NoteNextra-origin/content/Math4121/Math4121_L28.md

Math4121 Lecture 28

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Lebesgue Measure

Outer Measure

m_e(S) = \inf_{S \subseteq \bigcup_{j=1}^{\infty} I_j} \sum_{j=1}^{\infty} \ell(I_j)

If S\subseteq I is measurable, then m_i(S)=m_e(I)-m_e(I\setminus S)

Lebesgue criterion for measurability

S\subseteq I is measurable if and only if m_e(I)=m_e(S)+m_e(I\setminus S)

Caratheodory's criteria

Lebesgue criterion holds if and only if for any X of finite outer measure,

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

Local additivity

\{I_j\}_{j=1}^{\infty} is a collection of disjoint intervals, then

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

Proved on Friday

Proof

\implies If Lebesgue criterion holds for S, then for any X of finite outer measure,

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

First, we extend Lebesgue criterion to intervals I that may not contain S. Then we can find J,K intervals neighboring I such that S\subseteq \tilde{I}=J\cup I\cup K.

By Lebesgue criterion,

\begin{aligned}
m_e(\tilde{I})&=m_e(\tilde{I}\cap S)+m_e(\tilde{I}\setminus S)\\
&=m_e(S)+m_e(\tilde{I}\setminus S)\\
&=m_e(S^c\cap \tilde{I})+m_e(S\cap \tilde{I})\\
&=\sum_{L\in \{J,I,K\}}m_e(L\cap S^c)+m_e(L\cap S)\\
&\geq \sum_{L\in \{J,I,K\}}m_e(L)\\
&=m_e(\tilde{I})
\end{aligned}

Therefore, m_e(I)=m_e(S^c\cap I)+m_e(S\cap I).

Now, let X has finite outer measure, let \epsilon>0, we can find \{I_j\}_{j=1}^{\infty} covering X and

\sum_{j=1}^{\infty} \ell(I_j)<m_e(X)+\epsilon

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

Revisit Borel's criterion

1. m(I)=\ell(I)
2. If \{S_j\}_{j=1}^{\infty} is a sequence of disjoint measurable sets, then m\left(\bigcup_{j=1}^{\infty} S_j\right)=\sum_{j=1}^{\infty} m(S_j)
3. If R\subseteq S, then m(S\setminus R)=m(S)-m(R)

Theorem 5.8 (Countable additivity for Lebesgue measure)

If \{S_j\}_{j=1}^{\infty} is a sequence of disjoint measurable sets, whose union S=\bigcup_{j=1}^{\infty} S_j, has finite outer measure, then

m_e(S)=\sum_{j=1}^{\infty} m_e(S_j)

Proof

First we prove m_e(\bigcup_{j=1}^{\infty} S_j)=\sum_{j=1}^{\infty} m(S_j) by induction.

n=1 is trivial.

Let n>1 and suppose the statement holds for n-1. Take X=\bigcup_{j=1}^{n-1} S_j, then S_n\cap X=S_n, X\setminus S_n=\bigcup_{j=1}^{n-1} (S_j).

By Caratheodory's criteria,

\begin{aligned}
m_e(X)&=m_e(S_n)+m_e(\bigcup_{j=1}^{n-1} S_j)\\
m_e(\bigcup_{j=1}^{n} S_j)&=m(S_n)+\sum_{j=1}^{n-1} m(S_j)\\
&=\sum_{j=1}^{n} m(S_j)
\end{aligned}

Take the limit as n\to\infty, and justify this.

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

Since m_e(S) is finite and m(S_j) is monotone, the limit exists.

Therefore, \sum_{j=1}^{\infty} m(S_j)\leq m_e(S)\leq \sum_{j=1}^{\infty} m(S_j)

So S is measurable.

Proposition 5.9 (Preview)

Any finite union (and intersection) of measurable sets is measurable.

Proof

Let S_1, S_2 be measurable sets.

We prove by verifying the Caratheodory's criteria for S_1\cup S_2.