How can you prove that if $X$ is an integrable variable, then the $E(X \mid Y)$ is finite regardless of whether the variable $X$ is continuous or discrete?
The problem is: Proof that if $X$ is an integrable variable, then $E(X \mid Y)$ is finite. Additionally, proof $E\bigl( E(X \mid Y) \bigr) = E(X)$. The second part, I don't have problem, but I have very problems for proof that $E(X \mid Y)$ is finite.
If $X$ and $Y$ are discrete random variables:
$$E(X \mid Y=y) = \sum_{x}x\left(P(X=x \mid Y=y)\right) = \sum_{x}\left(x\left(\frac{P\left(X \;=\; x \;,\; Y\; =\; y\right)}{P(Y\; =\; y)}\right)\right)$$
If $X$ and $Y$ are continuous random variables:
$$E(X \mid Y=y) = \int_{-\infty}^{\infty} x\left(f_{X|Y=y}\left(y\right)\right)\; dy= \int_{-\infty}^{\infty}\left(x\left(\frac{f_{X,Y}\left(x,y\right)}{f_{X}\left(x\right)}\right)\right) dy$$