Area Formula (1st part)

Geometric Measure Theory has been a well-recognized field with plentiful natural connections to many other areas of mathematics. Over the past thirty years, this theory has contributed to major advances in geometry and analysis, including, for example, the original proof of the positive mass conjecture in Cosmology. In this paper, we will start with basic ideas in this field to end up with one of the main results, what is known as "Area Formula". \[ Let \: A \: \subseteq \mathbb{R}^{n}\ , 0 \leq s< \infty \ , 0 \leq \delta< \infty \] \[ H_{\delta}^{s}(A)= \inf \left\{ \sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^s : A \subseteq \bigcup_{i=1}^{\infty}F_{i} ,\, diam \, F_{i} \leq \delta \right\}\] \[\lim_{\delta \to 0}H_{\delta}^{s}(A)=\sup_{\delta}H_{\delta}^{s}(A) =H^{s}(A)\] Where \[ \alpha(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma(\frac{s}{2}+1)} \]

$H^{s}(A)$ is called Hausdorff Measure

$$\\$$ $\textbf{Lemma}$ $$\\$$ Hausdorff Measure is a Borel regular Measure $$\\$$ $\textbf{Proof}$ $$\\$$ $\textbf{First Step:}$ to prove the function we have defined is really a measure we need to prove the sub-additivity property of measures. It is obvious that $H^{s}(\emptyset)=0$. Let $A \subseteq \bigcup_{i=1}^{\infty}A_{i}$ and from then from (def 1.1), for each $i$ we have: \[H_{\delta}^{s}(A_{i})= \inf \left\{ \sum_{j=1}^{\infty} \alpha(s) \left( \frac{diam \, F_{i,j}}{2} \right)^s : A_{i} \subseteq \bigcup_{j=1}^{\infty}F_{i,j} ,\, diam \, F_{i,j} \leq \delta \right\},\] then we have for some $\{F_{i,j}\}_{j}$ such that $ A_{i} \subseteq \bigcup_{j=1}^{\infty}F_{i,j}$ \[ \sum_{j=0}^{\infty} \alpha(s) \left( \frac{diam \, F_{i,j}}{2} \right)^s \leq H_{\delta}^{s}(A_{i})+\frac{\epsilon}{2^{i}},\] Therefore, \[ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \alpha(s) \left( \frac{diam \, F_{i,j}}{2} \right)^s \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_{i})+\epsilon \] By rearrangement: \[\sum_{(i,j)\in \mathbb{N}^2} \alpha(s) \left( \frac{\mathrm{diam}(F_{i,j}) }{2} \right)^{s} \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_i) + \epsilon\] It follows that if we choose any bijection $\phi: \mathbb{N} \rightarrow \mathbb{N}^2 $, then \[\sum_{i=1}^\infty \alpha(s) \left( \frac{\mathrm{diam}(F_{\phi(i)})}{2} \right)^s \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_i) + \epsilon\] \[ \sum_{m=1}^{\infty} \; \sum_{i+j=m} \alpha(s) \left( \frac{diam \, F_{i,j}}{2} \right)^s \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_{i})+\epsilon \] Since \[A \subseteq \bigcup_{i=1}^{\infty} A_{i} \subseteq \bigcup_{j,i=1}^{\infty} F_{i,j}, \] then: \[ H_{\delta}^{s}(A) \leq \sum_{i=1}^\infty \alpha(s) \left( \frac{\mathrm{diam}(F_{\phi(i)})}{2} \right)^s \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_{i})+\epsilon. \] \[ H_{\delta}^{s}(A) \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_{i})+\epsilon \] We let $\epsilon \rightarrow 0$, to get \[ H_{\delta}^{s}(A) \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_{i}) \leq \sum_{i=1}^{\infty} H_{\delta}^{s}(A_i)\] By letting $\delta \rightarrow 0$ \[\ H^{s}(A) \leq \sum_{i=1}^{\infty} H^{s}(A_{i}) \] Thus $H^{s}$ is an measure. $$\\$$ $\textbf{Second Step:}$ We now prove it is a Borel measure. $$\\$$ We introduce Caratheodory's Criterion: $$\\$$ let $\mu$ be a measure on $R^n$, then $\mu$ is a Borel measure $\iff$ $\mu(A \cup B) = \mu(A)+\mu(B)$ whenever $dist(A,B)>0$ $$\\$$ So it is sufficient to prove $H^{s}(A \cup B) = H^{s}(A)+H^{s}(B)$ , whenever $dist(A,B)>0$ $$\\$$ Assume $dist(A,B)=\delta>0 $, and we have some covering of $A \cup B$ such that $A \cup B \subseteq \bigcup_{j=1}^{\infty} F_{i} $ such that $diam (F_{i})< \frac{\delta}{4}$ for all $i,j$ so $F_{i} \cap F_{j}=\phi$ for each $i \ne j$ and if $A \cap F_{i} \ne \phi \iff B \cap F_{i}=\phi $ $$\\$$ Let $\{A_{i}\} = \{F_{i}:A \cup F_{i} \ne \phi\}$ and $\{B_{i}\} = \{F_{i}:B \cup F_{i} \ne \phi\}$ \[\sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, F_{i}}{2} \right)^s = \sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, A_{i}}{2} \right)^s + \sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, B_{i}}{2} \right)^s \] \[ H_{\frac{\delta}{4}}^{s}(A \cup B) \geq H_{\frac{\delta}{4}}^{s}(A)+H_{\frac{\delta}{4}}^{s}(B) \] \[\delta \to 0 , \; H^{s}(A \cup B) \geq H^{s}(A)+H^{s}(B)\] So we finally have: \[ H^{s}(A \cup B) = H^{s}(A)+H^{s}(B)\] Thus $H^{s}$ is a Borel measure. $$\\$$ $\textbf{Third step:}$ The remaining part is to prove that it is Borel regular, so the next step is to prove the following statement: $$\\$$ For every $A \subseteq R^{n} $ there exists a Borel set (call it $B$) such that $A \subseteq B$ and $H^{s}(A)=H^{s}(B) $} $$\\$$ Assume we have $A \subseteq R^{n} $ such that $H^{s}(A)<\infty$ $$\\$$ It is straight forward that for any set $F \subseteq R^{n}$, $diam(F)=diam(\bar{F})$, where $\bar{F}$ is the closure of $F$. \[ H_{\frac{1}{k}}^{s}(A)= \inf \left\{ \sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, \bar{F}^{k}_{i}}{2} \right)^s : A \subseteq \bigcup_{i=1}^{\infty}F^{k}_{i} ,\, diam \, \bar{F}^{k}_{i} \leq \frac{1}{k} \right\}\] \[ H_{\frac{1}{k}}^{s}(\bigcup_{i=1}^{\infty} \bar{F}^{k}_{i}) \leq \sum_{i=1}^{\infty} \alpha(s) \left( \frac{diam \, \bar{F}^{k}_{i}}{2} \right)^s < H_{\frac{1}{k}}^{s}(A) + \frac{1}{k} \] Let $\bigcup_{i=1}^{\infty} \bar{F}^{k}_{i} = B_{k}$ \[ H_{\frac{1}{k}}^{s}(B_{k})< H_{\frac{1}{k}}^{s}(A) + \frac{1}{k} \] Taking $B= \bigcap_{k=1}^{\infty} B_{k}$ \[ H_{\frac{1}{k}}^{s}(B)\leq H_{\frac{1}{k}}^{s}(A) \] Taking $k \to \infty$ , $ H^{s}(B) \leq H^{s}(A)$ which implies directly that: \[ H^{s}(A)=H^{s}(B) \]

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$$\\$$ $\textbf{lemma}$ $$\\$$ $H^{0}(A)$ counts the number of elements in $A$ $$\\$$ $\textbf{proof}$ $$\\$$ If $A$ consists of $n$ elements $\{x_{i}\}$, we can take a finite disjoint covering such that each element is contained in a ball $B(x_{i}; \delta)$ $$\\$$ From the definition of Hausdorff Measure we have: $$\\$$ $H^{0}_{\delta} (A) \leq \sum_{i=1}^{n} \alpha(0) \left( \frac{diam \, F_{i}}{2} \right )^{0} = \sum_{i=1}^{n} 1 = n $ $$\\$$ We can see that the right side counts the number of covering sets we use, it's obvious that the elements for small $\delta$ covered by disjoint covering, so we can not have smaller than $n$. $$\\$$ If $A$ is an infinite set, we take some finite subsets of it $\{A_{i} \}$ such that $\#(A_{i})=i$ we obviously have for all $i \in \mathbb{N}$ \[ i=H^{0}(A_{i})< H^{0} (A) \] We get $H^{0}(A)=\infty$ $$\\$$

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$$\\$$ $\textbf{Definition}$ The function $f:E \subseteq R^{n} \to R^{m}$ is said to be $\textbf{Lipschitz}$ on $E$ if and only if it satisfies the following inequality: \[ |f(y)-f(x)| \leq C|y-x| \] For some constant $C \in R^{+}$ for all $x,y \in E$ $$\\$$ The infimum of the set of constants $C$ is the $\textbf{Lipschitz constant}$ and denoted by $\mathbf{lip(f;E)}$ $$\\$$ If $E=R^{n}$ we simply write $\mathbf{lip(f)}$ for the $\textbf{Lipschitz constant}$ $$\\$$

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$\textbf{lemma}$ $$\\$$ If $f:R^{n} \to R^{m}$ is a Lipschitz function on $R^{n}$ and $A \subseteq R^{n}$ then we have: \[ H^{s}(f(A)) \leq (lip(f))^{s} H^{s}(A) \] $\textbf{Proof}$ $$\\$$ For each cover $\{F_{i}\}_{i=1}^{\infty}$ for $A$, $f(A)$ is covered by $\{f(F_{i})\}_{i=1}^{\infty}$ and obviously $diam(f(F_{i})) \leq (lip(f)) diam(F_{i})$ For each $i$ , we get: \[H^{s}_{lip(f) \delta} (A) \leq \sum_{i=0}^{\infty} \alpha(s) \left( \frac{diam \, f(F_{i})}{2} \right)^s \leq \sum_{i=0}^{\infty} \alpha(s) \left( \frac{(lip(f))diam \, F_{i}}{2} \right)^s \] We get: \[ H_{lip(f) \delta}^{s}(f(A)) \leq (lip(f))^{s} H_{\delta}^{s}(A) \] Finally, take $\delta \to 0$ we get the desired inequality: \[ H^{s}(f(A)) \leq (lip(f))^{s} H^{s}(A) \]

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$\textbf{lemma}$ $$\\$$ $H^{1}=\mathfrak{L}^{1}$ on $R^{1}$ $$\\$$ $\textbf{Proof}$ $$\\$$ Assume $A \subset R$ \[\mathfrak{L}^{1}(A) = \inf \left\lbrace \sum_{i=1}^{\infty} diam (F_{i}) : A \subseteq \bigcup_{i=1}^{\infty} F_{i} \right\rbrace \leq \] \[ \inf \left\lbrace \sum_{i=1}^{\infty} diam(F_{i}) : A \subseteq \bigcup_{i=1}^{\infty} F_{i} \; diam(F_{i})<\delta \right\rbrace =H^{1}_{\delta}(A)\] So we have $\mathfrak{L}^{1}(A) \leq H^{1}(A) $ $$\\$$ For some cover $\{F_{i}\}_{i=1}^{\infty}$if we have some sets of the cover where $diam(F_{i})>\delta$, we can cut each set of these into finite number of sets such that $diam(F_{i})=\sum_{j} diam(F_{j,i}) , \; diam(F_{j,i})<\delta$ $$\\$$ So we can see that : \[\mathfrak{L}^{1}(A) = \inf \left\lbrace \sum_{i=1}^{\infty} diam(F_{i}) : A \subseteq \bigcup_{i=1}^{\infty} F_{i} \; \right\rbrace \] \[= \inf \left\lbrace \sum_{j,i=1}^{\infty} diam(F_{j,i}) : A \subseteq \bigcup_{j,i=1}^{\infty} F_{j,i} \; diam(F_{j,i})<\delta \right\rbrace = H^{1}_{\delta}(A) \]

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$\textbf{lemma}$ $$\\$$ $\mathfrak{L}^{n}=H^{n} $ on $R^{n}$ $$\\$$

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$\textbf{lemma}$ $$\\$$ If $\Gamma$ a curve in $R^n$ then $H^{1}(\Gamma)=Length(\Gamma)$ $$\\$$ $\textbf{Proof}$ $$\\$$ Let $\gamma:[0,a] \to R^{n}$ be a parametrization of the curve $\Gamma$. $$\\$$ $Length(\Gamma)= sup \{ \sum_{i=1}^{n} |\gamma(a_{i+1})-\gamma(a_{i})| :0=a_{1} < a_{2} < \cdots < a=a , n \in \mathcal{N} \} $ $$\\$$ Let $P:R^{n} \to R^{n}$ be the projection of $R^{n}$ onto the line connecting the two points $\gamma(0),\gamma(a)$ \\ Since $|P(x)-P(y)| \leq |x-y|$, it is Lipschitz. $$\\$$ So we have $H^{1} (P(A)) \leq H^{1} (A)$ for any set $A \subseteq R^{n}$ $$\\$$ Then $H^{1} (P(\Gamma)) \leq H^{1} (\Gamma)$. $$\\$$ And it also obvious that $P(\Gamma)$ contains the segment [$\gamma(0),\gamma(a)$] due to the continuity of $\gamma$: $$\\$$ $|\gamma(a)-\gamma(0)| \leq H^{1}([\gamma(0),\gamma(a)]) \leq H^{1} (P(\Gamma)) \leq H^{1} (\Gamma)$ $$\\$$ Also $\gamma([a_{i},a_{i+1}])$ is compact due to the continuity of $\gamma$, so it is $H^{1}$ measurable set. $$\\$$ Let $\Gamma_{i} = \gamma([a_{i},a_{i+1}])$, so $\Gamma = \bigcup_{i} \Gamma_{i}$ and $H^{1}(\Gamma_{i} \bigcap \Gamma_{i+1})=H^{1}(\{a_{i+1}\})=0$ $$\\$$ \[\sum^{n}_{i=1} |\gamma(a_{i+1})-\gamma(a_{i})| \leq \sum^{n}_{i=1} H^{1}(\Gamma_{i}) = H^{1}(\Gamma)\] Which means: \[Length(\Gamma) \leq H^{1}(\Gamma) \] We have now to prove the other side of the inequality. $$\\$$ There exists a continuous injective map $\gamma^{*}:[0,Length(\Gamma)] \to R^{n}$ which its existence is proven easily in Differential Geometry, this map $\gamma(l)$ determines where on the curve the length to that point equals $l$ $$\\$$ From the definition of $Length$ we can see that $|\gamma^{*}(l_{1})-\gamma^{*}(l_{2})| \leq |l_{1}-l_{2}|$ (it is lipschitz with $lip(\gamma^{*}) \leq 1$), so we get : $$\\$$ \[H^{1}(\Gamma) = H^{1}(\gamma^{*}([0,Length(\Gamma)])) \leq H^{1}([0,Length(\Gamma)]) = Length(\Gamma) \] Finally we have: \[Length(\Gamma) = H^{1}(\Gamma) \] $$\\$$

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