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Let $\mu(\cdot)$ be a probability measure on $W \subseteq \mathbb{R}^p$, so that $\int_W \mu(dw) = 1$.

Consider a function $f: X \times Y \times W \rightarrow \mathbb{R}_{\geq 0}$, with $X \subset \mathbb{R}^n$ compact, $Y \subset \mathbb{R}^m$ compact, such that: $\forall w$ $f(\cdot,\cdot,w)$ is continuous, $\forall (x,y)$ $f(x,y,\cdot)$ is measurable.

Assume that for any compact $\underline{W} \subset W$ we have

$$ \max_{y \in Y} \int_{\underline{W}} f(x,y,w) \mu(dw) \leq F(x) $$

On the whole $W$, we assume that

$$ \max_{y \in Y} \int_W \sup_{x \in X} f(x,y,w) \mu(dw) < \infty $$

This implies that for any $y$ the family $\{w \mapsto f(x,y,w)\}_{x \in X}$ is Uniformly Integrable.

Are the assumptions sufficient to say the following?

$$ \max_{y \in Y} \int_W f(x,y,w) \mu(dw) \leq F(x) $$

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I don't think so. Let $W = \mathbb{R}$. Consider $f(x,n,w) = n$ if $w \in [n,n+1]$, $0$ otherwise. When you integrate over $W$ you get $\infty$. –  Adam May 15 '12 at 3:13
    
I completely misread the question, it seems! o.O –  Sam May 15 '12 at 6:22
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up vote 2 down vote accepted

Suppose not. Then there are $x_0 \in X$, $y_0 \in Y$ such that $\int_W f(x_0,y_0,w) \mu(dw) > F(x_0)$. And from the inner regularity of the measure $f(x_0,y_0,w) \mu(dw)$ there is compact $K \subset W$ such that $\int_K f(x_0,y_0,w) \mu(dw) > F(x_0)$.
But $\int_K f(x_0,y_0,w)\mu(dw) \le \max_{y \in Y} \int_K f(x_0,y,w)\mu(dw) \le F(x_0)$, contradiction.

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I'm wondering on what happens if $f(x,y,w) := g(y,w) \ \forall x$ and $F(x) = F \in \mathbb{R}_{\geq 0} \ \forall x$. In this case, $\sup_x f(x,y,w) = g(y,w)$. Therefore we only have $\max_{y \in Y} \int_{\underline{W}} g(x,w) \mu(dw) \leq F$. So why should we have $\max_{y \in Y} \int_W g(y,w) \mu(dw) \leq F$? –  Adam May 15 '12 at 4:46
    
I meant "we only have $\max_{y \in Y}\int_{\underline{W}} g(y,w) \mu(dw) \leq F$"... Can you be more formal when you say "from the inner regularity"? –  Adam May 15 '12 at 5:00
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$\rho(dw) = f(x_0, y_0, w) \mu(dw)$ is a positive measure, defined by $\rho(A) = \int_A f(x_0,y_0,w)\mu(dw)$. According to your assumptions, $\rho(K) \le F(x_0) < \infty$ for every compact set $K$. There is a theorem (see e.g. Rudin, "Real and Complex Analysis", Theorem 2.18) that any positive Borel measure on a locally compact Hausdorff space in which every open set is $\sigma$-compact, such that every compact set has finite measure, is regular. –  Robert Israel May 15 '12 at 7:47
    
This means that for every Borel set $E$, $$\rho(E) = \inf \{\rho(V): E \subseteq V,\ V \text{ open}\} = \sup \{\rho(K): K \subseteq E,\ K \text{ compact}\}$$ –  Robert Israel May 15 '12 at 7:49
    
The part with the $V$ open is outer regularity, the part with the $K$ compact is inner regularity. –  Robert Israel May 15 '12 at 7:51
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