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)


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



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