Continuity of Expected Value Let $m(\cdot)$ be a probability measure on $Z$, so that $\int_Z m(dz) = 1$.
Consider a continuous function $f: X \times Y \times Z \rightarrow \mathbb{R}_{\geq 0}$, where $X \subseteq \mathbb{R}^n$, $Y \subset \mathbb{R}^m$ is compact, $Z \subseteq \mathbb{R}^p$ is closed.
Define 
$$ \hat{f}(x) \ := \ \inf_{y \in Y} \ \int_Z f(x,y,z) m(dz) $$


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*Under which conditions is function $\hat{f}(\cdot)$ continuous?

*Instead of assuming continuity of $f(\cdot)$, let us assume that for each $z \in Z$ the map $(x,y) \mapsto f(x,y,z)$ is continuous. Now under which conditions we have continuity of $\hat{f}(\cdot)$?
Note: in the non-probabilistic case, $Y$ compact and $(x,y) \mapsto f(x,y)$ continuous are sufficient to establish continuity of $x \mapsto \inf_{y \in Y} f(x,y) $.
 A: Suppose 
1) $f(x,y,\cdot)$ is $m$-measurable for all $x,y$, 
2) $f(\cdot,\cdot,z)$ is continuous on $X \times Y$ for all $z \in Z$,
3) for each $x_0 \in X$ there is $r > 0$ such that $\int_Z g(z)\ m(dz) < \infty$ where 
$g(z) = \sup \{f(x,y,z): |x-x_0| \le r, y \in Y\}$ 
Then I claim it works.  Suppose it didn't.  Then there would be $\epsilon > 0$ and a sequence $x_n \to x_0$ in $X$ such that $|\hat{f}(x_n) - \hat{f}(x_0)| > \epsilon$. We may assume $|x_n - x_0| < r$ as given in (3). 
Case 1: Suppose $\hat{f}(x_n) - \hat{f}(x_0) > \epsilon$ for infinitely many $n$.  We can assume WLOG this is true for all $n$.  There is some $y \in Y$ such that 
$ \int_Z f(x_0, y, z)\ m(dz) < \hat{f}(x_0) + \epsilon/2$, so 
$$\int_Z f(x_n,y,z)\ m(dz) - \int_Z f(x_0,y,z)\ m(dz) \ge \hat{f}(x_n) - \int_Z f(x_0,y,z)\ m(dz) > \epsilon/2$$
 But this contradicts the Lebesgue Dominated Convergence Theorem.
Case 2: $\hat{f}(x_0) - \hat{f}(x_n) > \epsilon$ for infinitely many $n$.  Again we can assume this is true for all $n$.  For each $n$ there is $y_n \in Y$ such that 
$\int_Z f(x_n, y_n, z)\ m(dz) < \hat{f}(x_n) + \epsilon/2$.  By compactness, some
subsequence of $y_n$ converges to some $y_0 \in Y$.  Again, we can assume this is
the whole sequence.  Then
$$ \int_Z f(x_0,y_0,z)\ m(dz) - \int_Z f(x_n,y_n,z)\ m(dz) \ge
\hat{f}(x_0) - \int_Z f(x_n,y_n,z)\ m(dz) > \epsilon/2$$
which again contradicts the Lebesgue Dominated Convergence Theorem.
