Math4121 Lecture 30

Lebesgue Measure

$\mathfrak{M}=\{S\subseteq\mathbb{R}:S\text{ is Lebesgue measurable}\}$ is a $\sigma$ -algebra on $\mathbb{R}$ (closed under complementation and countable unions).

Consequence of Lebesgue Measure

Every open set and closed set is Lebesgue measurable.

Inner and Outer Regularity of Lebesgue Measure

Outer regularity:

m_e(S)=\inf_{U\text{ open},S\subseteq U}m(U)

Inner regularity:

m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)

Proof

Inner regularity:

Since $m_i(S)=m(I)-m_e(I\setminus S)$ , $S\subseteq I$ for some closed interval $I$ . Let $\epsilon>0$ and $U$ be an open set such that $I\setminus S\subseteq U$ and $m(U)<m(I\setminus S)+\epsilon$ .

Take $K=I\setminus U$ . Then $K\subseteq S$ and $K$ is closed and

m(K)=m(I)-m(U)>m(I)-m(I\setminus S)-\epsilon

So $m_i(S)<m(K)+\epsilon$ . Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$ .

We can approximate $m(S)$ from outside by open sets. If we are just concerned with “approximating” $m(S)$ , we can use finite union of intervals.

Symmetric difference

The symmetric difference of two sets $S$ and $T$ is defined as

S\Delta T=(S\setminus T)\cup(T\setminus S)

The XOR operation on two sets.

Theorem

If $S\subset I$ is measurable, then for every $\epsilon>0$ , $\exists I_1,I_2,\cdots,I_n\subset I$ open intervals such that

m(S\Delta U)<\epsilon

where $U=\bigcup_{j =1}^n I_j$ .

Proof

Let $\epsilon>0$ and $m(V)<m(S)+\frac{\epsilon}{2}$ . Let $K\subseteq S$ be closed set such that $m(S)-\frac{\epsilon}{2}<m(K)$ . $V$ is an open cover of closed and bounded set $K$ . By Heine-Borel theorem, $K$ has a finite subcover. Let $I_1,I_2,\cdots,I_n$ be the open intervals in the subcover.

Check:

m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon

Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and

m(T)=\sum_{j=1}^\infty m(T_j)

Corollary (Better osgood’s theorem on Lebesgue measure)

If $S_1\subseteq S_2\subseteq S_3\subseteq\cdots$ are measruable (no need to be closed and bounded) and $S=\bigcup_{j=1}^\infty S_j$ , then

m(S)=\lim_{j\to\infty}m(S_j)

Proof:

Let $T_1=S_1$ and $T_i=S_i\setminus S_{i-1}$ for $i\geq 2$ . Still have $S=\bigcup_{j=1}^\infty T_j$ .

Where $T_i$ are disjoint measurable sets. So $m(S)=\sum_{j=1}^\infty m(T_j)$ .

So $\lim_{j\to\infty}m(S_j)=\sum_{j=1}^\infty m(T_j)=m(S)$ .