$\int f_kd\mu_k\to\int fd\mu$? If a measure $\mu_k$ converges weakly to a measure $\mu$, i.e. $\int fd\mu_k\to \int fd\mu$ for every $f\in C_b(\mathbb{R})$ (contiuous and bounded functions), and let $f_k\to f$ a.e. pointwise. Then I was wondering what extra condition we need so that $\int f_kd\mu_k\to\int fd\mu$. I have $$\int f_kd\mu_k=\underbrace{\int fd\mu_k}_{\to\int fd\mu}+\int (f_k-f)d\mu_k$$ but I am not sure when the second term goes to zero.
 A: Even if $\mu_k=\mu$, a.e. convergence of $f_k$ to $f$ is not sufficient to guarantee convergence of the integral. As an example take a non-zero positive function $\phi\in C_c(\mathbb{R})$ with support in $(0,1)$ and let $f_k(x)=k\phi(kx)$. Then $f_k\to 0$ pointwise, but
$$
\int_0^1 f_k(x)\,dx=\int_0^k \phi(x)\,dx=\int_0^1 \phi(x)\,dx>0.
$$
(The measure $\mu$ here is the uniform measure on $(0,1)$.)
The claimed convergence is true for all weakly convergent sequences $(\mu_k)$ if and only if $(f_k)$ is uniformly bounded and $f_k\to f$ uniformly on compact sets:
Every weakly convergent sequence of finite measures is bounded by the uniform boundedness principle. Moreover, it is tight, that is, for every $\epsilon>0$ there exist $R>0$ and $N\in \mathbb N$ such that $\mu_k(\mathbb R\setminus[-R,R])<\epsilon$ for all $k\geq N$. If $(f_k)$ is uniformly bounded and $f_k\to f$ uniformly on compact sets, then
\begin{align*}
\left\lvert\int (f-f_k)\,d\mu_k\right\rvert&\leq \left\lvert\int_{\mathbb R\setminus[-R,R]}(f-f_k)\,d\mu_k\right\rvert+\left\lvert\int_{[-R,R]}(f-f_k)\,d\mu_k\right\rvert\\
&\leq 2\epsilon\sup_j \|f_j\|_\infty+\sup_{|x|\leq R}|f(x)-f_k(x)|\cdot\sup_{j}\|\mu_j\|,
\end{align*}
which gives the desired convergence.
For the converse implication, the uniform boundedness follows from the uniform boundedness principle. Now assume $(f_k)$ does not converge uniformly on compact sets to $f$, that is, there exist $R>0$, $\epsilon>0$, a subsequence $(f_{k_n})$ of $(f_k)$ and a sequence $(x_n)$ in $[-R,R]$ such that $|f(x_n)-f_{k_n}(x_n)|\geq \epsilon$.
Since $[-R,R]$ is compact, there exists a subsequence of $(x_n)$ that converges to some $x\in [-R,R]$. To avoid even more indices, I'll just assume that $(x_n)$ itself converges. This means that also $\delta_{x_n}\to \delta_x$ weakly.
Since $f$ is continuous, there exists $N\in \mathbb N$ such that $|f(x)-f(x_n)|<\epsilon/2$ for $n\geq N$. Thus
$$
\left\lvert\int f\,d\delta(x)-\int f_n\,d\delta(x_n)\right\rvert\geq |f(x_n)-f_n(x_n)|-|f(x)-f(x_n)|\geq \epsilon/2
$$
in contradiction to our assumption.
A: Some equicontinuity property along with uniform integrability is enough. Here is a result by Mann, Wald, Prokhorov, Rubin (see, Kallenberg, O. Foundations of probability, second ed. Springer. 2001, pp. 76)

Suppose  $(S,d)$ and $(T,\rho)$ are  metric spaces, $\{X_n,X:n\in\mathbb{N}\}$ are random variables taking values in $S$ such that  $X_n\stackrel{law}{=}\mu_n\stackrel{n}{\Longrightarrow}\mu\stackrel{law}{=} X$. If $(f_n,f:n\in\mathbb{N})$ are measurable functions from $S$ into  $T$, and there is a measurable set $C\subset S$ such that

*

*$\operatorname{supp}\mu\subset C$, and

*$f_n(s_n)\xrightarrow{n\rightarrow\infty}f(s)$ for any $s_n\rightarrow s\in C$,
Then $f_n(X_n)\stackrel{n}{\Longrightarrow} f(X)$, that is $$\int g(f_n(X_n))dP=\int_S g(f_n(x))\,\mu_n(dx)\xrightarrow{n\rightarrow\infty}\int_S g(f(x))\,\mu(dx)=\int g(f(X))\,dP$$  for all $g\in\mathcal{C}_b(T)$.


If $T=\mathbb{R}$, and the sequence $(f_n(X_n):n\in\mathbb{N})$ is uniformly integrable, then you may use $g(t)=t$ to get the convergence you are looking for, that is $$\int f(x)\,\mu_n(dx)\xrightarrow{n\rightarrow\infty}\int f(x)\,\mu(dx)$$
(See Billingsley, Convergence of Probability measures, Wiley 1968, Theorem 5.4)

A slightly more general version of the result above can be found in van der Vaart, A.W. Asymptotic statistics, Cambridge 1998, pp. 258-259 where some measurability assumptions are dropped and the integrals are "upper" integrals (upper Daniell integrals).
In both references, the proof of the statements relay on the Portmanteau theorem.
