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?



Absolute continuity of limiting measures
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. 


