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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 $X$.

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    $\begingroup$ For every countable family $\mu_n$ of finite measures you find another one $\lambda$ which dominates them all in the sense that $\lambda(A)=0$ implies $\mu_n(A)=0$. Radon-Nikodym implies that $\mu_n$ have densities $\varphi_n$ so that you ask for $\int f_n\varphi_nd\lambda \to \int f\varphi d\lambda$. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 21, 2021 at 11:50
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    $\begingroup$ Without Radon-Nikodym: For non-negative $f$, $\lim\int_X f_n-fd\mu_n=\lim\int_0^\infty \mu_n \left (\{x \in X \mid f_n(x)-f(x)>t\} \right )dt$. $\endgroup$
    – vitamin d
    Commented Oct 21, 2021 at 11:52
  • $\begingroup$ Thanks to everyone for the input. @JochenWengenroth, so, one deduces that if $\mu_n = \varphi_n d\lambda$ and $\varphi_n \to \varphi$ pointwise, and $f_n$ is dominated by some $\lambda$-integrable function, then $\int_X|f_n-f|d\mu_n \to 0$ by the classical Dominated Convergence Theorem, right ? $\endgroup$
    – dohmatob
    Commented Oct 21, 2021 at 12:43

3 Answers 3

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$\newcommand\ep\varepsilon$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 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.

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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.

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  • $\begingroup$ Thanks for mentioning Cafiero's theorem (new to me). Upvoted.The notion of exhaustiveness eludes me. Are there any conditions which can be imposed separately on $(f_n)_n$ and $(\mu_n)_n)$ (in the spirit of those mentioned in the question), to ensure that $(f_n\mu_n)_n$ is "uniformly exhaustive" ? $\endgroup$
    – dohmatob
    Commented Oct 21, 2021 at 14:38
  • $\begingroup$ @dohmatob to my knowledge, no: however, as said in the linked Q&A, Gaetano Fichera proved a simpler necessary and sufficient condition for the case $\mu_n=\mu$ for all $n$. Perhaps you can have a look at this fact in order to find conditions on $\{f_n\}$ and $\{\mu_n\}$ to assure that the family of their product $\{f_n\mu_n\}$ is uniformly exaustive. $\endgroup$
    – Daniele Tampieri
    Commented Oct 21, 2021 at 14:52
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    $\begingroup$ When $\mu_n = \mu$ for all $n$, then we are back to the classical DCT (and avatars). The last part of your remark is precisely the question in my last comment ;) $\endgroup$
    – dohmatob
    Commented Oct 21, 2021 at 14:54
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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.

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  • $\begingroup$ Thanks for the input. The question to which you've reduced the situation doesn't seem very transparent / exploitable (due to the construction of $\mu$). $\endgroup$
    – dohmatob
    Commented Oct 21, 2021 at 12:23

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