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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 completely misread the question, it seems! o.O2012-05-15

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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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    Is the family $\{w \mapsto \sup_{x \in X} f(x,y,w) \}_{y \in Y}$ Uniformly Integrable?2012-05-16