# Outer regularity. Theorem.

This property connects the outer Lebesgue measure of an arbitrary set \( E \subset R^d \) to the outer Lebesgue measure of open sets which contain \( E \).

### Outer regularity. Theorem.

Let \( E \subset \mathbb{R}^d \) be an arbitrary set. Then one has:

The prove is illustrative of the \( \frac{\varepsilon}{2^n} \) trick.

#### Proof

From monotonicity of outer Lebesgue measure, we have:

So we are left to show the reverse:

Reading this second less-equal as "not greater than" can motivate intuition.

Tao points out specifically that the inequality is trivial if \( m^*(E) \) is infinite, and we can assume that \( m^*(E) \) is finite.

What follows is a exemplary use of the \( \frac{\varepsilon}{2^n} \) trick.

Let \( \varepsilon > 0 \) be a real. The definition of outer Lebesgue measure affords us the ability to assert the existance of a countable family of boxes covering \( E \) such that it's outer measure is slightly greater than the infimum:

We can enlarge each box \( B_i \) to become an open box \( B'_i \) such that \( |B_i'| \le |B_i| + \frac{\varepsilon}{2^i} \). The union of these open boxes is also open and also contains \( E \). In particular, we have:

Applying subadditivity (the outer measure of the union is less than the sum of the outer measures) we have:

Denote this union as \( B' = \cup_{i=1}^{\infty}B'_i \). We have found an open set \( B' \) that contains \( E \) and has outer measure \( m^*(B') \le m^*(E) + 2\varepsilon \). The infimum of measure over open covers of \( E \) must be less than this one instance of an open cover:

As \( \varepsilon \) was arbitrary, we have: