Why is that $\mathbb{E}[X|A]\mathbb{P}(A)=\int x \mathbb{I}_{A} dP$?

Question

I'd like to prove that:

$$ \mathbb{E}[X|A]\mathbb{P}(A)=\int x \mathbb{I}_{A} dP $$

This came to me because the "Law of total expectation" would be way easier to prove if I could argue that:

$$ \mathbb{E}[X] = \sum_{i=1}^\infty \int x \mathbb{I}_{(A_i)} dP=\sum_{i=1}^\infty\mathbb{E}[X|A_i]\mathbb{P}(A_i) $$

with $A_i$ a partition of $\Omega$.

Answer 1

Assuming that $A$ is a set, $Z=\mathbb{E}[X|A]$ is a constant and thus:

$$ZP(A)=Z\mathbb{E}[\mathbb{I}_A]=\mathbb{E}[Z\mathbb{I}_A]=\mathbb{E}[\mathbb{E}[X\mathbb{I}_A|A]]=\mathbb{E}[X\mathbb{I}_A]$$

where we used that $\mathbb{I}_A \in \sigma(A)$ (therefore $ \mathbb{E}[X\mathbb{I}_A|A]=\mathbb{I}_A\mathbb{E}[X|A]$) and the last equality is the tower law.