Continuity of $\max$ of Lebesgue integral Let $m$ be a probability measure on $Z \subseteq \mathbb{R}^p$, so that $m(Z)=1$.
Consider a locally bounded $f: X \times Y \times Z \rightarrow \mathbb{R}_{\geq 0}$, with $X \subseteq \mathbb{R}^n$, $Y \subseteq \mathbb{R}^m$ compact, such that $f(\cdot, \cdot, z)$ is continuous, and $f(x,y,\cdot)$ is measurable. Also, $f$ is uniformly ($\forall (x,y,z) \in X \times Y \times Z $) upper bounded by a constant.
Prove that the following mapping is continuous.
$$ x \mapsto \max_{y \in Y} \int_Z f(x,y,z) m(dz) $$
Comment. I think I proved the claim by invoking the Lebesgue-Vitali Theorem. I would like to get a "relatively-simple" proof invoking, perhaps, the Dominated Convergence Theorem.
 A: Denote 

$$F(x,y):=\int_Zf(x,y,z)dm(z)\quad\mbox{and}\quad G(x):=\sup_{y\in K}\;F(x,y).$$ 

First, we show that $F$ is continuous. It follows that $G$ is finite and $G(x)=\max_{y\in K}F(x,y)$. Then we show that $G$ is continuous.

Step 1: $F$ is continuous on $X\times Y$.

Proof: By assumption, $f(x,y,\cdot)$ is measurable and bounded on $Z$. Since $Z$ has finite measure, it follows that it is integrable. So $F$ is well-defined. Now since the spaces under consideration are metric, it suffices to prove sequential continuity. So let $(x_n,y_n)$ converge to $(x,y)$ in $X\times Y$.
The sequence of measurable functions $f(x_n,y_n,\cdot)$ is uniformly bounded by, say, $M$ on $Z$. Now the constant function $M$ is integrable over $Z$ as $Z$ has finite measure. On the other hand, $f(x_n,y_n,\cdot)$ converges pointwise to $f(x,y,\cdot)$ on $Z$ by continuity of each $f(\cdot,\cdot,z)$. By the Dominated Convergence Theorem
$$
\lim \int_Zf(x_n,y_n,z)dm(z)=\int_Z\lim f(x_n,y_n,z)dm(z)=\int_Zf(x,y,z)dm(z)
$$
i.e.$\lim F(x_n,y_n)=F(x,y)$. QED.

Step 2: $G$ is continuous on $X$.

Proof: take $x_0\in X$, fix $\epsilon>0$, and denote $B(x_0,1)$ the closed unit ball centered at $x_0$. Then note that the function $(x,y)\longmapsto F(x,y)$ is continuous on the compact $B(x_0,1)\times Y$. By Heine-Cantor, the latter is uniformly continuous. In particular there exists $1>\delta>0$ such that $|F(x,y)-F(x_0,y)|\leq \epsilon$ for every $\|x-x_0\|\leq\delta$ and every $y\in Y$. Now observe 
$$
F(x,y)=F(x_0,y)+F(x,y)-F(x_0,y)\leq F(x_0,y)+\epsilon\leq G(x_0)+\epsilon
$$
for every $y\in Y$ and every $\|x-x_0\|\leq\delta$. Taking the sup/max over $Y$ on the lhs yields
$$
G(x)\leq G(x_0)+\epsilon \qquad\forall \|x-x_0\|\leq\delta.
$$
So $G(x)-G(x_0)\leq \epsilon$ and, likewise, $G(x_0)-G(x)\leq \epsilon$, whence $|G(x)-G(x_0)|\leq \epsilon$  for every $ \|x-x_0\|\leq\delta$. So $G$ is continuous at $x_0$. QED.
