Math4121 Lecture 27

Lebesgue Measure

Outer Measure

m_e(S)=\inf\left\{\sum_{n=1}^\infty \ell(I_n): S\subset \bigcup_{n=1}^\infty I_n\right\}

where $I_j$ is an open interval

Properties:

$m_e(I)=\ell(I)$
Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
does not respect complementation (Build in to Borel measure)

Why does Jordan content respect complementation?

$(\text{Finite union of intervals })^C=\text{another finite union of intervals}$

We know this failed for countable unions.

Example:

\bigcup_{n=1}^\infty \left(q_n-\frac{\epsilon}{2^n},q_n+\frac{\epsilon}{2^n}\right)

Where $q_n$ is dense.

Inner Measure

Say $S\subset I$

m_i(S)=m(I)-m_e(I\setminus S)

where $m(I)=\ell(I)$

Say $S$ is (Lebesgue) measurable if $m_i(S)=m_e(S)$ , call this value $m(S)=m_e(S)=m_i(S)$ the (Lebesgue) measure of $S$ .

Corollary of measurability of subsets

If $S$ is measurable, and $S\subset T$ , then

m(S)=m_e(S)=m(I)-m_e(I\setminus S)

m(I\setminus S)=m(I)-m(S)

$I\setminus S$ is Lebesgue measurable and $m(I)=m(S)+m(I\setminus S)$

Proposition 5.8 (Countable additivity over measurable sets)

If $S_n$ are measurable, then

m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)

Proof

Let $\epsilon>0$ and for each $j$ , let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.

\sum_{i=1}^\infty \ell(I_{i,j})<m(S_j)+\frac{\epsilon}{2^j}

Then $\bigcup_{j=1}^\infty \bigcup_{i=1}^\infty I_{i,j}$ is a countable cover of $\bigcup_{j=1}^\infty S_j$ and

m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq \sum_{j=1}^\infty \sum_{i=1}^\infty \ell(I_{i,j})<\sum_{j=1}^\infty \left(m_e(S_j)+\frac{\epsilon}{2^j}\right)=\sum_{j=1}^\infty m_e(S_j)+\epsilon

Since $\epsilon$ is arbitrary, we have

m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq\sum_{j=1}^\infty m_e(S_j)=\sum_{j=1}^\infty m(S_j)

Corollary: inner measure is always less than or equal to outer measure

m_i(S)\leq m_e(S)

Proof

m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)

Caratheodory’s Criterion

Lemma 5.7 (Local additivity)

If $\{I_j\}_{j=1}^\infty$ are pairwise disjoint open intervals, then

m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)

Proof

For each $j$ , let $\{J_i\}_{i=1}^\infty$ be a cover of $S\cap \left(\bigcup_{j=1}^\infty I_j\right)$ such that $\sum_{i=1}^\infty \ell(J_i)<c_e(S\cap \left(\bigcup_{j=1}^\infty I_j\right))+\epsilon$ . Since $\{I_j\}_{j=1}^\infty$ are pairwise disjoint, so is $\{J_i\cap I_j\}_{j=1}^\infty$ for each $i$ .

\sum_{j=1}^\infty m_e(J_i\cap I_j)=m_e(J_i)

\begin{aligned} m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)&\leq \sum_{j=1}^\infty m_e(S\cap I_j)\\ &\leq \sum_{j=1}^\infty m_e(\bigcup_{i=1}^\infty J_i\cap I_j)\\ &= \sum_{j=1}^\infty m_e(J_i)+\epsilon \end{aligned}

Since $\epsilon$ is arbitrary, we have

m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)

Theorem 5.6 (Caratheodory’s Criterion)

A set $S$ is measurable if and only if for every set $X\in \mathbb{R}$ of finite outer measure,

m_e(X)=m_e(X\cap S)+m_e(X\setminus S)

Lebesgue: $X=I$ and $S\subset I$ we can cut any set by a measurable set to get a measurable set. (no matter how big the set is)