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:

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.

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.

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.


\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.

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