# Condition implying tightness of sequence of probability measures

A sequence of probability measures $\mu_n$ is said to be tight if for each $\epsilon$ there exists a finite interval $(a,b]$ such that $\mu((a,b])>1-\epsilon$ For all $n$.

With this information, prove that if $\sup_n\int f$ $d\mu_n<\infty$ for a nonnegative $f$ such that $f(x)\rightarrow\infty$ as $x\rightarrow\pm\infty$ then $\{\mu_n\}$ is tight.

This is the first I've worked with tight probability measure sequences so I'm not sure how to prove this. Any thoughts?

Note that for a fixed $t_0$, we have $$\mu_n(\mathbf R\setminus [-t_0,t_0])\cdot\inf_{x:|x|\geqslant t_0 } f(x)\leqslant \int_\mathbf R f(x) \mathrm d\mu_n,$$ hence defining $M:=\sup_n\int_\mathbf R f(x) \mathrm d\mu_n$, it follows that $$\mu_n(\mathbf R\setminus [-t_0,t_0])\cdot\inf_{x:|x|\geqslant t_0 } f(x)\leqslant M.$$ It remains to choose a good $t_0$ for a given $\varepsilon$.
The condition to ensure the tightness is that $\displaystyle \frac{f(x)}{x} \to \infty$ as $x \to \infty$. In fact if e.g. $f(x)=|x|$ then $$\sup_n \int f \, dμ_n<\infty$$ is only a necessary condition for the tightness.
• Actually, it looks to me like Davide's answer is correct - the given condition is sufficient. And I don't think what you said about a necessary condition is right. Consider a sequence of measures where $\mu_n$ puts mass $n^{-1/2}$ at $n$ and the rest at 0. Is it possible you are thinking of conditions for uniform integrability instead of tightness? – Nate Eldredge Jul 30 '15 at 13:43