Take the 2-minute tour ×
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.

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

Consider $f: X \times W \rightarrow \mathbb{R}_{\geq 0}$, $X \subseteq \mathbb{R}^n$ such that

$\forall w \in W$ $\ x \mapsto f(x,w)$ is continuous;

$\forall x \in X$ $\ w \mapsto f(x,w)$ is measurable.

Assume that ($f(x,\cdot)$ is integrable):

$$ \sum_{k=0}^{\infty} 2^k m(\{w \in W \mid 2^k \leq f(x,w) \leq 2^{k+1} \}) < \infty $$

Define

$$ \epsilon_n(\delta) := \sup_{\xi \in \{x\} + \delta \overline{\mathbb{B}} } m(\{ w \in W \mid f(\xi,w) \geq 2^n \}) $$

Assume that: $\lim_{n \rightarrow \infty} \epsilon_n(\delta) \rightarrow 0$ for all $\delta \in \mathbb{R}_{\geq 0}$.

Does there exist $\delta > 0$ such that the following holds true?

$$ \sup_{\xi \in \{x\} +\delta \overline{\mathbb{B}} } \ \sum_{k=0}^{\infty} 2^k m(\{w \in W \mid 2^k \leq f(\xi,w) \leq 2^{k+1} \}) < \infty $$

share|improve this question

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.