# Tightness of a family of probability measures.

Let $$\mathscr{A}$$ be a family of probability measures then this family is tight iff there exists a function $$f\in C(\mathbb{R^n})$$ such that $$f(x)\to\infty$$ as $$|x|\to\infty$$ and $$\sup_{\mu\in\mathscr{A}}\int f\,\mathrm{d}\mu<\infty.$$

I think the function needs to be nonnegative. If that is the case I can prove the tightness of the family but again I am unable to show the existence of such a function from the tightness. Any help will be appreciated.

If the family is tight there is a sequence $$n_k$$ increasing to $$\infty$$ such that $$\mu \{x:n_n \leq\|x\| for all $$\mu$$ in the family. . Let $$f(x)=g(\|x\|)$$ where $$g$$ is a piece wise linear function on $$[0,\infty)$$, linear on each of the intervals $$[n_k,n_{k+1})$$ taking the value $$k$$ at $$n_k$$. Note that $$g(x) \leq k+1$$ on $$[n_k, n_{k+1})$$. It follows that $$\int f d \mu \leq \sum_k \frac {k+1} {2^{k}}$$ for all $$\mu$$.
• @mudok Not necessary to assume that $f$ is non-negative. . If $f$ is continuous and $f(x) \to \infty$ as $|x| \to \infty$ tehn $f$ has an absolute minimum, say $a$. Then $f-a$ is non -negative and $\int (f-a)d \mu$ is bounded, so tightness follows. – Kavi Rama Murthy Oct 24 '18 at 11:49