This post is about a basic fact which is not very well known. Let ${{(\mu_i)}_{i\in I}}$ be a family of probability measures on a topological space ${E}$ equipped with its Borel sigma-field. The following two properties are equivalent, and when they hold we say that ${{(\mu_i)}_{i\in I}}$ is tight.

1. for any ${\varepsilon>0}$ there exists a compact set ${K_\varepsilon\subset E}$ such that

$\sup_{i\in I}\mu_i(K_\varepsilon^c)\leq\varepsilon;$

2. there exists a measurable ${f:E\rightarrow[0,\infty]}$ with compact level-sets such that

$\sup_{i\in I}\int\!f\,d\mu_i<\infty.$

Recall that the level-sets of ${f}$ are the sets ${\{x\in E:f(x)\leq r\}}$, ${r\geq0}$. If ${E}$ is not compact then it cannot be a level-set of ${f}$ and thus ${f}$ cannot be bounded, while if ${E}$ is compact then the property holds trivially with ${f}$ constant.

To deduce 1. from 2. we write, using the Markov inequality, for any ${r>0}$,

$\sup_{i\in I}\mu_i(\{x\in E:f(x)\leq r\}^c) \leq \frac{1}{r}\sup_{i\in I}\int\!f\,d\mu_i=\frac{C}{r},$

which leads to take ${r=r_\varepsilon=C/\varepsilon}$ and ${K_\varepsilon=\{x\in E:f(x)\leq r_\varepsilon\}}$, for any ${\varepsilon>0}$.

To deduce 2. from 1. we first extract from 1. a sequence of compact subsets ${{(K_{1/n^2})}_{n\geq1}}$. We can assume without loss of generality that it grows: ${K_{1/n^2}\subset K_{1/m^2}}$ if ${n\leq m}$. Now, for any ${x\in F=\cup_{n}K_n}$, there exists ${n_x}$ such that ${x\in K_{1/m^2}}$ for any ${m\geq n_x}$, and thus ${\sum_n \mathbf{1}_{K_{1/n^2}^c}(x)=n_x-1<\infty}$. As a consequence, if one defines

$f:=\sum_n\mathbf{1}_{K_{1/n^2}^c}$

then ${f<\infty}$ on ${F}$ while ${f=\infty}$ on ${F^c}$, and ${f}$ has compact level-sets since ${\{x\in E:f(x)\leq n-1\}=K_{1/n^2}}$ for any ${n\geq1}$. On the other hand, by definition of ${K_\varepsilon}$,

$\sup_{i\in I}\int\!f\,d\mu_i \leq\sum_n\frac{1}{n^2}<\infty.$

Tightness is an important concept of probability theory. A famous theorem of Prokhorov states that a family of probability measures is tight if and only if it is relatively compact for the topology of narrow convergence.

Taking ${X_i\sim\mu_i}$ for every ${i\in I}$, the second property reads

$\sup_{i\in I}\mathbb{E}(f(X_i))<\infty.$

It plays for tightness the role played by the famous de la Vallée Poussin criterion for uniform integrability.

## One Comment

1. student 2017-08-10

Nice characterization. Having a close look at the proof it seems sufficient to require the level sets of $f$ to be only relatively compact. The inclusion ${\{x\in E:f(x)\leq n-1\}\subset K_{1/n^2}}$ shows this (besides, I could not see why the other inclusion would hold).

Regards

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