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In my self study of weak convergence of measures I came across this statement:

Let $M$ be a non-empty set of probability measures on $R$ then $M$ is tight if and only if there exists a non decreasing function $\phi: [0, +\infty) \rightarrow [0, \infty)$ s.t.

  • $\phi(x) \rightarrow + \infty$ as $x \rightarrow +\infty$
  • $\sup_{\mu \in M} \int \phi(|x|) \, \mu(dx) < \infty$

anybody know how to prove it?

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closed as off-topic by Did, Namaste, Shailesh, Claude Leibovici, Ove Ahlman Nov 20 '17 at 7:37

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  • $\begingroup$ What if you put \phi(x)=x, then tightness implies the two conditions above ! $\endgroup$ – user3503589 Nov 19 '17 at 13:38
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    $\begingroup$ @user3503589 There exist probability measures that have no expectation. $\endgroup$ – Michael Greinecker Nov 19 '17 at 14:53
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Let $\phi$ be such a function and suppose that $M$ is not tight. Then there is some $\epsilon>0$ such that for each $n$ there is some $\mu_n\in M$ such that $\mu_n\big([-n,n]\big)<1-\epsilon$. Now $$\int \phi\big(|x|\big)~\mu_n(dx)\geq\int_{(-\infty,-n)\cup(n,\infty)} \phi\big(|x|\big)~\mu_n(dx)\geq\int_{(-\infty,-n)\cup(n,\infty)} \phi(n)~\mu_n(dx)\geq\phi(n)\epsilon.$$ Since $\lim_n\phi(n)\epsilon=\infty$, we obtain a contradiction.

For the other direction, assume that $M$ is tight. For each $n$, there is some $r_n>0$ such that $$\sup_{\mu\in M}\mu\big([-r_n,r_n]\big)>1-1/2^{2n}.$$ This gives us a sequence $\langle r_n\rangle$, which we can take without loss of generality to be strictly increasing. Let $r_0=0$. Define $\phi$ by letting $\phi(x)=2^{n-1}$ for the unique $n$ such that $x\in [r_{n-1},r_n)$. Then $\phi$ has the desired properties.

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