 Math and science::Analysis::Tao, measure::02. Lebesgue measure

# 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:

[$m^*(E) = \inf_{\text{over what?} } m^*(U).$ ]

The proof can be intuitively grasped by noticing that $$m^*(E)$$ can be used to specify an open cover of $$E$$ with measure arbitrarily close to $$m^*(E)$$.

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

#### Proof

From monotonicity of outer Lebesgue measure, we have:

[$m^*(E) \le \text{something}.$ ]

So we are left to show the reverse:

[$\text{something} \le m^*(E).$]

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:

$\sum_{i=1}^{\infty} |B_i| \le m^*(E) + \varepsilon$

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:

[$\sum_{i=1}^{\infty}|B'_i| \le m^*(E) + \varepsilon + \sum_{i=1}^{\infty} \frac{\varepsilon}{2^i} = m^*(E) + \text{what?}$]

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

$m^*\left( \bigcup_{i=1}^{\infty} B'_i \right) \le m^*(E) + 2\varepsilon.$

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:

$\inf_{E \subset U, U \text{is open}} m^*(U) \le m^*(B') \le m^*(E) + 2\varepsilon.$

As $$\varepsilon$$ was arbitrary, we have:

[$\inf_{E \subset U, U \text{is open} } m^*(U) \quad \text{?} \quad m^*(E),$]
and combined with our initial subadditivity claim, we have:
$\inf_{E \subset U, U \text{is open}} m^*(U) = m^*(E).$