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Prove: If there is a $\phi(X)\geq0$ such that $\phi(x)\rightarrow \infty$ for $|x|\rightarrow \infty$ and $\sup_n\int\phi(x)dF_n(x)<\infty$

Then $F_n$ is tight.

The definition of tightness of probability measures:

$F_n$ is called tight. If for every $\epsilon>0$ there is a compact set $K_{\epsilon}$ such that $\mu(K_{\epsilon})>1-\epsilon$

Can someone give me a tip?

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    $\begingroup$ Hint: For every $c$ there exists $x$ such that $\phi(t)\geqslant c$ for every $t\geqslant x$, then $K=[0,x]$ is compact and, for every $n$, $$c\mu_n((x,\infty))\leqslant\int_x^{\infty}\phi(t)dF_n(t)\leqslant\sup_k\int_0^{\infty}\phi(t)dF_k(t)\lt\infty.$$ $\endgroup$ – Did Feb 8 '15 at 18:29
  • $\begingroup$ So, we have $c\mu_n((x,\infty))<L<\infty$ which is equivalent to $1-\frac{L}{c}<\mu_n[0,x]$ but c was arbitrary, hence $\frac{L}{c}$ plays the role of our $\epsilon$ ? $\endgroup$ – Epsilondelta Feb 8 '15 at 18:59
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Following Did's hint, let $M:=\sup_k\int_{-\infty}^{\infty}\phi(t)dF_k(t)$, which is finite by assumption. Then for all $x$, $$ \mu_n\left(\mathbb R\setminus \left[-x,x\right]\right)\cdot \inf\left\{\phi(t), \left\lvert t\right\rvert\gt x\right\}\leqslant \int_x^{\infty}\phi(t)dF_n(t)+\int_{-\infty}^{-x}\phi(t)dF_n(t)\leqslant M$$ hence $$ \mu_n\left(\mathbb R\setminus \left[-x,x\right]\right)\leqslant M\left( \inf\left\{\phi(t), \left\lvert t\right\rvert\gt x\right\}\right)^{-1}. $$ Since for all positive $x$, the interval $\left[-x,x\right]$ is compact, it suffices to shows that for each positive $\varepsilon$, there exists $x$ such that $M\left( \inf\left\{\phi(t), \left\lvert t\right\rvert\gt x\right\}\right)^{-1}\lt\varepsilon$, which is possible thanks to the assumption $\phi(x)\to +\infty$ as $\left\lvert x\right\rvert\to +\infty$.

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