Conditional Expectation & Integration Theory

Question

Let $(\Omega, \mathcal{A}, P)$ be a measure space.

Easy question: can somebody explain to me why this holds?

$\int_B E(X|B) dP = P(B) E(X|B)$ (for some B taken from the sigma-algebra)

Is it the same as:

$\int_B E(X|B) dP = E(X|B) \int 1_{B} dP$

If yes, why can I pull out $E(X|B)$? Why do I not need the assumption of independence?

Thanks very much!

Answer 1

$\mathsf E(X\mid B)$ is an expectation over an event $B$, and thus a constant.

It is not to be confused with expectation over a sigma-algebra (usually generated by another random variable), which would be a random variable.

For any constant $C$, we have $\int_B C \mathrm d\mathsf P ~{=C\int_B \mathrm d\mathsf P\\= C~\mathsf P(B)}$

To be redundant and repeat again, $\mathsf E(X\mid B)$ is a constant.   So....