142 lines
3.4 KiB
Markdown
142 lines
3.4 KiB
Markdown
# Math4121 Lecture 34
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> Important:
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>
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> $\mathfrak{M}=\{S\subset \mathbb{R}: S \text{ satisfies the caratheodory condition}\}$, that is, for any $X$ of finite outer measure,
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>
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> $$m_e(X)=m_e(X\cap S)+m_e(X\cap S^c)$$
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>
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> In particular, the measure of sets can be infinite, not necessarily bounded. (We want to make the real line measurable.)
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## Lebesgue Integral
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### Simple Function
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A function $\phi$ is called a simple function if
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$$
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\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)
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$$
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where $a_i\in \mathbb{R}$ and $\chi_{S_i}=\begin{cases}1, & x\in S_i \\ 0, & x\notin S_i\end{cases}$
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where $\{S_i\}_{i=1}^{n}$ are pairwise disjoint each having finite measure.
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**constant function** is not simple ($\mathbb{R}$ is not finite measurable sets.)
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#### Theorem 6.6
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A function $f$ is measurable on $[a,b]$ if and only if there exists a sequence of simple functions $\{\phi_n\}$ such that $\lim_{n\to\infty} \phi_n(x)=f(x)$ almost everywhere on $[a,b]$.
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Proof:
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Partition $[-n,n]$ into $n2^{n+1}$ pieces.
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(These are just horizontal strips from $-n$ to $n$ with width $\frac{1}{2^n}$.)
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$$
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E_{n,k}=\{x\in[-n,n]:\frac{k}{2^n}\leq f(x)<\frac{k+1}{2^n}\}
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$$
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for $-n2^n<k<n2^n$
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$$
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E_{n,n2^n}=\{x\in[-n,n]:f(x)\geq n\}
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$$
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$$
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E_{n,-n2^n}=\{x\in[-n,n]:f(x)<\frac{-n2^n+1}{2^n}\}
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$$
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$$
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\phi_n(x)=\frac{k}{2^n}\chi_{E_{n,k}}(x)
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$$
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is a simple function.
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We need to justify that $\phi_n(x)\to f(x)$ for all $x\in\mathbb{R}$.
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Let $x\in\mathbb{R}$. And choose $n_0$ large such that $x\in [-n_0,n_0]$ and $f(x)\in [-n_0,n_0]$.
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Then, for $n\geq n_0$,
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$$
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|\phi_n(x)-f(x)|<\frac{1}{2^n}\to 0
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$$
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as $n\to\infty$.
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QED
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### Integration
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Given a measurable set $E$ and a simple function $\phi$, we define
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$$
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\int_E \phi dm=\sum_{i=1}^{n} a_i m(E\cap S_i)
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$$
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#### Properties 6.10
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Let $\phi$ and $\psi$ be simple functions, $c\in \mathbb{R}$, and $E=E_1\cup E_2$ where $E_1\cap E_2=\emptyset$ and $E_1,E_2\in \mathfrak{M}$. Then,
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1. $\int_E c\phi dm=c\int_E \phi dm$ (linearity)
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2. $\int_E (\phi+\psi)dm=\int_E \phi dm+\int_E \psi dm$ (additivity of simple functions)
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3. if $\phi(x)\leq \psi(x)$ for all $x\in E$, then $\int_E \phi dm\leq \int_E \psi dm$ (monotonicity)
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4. $\int_E \phi(x)dm=\int_{E_1} \phi(x)dm+\int_{E_2} \phi(x)dm$ (additivity over **disjoint** measurable sets)
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Proof:
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Let $\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)$ and $\psi(x)=\sum_{j=1}^{m} b_j \chi_{T_j}(x)$.
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2.
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$$
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\phi+\psi=\sum_{i=1}^{n} a_i \chi_{S_i}+\sum_{j=1}^{m} b_j \chi_{T_j}
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$$
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Without loss of generality, we may assume that $x\in E$, $\bigcup_{i=1}^{n} S_i=\bigcup_{j=1}^{m} T_j=E$.
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So
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$$
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\phi+\psi=\sum_{i,j=1}^{n,m}(a_i+b_j) \chi_{S_i\cup T_j}
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$$
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is a simple function.
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$$
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\begin{aligned}
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\int_E (\phi+\psi)dm&=\sum_{i,j=1}^{n,m}(a_i+b_j) m(E\cap S_i\cup T_j) \\
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&=\sum_{i=1}^{n} a_i \sum_{j=1}^{m} m(E\cap S_i\cup T_j)+\sum_{j=1}^{m} b_j \sum_{i=1}^{n} m(E\cap S_i\cup T_j) \\
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&=\sum_{i=1}^{n} a_i m(E\cap S_i)+\sum_{j=1}^{m} b_j m(E\cap T_j) \\
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&=\int_E \phi dm+\int_E \psi dm
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\end{aligned}
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$$
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3.
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$$
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\phi(x)=\sum_{i=1}^{n} a_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)
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$$
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$$
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\psi(x)=\sum_{i=1}^{n} b_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)
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$$
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If $x\in S_i\cap T_j$, then $\phi(x)=a_i$ and $\psi(x)=b_j$, therefore $a_i\leq b_j$.
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So,
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$$
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\begin{aligned}
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\int_E \phi dm&=\sum_{i=1}^{n} \sum_{j=1}^{m} a_i m(E\cap S_i\cap T_j) \\
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&\leq \sum_{i=1}^{n} \sum_{j=1}^{m} b_i m(E\cap S_i\cap T_j) \\
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&=\int_E \psi dm
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\end{aligned}
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$$
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QED
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Back on Wednesday.
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