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?

  • $\begingroup$ My assumption would be that it is still true due to the regularity of Lebesgue measure (as your linked answer showed for part a), but what is the motivation for working with non-measurable sets? Generally they are quite pathological and it is non-trivial to even construct such a beast. $\endgroup$ – Math1000 Nov 11 '19 at 19:57
  • $\begingroup$ (b) holds for inner Lebesgue measure $m_{*}.\;$ More generally, we have:$\;$ (A) $\;m^{*}(\liminf E_i) \leq \liminf m^{*}(E_i)\;$ and $\;$ (B) $\;m_{*}(\limsup E_i) \geq \limsup m_{*}(E_i)\;$ if $\;\bigcup_{i=n}^{\infty}E_i \;$ has finite inner Lebesgue measure for some $n.$ Here, the extreme limits on the sets are the usual set-theoretic limits and the other extreme limits are those for sequences of real numbers. $\endgroup$ – Dave L. Renfro Nov 11 '19 at 20:17

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 $$m^*(\bigcap_{n=1}^{\infty}A_n)=m^*(\emptyset)=0<m^*(V) \leq \liminf_n m^*(A_n)$$

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  • $\begingroup$ +1 How do we know $\bigcap_n A_n$ is empty? $\endgroup$ – Pascal's Wager Nov 11 '19 at 20:35
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    $\begingroup$ @Pascal'sWager the sets $V+q_n$ are disjoint so this will help you to understand $\endgroup$ – Marios Gretsas Nov 11 '19 at 20:37
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    $\begingroup$ I see. So the $V+q_n$ are disjoint since $V$ contains one element from each rational equivalence class. $\endgroup$ – Pascal's Wager Nov 11 '19 at 20:45

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