Let X and Y be independent random variables and let $f: \mathbb{R}^2\rightarrow \mathbb{R}$ be bounded measurable. How do you show that $E(f(X,Y)|Y) = h(Y)$ where $h(t) = E(f(X,t))$.
So, basically I have to show that $f(X,Y)$ is independent of $Y$. right? and how is it done?
As stated in the comments the actual statement that we should prove is, that $$\mathbb{E}[f(X,Y)|Y] = h(Y)$$ where $h(y)= \mathbb{E}[f(X,y)]$. Now clearly $h(Y)$ is $\sigma(Y)$-measurable and bounded, since $f$ is bounded. It remains to be proven, that
$$\int_A h(Y) \: dP = \int_A f(X,Y) \: dP$$ for any $A \in \sigma(Y)$. So let $A \in \sigma(Y)$ and write $A=\{Y \in B\}$ for some borel set $B \subseteq \mathbb{R}$ \begin{align*} \int_A f(X,Y) \: dP &= \int_\mathbb{R} \int_\mathbb{R} 1_{B}(y) f(x,y) P_X(dx)P_Y(dy) &\text{(by independence)} \\ &= \int_B \Big(\int_{\mathbb{R}}f(x,y)P_X(dx) \Big) P_Y(dy) \\ &=\int_B \mathbb{E}[f(X,y)] P_Y(dy) \\ &= \int_B h(y) P_Y(dy) \\ &=\int_A h(Y) \: dP \end{align*} And we have thus proven that $\mathbb{E}[f(X,Y)|Y] = h(Y)$.
It is very important to note, that in general $$h(Y) \neq \mathbb{E}[f(X,Y)]$$ and also in general $f(X,Y)$ is not independent of $Y$.