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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.

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If the family is tight there is a sequence $n_k$ increasing to $\infty$ such that $\mu \{x:n_n \leq\|x\| <n_{k+1}\} <\frac 1 {2^{k}}$ 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 $\mu$.

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  • $\begingroup$ Sir just to clarify a few things. Does it mean that my intuition about nonnegative function was correct? Or only continuous function will work? $\endgroup$ – mudok Oct 24 '18 at 10:54
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Oct 24 '18 at 11:49
  • $\begingroup$ Thanks a lot @Kavi Rama Murthy. $\endgroup$ – mudok Oct 24 '18 at 11:57

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