Doob–Dynkin lemma: Let $X$,$Y$ $:\Omega \rightarrow R^n$ be random elements and $\sigma(X)$ be the $\sigma$ algebra generated by $X$. Then $Y$ is $\sigma(X)$-measurable if and only if $Y=g(X)$ for some Borel measurable function $g:R^n\rightarrow R^n$.
Now I want to use this lemma to simplify $\sum_{n=0}^{\infty}E[f(X_{n+1}) \vert \sigma(X_n)] 1_{\{N=n\}}$ where $N<\infty$ is any discrete stopping time and $f$ is a bounded measurable function. I know that by definition, conditional expectation $E[f(X_{n+1}) \vert \sigma(X_n)]$ is $\sigma(X_n)$-measurable. So, by applying Doob–Dynkin lemma, $E[f(X_{n+1}) \vert \sigma(X_n)]= g(X_n)$ and then, \begin{equation} \sum_{n=0}^{\infty}E[f(X_{n+1}) \vert \sigma(X_n)] 1_{\{N=n\}} = \sum_{n=0}^{\infty} g(X_n) 1_{\{N=n\}} \hspace{5mm}(1) \end{equation}
But I am not sure about the next step. I think it is wrong to substitute $n$ in $g(X_n)$ by $N$ because in that case, it means that $E[f(X_{N+1}) \vert \sigma(X_N)]= g(X_N)$ which doesn't seem correct to me because without fixing the value of $N$, the definition of $\sigma(X_N)$ is not clear to me. If we do this seemingly wrong substitution, then
$\sum_{n=0}^{\infty}E[f(X_{n+1}) \vert \sigma(X_n)] 1_{\{N=n\}} = \sum_{n=0}^{\infty} g(X_n) 1_{\{N=n\}} = g(X_N) \sum_{n=0}^{\infty} 1_{\{N=n\}} = g(X_N). $
But As I mentioned above, I think this substitution is wrong and (1) at the end should also depend on $N$. Am I right? Any idea for simplifying (1)?
Better to write $E[f(X_{n+1})|\sigma(X_n)] = g(n,X_n)$ to acknowledge that the function representing the conditional expectation depends on the joint distribution of $X_n$ and $X_{n+1}$ (as well as on the function $f$). With his fix, your computation shows that $E[f(X_{N+1})|\sigma(X_N)]=g(N,X_N)$, a.s. on $\{N<\infty\}$.