# Convergence of a sequence of functions integrated over a sequence of measures

I have real-valued functions $\{f_n\},f$ on a subset $X\subset \mathbb R^n$ that are equicontinuous and I have Borel measures $\{\mu_n\},\mu$. I have that

1. For each fixed $m$, $\int f_m d\mu_n\to\int f_md\mu$ and $\int f d\mu_n\to\int fd\mu$.
2. $f_n\to f$ pointwise (so also uniformly since equicontinuous)
3. $|f_n|\leq g\forall n$, $|f|\leq g$ and $\int g d\mu_n<\infty\forall n$, $\int g d\mu<\infty$,$\int gd\mu_n\to\int gd\mu$.

The question is whether $\int f_nd\mu_n\to\int f d\mu$?

So for example if we take away equicontinuity then $f_n(x)=x^n$ on $[0,1]$ and $\mu_n=(\text{point mass at$1-1/n$})$ makes this break (LHS $\to1/e$, RHS=1 in the above question). But it's not equicontinuous. But the thing I worry about is that the integration doesn't have anything to do with the topology on $X$ so equicontinuity is a moot point. But I'm not sure about this.

So can you prove $\int f_nd\mu_n\to\int f d\mu$ given the assumptions including equicontinuity?

If it doesn't work, what sort of extra lax condition might fix it? (Without touching the convergence modes. I.e., without asking for the measures to converge in total variation, which would immediately fix it. Only stuff like $f_n$ have to be extra continuous somehow or $X$ has to be a product of intervals in $R^n$ etc.)

It works. By 3., you can restrict yourself to a compact subset $K$ of $X$, such that $\mu_n(K)$ is uniformly bounded and such that in every integral you only make a small error. You can do this by looking at the sets $\{g>\epsilon\}$ and let $\epsilon \to 0$. By 2., $f_n \to f$ uniformly on $K$. The fact that $K$ is compact is essential! The uniform convergence is not true otherwise, just take a continuous bump and let it wander off to infinity. Thus, in the integral $\int_K f_n d\mu_n$ you can replace $f_n$ by $f$ with a small error. Then use the second convergence in 1..
Edit 2: Ok, the above proof fails in the first step. One cannot choose $K$ uniformly in $n$. I do not see a counterexample at the moment, but you can easily fix the above proof by requiring that the sequence $\mu_n$ is tight, which ensures the existence of $K$.