Bayes-like conditional expectation

Question

For $X$ random variable on $\Omega$, $f$ a measurable function, and $B$ a measurable subset of $\Omega$, Can you please help rigorously proving that:

$\mathop{\mathbb{E}}[f(X)\cdot 1_B]=P(B)\cdot \mathop{\mathbb{E}}[f(X)| B]$

Answer 1

The towering property of conditional expectation can be used for this proof \begin{align*} \mathbb E[f(X) 1_B] &= \mathbb E[\mathbb E[f(X)1_B|1_B]]\\ &= \mathbb E[1_B \mathbb E[f(X)|1_B]]\\ &= \mathbb P[\{1_B=1\}]\cdot 1\cdot \mathbb E[f(X)|1_B=1] + \mathbb P(\{1_B=0\})\cdot 0 \cdot\mathbb E[f(X)|1_B=0]\\ &= \mathbb P[B] \mathbb E[f(X)|B] \end{align*} Since $1_B$ is discrete random variable and hence so is $\mathbb E[f(X)1_B|1_B]$