Let $(f\_n)\_n:X \to \mathbb R$ be a sequence of measurable functions on a measurable space $X$ converging pointwise to a function $f:X \to \mathbb R$, and let $(\mu\_n)\_n$ be a sequence of finite measures (e.g probability measures) on $X$ such that each $f\_n$ is integrable w.r.t $\mu\_n$.



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> **Question.** *Under what additional conditions do we have $\int\_X (f\_n-f)\,d\mu\_n \to 0$ ?*
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**Note.** In case it helps, it may be assumed that $\mu\_n$ converges (in some sense) to a measure $\mu$ on $X$.



Dominated convergence theorem when the measure space also varies with $n$
$\newcommand\ep\varepsilon$The conjunction of the following conditions is enough:


1. The $f\_n$'s are uniformly bounded: $|f\_n|\le M$ for some real $M>0$ and all $n$;
2. $X$ is Polish;
3. $f\_n\to f$ uniformly on every compact $K\subseteq X$;
4. $\mu\_n\to\mu$ weakly for some $\mu$.


Indeed, take any real $\ep>0$. By [Prokhorov's theorem](https://en.wikipedia.org/wiki/Prokhorov%27s_theorem#:%7E:text=In%20measure%20theory%20Prokhorov%27s%20theorem,on%20complete%20separable%20metric%20spaces.) and in view of conditions 2 and 4, $\mu\_n(X\setminus K)<\ep$ for some compact $K\subseteq X$ and all $n$. Write
$$I\_n:=\int\_X(f\_n-f)\,d\mu\_n=I\_{1,n}+I\_{2,n},$$
where
$$I\_{1,n}:=\int\_K(f\_n-f)\,d\mu\_n,\quad
I\_{2,n}:=\int\_{X\setminus K}(f\_n-f)\,d\mu\_n.$$
By condition 3, $I\_{1,n}\to0$. By conditions 1 and 3, $|f|\le M$ and hence $|f\_n-f|\le2M$. So, $|I\_{2,n}|\le2M\ep$. So, $\limsup\_n|I\_n|\le2M\ep$, for every real $\ep>0$. So, $I\_n\to0$, as desired.


