General property regarding outer measure for a nested sequence of sets (measurable or not).

Let $\bigcap_{n=1}^\infty E_n=∅$ and if $\mu^*(E_n) <\infty$ and $E_{n+1} \subseteq E_n $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) =0 $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is outer measure. Proof sketch please?

Solution 1:

It's false. (I was taking a shower when Michael posted his comment; what's below is a detailed exposition of a simpler version of what he said.)

Say $G$ is the group $[0,1)$, with addition modulo $1$. Note that Lebesgue outer measure is $G$-invariant. I'll be writing $a+b$ for the addition in $G$.

Let $H=[0,1)\cap\Bbb Q$, and let $C$ be a complete set of coset representatives for $H$ as a subgroup of $G$. Note that every $x\in G$ has a unique representation as $h+c$ with $h\in H$, $c\in C$.

Let $H=\{h_1,h_2,\dots\}$, and then define $H_n=\{h_n,h_{n+1},\dots\}$. Define $$E_n=H_n+C.$$

Then $\cap E_n=\emptyset$, although for every $n$ we have $$\mu^*(E_n)\ge\mu^*(h_n+C)=\mu^*(C)>0.$$