Area Formula (2st Part)

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Hausdorff Dimension

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Lemma

$$\\$$ For $s < t < \infty$, if $H^{s}(A) < \infty$ then $H^{t}=0$ $$\\$$

Proof

$$\\$$ As $H^{s}(A) < \infty$, there exists some covering $\{F_{i} \}^{\infty}_{i=1}$ where $diam(F_{i}) < \delta$ for each $i$ such that: \[ \sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^{s} < \infty \] Let $c=\sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^{s} $ Then we have: \[ H^{t}_{\delta} (A) \leq \sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^t =\sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^{s} \left( \frac{diam \, F_{i}}{2} \right)^{t-s}\] \[\leq \sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^{s} (\delta)^{t-s} = (\delta)^{t-s} c \] where $t-s>0$, let $\delta \to 0$ \[H^{t}(A)=0 \]

Corollary

For $s < t < \infty$, if $H^{s}(A)>0$ then $H^{t}(A)=\infty$

Definition

The Hausdorff Dimension for a set $A \subseteq R^{n}$ is defined by: \[H_{dim}(A)= \inf \{ s: H^{s}(A)=0 \} \] From the Corollary, we also have: \[ H_{dim}(A)= \sup \{ s: H^{s}(A)=\infty \} \]

Definition

$$\\$$ Cantor $$\\$$ A $\textit{Cantor set}$ is a set of points on a line segment made by removing the middle third of some interval (say we have [0,1] as our interval, we remove $ (\frac{1}{3},\frac{2}{3})) $ then removing the third middle of both of intervals resulted in the previous step, and continuing inductively in an infinite number of steps. $$\\$$

Lemma

Let $F$ be a cantor set, then $dim(F)=\frac{\log(2)}{\log(3)}$

Proof

Without loss of generality, assume our interval is $[0,1]$ $$\\$$ If $0 < \frac{1}H^{s}(F) < \infty$, $s$ is directly is $dim(F)$. Let $s=\log_{3}2$ we will prove that \[ \frac{1}{2^{s+1}} \leq \frac{1}{\alpha(s)} H^{s}(F)\leq \frac{1}{2^{s}} \] $\textbf{Step one}$: At any step of the steps illustrated in (Fig 1.) we can see that the intervals covers the set $F$. At step $C_{n}$, we have $2^{n}$ intervals of length $\frac{1}{3^{n}}$ $$\\$$ From the definition of Hausdorff measure we have: \[H_{3^{-n}}^{s}(F) \leq \alpha(s) 2^{n-s} 3^{-sn} = \alpha(s) 2^{-s} \] Let $n \to \infty$ we get: \[H^{s}(F) \leq \alpha(s) 2^{-s} \] $\textbf{Step two}$: Sue to the compactness of the Cantor set, every open cover of it has finite subcover. $$\\$$ Assume the the finite set $\{U_{i}\}$ of open intervals covers our set such that for each $i$ we have $\frac{1}{3^{i+1}} \leq |U_{i}| \leq \frac{1}{3^{i}} $. At $C_{j}$ step, we have $2^{j}$ intervals of length $\frac{1}{3^{j}}$ and the minimum space between two intervals is also $\frac{1}{3^{j}}$. $$\\$$ Assume $j$ is large such that $\frac{1}{3^{j}} \leq |U_{i}|$ for all $i$. The maximum number of intervals of step $C_{j}$ that $U_{i}$ for some $i$ covers is $2^{j-i} = 2^{j} 3^{-si} \leq 2^{j}3^{s} |U_{i}|^{s}$ $$\\$$ We can easily deduce from the previous inequality that if $\{U_{i}\}$ covers the $2^{j}$ interval at step $C{j}$, then: \[ 2^{j} \leq 2^{j}3^{s} \sum |U_{i}|^{s} \] \[\frac{\alpha(s)}{2^{s+1}} \leq \sum \alpha(s)\left( \frac{|U_{i}|}{2}\right) ^{s} \] And that is true for any cover as the inequality does not depend on j. \[\frac{\alpha(s)}{2^{s+1}} \leq H^{s}(F) \]

Measures Differentiation

$$\\$$ $B(x;r) = \{ y : |x-y| \leq r \}$ is the closed ball centred at $x$, $B_{R}$ denotes for $B(0;R)$ $$\\$$

Definition

$$\\$$ For $\mu$,$v$ Radon measures on $R^{n}$, the upper and lower derivatives of $\mu$ with respect to $v$ at some point $x \in R^{n}$ are defined by: \[\overline{D}(\mu,v,x)=\limsup_{r \to 0} \frac{\mu(B(x;r))}{v(B(x;r))} \] \[\underline{D}(\mu,v,x)=\liminf_{r \to 0} \frac{B(x;r))}{v(B(x;r))} \] If both upper and lower derivates are equal, we say $\mu$ is differentiable with respect to $v$ and we define the derivative by: \[D(\mu,v,x)= \overline{D}(\mu,v,x)=\underline{D}(\mu,v,x) = \lim_{r \to 0} \frac{\mu(B(x;r))}{v(B(x;r))} \]

Essential Theorems

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Lemma

$$\\$$ For $\mu$,$v$ Radon measures, if $D(\mu,v,x)$ exists, it is $\mu$- measurable function

Proof

$$\\$$ We will prove that $f(x)=\mu(B(x;r))$ is upper continuous function and the same will apply for $v(B(x;r))$. $$\\$$ By $\textbf{Baire's theorem on semi-continuous functions}$, $1_{B(x;r)}$ is upper continuos, therefore we have: \[\limsup_{n \to \infty} 1_{B(x_{n};r)} \leq 1_{B(x;r)}\] We have for such large $n$: $1_{B(x_{n};r)} < 1_{B(x;2r)} $, so by reverse Fatou's Theorem: \[\limsup_{n \to \infty} f(x_{n}) =\limsup_{n \to \infty} \int 1_{B(x_{n};r)} \leq \int \limsup_{n \to \infty} 1_{B(x_{n};r)} \leq \int 1_{B(x;r)} = f(x) \] Also from \textbf{Baire's theorem on semi-continuous functions} we can easily conclude that semi-continuous functions are measurable with respect to Borel measures. $$\\$$ Limit of measurable functions is measurable, so $D(\mu,v,x) = \lim_{r \to 0} \frac{\mu(B(x;r))}{v(B(x;r))} $ is measurable.
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Lemma

$$\\$$ Let $\mu$ and $v$ be Radon measures. if $0 < t < \infty$ and $A \in R^{n}$
  1. If $D(\mu,v,x) \leq t$ for all $x \in A$, then $\mu(A) \leq t v(A)$
  2. If $D(\mu,v,x) \geq t$ for all $x \in A$, then $\mu(A) \geq t v(A)$
  3. $D(\mu,v,x)$ exists and is finite $v$-a.e.
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Proof

$\textbf{First Step}$: $D(\mu,v,x)=\lim_{r \to 0} \frac{\mu(B(x;r))}{v(B(x;r))} \leq t$ $$\\$$ As $v$ are Radon measures, there exists open set $U$ such that $A \subseteq U$ and $v(U)<\epsilon + v(A)$ $$\\$$ For $x \in A$, $0 < r < r(x,\epsilon)$ we have $B(x;r) \subseteq U$ and : \[\frac{\mu(B(x_{i};r))}{v(B(x_{i};r))} \leq (t+\epsilon) \] \[\mu(B(x_{i};r)) \leq (t+\epsilon) v(B(x_{i};r)) \] From $\textbf{Vitali Covering Theorem}$, There exists a set of pairwise disjoint balls $\{B_{i}\}$ such that $A \subseteq \bigcup B_{i} \subseteq U$ and satisfying the previous relation: \[\mu(B_{i}) \leq (t+\epsilon) v(B_{i}) \] \[\mu(A) \leq \sum_{i} \mu(B_{i}) \leq (t+\epsilon) \sum_{i} v(B_{i}) \leq (t+\epsilon) v(U) \leq (t+\epsilon) (\epsilon + v(A)) \] Taking intersection of sets satisfying the previous relation for $\epsilon \to 0$ \[\mu(A) \leq t v(A)\] The same for the second inequality. $$\\$$ $\textbf{Second Step}$: For $0 < r < \infty$, $0 < s < t < \infty$ \[A_{s,t} = \{x \in B_{R} : \underline{D}(\mu,v,x) \leq s < t \leq \overline{D}(\mu,v,x) \} \] is the set where $D(\mu,v,x)$ does not exist. $$\\$$ We have the following two inequalities: \[ \mu(A_{s,t}) \leq s \textit{ } v(A_{s,t}) \] \[ t v(A_{s,t}) \leq \mu(A_{s,t}) \] Combining these two inequalities, we get: \[ t \textit{ } v(A_{s,t}) \leq s \textit{ } v(A_{s,t}) \] But that can happen only if $v(A_{s,t}) = 0$, which means $D(\mu,v,x)$ exists $v$ a.e. $$\\$$ Define $A_{r} = \{ x \in B_{R} : D(\mu,v,x) \geq r \}$ \[ \mu(A_{r}) \geq r \textit{ } v(A_{r}) \] As $A_{r}$ is bounded, $\mu(A_{r})$ can't equal $\infty$, therefore $v(A_{r}) \to 0$ as $r \to \infty$ $$\\$$ So it is finite $v$ a.e.

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