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I am trying to prove the well-known positivity property of conditional expectations. Namely, $X$ being a random variable on a measure space $(\Omega,\cal{F},P)$ and $\cal{A}$ being a sub-$\sigma$-algebra of $\cal{F}$, the following implication holds. $$ X \geq 0 \text{ a.s. } \Rightarrow E[X\mid\cal{A}] \geq 0 \text{ a.s. }$$

I try to prove this by contradiction. Suppose for some sub-$\sigma$-algebra $\cal{B}$, $P(\{E[X\mid\cal{B}] < 0 \}) >0$. Then I have $$\int_{\{E[X\mid\cal{B}] < 0 \}}E[X\mid\cal{B}]\,dP < 0$$ But then, $$\int_{\{E[X\mid\cal{B}] < 0 \}}E[X\mid\cal{B}]\,dP = \int_{\{E[X\mid\cal{B}] < 0 \}} X \,dP \geq 0$$ The last inequality holds due to the nonnegativity constraint on $X$. Here I have a contradiction so the proof is complete. But I feel like I missed something. Is this proof OK? Thanks.

Calculon
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    Looks okay. But have you a good argument for what you conclude after "Then we have"? – drhab Feb 04 '15 at 10:49
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    @drhab My argument was this: Since $EX\mid\cal{B}$ is negative over ${\omega : EX\mid\cal{B} < 0 }$, integrating it over these "events" should give something negative since probability measure is nonnegative. – Calculon Feb 04 '15 at 10:56
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    That's true but it needs some argumentation. If $P(Y<0)>0$ then it can be shown that some $\epsilon>0$ exists such that $P(Y<-\epsilon)>0$. That is a step in between. – drhab Feb 04 '15 at 11:00
  • I see it now. Thank you very much. – Calculon Feb 04 '15 at 11:01

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