I am trying to prove that $P(X=0)=1$ for a random variable $X$ which satisfies $P(X\ge 0 )=1$ and $E(X)=0$. I know that this seems intuitively clear, but I am having trouble with a formal proof (one that relies on Lebesgue integration).
I have tried to make a proof based on contradiction where I assume that $P(X>0) >0$ and show that this implies that $E(X)>0$, but I am not sure if my proof is correct. It looks something like this:
Consider the probability space $(\Omega,\Sigma,P)$. Let $A=\{\omega\in\Omega:X(\omega)=0\}$ and $ B=\{\omega\in\Omega:X(\omega)>0\}$. Clearly $A$ and $B$ are disjoint. Assume $P(B)>0$ and write
$E(X)=\int_\Omega X(\omega)P(d\omega)=\int_A 0 P(d\omega)+\int_B X(\omega)P(d\omega).$
Since $P(B)>0$, and $X(\omega)>0$ for all $\omega \in B$, this should imply that $E(X)>0$, which is a contradiction.
Is this correct? How do I know that my final statement holds? Can I make it more formal by introducing limits of some kind? For instance by using $P(X>0)=\lim_{n\rightarrow \infty} P(X\ge 1/n)$.
Your argument is right. Here's a direct approach, using essentially the same calculations. We partition the set $(X\geq 0)$ as the union of the sets $(X\geq 1)$ and $(2^{-n}\leq X<2^{-n+1})$, $n\in\mathbb{N}$:
\begin{align*} 0&=E[X]=\int XdP=\int_{X\geq 1}XdP+\sum_{n=1}^\infty\int_{2^{-n}\leq X<2^{-n+1}}XdP\\ &\geq P(X\geq 1)+\sum_{n=1}^\infty 2^{-n}P(2^{-n}\leq X<2^{n-1})\geq 0 \end{align*} so the inequalities are actually equalities. Since all numbers are nonnegative, they have to be zero, so $P(X\geq 1)=P(2^{-n}\leq X<2^{-n+1})=0$ for all $n$. Finally, $$1=P(X\geq 0)=P(X=0)+P(X\geq 1)+\sum_{n=1}^\infty P(2^{-n}\leq X<2^{-n+1})=P(X=0)$$