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A set $\Gamma$ of probability measures on $X$ is said to be tight if for every $\epsilon > 0$ there exists a compact set $K_\epsilon$ of $X$ such that

$$\mu(K) \geq 1 - \epsilon \text{ for all $\mu \in \Gamma$}.$$

Fine, now let $\mathcal M$ be the set of measurable mappings on the probability space $(\Omega, \mathcal F, P)$ if there exists $M > 0$ such that

$$\int_\Omega |f(\omega)| \, dP(\omega) \leq M \text{ for all $f \in \mathcal M$}.$$

How do I now show that the set $\{f_{\#}P : f \in \mathcal M\}$ is tight?

Please only give me a hint as this is "homework". What I tried to do was assume it is not tight and then try to construct a compact set that gives me a contradiction, but that seems to lead nowhere.

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1 Answer 1

up vote 3 down vote accepted

Try to use Markov's inequality; $P(|X|\geq a)\leq E(|X|)/a.$

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Ahh, it is so easy. Thanks! –  Jonas Teuwen Mar 10 '11 at 8:51

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