Let $X$ and $Y$ be real-valued random variables on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra.
Suppose $X$ is independent of $\mathcal F$ and $Y$ is $\mathcal F$-measurable. How can we show, that $$\operatorname P\left[X+Y\in A\mid Y=y\right]=\operatorname P\left[X+y\in A\right]\;\;\;\left(\operatorname P\circ Y^{-1}\right)\text{-almost surely}\tag 1$$ for all $A\in\mathcal B(\mathbb R)$ and $y\in\mathbb R$.
I was able to prove the similar result $$\operatorname P\left[X+Y\in A\mid\mathcal F\right]=\operatorname P\left[X+Y\in A\mid Y\right]\;\;\;\operatorname P\text{-almost surely}\;,\tag 2$$ but I've no idea how I need to start for $(1)$.
If $B_1$ and $B_2$ are Borel subsets of $\Bbb R$, then $$ \Bbb P[(X,Y)\in B_1\times B_2|\mathcal F](\omega)=\Bbb P[X\in B_1,Y\in B_2|\mathcal F](\omega) = \Bbb P[X\in B_1]\cdot 1_{\{Y\in B_2\}}(\omega), $$ $\Bbb P$-a.s., by the independence of $X$ from $\mathcal F$ and the $\mathcal F$-measurability of $Y$. The $\mathcal F$-measurable random variable on the right can be written as the Borel function $h(y):=\Bbb P[(X,y)\in B_1\times B_2]$ evaluated at $y=Y(\omega)$. A monotone class argument now shoes that $$ \Bbb P[(X,Y)\in C|\mathcal F](\omega)=\Bbb P[(X,y)\in C]\Big|_{y=Y(\omega)} $$ $\Bbb P$-a.s., for each Borel subset of $\Bbb R^2$. Your formula (1) now follows by taking $C=\{(x,y)\in\Bbb R^2: x+y\in A\}$.