Absolute continuity of limiting measures

Question

Let $A_n$, $B_n$ for $n \in \mathbb N$ be finte subsets of compact set $X$ in $\mathbb C$ such that $A_n \subset B_n$.

Let $\delta_{A_n}:= \frac{1}{|A_n|} \displaystyle\sum_{x\in A_n} \delta_x$ and $\delta_{B_n}:=\frac{1}{|B_n|} \displaystyle\sum_{x\in B_n} \delta_x$ be normalized dirac probability measures.

If $\delta_{A_n} \to \mu$ and $\delta_{B_n} \to \sigma$ as $n \to \infty$ in the weak* topology then can we say that $\mu$ is absolutely continuous with respect to $\sigma$?

If answer is no in general, can one suggest me reference for the weakest known sufficient condition on $A_n$ and $B_n$ (may be on $|A_n|$ and $|B_n|$) so that answer is yes?

Answer 1

Let $\nu:=\sigma$. This answer, based mainly on comments by Anthony Quas, provides a necessary and sufficient condition for $\mu\ll\nu$ (the absolute continuity of $\mu$ with respect to $\nu$) in terms of $|A_n|$ and $|B_n|$, assuming that $X$ contains at least two distinct non-isolated points.

More specifically, let $K:=(k_n)$ and $L:=(l_n)$ be two sequences of natural numbers such that $k_n\to\infty$ and $k_n\le l_n$ for all $n$. Let us say that the pair $(K,L)$ is good if for all sequences $(A_n)$ and $(B_n)$ of subsets of $X$ satisfying the conditions

1. $A_n\subseteq B_n$, $|A_n|=k_n$, and $|B_n|=l_n$ for all $n$ and
2. $\delta_{A_n}\to\mu$ and $\delta_{B_n}\to\nu$ for some probability measures $\mu$ and $\nu$

we have $\mu\ll\nu$, where the convergence $\lambda_n\to\lambda$ of probability measures means that $\int f\,d\lambda_n\to\int f\,d\lambda$ for all bounded continuous functions $f$.

Theorem. Suppose that $X$ contains at least two distinct non-isolated points. Then the pair $(K,L)$ is good if and only if $\limsup_n k_n/l_n>0$.

Proof. The "if" part: Suppose that $\limsup_n k_n/l_n>0$. Take any sequences $(A_n)$ and $(B_n)$ of subsets of $X$ satisfying the conditions 1 and 2. We then need to show that $\mu\ll\nu$. Passing to subsequences, without loss of generality we may assume that $|A_n|=k_n\ge al_n=a|B_n|$ for some $a\in(0,1)$ and all $n$. Then for all nonnegative bounded continuous functions $f$
$$a\int f\,d\mu\longleftarrow \frac a{|A_n|}\sum_{x\in A_n}f(x) \le\frac1{|B_n|}\sum_{x\in B_n}f(x)\longrightarrow\int f\,d\nu,\quad $$ whence $a\int f\,d\mu\le\int f\,d\nu$. So, by the regularity of the measures $\mu$ and $\nu$, it follows that $a\mu\le\nu$ and hence $\mu\ll\nu$, which completes the proof of the "if" part.

The "only if" part: Suppose that $\limsup_n k_n/l_n\not>0$, that is, $k_n/l_n\to0$. We then need to construct sequences $(A_n)$ and $(B_n)$ of subsets of $X$ satisfying the conditions 1 and 2 and such that $\mu\not\ll\nu$. Take any two distinct non-isolated points of $X$, say $x$ and $y$. Then there are sequences $(x_n)$ and $(y_n)$ of points in $X$ such that $x_n\to x$, $y_n\to y$, and the points $x_1,x_2,\dots,y_1,y_2,\dots$ are all pairwise distinct. Let $A_n:=\{x_1,\dots,x_{k_n}\}$ and $B_n:=A_n\cup\{y_1,\dots,y_{l_n-k_n}\}$. Then the sequences $(A_n)$ and $(B_n)$ of subsets of $X$ satisfy the conditions 1 and 2 with $\mu=\delta_x$ and $\nu=\delta_y$, so that $\mu\not\ll\nu$, which completes the proof of the "only if" part and thus the proof of the theorem.

Remark: The condition that $X$ contains at least two distinct non-isolated points was used only in the proof of the "only if" part of the theorem.