Let $\Delta[0,1]$ be the space of all signed Borel measures over $[0,1]$, ranging between -1 and 1. $[0,1]^{[0,1]}$ is the set of all functions with domain $[0,1]$ that takes values in $[0,1]$. Consider the functional $V: \Delta [0,1] \times [0,1]^{[0,1]}, V: (\mu,f) \mapsto \int fd\mu$. My question is, is $V$ continuous in $(\mu,f)$? Or, what additional conditions do we need for it to be?
Background: I'm trying to apply the Maximum theorem, where my objective function is of the form $\int fd\mu$.
If we put the weak$^*$-topology on $\Delta[0,1]$, then $V$ cannot even be jointly measurable, independently of the $\sigma$-algebra put on $[01]^{[0,1]}$. We can embed the unit interval $[0,1]$ as a subspace of $\Delta[0,1]$ via the embedding $x\mapsto \delta_x$. But the question has a negative answer on this smaller space too, the function $e:[0,1]\times [01]^{[0,1]}\to [0,1]$ given by $$e(x,f)=f(x)=\int f~\mathrm d\delta_x$$ is not jointly measurable, independently of the $\sigma$-algebra put on $[01]^{[0,1]}$. This follows from a deep result of Aumann.
There is a positive answer if we endow $\Delta[0,1]$ with the weakest topology such that the function $\mu\mapsto\mu(A)$ is continuous for every measurable set $A\subseteq [0,1]$ and $[01]^{[0,1]}$ with the uniform topology given by the metric $d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$. This follows from the fact that the simple functions, measurable functions that take on only finitely many values, are dense in $[01]^{[0,1]}$. However, $\Delta[0,1]$ is not metrizable and also not compact. The latter means one cannot apply the maximum theorem here. The topology on $\Delta[0,1]$ is also not metrizable, but the finer topology given by the total variation of signed measures is. But under this finer topology, $\Delta[0,1]$ is, of course, still not compact.