$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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proof

Content of Graphs

Let $f\in C([a, b], \mathbb{R})$ with $a

Concepts

Even though this says that $\mu(G)=\int_{\mathbb{R}^2}\chi d(x, y)=0$, we can't actually directly calculate the integral, so we take a different method for proving it.

Hypothesis

Coming soon

Results

Coming soon

Proof

We use the fact that the graph is uniformly continuous, and the if and only if condition for zero content. That is, we use the fact that $G$ is uniformly continuous to show that for all $\epsilon>0$, there exists a finite set of boxes $\{I_i\subseteq\mathbb{R}^n:1\leq i\leq m\}$ such that $$\begin{align*} G\subseteq\bigcup_{i=1}^n I_i \end{align*}$$ and $$\begin{align*} \sum_{i=1}^n \mu(I_i)< \epsilon. \end{align*}$$
Since $f$ is continuous and bounded on the closed domain $[a, b]$, then $f$ is uniformly continuous on $[a, b]$. That means for all $\epsilon'>0$, where $\epsilon'>2(b-a)\epsilon$, there exists some $\delta>0$ such that $|f(x)-f(y)|<\epsilon'$ for all $x, y$ that satisfies $|x-y|<\delta$.
We partition the interval $[a, b]$ into segments of length $\frac{n-a}{N}$ such that $\delta>\frac{b-a}{N}$ (ie. $N>\frac{b-a}{N}$).
Let $x_i=a+\frac{b-a}{N}$, and $I_j=x_i\times 2\epsilon'$. Being uniformly continuous, the graph is contained in $I_j$, and the volume of $I_j$ is $\frac{b-a}{N}2\epsilon'$ meaning $\mu(I_j)=2\frac{b-a}{N}\epsilon'$. Then $\sum_{j=1}^n\mu(I_j)=2(b-a)\epsilon'<\epsilon$. Hence by the condition for content zero, $G$ has content zero.

Content of Graphs

Let $f\in C([a, b], \mathbb{R})$ with $a

Concepts

Even though this says that $\mu(G)=\int_{\mathbb{R}^2}\chi d(x, y)=0$, we can't actually directly calculate the integral, so we take a different method for proving it.

Hypothesis

Coming soon

Results

Coming soon

Proof

We use the fact that the graph is uniformly continuous, and the if and only if condition for zero content. That is, we use the fact that $G$ is uniformly continuous to show that for all $\epsilon>0$, there exists a finite set of boxes $\{I_i\subseteq\mathbb{R}^n:1\leq i\leq m\}$ such that $$\begin{align*} G\subseteq\bigcup_{i=1}^n I_i \end{align*}$$ and $$\begin{align*} \sum_{i=1}^n \mu(I_i)< \epsilon. \end{align*}$$
Since $f$ is continuous and bounded on the closed domain $[a, b]$, then $f$ is uniformly continuous on $[a, b]$. That means for all $\epsilon'>0$, where $\epsilon'>2(b-a)\epsilon$, there exists some $\delta>0$ such that $|f(x)-f(y)|<\epsilon'$ for all $x, y$ that satisfies $|x-y|<\delta$.
We partition the interval $[a, b]$ into segments of length $\frac{n-a}{N}$ such that $\delta>\frac{b-a}{N}$ (ie. $N>\frac{b-a}{N}$).
Let $x_i=a+\frac{b-a}{N}$, and $I_j=x_i\times 2\epsilon'$. Being uniformly continuous, the graph is contained in $I_j$, and the volume of $I_j$ is $\frac{b-a}{N}2\epsilon'$ meaning $\mu(I_j)=2\frac{b-a}{N}\epsilon'$. Then $\sum_{j=1}^n\mu(I_j)=2(b-a)\epsilon'<\epsilon$. Hence by the condition for content zero, $G$ has content zero.
FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof

Content of Graphs

Let $f\in C([a, b], \mathbb{R})$ with $a

Concepts

Even though this says that $\mu(G)=\int_{\mathbb{R}^2}\chi d(x, y)=0$, we can't actually directly calculate the integral, so we take a different method for proving it.

Hypothesis

Coming soon

Results

Coming soon

Proof

We use the fact that the graph is uniformly continuous, and the if and only if condition for zero content. That is, we use the fact that $G$ is uniformly continuous to show that for all $\epsilon>0$, there exists a finite set of boxes $\{I_i\subseteq\mathbb{R}^n:1\leq i\leq m\}$ such that $$\begin{align*} G\subseteq\bigcup_{i=1}^n I_i \end{align*}$$ and $$\begin{align*} \sum_{i=1}^n \mu(I_i)< \epsilon. \end{align*}$$
Since $f$ is continuous and bounded on the closed domain $[a, b]$, then $f$ is uniformly continuous on $[a, b]$. That means for all $\epsilon'>0$, where $\epsilon'>2(b-a)\epsilon$, there exists some $\delta>0$ such that $|f(x)-f(y)|<\epsilon'$ for all $x, y$ that satisfies $|x-y|<\delta$.
We partition the interval $[a, b]$ into segments of length $\frac{n-a}{N}$ such that $\delta>\frac{b-a}{N}$ (ie. $N>\frac{b-a}{N}$).
Let $x_i=a+\frac{b-a}{N}$, and $I_j=x_i\times 2\epsilon'$. Being uniformly continuous, the graph is contained in $I_j$, and the volume of $I_j$ is $\frac{b-a}{N}2\epsilon'$ meaning $\mu(I_j)=2\frac{b-a}{N}\epsilon'$. Then $\sum_{j=1}^n\mu(I_j)=2(b-a)\epsilon'<\epsilon$. Hence by the condition for content zero, $G$ has content zero.

Content of Graphs

Let $f\in C([a, b], \mathbb{R})$ with $a

Concepts

Even though this says that $\mu(G)=\int_{\mathbb{R}^2}\chi d(x, y)=0$, we can't actually directly calculate the integral, so we take a different method for proving it.

Hypothesis

Coming soon

Results

Coming soon

Proof

We use the fact that the graph is uniformly continuous, and the if and only if condition for zero content. That is, we use the fact that $G$ is uniformly continuous to show that for all $\epsilon>0$, there exists a finite set of boxes $\{I_i\subseteq\mathbb{R}^n:1\leq i\leq m\}$ such that $$\begin{align*} G\subseteq\bigcup_{i=1}^n I_i \end{align*}$$ and $$\begin{align*} \sum_{i=1}^n \mu(I_i)< \epsilon. \end{align*}$$
Since $f$ is continuous and bounded on the closed domain $[a, b]$, then $f$ is uniformly continuous on $[a, b]$. That means for all $\epsilon'>0$, where $\epsilon'>2(b-a)\epsilon$, there exists some $\delta>0$ such that $|f(x)-f(y)|<\epsilon'$ for all $x, y$ that satisfies $|x-y|<\delta$.
We partition the interval $[a, b]$ into segments of length $\frac{n-a}{N}$ such that $\delta>\frac{b-a}{N}$ (ie. $N>\frac{b-a}{N}$).
Let $x_i=a+\frac{b-a}{N}$, and $I_j=x_i\times 2\epsilon'$. Being uniformly continuous, the graph is contained in $I_j$, and the volume of $I_j$ is $\frac{b-a}{N}2\epsilon'$ meaning $\mu(I_j)=2\frac{b-a}{N}\epsilon'$. Then $\sum_{j=1}^n\mu(I_j)=2(b-a)\epsilon'<\epsilon$. Hence by the condition for content zero, $G$ has content zero.
FullPage
result
concepts
hypothesis
implications
proof