Let $X$ be a compact, convex subset of $\mathbb R^n$ with a non-empty interior and endowed with the sup-norm and let $C:X \rightrightarrows X$ be a continuous (i.e. upper and lower hemicontinuous), nonempty-valued, compact-valued and convex-valued correspondence. Moreover, $C(x)$ is either a singleton or has a non-empty interior — though I think this is not necessary.
Let $\lambda$ denote the Lebesgue measure.
(1) Is it possible to show that for all sequences $(x_n)_{n \geq 1} \subset X$ such that $x_n \to x$, $\lambda(C(x_n)) \to \lambda(C(x))$?
(2) Is continuity of the correspondence enough?
Best Answer
Upper and lower hemicontinuity of $C$ implies that $C$ is continuous in the Hausdorff metric, $d_H$, considering $C: X \to P(X)$, where $P(X)$ denotes the set of all closed subsets of $X$. As such, $C(x_n) \to C(x)$ where convergence is wrt to $d_H$.
This answer provides a partial answer: $\lambda$ is upper semicontinuous wrt $d_H$, that is $\lim_{x_n \to x}\lambda(C(x_n))\leq \lambda(C(x))$. If $C$ is convex-valued we have that the inequality holds with equality. A sketch of the proof is as follows:
If $C(x)$ is a singleton, we are done, as $0=\lambda(C(x))$ and by upper semicontinuity, $\lambda(C(x))=0\geq \lim_{x_n \to x}\lambda(C(x_n))\geq 0$.
If $C(x)$ has a non-empty interior, then $\lambda(C(x))>0$. Thus, recalling that $C$ is convex-valued, this implies that $\exists N$ such that $\forall n\geq N$, $C(x_n)$ also has a nonempty interior.
Let $f_n(y):=1_{y \in \text{int}(C(x_n))}$. Note that $f_n$ is measurable with respect to the Lebesgue measure on $(X,\mathcal B)$ and dominated by 1 for all $n$. To see that $f_n$ converges pointwise to $f$, where $f(y)=1_{y \in \text{int}(C(x))}$, take any $y \in \text{int}(C(x))$. Let $\epsilon:=\min_{z \in \partial C(x)}\lVert z-y\rVert_\infty>0$. Let $B_{\delta}(z)$ be the $\delta$ open-ball around $z\in X$ wrt the sup-norm. Then, $B_{\epsilon/2}(y)\subset \text{int}(C(x))$ and as $d_H(C(x_n),C(x))\to 0 \implies d_H(\partial C(x_n),\partial C(x))\to 0$, we have that $\exists N'\geq N$ such that $\forall n \geq N'$ where $B_{\epsilon/2}(y) \subset co(\partial C(x_n))=C(x_n)$ (as $C$ is convex-valued), where $\partial A$ denotes the boundary of set $A$ and $co(A)$ its convex hull.
Then, by Lebesgue's dominated convergence theorem we have that $\lambda(C(x_n))=\lambda(\text{int}(C(x_n)))=\lim_{n\to \infty}\int_X f_n d\lambda = \int_X \lim_{n\to \infty} f_n d\lambda = \int_X f d\lambda = \lambda(\text{int}(C(x))) = \lambda(C(x)).$