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 $E$.

The prove is illustrative of the $\frac{\epsilon }{2^n}$ 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 $m^{\ast }(E)$ is infinite, and we can assume that $m^{\ast }(E)$ is finite.

What follows is a exemplary use of the $\frac{\epsilon }{2^n}$ trick.

Let $\epsilon >0$ 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 $B_i^{\prime }$ such that $|B_i^{\prime }|\le |B_i|+\frac{\epsilon }{2^i}$. 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 $B^{\prime }={\cup }_{i=1}^{\mathrm{\infty }}{B}_i^{\prime }$. We have found an open set $B^{\prime }$ that contains $E$ and has outer measure $m^{\ast }(B^{\prime })\le m^{\ast }(E)+2\epsilon$. The infimum of measure over open covers of $E$ must be less than this one instance of an open cover:

As $\epsilon$ was arbitrary, we have:

and combined with our initial subadditivity claim, we have: