If $X,Y,Z:\Omega\to\mathbb{R}$ are random variables, do we have $$\mathrm{E}[(X-\mathrm{E}[X|Y])^2]=\mathrm{E}[(X-\mathrm{E}[X|Y,Z])^2]+\mathrm{E}[(\mathrm{E}[X|Y,Z]-\mathrm{E}[X|Y])^2]?$$
Attempt
(1) $$\begin{aligned} \mathrm{E}[(X-\mathrm{E}[X|Y])^2] &= \mathrm{E}[(X-\mathrm{E}[X|Y,Z]+\mathrm{E}[X|Y,Z]-\mathrm{E}[X|Y])^2]\\ &= \mathrm{E}[(X-\mathrm{E}[X|Y,Z])^2]+\mathrm{E}[(X-\mathrm{E}[X|Y])^2]\\ &\phantom{{}={}}+2\mathrm{E}[(X-\mathrm{E}[X|Y,Z])(\mathrm{E}[X|Y,Z]-\mathrm{E}[X|Y])]. \end{aligned}$$
(2) We only need $\mathrm{E}[(X-\mathrm{E}[X|Y,Z])(\mathrm{E}[X|Y,Z]-\mathrm{E}[X|Y])]$ to be zero. By the property of conditional exception, I think we have $\mathrm{E}[(X-\mathrm{E}[X|Y,Z])(\mathrm{E}[X|Y,Z])]=0$. I don't know whether it is correct, since $\mathrm{E}[X|Y,Z]$ is function on product space $\Omega\times\Omega$, but $X$ is on $ \Omega$.
But i don't know hot to deal with $\mathrm{E}[X|Y]$. Is it measurable with respect to $\sigma(Y,Z)$?
The very first line says $X,Y,Z$ are all defined on $\Omega$. $\Omega\times \Omega$ does not come into the picture at all.
$\mathrm{E}[(X-\mathrm{E}[X|Y,Z])(\mathrm{E}[X|Y,Z])]=0$ is correct. In general $E[(X-E(X|\mathcal G))E(X|\mathcal G)]=0$.
$E(X|Y)$ is measurable w.r.t. $\sigma (Y)$ and $\sigma (Y)\subseteq \sigma (Y,Z)$ so $E(X|Y)$ is measurable w.r.t $\sigma (Y,Z)$