Prove $E[f(X,Y)\mid Y]=E[f(X,Y)\mid Y,Z]$ if $X$ is independent of $Y$ and $Z$ I would like to prove the following claim (which I think is true):
Suppose $\mathcal{F}_1 \subset \mathcal{F}_2$ are two $\sigma$-fields, $X,Y$ are random variables, and $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is a measurable function. If $X$ is independent of $\mathcal{F}_2$, $Y\in\mathcal{F}_1$, then $E[f(X,Y)\mid \mathcal{F}_1] = E[f(X,Y)\mid \mathcal{F}_2]$.
Intuitively, this is just saying $E[f(X,Y)\mid Y]=E[f(X,Y)\mid Y,Z]$ if $X$ is independent of the pair $(Y,Z)$, but I don't know how to prove the above claim rigorously.
Thanks in advance!
 A: Let $F$ be an element of $\mathcal F_2$. Since $X$ is independent of $\mathcal F_2$, we have
$$\tag{*}     \mathbb E\left[f\left(X,Y\right)\mathbf 1_F   \right]=\int_{\mathbb R} \mathbb E\left[f\left(x,Y\right)\mathbf 1_F   \right]  \mathrm dP_X(x)         $$
and for all real number $x$, $\mathbb E\left[f\left(x,Y\right)\mathbf 1_F   \right] =\mathbb E\left[f\left(x,Y\right)\mathbb E\left[   \mathbf 1_F   \mid \mathcal F_1  \right]\right]  $ hence 
$$\mathbb E\left[f\left(X,Y\right)\mathbf 1_F   \right]=\mathbb E\left[f\left(X,Y\right)\mathbb E\left[   \mathbf 1_F   \mid \mathcal F_1  \right]  \right] .$$
Let $Z:=f\left(X,Y\right)$. Since 
$$\mathbb E\left[Z\mathbb E\left[   \mathbf 1_F   \mid \mathcal F_1  \right] \mid\mathcal F_1    \right]= \mathbb E\left[   \mathbf 1_F   \mid \mathcal F_1  \right]\mathbb E\left[Z \mid\mathcal F_1    \right]= \mathbb E\left[ \mathbf 1_F \mathbb E\left[  Z  \mid \mathcal F_1  \right] \mid\mathcal F_1    \right],    $$
we have 
$$\mathbb E\left[f\left(X,Y\right)\mathbb E\left[   \mathbf 1_F   \mid \mathcal F_1  \right]  \right]=\mathbb E\left[\mathbb E\left[   f\left(X,Y\right)\mid\mathcal F_1  \right]    \mathbf 1_F      \right].$$
We showed that for all $F\in\mathcal F_2$,
$$
\mathbb E\left[f\left(X,Y\right)\mathbf 1_F   \right]=\mathbb E\left[\mathbb E\left[   f\left(X,Y\right)\mid\mathcal F_1  \right]    \mathbf 1_F      \right],
$$
which is what we wanted.
Proof of (*): we use the following fact: if $X$ is independent of a vector $(V_1,V_2)$, then for any function $g\colon \mathbb R^3\to\mathbb R$, 
$$\mathbb E\left[g \left(X,V_1,V_2\right) \right]  =\int_{\mathbb R} \mathbb E\left[g\left(x,V_1,V_2\right)  \right]  \mathrm dP_X(x),$$
which follows from an application of Fubini's theorem and the fact that the law of $    \left(X,V_1,V_2\right)$ is the product measure between the laws of $X$ and $\left(V_1,V_2\right)$. We then apply this to $V_1=Y$, $V_2=\mathbf 1_F$ and $g \colon (x,v_1,v_2) \mapsto f(x,v_1)v_2$.
