Show that $\Omega \setminus N_A \ni x \mapsto 1_A(x) \cdot \int\limits_{\Upsilon} k(x,t)f(t)d\nu(t)$ is measurable Assume $(\Omega,\mathcal{A},\mu)$ and $(\Upsilon,\mathcal{B},\nu)$ are measure spaces with $\sigma$-finite measures $\mu, \nu$ and $k \in L^2(\Omega \times \Upsilon, \mathcal{A} \otimes \mathcal{B}, \mu \times \nu)$. I have shown for all $A \in \mathcal{A}$ with $\mu(A) < \+ \infty$ and for all $f \in L^2(\Upsilon,\mathcal{B},\nu,\mathbb{C})$ that the function $(x,t) \mapsto 1_A(x) \cdot \int\limits_{\Upsilon} k(x,t)f(t) d\nu(t)$ is integrable. What I want to do now is to follow that for a null set $N_A \in \mathcal{A}$ the function $g: \Omega \setminus N_A \ni x \mapsto 1_A(x) \cdot \int\limits_{\Upsilon} k(x,t)f(t)d\nu(t)$ is measurable. My idea is that since the the integral of $g$ is finite it is finite almost everywhere. Now I define the set $N_A$ as the points $x$ for which $g(x)$ is not finite. Is it now guaranteed that $g$ is measurable?

$\int\limits_{\Omega\times\Upsilon} 1_A(x) k(x,t)f(t) d(\mu\times\
u)(x,t)=\int\limits_{\Omega}1_A(x) \int\limits_{\Upsilon} k(x,t)f(t) d\
u(t) d\mu(x)$

Follows from Fubini Tonelli because the following function is measurable and integrable.

$\Omega \times \Upsilon \
i(x,t) \mapsto 1_A(x) k(x,t)f(t)$

And since the first integral is finite $g$ as the inner integrand of the right hand side is also integrable and therefore measurable.

BUT the inner integrand doesn't necessarily exist everywhere, it only has to be defined almost everywhere and we therefore have to exclude the set $N_A$ as the set of all $x$ for which $1_A(x) \int\limits_{\Upsilon} k(x,t)f(t) d\
u(t)$ doesn't exist.