Conditional identity of conditional expectations

Question

Suppose that random variables $X_1,\ldots,X_{n+1}$ are such that, conditionally on an event $A$ (assumed $\mathcal{F}_j$-measurable for some $j\in\{1,\ldots,n\}$, where $\mathcal{F}_j$ is the natural filtration), $X_{n+1}= X$ where $X$ is some other given independent random variable. Intuitively it should hold that conditionally on $A$ then $\mathbb{E}_{\mathcal{F}_n}X_{n+1}=\mathbb{E}X$. How does one justifies this rigorously? I was thinking of removing the conditioning on the filtration using independence of $X$, but before doing that, I should show that conditionally on $A$, $\mathbb{E}_{\mathcal{F}_n}X_{n+1}=\mathbb{E}_{\mathcal{F}_n}X$. But how does one show a conditional (w.r.t. $A$) identity between random variables? My attempt (EDIT): to show that $\mathbb{E}_{\mathcal{F}_n}X_{n+1}=\mathbb{E}_{\mathcal{F}_n}X$ in general, one usually considers an arbitrary event $F\in\mathcal{F}_n$ and and proves that $\int_F\mathbb{E}_{\mathcal{F}_n}X_{n+1}\mathbb{P}(d\omega)=\int_F\mathbb{E}_{\mathcal{F}_n}X\mathbb{P}(d\omega)$, so to show that they are (almost surely) equal conditionally on the event (assumed nonnegligible) A, perhaps (claim 1) we should show that $\int_F\mathbb{E}_{\mathcal{F}_n}X_{n+1}\mathbb{P}(d\omega|A)=\int_F\mathbb{E}_{\mathcal{F}_n}X\mathbb{P}(d\omega|A)$ where $\mathbb{P}(d\omega|A)$ is the regular conditional probability measure. If this is a correct starting point, it should follow, by definition of the conditional expectation, that $\int_F\mathbb{E}_{\mathcal{F}_n}X_{n+1}\mathbb{P}(d\omega|A)=\int_FX_{n+1}\mathbb{P}(d\omega|A)$ and $\int_F\mathbb{E}_{\mathcal{F}_n}X\mathbb{P}(d\omega|A)=\int_FX\mathbb{P}(d\omega|A)$. Furthermore being $X_{n+1}=X$ conditionally on $A$ should imply (claim 2) that for any such $F$, $\int_FX_{n+1}\mathbb{P}(d\omega|A)= \int_FX\mathbb{P}(d\omega|A)$, yielding the result. The things I am not sure how to show rigorously is whether the claims 1,2 are correct. Thank you for any help.