150 lines
3.5 KiB
Markdown
150 lines
3.5 KiB
Markdown
# Math4121 Lecture 27
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## Lebesgue Measure
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### Outer Measure
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$$
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m_e(S)=\inf\left\{\sum_{n=1}^\infty \ell(I_n): S\subset \bigcup_{n=1}^\infty I_n\right\}
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$$
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where $I_j$ is an open interval
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**Properties:**
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1. $m_e(I)=\ell(I)$
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2. Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
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3. does not respect complementation (Build in to Borel measure)
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Why does Jordan content respect complementation?
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$(\text{Finite union of intervals })^C=\text{another finite union of intervals}$
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We know this failed for countable unions.
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Example:
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$$
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\bigcup_{n=1}^\infty \left(q_n-\frac{\epsilon}{2^n},q_n+\frac{\epsilon}{2^n}\right)
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$$
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Where $q_n$ is dense.
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### Inner Measure
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Say $S\subset I$
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$$
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m_i(S)=m(I)-m_e(I\setminus S)
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$$
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where $m(I)=\ell(I)$
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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 $S$.
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#### Corollary of measurability of subsets
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If $S$ is measurable, and $S\subset T$, then
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$$
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m(S)=m_e(S)=m(I)-m_e(I\setminus S)
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$$
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$$
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m(I\setminus S)=m(I)-m(S)
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$$
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$I\setminus S$ is Lebesgue measurable and $m(I)=m(S)+m(I\setminus S)$
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#### Proposition 5.8 (Countable additivity over measurable sets)
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If $S_n$ are measurable, then
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$$
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m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)
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$$
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<details>
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<summary>Proof</summary>
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Let $\epsilon>0$ and for each $j$, let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.
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$$
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\sum_{i=1}^\infty \ell(I_{i,j})<m(S_j)+\frac{\epsilon}{2^j}
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$$
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Then $\bigcup_{j=1}^\infty \bigcup_{i=1}^\infty I_{i,j}$ is a countable cover of $\bigcup_{j=1}^\infty S_j$ and
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$$
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m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq \sum_{j=1}^\infty \sum_{i=1}^\infty \ell(I_{i,j})<\sum_{j=1}^\infty \left(m_e(S_j)+\frac{\epsilon}{2^j}\right)=\sum_{j=1}^\infty m_e(S_j)+\epsilon
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$$
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Since $\epsilon$ is arbitrary, we have
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$$
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m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq\sum_{j=1}^\infty m_e(S_j)=\sum_{j=1}^\infty m(S_j)
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$$
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</details>
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#### Corollary: inner measure is always less than or equal to outer measure
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$$
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m_i(S)\leq m_e(S)
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$$
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<details>
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<summary>Proof</summary>
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$$
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m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)
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$$
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</details>
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### Caratheodory's Criterion
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#### Lemma 5.7 (Local additivity)
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If $\{I_j\}_{j=1}^\infty$ are pairwise disjoint open intervals, then
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$$
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m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)
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$$
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<details>
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<summary>Proof</summary>
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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 $i$.
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$$
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\sum_{j=1}^\infty m_e(J_i\cap I_j)=m_e(J_i)
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$$
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$$
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\begin{aligned}
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m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)&\leq \sum_{j=1}^\infty m_e(S\cap I_j)\\
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&\leq \sum_{j=1}^\infty m_e(\bigcup_{i=1}^\infty J_i\cap I_j)\\
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&= \sum_{j=1}^\infty m_e(J_i)+\epsilon
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\end{aligned}
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$$
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Since $\epsilon$ is arbitrary, we have
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$$
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m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)
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$$
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</details>
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#### Theorem 5.6 (Caratheodory's Criterion)
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A set $S$ is measurable if and only if for every set $X\in \mathbb{R}$ of finite outer measure,
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$$
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m_e(X)=m_e(X\cap S)+m_e(X\setminus S)
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$$
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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)
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