Conditional expectation and density function

Question

I am kinda new to probabilities and I have a question.

Let $X,Y$ be random variables.

With $f_Y(y) = \int_\mathbb{R} f_{(X,Y)} (x,y) \mu(dx)$ we define the density of $Y$ and let

$f_{X|_y}(x) := \frac{f_{(X,Y)} (x,y)}{f_Y(y)}$ if the quotient is well defined, otherwise its $0$.

Furthermore let $h(y) := \int_\mathbb R x f_{X\mid_y} (x)\mu(dx) $

Why is it that $\mathbb E[X \mid Y] := \mathbb E[X \mid \sigma(Y)] = h(Y)$ ?

Answer 1

Hint: $\mathbb E[X|Y]$ is the unique random variable $Z$ such that $$ \mathbb E(Z\mathbf{1}_F)=\mathbb E(X\mathbf 1_F) $$ for every $\sigma(Y)$-measurable set $F$. Try to check that $h(Y)$ satisfies the above condition.