This is Theorem 4.1.8 of Durrett $3^{rd}$, stating as follows:
Suppose $\mathbb{E}X^{2}<\infty$. Then $\mathbb{E}(X|\mathcal{F})$ is the variable $Y\in\mathcal{F}$ that minimizes the ``mean square error'' $\mathbb{E}(X-Y)^{2}$.
In the proof, he let $Y\in L_{2}(\mathcal{F})$ and $Z:=\mathbb{E}(X|\mathcal{F})-Y$, then he get following computation:
$$\mathbb{E}(X-Y)^{2}=\mathbb{E}(X-\mathbb{E}(X|\mathcal{F})+Z)^{2}=\mathbb{E}(X-\mathbb{E}(X|\mathcal{F})^{2}+\mathbb{E}Z^{2},$$ and he said, which confused me, that
from this formula, it is easy to see that $\mathbb{E}(X-Y)^{2}$ minimizes when $Z=0$.
...Well I did not see why this is true even thought he claims that this is easy.
How did he know when $Z=0$, $\mathbb{E}(X-Y)^{2}$ minimizes? Taking derivative or something else?
I am really confused here, thank you so much in advance for any explanations!
Since $\mathbb EZ^2\geq 0$, it follows that for all $Y$ whatsoever, $$ \mathbb E(X-Y)^2\geq \mathbb E\bigl(X-\mathbb E(X\mid \mathcal F)\bigr)^2. $$ Now, choosing $Y=\mathbb E(X\mid \mathcal F)$ shows that equality is attained, so that $\mathbb E(X\mid F)$ is a minimizer.