Dominated convergence theorem when the measure space also varies with $n$ 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 $\mu_n$.

Question. Under what additional conditions do we have $\int_X (f_n-f)\,d\mu_n \to 0$ ?

Note. In case it helps, it may be assumed that $\mu_n$ converges (in some sense) to a measure $\mu$ on $X$.
 A: $\newcommand\ep\varepsilon$The conjunction of the following conditions is enough:

*

*The $f_n$'s are uniformly bounded: $|f_n|\le M$ for some real $M>0$ and all $n$;

*$X$ is Polish;

*$f_n\to f$ uniformly on every compact $K\subseteq X$;

*$\mu_n\to\mu$ weakly for some $\mu$.

Indeed, take any real $\ep>0$. By Prokhorov's theorem 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.
A: This answer should perhaps be posted as a comment, since the whole topic was already debated in this MathOverflow Q&A. However i briefly restate the main result here and leave the above link for further detals: a definitive answer is Cafiero's convergence theorem (see [1]) which, roughly states that$\DeclareMathOperator{\Dm}{\operatorname{d\!}}$
$$
\int_X f_n\Dm\mu_n - \int_X f\Dm\mu_n \to 0 \iff \text{$(f_n\cdot\mu_n)_{n\geq 1}$ is uniformly exaustive.}
$$
Note that this necessary and sufficient condition is not very well known, even in the circles of measure theorists.
Reference
[1] Cafiero, F. (1953), "Sul passaggio al limite sotto il segno d'integrale per successioni d'integrali di Stieltjes-Lebesgue negli spazi astratti, con masse variabili con gli integrandi" [On the passage to the limit under the sign of integral for sequences of Stieltjes–Lebesgue integrals in abstract spaces, with masses varying jointly with integrands], Rendiconti del Seminario Matematico della Università di Padova (in Italian), 22: 223–245, MR0057951, Zbl 0052.05003.
A: Using the Radon-Nikodym Theorem this problem can be reduced to the case for a fixed measure.
Suppose none of the $\mu_n$ is the zero-measure, i.e. $\mu_n(X)>0$ for all $n \in \mathbb{N}$.
For a series $(\alpha_n)_{n \in \mathbb{N}} \subset (0,1)$ with $\sum\limits_{n=1}^\infty \alpha_n = 1$ define the probability measure
$$
\mu = \sum\limits_{n=1}^\infty \alpha_n \, \frac{\mu_n}{\mu_n(X)} \ .
$$
Now for any set $A \in \mathcal{B}(X)$ we have the implication
$$
0= \mu(A) = \sum\limits_{n=1}^\infty \alpha_n \, \frac{\mu_n(A)}{\mu_n(X)} \ \Rightarrow \ 0 = \mu_n
$$
for arbitrary $n \in \mathbb{N}$. Radon-Nikodym then implies the existance of measurable functions $g_n : X \rightarrow [0,\infty)$ such that $\mu_n = g_n \cdot \mu$ or equvalently
$$
\mu_n(A) = \int_A g_n \, d\mu \ , \ \forall A \in \mathcal{B}(X) \ .
$$
The question whether $\int_X f_n-f \, d\mu_n$ converges to zero now becomes the question whether $\int_X (f_n-f)g_n \, d\mu$ converges to zero.
