Counter examples in measures

Question

If $$\lim\mu (A_n)=0$$ is it true that $$\mu(\lim\sup A_n)=0 \quad ?$$

We know that if $(A_n)_n$ is an increasing sequence of sets then it is true. But for an arbitrary sequence of sets I believe it doesn't hold, though I cannot think of a counter-example.

I came up with this answer, but I am not sure if it is correct: $\mu(A)=0$, if $A$ is finite or $\mu(A)=\infty$, if $A$ is infinite. $A_n=\{1,2,...,n\}$. Then $\mu(A_n)=0$ while $\mu(limsupA_n)=\mu(\mathbb{N})=\infty$

Answer 1

No. Let $\mu$ be the Lebesgue measure on $[0,1]$, and consider the "typewriter sequence," i.e. let $A_n:=\displaystyle \Big[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\Big]$ whenever ${2^k \leq n < 2^{k+1}}$. Clearly $\mu(A_n)\to 0$, but $\bigcup_{n=2^k}^{2^{k+1}-1} A_n = [0,1]$ for each $k$; i.e. $$ \mu\big(\limsup A_n\big) = \lim_{m\to\infty} \mu\Big(\bigcup_{k\geq m}A_k\Big) = 1 $$

Answer 2

If you are familiar with Borel Cantelli Lemma then you know that for indpendent events $(A_n)$, the condition $\mu (\lim \sup A_n)=0$ holds iff $\sum \mu (A_n) <\infty$. So $\lim \mu(A_n)=0$ does not guarantee that $\mu (\lim \sup A_n)=0$.