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Let $m$ be a probability measure on $W \subseteq \mathbb{R}^m$ so that $m(W)=1$.

For all $k \in \mathbb{Z}_{\geq 0}$ let $f_k: X \times W \rightarrow \mathbb{R}_{\geq 0}$ be measurable and locally bounded, where $X \subseteq \mathbb{R}^n$.

Consider a sequence $\{X_k\}_{k=0}^{\infty}$ of compact sets $X_k \subset X$ such that $X_{k} \supseteq X_{k+1} \supseteq \cdots \supseteq X_\infty = \{\bar x\}$.

Assume that there exists and integrable function $\bar{f}: W \rightarrow \mathbb{R}_{\geq 0}$ such that $f_k(x,w) \leq \bar f(w)$ for all $x \in \mathbb{R}^n$ and $w \in W$.

Say if the following is true.

$$ \limsup_{k \rightarrow \infty} \ \max_{x \in X_k} \int_{W} f_k(x,w) m(dw) \leq \int_W \limsup_{k \rightarrow \infty} f_k(\bar{x}, w) m(dw) $$

Note: if $X_k = \{\bar{x}\}$ for all $k$, then the claim is true because of the Fatou's Lemma.

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1 Answer 1

up vote 2 down vote accepted

Try $f_k(x,w)=\bar f(w)\cdot\mathbf 1_{x\ne\bar x}$, assuming that $X_k\ne\{\bar x\}$ for every $k$. Then each maximum on the LHS is the integral of $\bar f$ and each limsup on the RHS is zero, hence LHS $\gt0$ and RHS $=0$ except in the trivial case where $m(\{\bar f\ne0\})=0$.

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