# Outer Measure on $\mathbb{R}$

Let $p\ge0$, $d>0$ and $X \subset \mathbb{R}$.

(i) Define:

$$H_{p,d}(X) := \inf\left\{\sum_{i=1}^{\infty}\operatorname{diam} \left(\frac{A_i}{2}\right)^p : X \subset \bigcup_{i=1}^{\infty}A_i, 0 < \operatorname{diam}A_i \leq d \right\}.$$ If $X$ is empty, $H_{p,d}(X)=0$ .

(ii) Now, define $$H_p(X) := \lim_{d\rightarrow 0} H_{p,d}(X).$$

Prove that $H_p$ is an outer measure on $\mathbb{R}$ .

I have proved that $H_{p,d}$ is an outer measure on $\mathbb{R}$. Using that I was trying to show that $H_p$ is an outer measure on $\mathbb{R}$ , but I cannot prove the countable subadditivity of $H_p$.

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Please edit the equations so that it mathjax can display them properly. – UnadulteratedImagination Jan 20 '13 at 8:44

You already proved that $H_{p,d}$ is an outer measure, so for a sequence of subsets $X_i$ of $\mathbb{R}$, we have $$H_{p,d}\left(\bigcup_{i\in\mathbb{N}} X_i\right)\leq \sum_{i=1}^{\infty}H_{p,d}(X_i).$$ Taking limits yields $$H_{p}\left(\bigcup_{i\in\mathbb{N}} X_i\right)\leq \lim_{d\rightarrow 0}\lim_{n\rightarrow\infty}\sum_{i=1}^{n}H_{p,d}(X_i),$$ so showing that the two limits on the right can be interchanged would show the countable subadditivity.

(1) We may assume that $$L:=\sum_{i=1}^{\infty}H_{p}(X_i) < \infty$$ as otherwise the required inequality holds trivially.

(2) Note that $H_{p,d}$ is decreasing as a function of $d$ as the set $$\left\{A\supset X: 0< \operatorname{diam} A \leq d \right\}$$ grows as $d$ grows, so the infimum in the definition of $H_{p,d}$ can only get smaller.

(3) Next, consider the sequence of functions defined by $$f_n(d) := \sum_{i=1}^n H_{p,d}(X_i).$$ By monotony, and since $L$ is required to be finite, the limit $$H_{p,0}(X_i) := \lim_{d\rightarrow 0} H_{p,d}(X_i)$$ exists and is the maximum value of $H_{p,d}$. We thus observe that $$\sum_{i=1}^n \left\|H_{p,d}(X_i)\right\|_{\infty} = \sum_{i=1}^n H_{p,0}(X_i) = \sum_{i=1}^n H_{p}(X_i) \leq L <\infty$$ for all $n\in\mathbb{N}$, where $\left\|\cdot\right\|_{\infty}$ denotes the supremum norm, i.e. the series $$\sum_{i=1}^{\infty} \left\|H_{p,d}(X_i)\right\|_{\infty}$$ converges. This shows that the sequence $f_n$ is uniformly convergent with respect to $d$, hence the limit function $f$ is continuous (at least) at $d=0$.

Using continuity, we finally get $$\lim_{d\rightarrow 0}\lim_{n\rightarrow\infty}\sum_{i=1}^{n}H_{p,d}(X_i) = \lim_{d\rightarrow 0 } f(d) = f(0) = \lim_{n\rightarrow\infty}f_n(0) = \sum_{i=1}^{\infty}H_{p}(X_i)$$ as was to be shown.

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Thanks very much for your help. I have a little confusion.You have written 'We thus observe that....'In the 2nd and 3rd term where does infinity come from? – Ester Jan 20 '13 at 10:39
These weren't supposed to be there, fixed it! :) – Andy Brandi Jan 20 '13 at 10:42