Problem on moment convergence for sufficiently tight measures. 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. 
 A: 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 $n$.
A: 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.
