# Definitions:

An open box in $$\mathbb{R}^d$$ is defined to be either the empty set, or a set of the form $$(a_1,b_1) \times \cdots \times (a_d,b_d)$$ where $$a_j < b_j$$ for each $$j\in\{1,\ldots,d\}$$. The volume of an open box $$B$$ in $$\mathbb{R}^d$$ is defined to be zero if $$B = \varnothing$$, and $$(b_1-a_1)\cdots(b_d-a_d)$$ otherwise. In either case it is denoted by $$\text{vol}_d(B)$$. In abbreviation of the statement that $$\mathcal{B}$$ is a finite or countably infinite collection of open boxes in $$\mathbb{R}^d$$, let us write $$\mathcal{B} \in \mathscr{B}_d$$. The outer measure of a subset $$\Omega$$ of $$\mathbb{R}^d$$ is defined to be the extended real number $$m_d^*(\Omega) := \inf\left\{ a \in\mathbb{R} \cup \{\pm\infty\} \hspace{1mm}:\hspace{1mm} \text{there exists a } \mathcal{B} \in \mathscr{B}_d \text{ such that both} \hspace{2mm} a = \sum_{B \in \mathcal{B}} \text{vol}_d(B) \hspace{2mm}\text{and}\hspace{2mm} \Omega \subseteq \bigcup_{B \in \mathcal{B}} B \right\}.$$

# Context

I'm following Terence Tao's Analysis II. Assume that some the theory of the outer measure, and of the Lebesgue measurable sets has already been developed.

# Question

Recently, I proved as an exercise that for any set $$A \subseteq \mathbb{R}^d$$ (which need not be Lebesgue measurable), and any $$\varepsilon>0$$, there exists an open set $$G$$ such that both $$A \subseteq G$$ and $$m^*(G) \leq m^*(A) + \varepsilon$$.

Furthermore, I see by this theorem that if $$A$$ is Lebesgue measurable, then for any $$\varepsilon>0$$, there exists a closed set $$F$$ such that both $$F \subseteq A$$ and $$m^*(A) \leq m^*(F) + \varepsilon$$. But is it necessary to assume that $$A$$ is Lebesgue measurable? That is, are either of the following statements true:

1. For every $$A \subseteq \mathbb{R}^d$$ (which need not be Lebesgue measurable), and any $$\varepsilon>0$$, there exists a closed set $$F$$ such that both $$F \subseteq A$$ and $$m^*(A) \leq m^*(F) + \varepsilon$$.
2. For every $$A \subseteq \mathbb{R}^d$$ (which need not be Lebesgue measurable), and any $$\varepsilon>0$$, there exists an open set $$F$$ such that both $$F \subseteq A$$ and $$m^*(A) \leq m^*(F) + \varepsilon$$.

Side note: proof of the claim concerning approximation by open sets from above

Basically, this falls out of the presence of the infimum in the definition of $$m^*(A)$$. Indeed, $$\sum_{B \in \mathcal{B}} \text{vol}_d(B) \leq M^*(A) + \varepsilon$$ for some $$\mathcal{B} \in \mathscr{B}_d$$ with $$A \subseteq \bigcup \mathcal{B}$$. Then, because $$m^*(B) = \text{vol}_d(B)$$ for any box $$B$$, and using sub-additivity, we have $$m^*(\bigcup \mathcal{B}) \leq \sum_{B \in \mathcal{B}} m^*(B) = \sum_{B \in \mathcal{B}} \text{vol}_d(B).$$ Finally, calling $$G : = \bigcup \mathcal{B}$$, the claim is true.

For 1., the answer is No. You are asking if a (measurable) set can be approximated from above with closed sets. Assume we are dealing with a measurable set. Passing to complements, the equivalent thing is whether a measurable set can be approximated from the inside with open subsets. The answer is no. One reason can be that the set has positive measure, but void interior (think of some Cantor type subset of $$[0,1]$$)
For 2., the answer is again no. Take the standard example of a non-measurable subset. Say $$F\subset [0,1]$$ is a system of representatives for $$\mathbb{R}\mod \mathbb{Q}$$. Since a countable family of translates of $$F$$ covers $$[0,1]$$, it must have a positive exterior measure. But since these translates are all disjoint, you cannot fit inside a measurable subset of measure $$>0$$.
In general, for any subset $$M$$, the supremum of the measure of all closed subsets of $$M$$ will give you the interior measure $$m_{*}(M)$$, while the infimum of measures of open suprasets will give the exterior measure $$m^{*}(M)$$. These will be equal if and only if the subset if measurable ( assuming they are finite).
No. Open subsets of $$A$$, and closed subsets of $$A$$, and more generally Borel subsets of $$A$$, and even more generally Lebesgue measurable subsets of $$A$$ cannot have measure larger than the inner measure of $$A$$.
• Hm, thank you for the info. But I'm not sure I understand the relevance. I didn't think that I was asking for $F$ to have a a measure larger than the inner measure of $A$. I though I was seeking $F$ with a measure just barely smaller than the outer measure of $A$, and I'm afraid I can't see why these would be the same ask. Is this because if $A$ is not measurable, then the difference between the outer and inner measures of $A$ is positive? – Thomas Winckelman May 23 at 2:30
• @ThomasWinckelman As long as you're talking about sets of finite measure (which seems to be the case as you're adding $\varepsilon$ to measures), non-measurability means exactly that the inner measure is strictly smaller than the outer measure. (I took that for granted in my answer, since non-measurability was assumed in the question.) – Andreas Blass May 23 at 13:52