Increasing sigma algebras and convergence of random variables Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space, $\mathscr{F}_n\subset \mathscr{F}$ an increasing subsequence of $\sigma$-fields, and $\mathscr{F}_\infty$ the $\sigma$-field generated by $\bigcup_n\mathscr{F}_n$. For any $X\in \mathrm{L}^2(\mathscr{F}_\infty, \mathrm{P})$, show that there exist $X_n\in \mathrm{L}^2(\mathscr{F}_n, \mathrm{P})$ such that $\lim_{n \to \infty}\mathrm{E}[\mid X_n-X\mid^2]=0$.
I do not know how to approach this. I cannot use martingles. My knowledge about conditional expectation is in the basic level of undergraduate course on probability. 
 A: Ok, let's take $X_{n}=E(X|\mathscr{F}_{n})$. By definition, $X_{n}$ is $\mathscr{F}_{n}$-measurable.
Using that $(\mathscr{F}_{n})_{n}$ is increasing and the definition of conditional expectation, it's easy to see that $E(X_{m}|\mathscr{F}_{n})=X_{n}$ for $m\geq n$. With this property, you can prove that the sequence of increments $(X_{n+1}-X_{n})_{n}$ is orthogonal in $L^{2}(\Omega,\mathscr{F}_{\infty})$. By making the telescopic sum for $X_{n}-X_{1}$ and taking norms, you obtain $\|X_{n}-X_{1}\|_{2}=\displaystyle\sum_{i=1}^{n-1}\|X_{i+1}-X_{i}\|_{2}$ and similarly, if $m\geq n$, $\|X_{m}-X_{n}\|_{2}=\displaystyle\sum_{i=m}^{n-1}\|X_{i+1}-X_{i}\|_{2}$
Since that $(X_{n})$ is a bounded sequence in $L^{2}(\Omega,\mathscr{F}_{\infty})$, the first sum is convergent, and then, by looking at the second sum, we see that $(X_{n})$ is a Cauchy sequence in $L^{2}(\Omega,\mathscr{F}_{\infty})$.
So, there exists $Z\in L^{2}(\Omega,\mathscr{F}_{\infty})$ such that $X_{n}\longrightarrow Z$ in $L^{2}(\Omega,\mathscr{F}_{\infty})$
Now, since that $E(-|\mathscr{F}_{n})$ is a continuous operator in $L^{2}$ and $E(X_{m}|\mathscr{F}_{n})=X_{n}$ for $m\geq n$, when $m$ goes to infinity, we have $E(Z|\mathscr{F}_{n})=X_{n}=E(X|\mathscr{F}_{n})$. Therefore $\int_{A}X\;dP=\int_{A}Z\;dP\;\forall A\in\bigcup_{n}\mathscr{F}_{n}$. By the $\pi-\lambda$-theorem, we have $\int_{A}X\;dP=\int_{A}Z\;dP\;\forall A\in\mathscr{F}_{\infty}$. This implies that $Z=X$ almost surely.
