Assume that we have a metric space $(A,\rho)$ and a sequence of probabilistic Borel measures $\mu_{n}$ on $A$ that converges weakly to the probabilistic Borel measure $\mu$. Assume also that one is given a sequence $A_{j}$ of closed subsets of $A$. Let $\limsup_{n\to\infty} A_{n}=\bigcap_{j=1}^{\infty}\bigcup_{i=j}^{\infty} A_{i}$ be the usual set-theoretic upper limit of the sequence $A_{j}$. An easy combination of Portmanteau theorem and Fatou lemma gives one that $$\limsup_{m\to\infty}\limsup_{n\to\infty}\mu_{n}(A_m)\leq \mu(\limsup_{m\to\infty} A_{m}).$$

I am interested under what additional assumptions on the measures $\mu_n$ one can get the stronger statement:

$$\limsup_{n\to\infty}\mu_{n}(A_n)\leq \mu(\limsup_{n\to\infty} A_{n})?$$

Note that this is true if $\mu_{n}$ converges strongly to $\mu$ and in this case the assumption that $A_{n}$ are closed is unnecessary. The inequality also trivially holds if we assume that $A_{n}$ is a nested family of closed sets, that is $A_{n+1}\subseteq A_{n}$. On the other hand the inequality does not hold for an arbitrary weakly convergent sequence of probability measures as the following trivial counterexample shows:

$$A=\mathbb R, \rho(x,y)=|x-y|, \mu_{n}=\delta_{\frac{1}{n}}, \mu=\delta_{0}, A_{n}=\{\frac{1}{n}\}.$$


2 Answers 2


There is a version of a generalized Fatou lemma, under the condition that, for all measurable set $E$, $\liminf_n \mu_n(E)\leq\mu(E)$, see

O. Hernández-Lerma, J-B. Lasserre, Fatou's lemma and Lebesgue's convergence theorem for measures, J. Appl. Math. Stochastic Anal. 13 (2000), no. 2, 137–146.

There is also the possibility to modify the right-hand side of the seeked inequality as shown in

E. Feinberg, P. Kasyanov, N. Zadoianchuk, Fatou's lemma for weakly converging probabilities, Theory Probab. Appl. 58 (2014), no. 4, 683-689 (Arxiv1206.4073).

Theorem 1.1 p.2, stated for measurable functions instead of sets, says

Let $\{\mu_{n}\}\subset\mathbb{P}(A)$ converge weakly to $\mu\in\mathbb{P}(A)$, and let $\{f_{n}\}_{n\geq1}$ be a sequence of measurable nonnegative $\overline{\mathbb{R}}$-valued functions on $A$. Then $$ \int \liminf_{n\to\infty,~s'\to s}f _ { n } ( s ^ { \prime } ) \mu ( d s ) \leq \liminf _ { n \rightarrow \infty } \int f _ { n } ( s ) \mu _ { n } ( d s ). $$ or, equivalently, with $\limsup$ instead of $\liminf$, and a sequence $\{f_{n}\}$ of measurable functions uniformly bounded above, $$ \limsup_ { n \rightarrow \infty } \int f _ { n } ( s ) \mu _ { n } ( d s )\leq \int \limsup_{n\to\infty,~s'\to s}f _ { n } ( s ^ { \prime } ) \mu ( d s ). $$


This is a comment, not an answer, but I am not entitled. The short answer, if not directly to your question, is that this works fine if the limit set has the continuity property that its boundary has zero $\mu$ measure. This follows from early work of Grothendieck which can easily found with a search machine.

Added on request: The wikipedia article „Weak convergence of measures“ gives the precise result and some references, though not specifically to Grothendieck. I am fairly sure that it is in his text on topological vector spaces but I don‘t have access to a copy. The classical text by Billingsley which is referenced in wiki is probably the best place to start.

  • $\begingroup$ Can you be a little more specific about the reference? Grothendieck has quite a few early papers on analysis. $\endgroup$
    – Nik Weaver
    Nov 26, 2018 at 20:52
  • $\begingroup$ I'm pretty sure the characterization you refer to (the Portmanteau theorem) is due to Alexandroff. $\endgroup$ Nov 27, 2018 at 12:25
  • $\begingroup$ If you give me a reference, I will check. $\endgroup$
    – user131781
    Nov 27, 2018 at 12:50
  • $\begingroup$ @user131781 See Theorem 3 at mathnet.ru/php/… (warning: the PDF takes some time to download.) $\endgroup$ Nov 27, 2018 at 21:10

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