Let $(\Omega, \mathcal{A},P)$ be a probability space, and let $(\mathcal{F}\_k)\_{k \geq 1}$ be a filtration which converges to $\mathcal{A}$. I suppose it is true that 
$$
E \left( \big(E \left( X | \mathcal{F}\_k \right) \big)^2 \right) \to E \left(X^2 \right).
$$
How to prove this? I guess one will need Jensen's inequality and a convergence theorem for submartingales, but cannot find the right reference (I am working on number theory, not probability).


(If you know an answer, please also provide a citable reference.)



Convergence of conditional second moments
Let us state Corollary 2.1 of [these notes](http://www.statslab.cam.ac.uk/~ps422/mynotes.pdf).



> 
> Let $p>1$, $X\in\mathbb L^p$ and let $\left(\mathcal F\_n\right)\_{n\geqslant 1}$ be a filtration. Denote by $\mathcal F$ the $\sigma$-algebra generated by $\bigcup\_{n\geqslant 1}\mathcal F\_n$. Then the convergence 
>  $$\lim\_{n\to +\infty} \mathbb E\left[X\mid\mathcal F\_n\right]=\mathbb E\left[X\mid\mathcal F\right],$$
>  where the convergence holds almost everywhere and in $\mathbb L^p$.
> 
> 
> 


The wanted result follows from this corollary and the elementary fact that if $\left\lVert Y\_n-Y\right\rVert\_2\to 0$ then $\left\lVert Y\_n\right\rVert\_2^2\to \left\lVert Y\right\rVert\_2$.


