Is this function measurable? It is defined by an inner product

Question

Suppose $f\in L^2(X\times Y)$ where $X$ and $Y$ are finite measure spaces and we have a product measure $\mu$.

Suppose further that $g\in L^2(X)$, and $h:Y\to \mathbb{R}$ is defined by $h(y)=\int f(x,y)\overline{g(x)}dx$.

Can we say that $h$ is measurable? I want to show that $h\in L^2(Y)$. It is not too hard to show that the $L^2$ norm is finite. But of course this only makes sense if $h$ is measurable...

Answer 1

A part of the proof of Tonelli Theprem says that if we integrate a non-negative jointly measurable function of two variables w.r.t. one variable the result is a measurable function of the second variable. Now write $f$ and $g$ as linear combinations of non-negative measurable functions. [ $a+ib=a^{+}-a^{-}+i(b^{+}-b^{-})$ for $a, b \in \mathbb R$]. Since linear combinations of measurable functions are measurable we are done.

Answer 2

Let $\mu\times \nu$ be the product measure on $\mathscr A\times \mathscr B.$

Lemma: If $E\in \mathscr A\times \mathscr B,$ then the map $y\mapsto \mu(E^y)$ is $\nu$-measurable.

Sketch of proof: Let $\mathscr F$ be the set of all $E\in \mathscr A\times \mathscr B$ for which the map is measurable. Then it's easy to show that finite disjoint unions of measurable rectangles in $\mathscr A\times \mathscr B$ belong to $\mathscr F,$ so $\mathscr F$ is an algebra of subsets of $\mathscr A\times \mathscr B.$ Now, if $\{E_n\}\subseteq \mathscr F$ is a sequence that increases to $E$ then the functions $\mu(E_n^y)$ are measurable and increase pointwise to $\mu(E^y)$ so $\mu(E^y)$ is measurable and $E\in \mathscr F$. Finally, if $\{E_n\}\subseteq \mathscr F$ is a decreasing sequence, then $\mu(\bigcap E^y_n)=\lim \mu(E_n^y)$ because $\mu$ is finite and so $\mu(\bigcap E^y_n)$ is measurable which implies that $\bigcap E_n\in \mathscr F$ and the lemma follows from the monotone class theorem.

Now, you have a measurable function $f$ on the product space and want to show that $y\mapsto \int_Xf^yd\mu$ is measurable. Without loss of generality, $f>0.$ First suppose that $E \in \mathscr A\times \mathscr B$ and that $f=1_E$. Then, $\int_Xf^yd\mu=\mu(E^y)$, which is measurable, by the lemma. And so now the result is true for simple functions. So, take any sequence $f_n$ of simple functions increasing pointwise to $f$. Then, $g_n:=\int_Xf_n^yd\mu$ are measurable and an application of the monotone convergence theorem allows us to conclude.