I'm wondering if the classic dominated convergence theorem can be generalized to the following uniform version.
Let $\{\mu_n:n=1,2,...\}$ be a countable set of probability measures. Let $\{f_k: k=1,2,\cdots\}$ be a sequence of measurable functions. Suppose for each $n$, $f_k(x)\to 0$ holds $\mu_n$-almost surely. Suppose there is a measurable function $g$ such that (i) $\sup_k |f(x)|\le g(x)$ for all $x$; (ii) $\sup_n \int g(x) \text{d}\mu_n < \infty$. Then, do we have $\sup_n |\int f_k(x)\text{d}\mu_n| \to 0$ as $k\to\infty$?
Based on the classic dominated convergence theorem, we know $\int f_k(x)\text{d}\mu_n\to0$ for each $n$. The question is that does it convergences uniformly in $n$ as stated above? If not, what extra conditions do we need?
Unfortunately not. For a simple counterexample consider the measurable space $[0,1]$ with the Lebesgue $\sigma$-algebra, and define $f_k=\mathbb 1_{[0,1/k]}$, and $\mu_n=n\lambda|_{[0,1/n]}$, where $\lambda$ is the usual Lebesgue measure. We thus have, for any $k\in \mathbb N$, that $$\sup_{n\in\mathbb N}|\int f_kd\mu_n|=\sup_{n\in\mathbb N}|n\int f_k\mathbb 1_{[0,1/n]}d\lambda|=1,$$ because for any $n>k$ we have $\int f_kd\mu_n=1$.