3.5 KiB
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)