## Small things about Hausdorff measure

Suppose ${X}$ is a metric space, ${\mathscr{F}}$ is a families of subsets of ${X}$, and ${\phi}$ is a function on ${\mathscr{F}}$ such that

$\displaystyle 0\leq \phi(A)\leq \infty\quad\text{ whenever }A\in \mathscr{F}$

we can obtain a measure through a prodecure called Caratheodory’s construction. Define

$\displaystyle \tilde{\phi}_\delta(E)=\inf\left\{\sum_i\phi(A_i)\big| E\subset\cup_i A_i\text{ and diam}A_i<\delta\right\}$

for any set ${E\subset X}$. Then

$\displaystyle \tilde{\phi}_\delta\geq\tilde{\phi}_\sigma \text{ if }0<\delta<\sigma<\infty$

$\displaystyle \tilde{\phi}_\delta(E\cup F)\geq \tilde{\phi}_\delta(E)+\tilde{\phi}_\delta(F)\text{ if dist}(E,F)>\delta>0$

This means ${\tilde{\phi}_\delta(E)}$ is monotone, then it is reasonable to define

$\displaystyle \lambda(E)=\lim\limits_{\delta\rightarrow 0+}\tilde{\phi}_\delta(E)=\sup_{\delta>0}\tilde{\phi}_\delta(E)$

${\tilde{\phi}_\delta}$ and ${\lambda}$ are (outer) measures. Moreover

$\displaystyle \lambda(E\cup F)\geq \lambda(E)+\lambda(F)\text{ if dist}(E,F)>0$

This means all open set are measurable in ${\lambda}$. If ${\mathscr{F}}$ contains Borel sets of ${X}$, then ${\lambda}$ is Borel regular measure.

Suppose ${X=\mathbb{R}^n}$, ${\mathscr{F}}$ is the Borel set of ${\mathbb{R}^n}$, ${m\geq0}$

$\displaystyle \phi(B)=w_m\left(\frac{\text{diam }B}{2}\right)^m, \quad B\in \mathscr{F}$

Then the Caratheodory’s construction will yield the ${m-}$dimensional Hausdorff measure.

Lemma: ${X}$ is a (separable?) metric space, ${\mu}$ is a measure on ${Y}$, ${f:X\rightarrow Y}$, ${f(A)}$ is measurable whenever ${A}$ is a Borel subset of ${X}$, define

$\displaystyle \phi(A)=\mu(f(A))$

${\lambda}$ is obtained from the Caratheodory’s construction from ${\phi}$ on all Borel subsets of ${X}$. Then

$\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)\text{ whenever }A\text{ is a Borel set }$

where ${N(f|A,y)=card\{x\in A|f(x)=y\}}$. Proof: Find Borel partitions ${\mathscr{P}_i}$ of ${A}$ such that ${\mathscr{P}_{i+1}}$ is a refinement of ${\mathscr{P}_i}$, or all subsets in ${\mathscr{P}_i}$ are unions of that of ${\mathscr{P}_{i+1}}$. And also assume

$\displaystyle \lim_{i\rightarrow \infty} \sup_{E\subset \mathscr{P}_i} \text{diam}(E)=0$

Let ${\chi^i_{f(E)}}$ be the characteristic function of ${f(E)}$, ${E}$ is from partition ${\mathscr{P}_i}$.

$\displaystyle \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y)\nearrow N(f|A,y)$

From Levy’s monotone convergence thm,

$\displaystyle \int_Y N(f|A,y)d\mu(y)=\lim_{i\rightarrow \infty}\int_Y \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y) d\mu(y)=\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\mu(f(E))$

$\displaystyle =\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\phi(E)\geq \lambda(A)$

by the defintion of ${\lambda(A)}$.

Note that the defition of ${\phi(A)}$ implies ${\tilde{\phi}_\delta\geq \phi(A)}$ for any ${\delta>0}$, hence ${\lambda(A)\geq \phi(A)}$ for very Borel set ${A}$. Since all Borel set are ${\lambda}$ measurable, for any partition ${\mathscr{P}}$

$\displaystyle \lambda(A)=\sum_{E\in \mathscr{P}}\lambda(E)\geq \sum_{E\in \mathscr{P}}\phi(E)=\sum_{E\in \mathscr{P}}\mu(f(E))$

combining with the result before,

$\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)$

$\Box$

Theorem: Suppose ${f:\mathbb{R}^l\rightarrow \mathbb{R}^n}$ is a Lipschitzian map. ${m}$ is nonegative number, then

$\displaystyle \int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)\leq (Lip f)^m\mathscr{H}^m(A)$

whenever ${A\subset\mathbb{R}^l}$ is a Borel subset. Proof: Let us apply the above lemma with ${\mu=\mathscr{H}^m}$ on ${\mathbb{R}^n}$, then

$\displaystyle \lambda(A)=\int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)$

For any Borel set ${B}$, any covering of ${B}$ by sets diameter ${\leq \delta}$ yields a covering of ${f(B)}$ of diameter ${\leq (Lip f)\delta}$, then

$\displaystyle \phi(B)=\mathscr{H}^m(f(B))\leq (Lip f)^m\mathscr{H}^m(B)$

$\displaystyle \tilde{\phi}_\delta(A)\leq \sum_{\text{diam}B_i<\delta, B_i\in \mathscr{P}}\phi(B_i)\leq (Lip f)^m\mathscr{H}^m(A)$

By taking the limit, ${\lambda(A)\leq (Lip f)^m\mathscr{H}^m(A)}$. $\Box$

Corollary: For any connected set ${E\subset \mathbb{R}^n}$

$\displaystyle \mathscr{H}^1(E)\geq \text{diam}\,E$

Proof: Since ${\mathscr{H}^1}$ is a regular measure on ${\mathbb{R}^n}$, we can find Borel set ${B}$ containing ${E}$ with the same ${\mathscr{H}^1}$ measure. Choose ${a\in B}$, define ${f:\mathbb{R}^n\rightarrow \mathbb{R}^1}$, ${f(x)=|x-a|}$. Then ${f}$ is Lipschitzian map and ${Lip\, f=1}$. By the previous theorem

$\displaystyle \mathscr{H}^1(B)\geq\int_{\mathbb{R}^1}N(f|B,y)d\mathscr{H}^1(y)\geq \mathscr{H}^1(f(E))=\mathscr{L}^1(f(E))\geq\text{diam}\,E$

$\Box$