Let $(X_i,\mathcal{F}_i,\mu_i)$, $i\in \{1,2\}$, be a $\sigma$-finite measure space and $E_i$ an ideal of measurable functions on $X_i$ with full carrier (for example $E_i=L^p(X_i,\mu_i)$).
A positive linear operator $T\colon E_1\to E_2$ has an integral kernel if $$Tf(x)=\int k(x,y) f(y)\,d\mu_1(y)$$ for some measurable function $k\colon E_2\times E_1\to\mathbb{R}_+$.
It is known* that $T$ has an integral kernel if and only if for every sequence $(u_n)$ in $E_1$ such that $0\leq u_n\leq u$ for some $u\in E_1$ and $u_n\to 0$ locally in measure the sequence $(Tu_n)$ converges to $0$ almost everywhere.
I am wondering if there is a similar result for the existence of a continuous integral kernel. To be more precise:
If the spaces $X_i$ carry some reasonable topology (e.g. locally compact Polish), $\mathcal{F}_i$ is the Borel $\sigma$-algebra and $\mu_i$ is a Radon measure, is there a similar criterion for a linear operator between spaces of measurable functions to have a continuous integral kernel?
*See for example A.R. Schep, Kernel Operators, Nederl. Akad. Wetensch. Indag. Math. (1979), although this result seems to go back to Buhvalov.
