Let $\mu_n$ be a sequence of probability measures on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ such that $\mu_n \to^d \mu$. Let $\{f_a\}_{a\in\mathbb{R}}$ be a colection of bounded real functions, each one continuous $\mu$-a.e.
I want to show that if exists an $\mu$-integrable function $h : \mathbb{R} \to (0,\infty)$, continuous $\mu$-a.e., such that $|f_a(x) - f_b(x)| \leq |b-a|h(x)$ for all $a,b,x\in\mathbb{R}$, then:
$$ \lim_{n\to\infty} \sup_{a\in\mathbb{R}} \left| \int f_a d\mu_n - \int f_a d\mu \right| = 0$$
I got the easiest part. Constructing $X_n \sim \mu_n$ and $X \sim \mu$ we have, by the continuous mapping theorem, that $f_a(X_n) \to^d f_a(X)$ for all $a\in\mathbb{R}$. Since the functions are bounded their expectations are well defined and therefore $$ \lim_{n\to\infty} \left| \int f_a d\mu_n - \int f_a d\mu \right| = 0\text{ for all }a\in\mathbb{R}.$$
By the triangle inequality we also have $$ \left| \int f_b d\mu_n - \int f_b d\mu \right| \leq \left| \int (f_b-f_a) d\mu_n - \int (f_b-f_a) d\mu \right| + \left| \int f_a d\mu_n - \int f_a d\mu \right| $$ and then $$ \left| \int f_b d\mu_n - \int f_b d\mu \right| \leq \left| \int f_a d\mu_n - \int f_a d\mu \right| + |b-a|\int h d\mu_n + |b-a|\int h d\mu .$$
Let $A_k = [-k,k]$ and let $C_N = \{ -k, -k+\frac{1}{N}, -k+\frac{1}{N}, \cdots, k-\frac{1}{N} , k\}$, then:
$$ \sup_{a\in A_k} \left| \int f_a d\mu_n - \int f_a d\mu \right| \leq \max_{a_i\in C_N} \left| \int f_{a_i} d\mu_n - \int f_{a_i} d\mu \right| + \frac{1}{2N}\int h d\mu_n + \frac{1}{2N}\int h d\mu $$
For a sufficiently large $n$ we can say, by the continuous mapping theorem on $h$, that: $$ \sup_{a\in A_k} \left| \int f_a d\mu_n - \int f_a d\mu \right| \leq \max_{a_i\in C_N} \left| \int f_{a_i} d\mu_n - \int f_{a_i} d\mu \right| + \frac{2}{N}\int h d\mu $$
Since $h$ is integrable, given $\epsilon > 0$, we can take $N$ such that $\frac{2}{N}\int h d\mu < \frac{\epsilon}{2}$ and then we can find $n$ such that $\max_{a_i} \left| \int f_{a_i\in C_N} d\mu_n - \int f_{a_i} d\mu \right| < \frac{\epsilon}{2}$, therefore:
$$ \lim_{n\to\infty} \sup_{a\in A_k} \left| \int f_a d\mu_n - \int f_a d\mu \right| = 0\text{ for all }k>0$$
And now I'm stuck and I believe I'm missing where to use the fact that functions $f_a$ are bounded (for more then have a well defined expectation). Suggestions?
I think this question might be wrong. Here is a possible counter-example.
Let $f_a= a \mathbb I_{[0,1]}$, and $h=\mathbb I_{[0,1]}+\phi(x)\mathbb I_{[0,1]^c}$, where $\phi(\cdot)$ is the density of a Normal distribution (or any other distribution that never hits zero and has first moment). From this, it’s clear that $$\mid f_b(x) - f_a(x) \mid \leq \mid b-a\mid h(x)$$
Now, let $\mu$ be the measure of a Uniform $(0,1)$ and $\mu_n$ be a measure for a $U(-1/n, 1- 1/n)$. It’s clear that $\mu_n \rightarrow_d \mu$.
But, note that for every $n \in \mathbb N$ and any $a \in \mathbb R$, you get
$$\left| \int f_a d\mu_n - \int f_a d\mu\right|= \left| a(1-\frac{1}{n}) - a\right |= \left | \frac{a}{n}\right| $$
Therefore, $$\lim_{n \to \infty} \sup_{a \in \mathbb R}\mid \frac{a}{n} \mid =+ \infty$$