97 lines
2.5 KiB
Markdown
97 lines
2.5 KiB
Markdown
# Math4121 Lecture 30
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## Lebesgue Measure
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$\mathfrak{M}=\{S\subseteq\mathbb{R}:S\text{ is Lebesgue measurable}\}$ is a $\sigma$-algebra on $\mathbb{R}$ (closed under complementation and countable unions).
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### Consequence of Lebesgue Measure
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Every open set and closed set is Lebesgue measurable.
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#### Inner and Outer Regularity of Lebesgue Measure
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Outer regularity:
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$$
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m_e(S)=\inf_{U\text{ open},S\subseteq U}m(U)
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$$
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Inner regularity:
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$$
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m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
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$$
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<details>
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<summary>Proof</summary>
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Inner regularity:
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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 $m(U)<m(I\setminus S)+\epsilon$.
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Take $K=I\setminus U$. Then $K\subseteq S$ and $K$ is closed and
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$$
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m(K)=m(I)-m(U)>m(I)-m(I\setminus S)-\epsilon
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$$
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So $m_i(S)<m(K)+\epsilon$. Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$.
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</details>
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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.
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#### Symmetric difference
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The symmetric difference of two sets $S$ and $T$ is defined as
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$$
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S\Delta T=(S\setminus T)\cup(T\setminus S)
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$$
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_The XOR operation on two sets._
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#### Theorem
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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
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$$
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m(S\Delta U)<\epsilon
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$$
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where $U=\bigcup_{j =1}^n I_j$.
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<details>
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<summary>Proof</summary>
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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.
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Check:
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$$
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m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
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$$
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</details>
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Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and
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$$
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m(T)=\sum_{j=1}^\infty m(T_j)
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$$
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#### Corollary (Better osgood's theorem on Lebesgue measure)
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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
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$$
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m(S)=\lim_{j\to\infty}m(S_j)
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$$
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Proof:
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Let $T_1=S_1$ and $T_i=S_i\setminus S_{i-1}$ for $i\geq 2$. Still have $S=\bigcup_{j=1}^\infty T_j$.
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Where $T_i$ are disjoint measurable sets. So $m(S)=\sum_{j=1}^\infty m(T_j)$.
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So $\lim_{j\to\infty}m(S_j)=\sum_{j=1}^\infty m(T_j)=m(S)$.
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