We have two theorems which are sometimes called the continuity properties of Lebesgue Measure. Notational note: $m^*$ is Lebesgue outer measure defined by $m^*(E)=\inf\{ \sum l(I_j): E \subseteq \bigcup_jI_j\}$.
a. Let $E_1 \subseteq E_2 \subseteq \cdots \subseteq \mathbb{R}$, where each $E_i$ is (Lebesgue) measurable. Then $m^*\left(\bigcup_i E_i\right)=\lim_{i \to \infty} m^*(E_i)$.
b. Let $ \mathbb{R} \supseteq E_1 \supseteq E_2 \supseteq \cdots$ where each $E_i$ is (Lebesgue) measurable and $m^*(E_1)< \infty$. Then $m^*\left(\bigcap _i E_i\right)=\lim_{i \to \infty} m^*(E_i)$.
I wanted to ask if properties a. and b. above still hold if we relax the requirement that each $E_i$ be measurable.
My initial guess was "no," as the proofs I am familar with utilize the excision property (which is not true for non-measuable sets in general).
However, according to Outer measure of a nested sequence of non-measurable sets
, property a. is actually true even if we don't assume that each $E_i$ is measurable.
What about property b.? Is it true that any decreasing sequence of subsets of $\mathbb{R}$ has $\bigcap _i m^*(E_i)=\lim_{i \to \infty} m^*(E_i)$ as long as we assume that the outer measures are finite? Or is there a quick counterexample?
Best Answer
Take as a counterexeample $A_n=\bigcup_{k=n}^{\infty}(V+q_k)$ where $V$ is a non measurable Vitali set of $[0,1]$ and $\{q_1,...,q_n....\}$ is an enumeration of the rationals on $[-1,1]$
We have that $\bigcup_{k=1}^{\infty}(V+q_k) \subseteq [-1,2]$ and $A_{n+1} \subseteq A_n$
We have that $$m^*(\bigcap_{n=1}^{\infty}A_n)=m^*(\emptyset)=0<m^*(V) \leq \liminf_n m^*(A_n)$$