Conditional expectation: I want to prove $E[E(X\mid Y)|Y]=E(X\mid Y)$
I attempted the following. Is it correct? $$\begin{align*}E[E(X\mid Y)|Y=y]&=\int_{-\infty}^\infty E(X\mid Y=y)f_{X\mid Y}(x\mid y)~dx\\ &=E(X\mid Y=y)\int_{-\infty}^\infty f_{X\mid Y}(x\mid y)~dx \\&=E(X\mid Y=y)\end{align*}$$ $$\text{Hence, } E[E(X\mid Y)|Y]=E(X\mid Y)$$
On
It might be easier to see what's going on if you also expand the conditional expectation and write its integral out. $$ E(E(X|Y)|Y=y)=\int_{-\infty}^\infty E(X|Y=y)f_{X|Y=y}(x)dx=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty x f_{X|Y=y}(x)dx\right)f_{X|Y=y}(x)dx\\ =\left(\int_{-\infty}^\infty x f_{X|Y=y}(x)dx\right)\int_{-\infty}^\infty f_{X|Y=y}(x)dx\\ =\left(\int_{-\infty}^\infty x f_{X|Y=y}(x)dx\right)\\ =E(X|Y) $$
You haven't defined your symbols for much of this; so, I don't want to say that I'm quite comfortable with what you've written there. Plus - assuming that $f_{X\vert Y}$ is a density function - you've made an inherent assumption that your variable is continuously distributed.
However, let me present you with a hint for a much more general proof. Remember that $\mathbb{E}[X\mid Y]$ is itself a random variable - and in particular, is a random variable that is measurable with respect to the $\sigma$-algebra generated by $Y$.
You can prove that if $Z$ is a random variable that is measurable with respect to the $\sigma$-algebra $\mathcal{F}$, then $\mathbb{E}[Z\mid \mathcal{F}]=Z$.
(To make logical sense of this: remember, we can think of conditional expectation as "If we know only the information encoded in $\mathcal{F}$, what can we say about the value of $Z$?" If $Z$ is measurable with respect to $\mathcal{F}$, then of course we know exactly the value of $Z$ given that information.)