115 lines
3.1 KiB
Markdown
115 lines
3.1 KiB
Markdown
# Math4121 Lecture 35
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## Continue on Lebesgue Integration
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### Lebesgue Integration
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#### Definition of Lebesgue Integral
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For simple functions $\phi = \sum_{i=1}^{n} a_i \chi_{S_i}$, given a measure $E$, the Lebesgue integral is defined as:
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$$
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\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)
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$$
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Given a non-negative measurable function $f$ and a measurable set $E$.
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Define $\int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}$
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(**We do allows $\int_E f \, dm = \infty$**)
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For general measurable function $f$, we can define $f^-(x)=\max\{0,-f(x)\}$, $f^+(x)=\max\{0,f(x)\}$. (The positive part of the function and the negative part of the function, both non-negative)
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Then $f=f^+-f^-$.
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We say $f$ is integrable if $\int_E f^+ \, dm < \infty$ and $\int_E f^- \, dm < \infty$. (both finite) If at least one is finite, define
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$$
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\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm
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$$
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We allow for $A-\infty = -\infty$ and $A+\infty = \infty$ for any $A\in \mathbb{R}$. But not $\infty-\infty$.
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#### Immediate Properties of Lebesgue Integral
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If $f$ is measurable and $m(E)=0$, then $\int_E f \, dm = 0$.
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If $E=E_1\cup E_2$ and $E_1\cap E_2=\emptyset$, then $\int_E f \, dm = \int_{E_1} f \, dm + \int_{E_2} f \, dm$.
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#### Corollary
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If $f\leq g$ almost everywhere, ($f\leq g$ except for a set of measure 0), then $\int_E f \, dm \leq \int_E g \, dm$.
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Proof:
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Let $F=\{x\in E: f(x)>g(x)\}$. Then $m(F)=0$.
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$$
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\begin{aligned}
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\int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
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&\leq \int_{E\setminus F} g \, dm+\int_E g \, dm\\
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&= \int_E g \, dm
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\end{aligned}
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$$
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QED
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#### Proposition 6.13
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If $f$ is non-negative and $\int_E f \, dm =0$, then $f=0$ almost everywhere on $E$, $f(x)=0$ $\forall x\in E\setminus F$, where $m(F)=0$.
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Proof:
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Let $E_n=\{x\in E: f(x)\geq \frac{1}{n}\}$. Then $\frac{1}{n}\chi_{E_n}(x)\leq f(x)$ for all $x\in E$.
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By definition $\frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E f \, dm =0$.
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Therefore, $m(E_n)=0$ for all $n$.
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Now $U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n$, and $E_n\subseteq E_{n+1}$ for all $n$.
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Therefore, $m(U)=m(\bigcup_{n=1}^{\infty} E_n)=\lim_{n\to\infty} m(E_n)=0$.
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QED
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### Convergence Theorems
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When does $\lim_{n\to\infty} \int_E f_n \, dm = \int_E \lim_{n\to\infty} f_n \, dm$?
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#### Theorem 6.14 Monotone Convergence Theorem
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Let $\{f_n\}$ be a monotone increasing sequence of measurable functions on $E$ and $f_n\to f$ almost everywhere on $E$. ($f_n(x)\leq f_{n+1}(x)$ for all $x\in E$ and $n$)
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If there exists $A>0$ such that $\left|\int_E f_n \, dm\right|\leq A$ for all $n\in \mathbb{N}$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists for almost every $x\in E$ and it is integrable on $E$ and
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$$
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\int_E f \, dm = \lim_{n\to\infty} \int_E f_n \, dm
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$$
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Proof:
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First to show the limit exists almost everywhere. It suffices to show
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$$
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\mathcal{U}=\{x\in E: f_n(x) \text{ is unbounded}\}
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$$
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has measure 0.
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Let $\epsilon>0$ and write
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$$
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U=\bigcup_{n=1}^{\infty} E_n
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$$
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where $E_n=\{x\in E: |f_n(x)|\geq \epsilon\}$.
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Then $U\subseteq \mathcal{U}$ and $m(U)<\epsilon$.
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CONTINUE NEXT TIME.
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QED
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