conditional expectation wrt sigma field identity I'm reading a proof which seems to make use of an identity, below, that I can't prove.
Let $f$ be a Borel measurable function, assume that $X$ is measurable with respect to $\mathcal{F}$ and $Y$ is independent of $\mathcal{F}$.
Then $\mathbb{E}[f(X+Y) |\mathcal{F}] = \int f(y+X) P_Y(dy)$. Can some one verify if this is true? Thanks.
attempt: take limit of indicator functions. in the case where $f$ is the indicator function on the set $(\infty,z]$ this reduces to showing $\mathbb{P} [X+Y \leq z | \mathcal{F}] = \int_{-\infty}^{z-X} P_Y(dy)$ but I cannot prove this.
 A: Let $\phi:x\mapsto \int_{\mathbb R} f(x+y) dP_Y(y)$.
For any $A\in \mathcal F$ we show that $E[1_A f(X+Y)]=E[1_A \phi(X)]$. This will show that $E[f(X+Y)|\mathcal F]=\phi(X)$ a.s.
Let $Z:(\Omega,\mathcal A)\to (\Omega,\mathcal F)$, $\omega\mapsto \omega$ and note that $\mathcal F = \sigma(Z)$. Since $X$ is $\mathcal F$-measurable, by the Doob–Dynkin lemma, there is some measurable $g:(\Omega,\mathcal F)\to (\mathbb R, \mathcal B(\mathbb R))$ such that $X = g(Z)$.
Note that $A=Z^{-1}(A)$ thus
$$\begin{align}
E[1_A \phi(X)] 
&= E[1_{A}(Z) \phi(g(Z))] \\
&= \int_\Omega 1_A(z) \Big(\int_{\mathbb R} f\big(g(z)+y\big) dP_Y(y) \Big)dP_Z(z) \tag {1}\\
&= \int_{\mathbb R \times \Omega} 1_A(z)  f\big(g(z)+y\big) d(P_{Y}\otimes P_{Z})(y,z) \tag {2}\\
&= \int_{\mathbb R \times \Omega} 1_A(z)  f\big(g(z)+y\big) dP_{(Y,Z)}(y,z) \tag {3}\\
&= E[1_A(Z)f\big(g(Z)+Y \big)] \tag{4}\\
&= E[1_Af\big(X+Y \big)] 
\end{align}$$
$(1)$: Law of the unconscious statistician (or integration w.r.t. pushforward measure)
$(2)$: Fubini's theorem
$(3)$: $Y$ is independent of $\mathcal F$, hence $Y$ and $Z$ are independent
$(4)$: Law of the unconscious statistician (or integration w.r.t. pushforward measure)
