Martingales application Let $X$ be a random variable, $X\in\mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$, such that: $\mathbb{E}|X|<\infty$ and consider $\mathbb{F}$ a filtration.
Define: $M_n=\mathbb{E}[X|\mathcal{F}_n]$, $n\geq 0$. Observe using the expectations tower property that: $M_n=\mathbb{E}[M_{n+1}|\mathcal{F}_n]$ and therefore the process $M$ is a martingale.
Now define: $\nu_n:=\mathbb{E}(M_n-X)^2$. 
Show that $\nu_n$ is decreasing.
My idea is to use the following proposition about conditional expectations:
Let $X\in\mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G}$ a sub-$\sigma$-algebra of $\mathcal{F}$.
If $X'$ is a version of $\mathbb{E}[X|\mathcal{G}]$, then $X'\in\mathcal{L}^2(\Omega,\mathcal{G},\mathbb{P})$ and the following holds:
$\mathbb{E}(X-Y)^2=\mathbb{E}(X-X')^2 +\mathbb{E}(X'-Y)^2$, for all $Y\in \mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$.
Any help would be really welcome.
 A: As @PetiteEtincelle pointed out, the mentioned equality holds only for $Y \in L^2(\mathcal{G})$. The proposition can rephrased as follows:

Let $X \in L^2(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G}$ a sub-$\sigma$-algebra of $\mathcal{F}$. Then $$\mathbb{E} \big(|X-\mathbb{E}(X \mid \mathcal{G})|^2 \big)= \inf\{\mathbb{E}(|X-Y|^2); Y \in L^2(\mathcal{G})\}.$$

For two sub-$\sigma$-algebras $\mathcal{G}$, $\mathcal{H}$ satisfying $\mathcal{G} \subseteq \mathcal{H}$, it follows easily that
$$\begin{align}\mathbb{E} \big(|X-\mathbb{E}(X \mid \mathcal{G})|^2 \big) &= \inf\{\mathbb{E}(|X-Y|^2); Y \in L^2(\mathcal{G})\} \\ &\geq \inf\{\mathbb{E}(|X-Y|^2); Y \in L^2(\mathcal{H})\} \\ &= \mathbb{E} \big(|X-\mathbb{E}(X \mid \mathcal{H})|^2 \big) \end{align}$$
as $L^2(\mathcal{G}) \subseteq L^2(\mathcal{H})$. Applying this for $\mathcal{G} := \mathcal{F}_n$ and $\mathcal{H} := \mathcal{F}_{n+1}$ yields the claim.
If we interpret $\sigma$-algebras as information, then $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$ means that $\mathcal{F}_{n+1}$ does contain more information. The monotonicity of $\nu_n$ therefore shows that the approximation $\mathbb{E}(X \mid \mathcal{F}_n)$ becomes better (in mean-square sense) the more information we have.
A: Your idea is good, except that $\mathbb{E}(X-Y)^2=\mathbb{E}(X-X')^2 +\mathbb{E}(X'-Y)^2$ is true for all $Y\in \mathcal{L}^2(\Omega,\mathcal{G},\mathbb{P})$.
So we have $$\mathbb{E}(X-M_{n})^2=\mathbb{E}(X-M_{n+1})^2 +\mathbb{E}(M_{n+1}-M_{n})^2 \geq \mathbb{E}(X-M_{n+1})^2$$ since $M_n$ is $\mathcal{F}_{n+1}$ measurable. 
It is just your idea applied for $\mathcal{G} = \mathcal{F}_{n+1}$ and $Y = X_n$
