# On why a set of probability measures is tight iff there exists a non decreasing function. [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Namaste, Shailesh, Claude Leibovici, Ove Ahlman
If this question can be reworded to fit the rules in the help center, please edit the question.

• What if you put \phi(x)=x, then tightness implies the two conditions above ! – user3503589 Nov 19 '17 at 13:38
• @user3503589 There exist probability measures that have no expectation. – Michael Greinecker Nov 19 '17 at 14:53

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.