Why do we need square integrability in showing $E(Y|X)$ minimizes expected quadratic loss?

Question

I've read about that $E(Y|X)=\underset{f(x)\in \mathcal{F}}{\arg\min} E(Y-f(X))^2$, where $\mathcal{F}$ is the set of all square integrable functions in $x$. The proof of this result is simple and does not seem to use square integrability of function $f(x)$. I'm wondering why do we restrict $f(x)$ to be square integrable? If this restriction is relaxed for $\mathcal{F}$, does $E(Y|X)$ fail to be the argmin?

Answer 1

The correct restriction to make is to ensure square-integrability of $f(X)$, not necessarily $f$; i.e. $E(f(X)^2) < \infty$. Eventually in the proof of this you want to show that $$E[(Y- E(Y|X))(E(Y|X) - f(X)) ] = 0$$ for all reasonably defined $f(X)$. Before you prove that expectation is zero, you should prove it exists, something which the Cauchy-Schwartz inequality affords you. That requires the square-integrability of each of the factors, hence of $f(X)$.