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$\textbf{Motivation: }$ I was thinking about what happens when $\text{Var}(X)= 0$. It is known that this happens if and only if $X=a$ for some constant $a$. The proof in one direction (assuming $X=a$) is trivial. To do the other I found some argument, which basically boils down to the proposition below. Problem is that I don't know if the proposition is correct, and I don't know whether my proof for it is correct.

So the question is whether the following is sound:

$\textbf{Proposition: }$Suppose $g\geq0$ and $E[g(X)] = 0$. Then $P(g(X)>0) = 0$. Hence $g(X) = 0$ a.s.

$\textbf{Proof: }$ Suppose $g\geq0$ and $E[g(X)] = 0$ but now suppose $P(g(X)>0)\neq 0$. Hence $g(X)$ is non-zero somewhere with non-zero probability. Suppose it is non-zero on some $A\subseteq \mathbb{R}$. Then $$0=E[g(X)] = \int_\mathbb{R}g(X)dF(x) = \int_Ag(X)dF(x).$$ Since $g>0$ on $A$ this is a contradiction.

$\textbf{Application: }$ Now take $g(X) = (X-E(X))^2$. Then $E[g(X)] = 0$ implies $X = E(X)$ almost everywhere since $P(X-E(X) = 0) = 1$.

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  • $\begingroup$ Do note: I have not done measure theory yet (beside when talking about Lebesgue integration), and I strongly suspect this has something more to do with measure theory. $\endgroup$ Mar 5 '20 at 15:35
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    $\begingroup$ Looks right to me. I don't know any measure theory either, but I don't think you need that much machinery. $\endgroup$ Mar 5 '20 at 15:50
  • $\begingroup$ It feels like Markov's inequality could be relevant here: if $Z$ is a nonnegative RV, then for some $a >0$, $Pr(Z\geq a) \leq E[Z]/a$. So what if you take $Z = g(X) + \sqrt{a}$, and look at the limit as $a\to 0$? (Also your proof looks right, but I'm also terrified of measure theory, so thinking of an alternative...) $\endgroup$
    – Y. S.
    Mar 5 '20 at 15:54
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    $\begingroup$ The Proposition is correct but the formulation is unnecessarily cumbersome in my opinion. This formulation might be a little bit better: „Let $Y$ be a random variable that is non-negative almost everywhere and $\mathsf E(Y)=0$. Then $Y=0$ almost everywhere.“ $\endgroup$ Mar 5 '20 at 16:52
  • $\begingroup$ If $g$ is non-zero on some $A\subseteq \mathbb R$, it cannot contradicts to $\mathbb Eg(X)=0$. Say, $g(x)=x^2\mathbb 1_{x\leq 0}$ is positive on whole halfline and for $X=3$ a.s. $\mathbb E[g(X)]=0$. None contradiction arises. $\endgroup$
    – NCh
    Mar 5 '20 at 17:35

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