Let $\mu_n$ and $\mu$ Borel probability measures on $\mathbb{R}_+$ such that $\mu_n \Rightarrow \mu$ (converges weakly). Show that if

$$ \sup\limits_{n \geq 1} \int x^a d \mu_n(x) <\infty $$

for some $a\geq 1$, then for every $0 < b< a$

$$ \lim\limits_{n\to \infty} \int x^b d \mu_n(x) = \int x^b d \mu(x) $$

My attempt at proof.

I can show that measures $\mu_n$ are tight, I can also show that
$$ \sup_{n}\int_{x}^{\infty} d\mu_n \text{ decreases at least as } \frac{1}{x^a} \text{ when } x \to \infty $$

The above implies that for $x$ large enough $$ \int_{x}^{\infty} x^{b} d\mu_n(x) < \varepsilon \quad \quad \forall n $$ where $\varepsilon$ is arbitrary small.

I was not able to proceed and complete the proof from here. Comments, hints or complete proof would be very appreciated. Thanks.


Fix $0<b<a$, and for each $m\in \mathbb{N}$ let $f_m(x)=\min\{x^b,m^b\}$. Then $f_m(x)$ is a bounded continuous function on $(0,\infty)$, hence $$ \lim_{n\to\infty}\int_0^{\infty}f_m(x)\;d\mu_n(x)=\int_0^{\infty}f_m(x)\;d\mu(x)$$ by the definition of weak convergence.

Moreover, as you noted in your question, the "tails" $$ \int_m^{\infty}x^b\;d\mu_n(x) $$ can be made small uniformly in $n$ if $m$ is chosen large enough. The corresponding integral for $\mu$ is also small if $m$ is large enough, hence there exists some $m$ such that $$ \Big|\int_0^{\infty}x^b\;d\mu_n(x)-\int_0^{\infty}x^b\;d\mu(x)\Big|$$ $$\leq \int_0^{\infty}|x^b-f_m(x)|\;d\mu_n(x)+\Big|\int_0^{\infty}f_m(x)\;d\mu_n(x)-\int_0^{\infty}f_m(x)\;d\mu(x)\Big|+\int_0^{\infty}|x^b-f_m(x)|\;d\mu(x)$$ $$\leq \int_m^{\infty}x^b\;d\mu_n(x)+\Big|\int_0^{\infty}f_m(x)\;d\mu_n(x)-\int_0^{\infty}f_m(x)\;d\mu(x)\Big|+\int_m^{\infty}x^b\;d\mu(x)$$ $$<3\varepsilon$$ for all sufficiently large $n$.

  • $\begingroup$ How do you know that for sufficiently large $m$ the integral $\int_{m}^{\infty} x^b d\mu $ is arbitrary small? $\endgroup$ – them Mar 15 '16 at 1:53
  • $\begingroup$ The condition on the $\mu_n$, together with the weak convergence, implies that $x^b\in L^1(\mu)$. Then the tails can be made small by the dominated convergence theorem. $\endgroup$ – carmichael561 Mar 15 '16 at 1:57
  • $\begingroup$ Could you please elaborate. Why $x^b$ is in $L^1(\mu)$ ? This is the part I am missing. If $x^b$ is in $L^1(\mu)$ then I follow, as you say by DCT I get convergence of the tail ($x^b$ dominates $x^b{1}([m, \infty])(x)$, by pointwise convergence of the later to zero and DCT I get the result.) $\endgroup$ – them Mar 15 '16 at 2:02
  • 1
    $\begingroup$ If it's not, then $\int x^b\;d\mu=\infty$ since $x^b$ is non-negative. So $\int f_m(x)\;d\mu$ can be made arbitrarily large, by the monotone convergence theorem. Since $\int f_m\;d\mu_n\to \int f_m\;d\mu$ for each $m$, it follows that $\int f_m\;d\mu_n$ can be made arbitrarily large, which (since $x^a>x^b>f_m(x)$ for all large $x$) violates the hypothesis that $\sup_n \int x^a\;d\mu_n<\infty$. $\endgroup$ – carmichael561 Mar 15 '16 at 2:08
  • $\begingroup$ got it thanks a lot! $\endgroup$ – them Mar 15 '16 at 2:13

Suggestion: Decompose $\int x^b\,d\mu_n(x)$ as $$ \int(x\wedge N)^b\,d\mu_n(x)+\int[x^b-(x\wedge N)^b]\,d\mu_n(x), $$ and likewise for $\int x^b\,d\mu(x)$. In the first of these two integrals, the integrand $(x\wedge N)^b$ is bounded and continuous, hence subject to the weak convergence of $\mu_n$ to $\mu$. The second integral is smaller than $\int_N^\infty x^b\,d\mu_n(x)$ and can be made small (uniformly in $n$) as you have already noted.

  • $\begingroup$ Could you please elaborate: I see that on the LHS, the partition gives $\int_N^\infty x^b d \mu_n$ uniformly small, for large N. Why does it still hold when I pass to the limit, that it why for $\int_N^\infty x^b d \mu$ is small? thanks. $\endgroup$ – them Mar 15 '16 at 1:57

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