I just stumbled upon seeing that being-zero of expectation and conditional expectation on random variables is equivalent, more exactly:
Let $X$,$Y$ be random variables. Then $E(X)=0$ if and only if $E(X|Y)=0$ a.s.
This is a bit surprising, one implication I have already seen here, but now it's even equivalent. Q: Is my argument right? It is quite straightforward, but because the result is counter-intuitive I'd like to have a look that I didnt miss something.
Proof. By definition of conditional expectation, we have $$E[E(X|Y) 1_A]=E[X 1_A]$$ We condition on $Y$, that means on the generated $\sigma$-algebra $\sigma(Y)$. Choose $A=\Omega$: $$E[ E(X|Y)] =E(X)$$ Now, if rhs$=0$ then the integrand, $E(X|Y)=0$ a.s. If, on the other hand, $E(X|Y)=0$, the expectation on the lhs is $0$ and therefore $E(X)=0$.