A nonnegative random variable has zero expectation if and only if it is zero almost surely Let $Y$ be a non-negative random variable. Prove that $E(Y) = 0$ if and only if $P(Y=0)=1$.
My understanding is that while you can prove it for discrete $Y$, the result is true for all $Y$.

If $Y$ is a non-negative random variable defined on a probability space $\Omega$ and
$$E(Y) = \int_{\Omega} Y dP=0$$
then $Y = 0$ almost surely and $P(\{\omega \in \Omega: Y(\omega)=0\})=1$.
Proof: For any $m \in \mathbf{N}$, let
$$E_m = \{\omega \in \Omega:Y(\omega) > 1/m\}$$
then, since $Y$ is non-negative, we have $Y = Y1_{\Omega} \geq Y 1_{E_m}$ and then
$$0 = \int_{\Omega} Y dP \geq \int_{E_m} Y dP\geq \frac1{m}P(E_m) \geq 0,$$
and $P(E_m) = 0$.
So
$$0 \leq P(\{\omega \in \Omega:Y(\omega) \neq 0\})= P\left(\bigcup E_m\right) = \lim_{m \rightarrow \infty}P(E_m)=0$$
Hence
$$P(\{\omega \in \Omega:Y(\omega) \neq 0\})=0 \implies P(\{\omega \in \Omega:Y(\omega) = 0\})=1 $$ QED
Conversely, if $Y=0$ a.s. then $E(Y) = \int Y dP = 0$
