Conditional expectation of Borel function of two independent variables I am trying to solve the following problem:
Let $X$, $Y$ be two independent random variables, and let $f$ be a Borel function. Show that
$$
\mathbb{E}[f(X,Y)|Y=t]=\mathbb{E}[f(X,t)].
$$
I have no clear idea where to start. By the definition of conditional expectation, we should have that
$$
\int_{\{Y=t\}}f(X,Y)dP=\int_{\{Y=t\}}\mathbb{E}[f(X,Y)|Y].
$$
On the LHS I have what I wanted, i.e. $\mathbb{E}[f(X,t)]$. But how do I proceed from here? Also, why is independence of variables important?
 A: We want to show $\mathbb E[f(X,Y) | Y] = h(Y)$, where $h(t) = \mathbb E[f(X,t)]$ (Assume that $\mathbb E[|f(X,Y)|] < \infty $)
So we need to show that $h(Y)$ is $\sigma(Y)$ measurable, but $$h(Y) = \int_{\Omega_{1}}f(X(\omega_1),Y)d\mathbb P(\omega_1),$$ so it follows from Fubini-Theorem (Y is $\sigma(Y)$ measurable, and $h$ as a integral from measurable function, too.)
Next, take any $A \in \sigma(Y)$
We have to show $\mathbb E[h(Y)\chi_A] = \mathbb E[f(X,Y)\chi_A] $
We know that $(Y,\chi_A)$ is independent of $X$, let's form vector $(X,Y,\chi_A)$ with distribution $\mu_{X} \otimes \mu_{(Y,A)}$ (due to independence).
Then, we have:
$$\mathbb E[f(X,Y)\chi_A] = \int_{R^3}f(x,y)z \cdot d(\mu_X \otimes \mu_{(Y,A)})(x,y,z) = \int_{R^2}\int_R f(x,y)z \cdot d\mu_X(x)d\mu_{(Y,A)}(y,z) = \int_{R} \mathbb E[f(x,Y) \chi_A] d\mu_X(x) = \mathbb E[\int_{R} f(x,Y) d\mu_X(x) \chi_A] = \mathbb E[ H(Y) \chi_A]$$
All steps with interchanging due to Fubini
A: Theorem:
In general if $X,Y$ are independent random variables and $u,v:\mathbb R\to\mathbb R$ denote Borel-measurable functions then also $u(X),v(Y)$ are independent random variables.

It is evident that: $$\mathbb E[f(X,Y)\mid Y=t]=\mathbb E[f(X,t)\mid Y=t]$$Now observe that on base of the theorem we can conclude that $f(X,t)$ and $Y$ are independent so that:$$\mathbb E[f(X,t)\mid Y=t]=\mathbb E[f(X,t)]$$

edit: 
Writing $g\left(Y\right)=\mathbb{E}\left[f\left(X,Y\right)\mid Y\right]$
you are asked to prove that $g\left(t\right)=\mathbb{E}\left[f\left(X,t\right)\right]$.
We have: $$\begin{aligned}g\left(t\right)P\left(Y=t\right) & =\int_{\left\{ Y=t\right\} }g\left(Y\left(\omega\right)\right)P\left(d\omega\right)\\
 & =\int_{\left\{ Y=t\right\} }f\left(X\left(\omega\right),Y\left(\omega\right)\right)P\left(d\omega\right)\\
 & =\int f\left(X\left(\omega\right),t\right)\mathbf{1}_{Y=t}\left(\omega\right)P\left(d\omega\right)\\
 & =\int f\left(X\left(\omega\right),t\right)P\left(d\omega\right)\times\int\mathbf{1}_{Y=t}\left(\omega\right)P\left(d\omega\right)\\
 & =\mathbb{E}f\left(X,t\right)P\left(Y=t\right)
\end{aligned}
$$where the second equality is by definition.
From this it can be deduced that $g\left(t\right)=\mathbb{E}f\left(X,t\right)$
but this only under the extra condition that $P\left(Y=t\right)\neq0$.

edit2
Claim: if $g\left(t\right):=\mathbb{E}f\left(X,t\right)$ then: $$\mathbb{E}\left[v\left(Y\right)f\left(X,Y\right)\right]=\mathbb{E}\left[v\left(Y\right)g\left(Y\right)\right]\tag1$$
for any suitable Borelfunction $v:\mathbb{R}\to\mathbb{R}$ (especially the measurable indicator functions).
Proof:
$$\begin{aligned}\mathbb{E}\left[v\left(Y\right)g\left(Y\right)\right] & =\int v\left(y\right)\mathbb{E}f\left(X,y\right)\;dF_{Y}\left(y\right)\\
 & =\int v\left(y\right)\int f\left(x,y\right)\;dF_{X}\left(x\right)\;dF_{Y}\left(y\right)\\
 & =\int\int v\left(y\right)f\left(x,y\right)\;dF_{X}\left(x\right)\;dF_{Y}\left(y\right)\\
 & =\int v\left(y\right)f\left(x,y\right)\;dF_{\left(X,Y\right)}\left(x,y\right)\\
 & =\mathbb{E}\left[v\left(Y\right)f\left(X,Y\right)\right]
\end{aligned}
$$
Here the fourth equality rests on independence.
From $(1)$ you are allowed to conclude that $g\left(Y\right)=\mathbb{E}\left[f\left(X,Y\right)\mid Y\right]$
hence: $$\mathbb{E}f\left(X,t\right)=g\left(t\right)=\mathbb{E}\left[f\left(X,Y\right)\mid Y=t\right]$$
