Durrett exercise 3.2.5: how to apply dominated convergence? 
Suppose $g,h:\Bbb R\to\Bbb R$ are continuous with $g(x)>0$ and $|h(x)|/g(x)\to 0$ as $|x|\to\infty$. If $F_n\implies F$ (indicating weak convergence) and $\int g(x)dF_n(x)\le C<\infty$ then 
  $$\int h(x) dF_n(x)\to \int h(x) dF(x).$$

What is known is the embedding theorem: we can have $Y_n$ equal in distribution to $X_n\sim F_n$ and $Y$ equal in distribution to $X\sim F$ such that $Y_n\to Y$ a.s. Obviously this tells us to apply Dominated Convergence Thm. In particular, we must find another random variable $Z$ on the same probability space that dominates all $h(Y_n)$ and is integrable itself. However, I don't know how to find it, although I believe we must start from $g(Y_n)$ which dominate $h(Y_n)$ in a sense, but I have no idea how to formally write it out. 
 A: Using the embedding theorem, we are reduced to prove that if $C :=\sup_n\mathbb E\left[g\left(Y_n\right)\right]< +\infty$ and $Y_n\to Y$ almost surely, then $\mathbb E\left[h\left(Y_n\right)\right]\to \mathbb E\left[h\left(Y\right)\right]$. This can be done by dominated convergence. Fix a positive $R$ and define a continuous function $\phi_R$ which takes the value $1$ on $[-R,R]$ and $0$ outside $\mathbb R\setminus[-R-1,R+1]$, and $0\leqslant \phi\leqslant 1$. Now, we use the inequalities 
$$\left|\mathbb E\left[h\left(Y_n\right)\right] -\mathbb E\left[h\left(Y\right)\right]\right|\leqslant \left|\mathbb E\left[h\left(Y_n\right)\phi_R\left( h\left(Y_n\right)\right) \right] -\mathbb E\left[h\left(Y\right)\phi_R\left(h\left(Y\right)\right )\right]\right|+ \left|\mathbb E\left[h\left(Y_n\right)\left(1-\phi_R\right)\left( h\left(Y_n\right)\right) \right] -\mathbb E\left[h\left(Y\right)\left(1-\phi_R\right)\left(h\left(Y\right)\right )\right]\right|\\ \leqslant \left|\mathbb E\left[h\left(Y_n\right)\phi_R\left( h\left(Y_n\right)\right) \right] -\mathbb E\left[h\left(Y\right)\phi_R\left(h\left(Y\right)\right )\right]\right|+ \left|\mathbb E\left[h\left(Y_n\right)\left(1-\phi_R\right)\left( h\left(Y_n\right)\right) \right] \right| +\left|\mathbb E\left[h\left(Y\right)\left(1-\phi_R\right)\left(h\left(Y\right)\right )\right]\right|.$$
Since for any real number $t$, $\left|h(t)\right|\cdot\left(1-\phi_R\right)(h(t))\leqslant \left|h(t)\right| \mathbf 1_{\mathbb R\setminus  [-R,R] }(h(t))\leqslant g(t)\sup_{|x|\geqslant R}\left|h(x)\right| /g(x)$ and $\mathbb E \left[g\left(Y\right) \right] \leqslant \liminf_{ n\to +\infty} \mathbb E \left[g\left(Y_n\right) \right]\leqslant C $, we derive that 
$$\tag{*}  \left|\mathbb E\left[h\left(Y_n\right)\right] -\mathbb E\left[h\left(Y\right)\right]\right|\leqslant \left|\mathbb E\left[h\left(Y_n\right)\phi_R\left( h\left(Y_n\right)\right) \right] -\mathbb E\left[h\left(Y\right)\phi_R\left(h\left(Y\right)\right )\right]\right| +2C\sup_{|x|\geqslant R}\left|h(x)\right| /g(x).$$
From continuity of $h_R$, we infer that  $h\left(Y_n\right)\phi_R\left( h\left(Y_n\right)\right)\to h\left(Y\right)\phi_R\left( h\left(Y\right)\right)$ almost surely and $\left| h\left(Y_n\right)\phi_R\left( h\left(Y_n\right)\right)\right|\leqslant R$,  hence by (*), 
$$\limsup_{n\to  +\infty} \left|\mathbb E\left[h\left(Y_n\right)\right] -\mathbb E\left[h\left(Y\right)\right]\right|\leqslant 2C\sup_{|x|\geqslant R}\left|h(x)\right| /g(x).$$   
