# Is the Hausdorff outer measure regular?

An outer measure $\mu^*$ is said to be regular if for every set $A \subset X$

$$\mu^\ast (A)=\inf\{\mu^*(E) : E\supset A \text{ is } \mu^\ast\text{-measurable} \}$$

To check that an outer measure is regular, we just have to check whether

$$\mu^\ast(A)\geq\inf\{\mu^*(E) : E\supset A \text{ is } \mu^\ast\text{-measurable} \}$$

since the other inequality follows from the outer measure axioms.

I have to find out whether the $s$-dimensional Hausdorff outer measure

For any $s \geq 0$ and $\delta\gt0$ we define the $\delta$-approximating $s$-dimensional Hausdorff outer measure,

$$\mathfrak{h}_{s,\delta}^\ast (A)= \alpha_s \inf\left\{\sum_{i=1}^\infty \text{diam}^s(A_i): A \subset \bigcup_{i=1}^\infty A_i, \text{diam}(A_i)\lt\delta\right\}$$

and the $s$-dimensional Hausdorff measure,

$$\mathfrak{h}_{s}^\ast(A)=\sup_{ \delta\gt0} \mathfrak{h}_{s,\delta}^\ast(A)= \lim_{\delta \downarrow 0} \mathfrak{h}_{s, \delta}^\ast(A)$$

Here $0\lt\alpha_s\lt\infty$ is chosen so that for $s \in \mathbb{N}$ the $s$-dimensional Hausdorff measure of the $s$-dimensional unit cube is one.

is regular. Since the Lebesgue outer measure is one of these measures, and it is regular, I'm trying to prove the Hausdorff outer measure is regular. So far I've got to

for any $A \subset X$ and $\delta\gt0$

$$\mathfrak{h}_s^\ast(A) \geq \inf\{\mathfrak{h}_{s,\delta}^\ast(E):A \subset E \text{ is } \mathfrak{h}_s^\ast\text{-measurable}\}$$

but I don't know how to show I'm allowed to swap my sup and my inf. If indeed I am.

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You only need to consider the case $\mathfrak{h}_{s}^\ast(A) \lt \infty$, but you need to be a bit careful in choosing the outer approximations since swapping $\inf$ and $\sup$ certainly isn't allowed without some thinking. If you knew that you can always take the same set $E$ in the $\inf$ (which I will show in \eqref{eqn:ast} below) then you'd be essentially done, but it is impossible to say if you got so far or not.

I'm going to ignore $\alpha_s$ since it only is a scalar factor that doesn't play a rôle in the argument and the formulas will already get cumbersome enough without it.

The following argument is essentially the one from Fremlin's measure theory, Volume 4II, 471D, page 102. $\DeclareMathOperator{\diam}{diam}$ $\newcommand{\hsd}[1]{\mathfrak{h}_{s,#1}^\ast}$

Set $\delta_n = 2^{-n}$ and choose $(A_{i}^n)_{i \in \mathbb{N}}$ of diameter $\leq 2^{-n}$ such that $A \subset \bigcup_{i=1}^\infty A_{i}^n$ and that

$$\tag{1}\label{eqn:eq1} \sum_{i=1}^\infty (\diam{A_{i}^n})^s \leq \hsd{2^{-n}}(A) +2^{-n},$$ which is possible because of the definition of $\hsd{2^{-n}}$ as infimum.

Observe that there's no reason for the $A_{i}^n$ to be $\hsd{2^{-n}}$-measurable, let alone Borel, so we would like to “blow them up” slightly, so as to get open sets still approximating the $\hsd{2^{-n}}$-measure of $A$ well. The problem is that by doing so we will lose the diameter condition which appears in the definition of $\hsd{2^{-n}}$, but this isn't a serious problem: we can simply choose a larger $n$ and work with the sets we obtain from there. Here are the gory details:

Choose $0 \lt r_{i}^n \lt 2^{-n}$ so small that $$(\diam{(A_{i}^n)} + 2r_{i}^n)^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.$$ Now put $U_{i}^n = \{x \in X\,:\,d(x,A_{i}^n) \lt r_{i}^n\}$ and note that $U_{i}^n \supset A_{i}^n$ is an open set whose diameter satisfies $$\tag{2}\label{eqn:eq2} (\diam{U_{i}^n})^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.$$ Let $$B = \bigcap_{n=1}^\infty \bigcup_{i=1}^\infty U_{i}^n$$ and observe that $B$ is a $G_{\delta}$-set (countable intersection of open sets) containing $A$.

Given any $\delta \gt 0$, we can find $N$ such that $3 \cdot 2^{-N} \leq \delta$ so that for all $n \geq N$ we have $\diam{U_{i}^n} \leq \delta$. As $B \subset \bigcup_{i = 1}^\infty U_{i}^n$ we see from \eqref{eqn:eq1} and \eqref{eqn:eq2} that for $n \geq N$ $$\hsd{\delta}(B) \leq \sum_{i=1}^\infty (\diam{U_{i}^n})^s \leq \sum_{i=1}^\infty [(\diam{A_{i}^n})^s + 2^{-n-i}] \leq \hsd{2^{-n}}{(A)} + 2 \cdot 2^{-n}$$ Since $\hsd{2^{-n}}(A) \leq \mathfrak{h}_{s}^\ast(A)$ we get for $n \geq N$ $$\hsd{\delta}(B) \leq \mathfrak{h}_{s}^\ast(A) + 2^{-n}$$ and letting $n \to \infty$ this gives

$$\tag{\ast}\label{eqn:ast} \hsd{\delta}(B) \leq \mathfrak{h}_{s}^\ast(A)$$

for every $\delta \gt 0$. [Note: this is stronger than your condition involving $\inf$ since we can specify the set $E = B$ and thus avoid the infimum]

Taking the $\sup$ over all $\delta$ in \eqref{eqn:ast} this yields $$\mathfrak{h}_{s}^\ast(B) \leq \mathfrak{h}_{s}^\ast(A)$$ and since $B \supset A$ and $B$ is $\mathfrak{h}_{s}$-measurable (being Borel) we can finally conclude that $$\mathfrak{h}_{s}(B) = \mathfrak{h}_{s}^\ast(A),$$ as desired.

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(+1) for the use of \eqref :-) –  robjohn Apr 8 '12 at 12:51
Thanks. Nice, isn't it? :) –  t.b. Apr 8 '12 at 12:54