Given two compact Hausdorff spaces $K$ and $L$, a bounded and separately continuous function $f:K\times L\to \mathbb C$, and a complex measure with finite variation $\mu$ on $L$ endowed with the Borel $\sigma$-algebra, the parameter integral $$F(x)= \int_L f(x,y) d\mu(y)$$ defines a continuous function on $K$.
That $F$ is sequentially continuous (and hence continuous if $K$ is metrizable) is an immediate consequence of Lebegue'sLebesgue's theorem on dominated convergence but the proof (that I know) for the general case is much harder and uses a deep functional analytic result of Grothendieck that a uniformly bounded subset of $C(L)$ is compact in the weak topology if (and only if) it is compact with repect to the topology of pointwise convergence. (The map $\phi:K\to C(L)$, $x\mapsto f(x,\cdot)$ is continuous if $C(L)$ is endowed with the pointwise topology and hence has a compact image $A$ which, by Grothendieck's theorem, is thus weakly compact so that on $A$ both topologies coincide; hence $\phi: K\to C(L)$ is weakly continuous and thus $F$ is continuous as the composition of $\phi$ with the (weakly) continuous linear functional $g\mapsto \int_Lg d\mu$).
Question: Is there a purely measure theoretic proof?