d}P.$$
(The second equality follows from the fact that $X(\omega)=0$ for $\omega\in U$.)
Now, suppose that $P[\Omega\setminus U]>0$. Since $$\forall \omega\in (\Omega\setminus U)\, \big(X(\omega)>0\big)$$ 
there must be some $\epsilon>0$ and some $V\subseteq(\Omega\setminus U)$ such that $P[V]>0$ and $$\forall \omega\in V \, \big(X(\omega)>\epsilon\big).$$
Thus:
$$E[X]=\int_{\Omega\setminus U}X\, \textrm{d}P \geq \int_{V} X\, \textrm{d}P \geq \int_{V}\epsilon\,\textrm{d}P = \epsilon P[V]>0.$$
For the converse direction, suppose that $P[U]=1$.  Then $P[\Omega\setminus U]=0$ and so we get:
$$E[X]=\int_{\Omega\setminus U}X\, \textrm{d}P = 0.$$
