# Math4121 Lecture 24

## Chapter 5: Measure Theory

### Jordan Measurable

#### Proposition 5.1

A bounded set $S\subseteq \mathbb{R}^n$ is Jordan measurable if

$$
c_e(S)=c_i(S)+c_e(\partial S)
$$

where $\partial S$ is the boundary of $S$ and $c_e(\partial S)=0$.

Example:

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, $c_i(SVC(n))=0$.

### Additivity of Content

Recall that outer content is sub-additive. Let $S,T\subseteq \mathbb{R}^n$ be disjoint.

$$
c_e(S\cup T)\leq c_e(S)+c_e(T)
$$

The inner content is super-additive. Let $S,T\subseteq \mathbb{R}^n$ be disjoint.

$$
c_i(S\cup T)\geq c_i(S)+c_i(T)
$$

#### Proposition 5.2

Finite additivity of Jordan content:

Let $S_1,\ldots,S_N\subseteq \mathbb{R}^n$ are pairwise disjoint Jordan measurable sets, then

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

Proof:

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

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

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

QED

##### Failure for countable additivity for Jordan content

Notice that each singleton $\{q\}$ is Jordan measurable and $c(\{q\})=0$. But take $a\in \mathbb{Q}\cap [0,1]$, $Q\cap [0,1]=\bigcup_{q\in Q\cap [0,1]} \{q\}$, but $\mathbb{Q}\cap [0,1]$ is not Jordan  measurable.

Issue is a countable union of Jordan measurable sets is not necessarily Jordan measurable.
