@dohmatob: I am a bit surprised by the formulation of Theorem 3.1, because it is not defined what a convex correspondence is. I assume that it should be convex-valued and measurable in some previously defined sense, but there is no mentioning of any measurability hypothesis. But Diestel & Uhl have proved various theorems of this kind, maybe their monograph contains clearer formulations.

If you assume that $A$ is convex (and compact), it suffices to apply the earlier mentioned theorem: The set of measurable selections of an integrably bounded measurable multivalued function with compact convex values is weakly compact in $L_1$, Indeed if $a_n=\int f_n$ is a sequence in the image then the weak (sequential) compactness of $f_n$ implies that there is a subsequence with $\int f_{n_k}\to\int f$ weakly for some selection $f$. Because we are in the space $\mathbb R^n$, this implies strong convergence.

Maharam's decomposition you mention holds only if the measure space is separable. Otherwise an uncountable number of copies of intervals is involved, though I do not know the details by heart. (AFAIK, only this non-separable version is due to D. Maharam; the separable version had been known long before.)

(continuing comment): Note that the Debreu integral is defined by means of approximation of the mutlivalued functions in the hyperspace of nonempty compact (convex?) sets with the Hausdorff metric, hence by definition is compact (and convex). All result establishing this equality are one sense or another based on the weak compactness in $L_1$ of the set of selections of an integrably bounded multivalued map with compact convex values. I am afraid that for this result convexity is crucial.

As mentioned in a comment to my reply, I had a direct application of the measure isomorphism theorem in mind (which requires separability). However, I think that simply the proof of the convexity result by Aumann requires only a non-atomaic measure space, that is without loss of generality you can assume in a non-atomic measure space that $A$ is convex, and then again Aumann's proof for compactness should directly hold. Alternatively, once you assume that $A$ is convex, you can apply some of many results which establish equality of the Aumann and the Debreu integral.

By $v(\cdot)A$ I mean the multivalued function (which assumes values in the powerset of $\mathbb R^n$); the Aumann integral is defined as the set of integrals over all measurable selections of that multivalued function. Concerning going to Lebesgue measure: I had the theorem in mind that every atomless separable measure space is isomorphic to Lebesgue measure. I forgot to mention "separable". (One might still try to apply Maharam's results for the nonseparable case, but I am not sure whether this will lead to something.)

@TobiasDiez By shrinking $U$ (and then $V$), you can assume that $U$ is an open ball. But then you can assume $U=\mathbb R^n$ as well, because an open ball is homeomorphic to $\mathbb R^n$.

A simple observation: If you restrict $f$ to an arbitrarily large closed ball, then the inverse is continuous (because preimages of closed=compact sets are compact and thus closed). And at least the construction of obvious counterexamples is then prevented by the fact that images of points from outside large balls have to be covered by other charts as well. Of course, not being able to construct obvious counterexamples does not mean that it can be proved easily.

@leomonsaingeon: Even in 1d, I do not see why the result should be true; you are using that the image is homeomorphic to an interval. Or did you mean $m=1$ (and not merily $n=1$)? In that case the answer is trivially positive, because $n=m$: In the case $n=m$ you can just use the domain invariance theorem for continuous injective functions.

I do not have the references here, so I cannot provide them. Anyway, Matthew's answer is very similar to the example I have seen at that time.