Let $a=(a_{i})_{i\in\mathbb{N}}$ be a sequence of i.i.d. random variables taking values in $[\alpha,\beta]$ ($0<\alpha\le\beta$).
Let $X$ be a Borel-measurable function of $a\in\mathbb{R}^{\mathbb{N}}$ (measurable w.r.t. the smallest $\sigma$-algebra on $\mathbb{R}^{\mathbb{N}}$ for which all coordinate functions $a\mapsto a_{i}$ are Borel-measurable). Define $X_{n}(a_{1},\dots,a_{n}):=\mathbb{E}[X\mid a_{1},\dots,a_{n}]$ for any such function $X$.
Suppose we have $X_{n}\overset{L^{2}}{\rightarrow}Y$ where the $L^{2}$ convergence is taken with respect to the probability measure, i.e.
$$X_{n}\overset{L^{2}}{\rightarrow}Y\iff \lim_{n\to\infty}\mathbb{E}\left[(X_{n}-Y)^{2}\right]=0$$
It is said that, in this case, it implies
$$X_{n}=Y_{n}\text{ for a.e. }(a_{1},\dots,a_{n})\in\mathbb{R}^{n} \text{ and all } n\in\mathbb{N}$$
Why is that? I tried to use the fact that the conditional expectation minimizes the $L^{2}$ distance:
$$\mathbb{E}\left[(Y-Y_{n})^{2}\right]\le \mathbb{E}\left[(Y-X_{n})^{2}\right]$$
for all $n\in\mathbb{N}$ but it leads me nowhere since we can only bound the R.H.S. by $\epsilon$ for $n\geq N_{\epsilon}$.
(I'm assuming that $X\in L^2$ and that $Y_n:=\Bbb E[Y\,|\,a_1,\ldots,a_n]$.)
Method 1: Conditional expectation contracts the $L^2$-norm, so the $L^2$-distance between $Y_n$ and $\Bbb E[X_m\,|\,a_i,\ldots,a_n]$ is no larger than the $L^2$-distance between $Y$ and $X_m$; here $n$ is fixed and $m> n$. The latter distance converges to $0$ as $m\to\infty$ (by hypothesis). Thus $\Bbb E[X_m\,|\,a_i,\ldots,a_n]$ converges in $L^2$ to $Y_n$ as $m\to\infty$. But by the tower property of conditional expectation, $$ \Bbb E[X_m\,|\,a_1,\ldots,a_n]=\Bbb E[\Bbb E[X\,|\,a_1,\ldots,a_m]\,|\,a_1,\ldots a_n]=\Bbb E[X\,|\,a_1,\ldots,a_n]=X_n, $$ because $m>n$. It follows that $X_n=Y_n$ a.s., for each $n$.
Method 2: By the Martingale Convergence Theorem, $X_n$ converges a.s. and in $L^2$ to $X$. The hypothesis that $X_n$ converges in $L^2$ to $Y$ thus implies that $Y=X$ a.s., hence that $Y_n=\Bbb E[Y\,|\, a_1,\ldots,a_n]=\Bbb E[X\,|\, a_1,\ldots,a_n]=X_n$, a.s.