Math4121 Lecture 24

1. $S=\mathbb{Q}\cap [0,1]$ is not Jordan measurable.

Since $c_e(S)=0$ and $\partial S=[0,1]$, $c_i(S)=1$.

So $c_e(\partial S)=1\neq 0$.

2. $SVC(3)$ is Jordan measurable.

Since $c_e(S)=0$ and $\partial S=0$, $c_i(S)=0$. The outer content of the cantor set is $0$.

Any set or subset of a set with $c_e(S)=0$ is Jordan measurable.

3. $SVC(4)$

At each step, we remove $2^n$ intervals of length $\frac{1}{4^n}$.

So $S=\bigcap_{n=1}^{\infty} C_i$ and $c_e(C_k)=c_e(C_{k-1})-\frac{2^{k-1}}{4^k}$. $c_e(C_0)=1$.

So

$$\begin{aligned} c_e(S)&\leq \lim_{k\to\infty} c_e(C_k)\\ &=1-\sum_{k=1}^{\infty} \frac{2^{k-1}}{4^k}\\ &=1-\frac{1}{4}\sum_{k=0}^{\infty} \left(\frac{2}{4}\right)^k\\ &=1-\frac{1}{4}\cdot \frac{1}{1-\frac{2}{4}}\\ &=1-\frac{1}{4}\cdot \frac{1}{\frac{1}{2}}\\ &=1-\frac{1}{2}\\ &=\frac{1}{2}. \end{aligned}$$

And we can also claim that $c_i(S)\geq \frac{1}{2}$. Suppose not, then $\exists \{I_j\}_{j=1}^{\infty}$ such that $S\subseteq \bigcup_{j=1}^{\infty} I_j$ and $\sum_{j=1}^{\infty} \ell(I_j)< \frac{1}{2}$.

Then $S$ would have gaps with lengths summing to greater than $\frac{1}{2}$. This contradicts with what we just proved.

So $c_e(SVC(4))=\frac{1}{2}$.

General formula for $c_e(SVC(n))=\frac{n-3}{n-2}$, and since $SVC(n)$ is nowhere dense, $c_i(SVC(n))=0$.

$$\begin{aligned} \sum_{i=1}^N c_i(S_i)&\leq c_i(\bigcup_{i=1}^N S_i)\\ &\leq c_e(\bigcup_{i=1}^N S_i)\\ &\leq \sum_{i=1}^N c_e(S_i)\\ \end{aligned}$$

Since $\sum_{i=1}^N c(S_i)=\sum_{i=1}^N c_e(S_i)=\sum_{i=1}^N c_i(S_i)$, we have

$$c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)$$