Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $m(\{\bar f\ne0\})=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.