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.