Let $X^1, X^2, X^3,\ldots $ be $L^2$-integrable martingales. Assume that for each $t\in[0,T]$ there exists an $X_t\in L^2(\Omega)$, such that $X_t^n\rightarrow X_t$ in $L^2(\Omega)$ for $n\rightarrow\infty$.
I want to show, that $(X_t)_{t\in[0,T]}$ is a martingale. This is the first part of Exercise 21.4.3 taken from Achim Klenke, Probability Theory.
The objective is to find, that $E[X_t|\mathcal{F}_s]=X_s$ for any $s\le t$. The confusing part for me is, that this should hold almost surely, right? My rather foolish attempt would be to use the dominated convergence theorem for conditional expectations, but in this case I would need a $Y_t\in L^1(\Omega)$ with $|X^n_t|\le Y_t$ and that $X_t^n\rightarrow X_t$ almost surely for $n\rightarrow\infty$, such that each equality of $$E[X_t|\mathcal{F}_s]=\lim_{n\rightarrow\infty}E[X_t^n|\mathcal{F}_s]=\lim_{n\rightarrow\infty}X_s^n=X_s$$ holds almost surely and not only in $L^1(\Omega)$. Or am I overcomplicating things here?
Any hint or help is appreciated. Thank you in advance.
Dominated convergence theorem cannot be used here since there is no dominating function.
By Jensen's inequality for conditional expectations we get $E(E[(X_t^{n}-X_t)|\mathcal F_s])^{2}\le E(E[(X_t^{n}-X_t)^{2}|\mathcal F_s])=E(X_t^{n}-X_t)^{2} \to 0$. Thus, $E[X_t^{n}|\mathcal F_s) \to E[X_t|\mathcal F_s)$ in $L^{2}$. This implies almost sure convergence along a subsequence. Taking limits in $E[X_t^{n}|\mathcal F_s) =X_s^{n}$ we get $E[X_t|\mathcal F_s) =X_s$.