Discrete random variable-expected value

Question

For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$, show that $E [X | B] = \frac{E [1_{B}X]} {P [B]}$. I have $E [X | B] = \frac{\sum x_{i} * P(\{X=x_{i}\} \cap B) } {P [B]}$ and I don't know what next.

Answer 1

$\frac{E[X 1_B]}{p[B]}=\frac{\sum_{\omega \in \Omega} x \times 1_B\times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B\cup B^{c}} x \times 1_B\times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B} x \times 1_B\times P(\{X=x\} )+0 } {P [B]}=\frac{\sum_{\omega \in B} x \times P(\{X=x\} ) } {P [B]}=\frac{\sum_{\omega \in B} x \times P(\{X=x\} \cap \{ \omega \in B\} ) } {P [B]}=\frac{\sum_{\omega \in \Omega} x \times P(\{X=x\} \cap \{ \omega \in B\} ) } {P [B]}=\sum_{\omega \in \Omega} x \times \frac{P(\{X=x\} \cap \{ \omega \in B\}}{P [B]} )=\sum_{\omega \in \Omega} x \times P(\{X=x\}|B )=E(X|B)$

another way

$\frac{E[X 1_B]}{p[B]}=\frac{EE[X 1_B|X]}{p[B]}$

$=\frac{E(XE[1_B|X])}{p[B]}=\frac{E(Xg(X))}{p[B]}$

$=\frac{\sum_x x g(x) p(X=x)}{p[B]}=\frac{\sum_x x E(1_B|X=x) p(X=x)}{p[B]}$

$=\frac{\sum_x x p(B|X=x) p(X=x)}{p[B]}$

$=\frac{\sum_x x p(X=x|B) p(B)}{p[B]}$

$=\frac{p(B) \sum_x x p(X=x|B) }{p[B]}$

$=\sum_x x p(X=x|B)=E(X|B)$

note that $E[1_B|X]$ is a random variable that it is a function of $X$.