I have a basic question about weak limits that I hope someone can clarify. Let $(\Omega,\mathcal{F},P)$ where $\Omega \subseteq \mathbb{R}^k$ be a probability space and let $\{f_n\}$ be a sequence of functions in $L^2(\Omega,\mathcal{F},P).$ Suppose $\{f_n\}$ is bounded.
Then I think it is true that there is a subsequence $\{f_{n_m}\}$ that converges weakly to an $f \in L^2(\Omega,\mathcal{F},P)$ (please correct me if I am wrong here). Let $\mathcal{G}$ be a sub-sigma algebra of $\mathcal{F}$ and suppose each $f_n$ is in fact measurable with respect to $\mathcal{G}$. Then is it true that $f$ is also measurable with respect to $\mathcal{G}$?
To put it differently, it is as if I really have a sequence $\{f_{n_m}\}$ in $L^2(\Omega,\mathcal{G},P)$ with a weak limit say $\bar{f} \in L^2(\Omega,\mathcal{F},P)$. But if $\{f_{n_m}\}$ is viewed as a sequence in $L^2(\Omega,\mathcal{F},P)$, does it have the same weak limit $\bar{f}$?
To save on subscripts, let's just assume that the original subsequence converges weakly in $L^2(\Omega,\mathcal F,P)$ to $f$. I claim that $f\in L^2(\Omega,\mathcal G,P)$. To show this I use the fact that $[L^2(\Omega,\mathcal G,P)]^{\perp\,\perp}$ is equal to $L^2(\Omega,\mathcal G,P)$ because the latter is closed in $L^2(\Omega,\mathcal F,P)$. (The superscript ${}^\perp$ means orthogonal complement relative to the ambient space $L^2(\Omega,\mathcal F,P)$.) So let $h$ be an element of $[L^2(\Omega,\mathcal G,P)]^\perp$. Then $$ \langle f,h\rangle=\lim_n\langle f_n,h\rangle=0, $$ because each $f_n$ is an element of $L^2(\Omega,\mathcal G,P)$, and is therefore orthogonal to $h$. Thus $f\perp h$, so (as $h$ was arbitrary) $f\perp [L^2(\Omega,\mathcal G,P)]^\perp$. That is, $f\in [L^2(\Omega,\mathcal G,P)]^{\perp\,\perp}=L^2(\Omega,\mathcal G,P)$.
Of course, $\{f_n\}$ also converges weakly in $L^2(\Omega,\mathcal G,P)$ to $f$.