Proof that $\int_X K(\cdot,y)f(y) \, d\mu(y)$ is measurable if $K, f\in L^2$? Let $(X,\mathscr A,\mu)$ be a $\sigma$-finite measure space, $K\in L^2(X^2,\mathscr A \otimes \mathscr A,\mu\otimes\mu),$ and $f\in L^2(X,\mathscr A,\mu).$
I am trying to prove that $F(x)=\int_X K(x,y)f(y) \, d\mu(y)$ is $\mathscr A$-measurable. The book I am reading claims it is, without proof.*
Here is my current progress:
If we can show that $\int_{X^2}|K(x,y)|\,|f(y)|\,d(\mu\otimes\mu)<\infty,$ then Fubini's theorem immediately guarantees that $F$ is measurable. Using Fubini's theorem on $K$ gives $|K(x,\cdot)|^2\in L^1$ a.e., which is equivalent to $K(x,\cdot)\in L^2$ a.e. So Cauchy-Schwartz gives:
$$\int_X |K(x,y)| |f(y)|\, d\mu(y)\leq \| K(x,\cdot)\|_2 \|f\|_2\qquad \text{a.e.}$$
In the comments, Kavi Rama Murthy suggests squaring both sides of this inequality and integrating in $x$. This gives
$$\int_X\left(\int_X |K(x,y)| |f(y)|\, d\mu(y)\right)^2 d\mu(x)\leq \| K(x,\cdot)\|_2^2 \|f\|_2^2.$$
So perhaps the left side here simplifies to the expression $\int_{X^2}|K||f|d(\mu\otimes\mu)?$

*V. Moretti, Spectral Theory and Quantum Mechanics. Example 4.16 (4).
 A: Some versions of Fubini's theorem include the following. Let $(X,\mathcal{F}_X,\mu)$ and $(Y,\mathcal{F}_Y,\nu)$ are $\sigma-$finite. If $g:X\times Y\to\mathbb{R}$ is $\mathcal{F}_X\times\mathcal{F}_Y-$measurable and $g \in L^1(\mu\times\nu)$, then
$$x\to \int_{Y}g(x,y)d\nu(y) $$
is not just $\mathcal{F}_X-$measurable, but is also a function in $L^1(\mu)$. In your example you have $g(x,y)=K(x,y)f(y)$.
I actually don't think that $g$ will necessarily be integrable with respect to the product measure, but I think we can get the result using $\sigma-$finiteness. Let $\{A_n\}$ be an increasing sequence of sets of finite measure which expand to $X$. Then consider the functions $g_n(x,y)=1_{A_n}(x)K(x,y)f(y)$ and call $F_n(x)=\int_X 1_{A_n}(x)K(x,y)f(y)d\mu(y)=F(x)1_{A_n}(x)$. Then for any $n$ we have that
\begin{align*}\int_X\int_X |g_n(x,y)|d\mu(y)d\mu(x) &=\int_X\int_X |1_{A_n}(x)K(x,y)f(y)|d\mu(y)d\mu(x)\\
&= \int_{A_n}\int_X |K(x,y)f(y)|d\mu(y)d\mu(x)\\
&\leq \int_{A_n}||K(x,\cdot)||_2 ||f||_2 d\mu(x)\\
&= ||f||_2\int_{A_n}||K(x,\cdot)||_2 d\mu(x)
\end{align*}
Now the function $x\to ||K(x,\cdot)||_2^2$ is measurable by Fubini's Theorem and the assumption that $K\in L^2(X^2)$. Since $z\to\sqrt{z}$ is a Borel function, the composition $x\to \sqrt{||K(x,\cdot)||_2^2}=||K(x,\cdot)||_2$ is measurable. Hence the set $B=\{x: ||K(x,\cdot)||_2\leq 1\}$ is measurable. Hence
\begin{align*}
\int_{A_n}||K(x,\cdot)||_2 d\mu(x) &= \int_{A_n\cap B}||K(x,\cdot)||_2 d\mu(x) + \int_{A_n\cap B^c}||K(x,\cdot)||_2 d\mu(x)\\
&\leq  \int_{A_n\cap B}1 d\mu(x) + \int_{A_n\cap B^c}||K(x,\cdot)||_2^2 d\mu(x)\\
&= \mu(A_n\cap B) + \int_{A_n\cap B^c}\int_X |K(x,y)|^2 d\mu(y)d\mu(x)\\
&\leq \mu(A_n\cap B) +||K||_2
\end{align*}
Now both terms on the right are finite since $\mu(A_n)<\infty$ and $K\in L^2$ by assumption. This then shows that $\int_X\int_X |g_n(x,y)|d\mu(y)d\mu(x) < \infty$ and by Fubini's Theorem we may conclude that $F_n(x)=F(x)1_{A_n}(x)$ is a measurable function. Well $F_n(x)\to F(x)$ almost everywhere, so $F$ is the pointwise limit a.e. of measurable functions, hence $F$ is a measurable function. Do note that $F$ may not be integrable.
