134 lines
4.2 KiB
Markdown
134 lines
4.2 KiB
Markdown
# Math 4121 Lecture 36
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## Random visit for Lebesgue Integration
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### Convergence Theorem
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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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To show the limit exists almost everywhere, let $\epsilon>0$, set $E_n=\{x\in E: f_n(x)>\frac{A}{\epsilon}\}$. We will show $U=\bigcup_{n=1}^{\infty} E_n$ has measure $<\epsilon$. $f_n(x)\geq \frac{A}{\epsilon}\chi_{E_n}(x)$, so
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$$
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\frac{A}{\epsilon}m(E_n)=\int_E \frac{A}{\epsilon}\chi_{E_n}dm\leq \int_E f_n dm\leq A
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$$
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In particular, $m(E_n)<\epsilon$. Since $E_n\subset E_{n+1}$ for all $n$, $m(U)=\lim_{n\to\infty} m(E_n)<\epsilon$.
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$\lim_{n\to\infty} \int_E f_n dm\leq \int_E f dm\leq \lim_{n\to\infty}$. To show the reverse inequality, let $\phi$ be a simple function $\leq f$ of the form $\phi=\sum_{i=1}^{k} a_i\chi_{S_i}$ where $S_i$ is sidjoint and $\bigcup_{i=1}^{k} S_i\subseteq E$.
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Let $\alpha\in (0,1)$ and set $A_n=\{x\in S:f_n(x)-\alpha\phi(x)>0\}$. This ensures that $f_n(x)\geq \alpha\phi(x)$ for all $x\in A_n$.
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Notice that $A_n\subset A_{n+1}$ for all $n$ and $U=\bigcup_{n=1}^{\infty} A_n$. $lim_{n\to\infty} m(A_n\cap S_i)=m(S_i)$ for all $i$.
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$\int_{A_n} \phi dm=\sum_{i=1}^{k} a_i m(S_i\cap A_n)$.
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As $n\to\infty$, $m(A_n\cap S_i)\to m(S_i)$ and $\sum_{i=1}^{k} a_i m(S_i\cap A_n)\to \int_E \phi dm$.
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Let $\epsilon>0$. There exists $n_0$ large such that $\int_{A_n} \phi dm>\int_E \phi dm-\epsilon$ for all $n\geq n_0$.
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Then for such $n\geq n_0$,
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$$
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\int_E f_n dm\geq \int_{A_n} f_n dm\geq \int_{A_n} \alpha \phi dm>\alpha(\int_E \phi dm-\epsilon)
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$$
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So, $\lim_{n\to\infty} \int_E f_n dm\geq \alpha(\int_E \phi dm-\epsilon)$.
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Since $\alpha,\epsilon$ are arbitrary, set $\alpha\to 1$ and $\epsilon\to 0$ to get $\lim_{n\to\infty} \int_E f_n dm\geq \int_E \phi dm$.
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For any simple function $\phi\leq f$, taking sup over all simple functions $\phi\leq f$ gives $\lim_{n\to\infty} \int_E f_n dm\geq \int_E f dm$.
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QED
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#### Lemma Absolute Integrability
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$f$ is integrable on $E$ if and only if $|f|$ is integrable on $E$ and $\left|\int_E f \, dm\right|\leq \int_E |f| \, dm$.
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Proof:
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If $f^+$ and $f^-$ are integrable and $|f|=f^+-f^-$. So setting $E_1=\{x\in E: f(x)\geq 0\}$ and $E_2=\{x\in E: f(x)<0\}$, these are disjoint and $E=E_1\cup E_2$.
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$$
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\int_E |f| dm=\int_{E_1} f^+ dm+\int_{E_2} f^- dm
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$$
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For the reverse inequality, note that
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$$
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\int_E f^+ dm\leq \int_E |f| dm
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$$
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and
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$$
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\int_E f^- dm\leq \int_E |f| dm
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$$
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QED
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#### Corollary Properties of Integrals
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Let $f$ and $g$ be integrable on $E$, and $c\in \mathbb{R}$.
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1. $\int_E (cf) dm=c\int_E f dm$
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2. $\int_E (f+g) dm=\int_E f dm+\int_E g dm$
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Proof:
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First we prove it for $f,g$ nonnegative and $c\geq 0$.
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Take simple functions $\phi_n\to f$ and $\psi_n\to g$ pointwise. Then $c\phi_n\to cf$ and $\phi_n+\psi_n\to f+g$ pointwise.
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By Monotone Convergence Theorem,
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$$
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\int_E cf dm=\lim_{n\to\infty} \int_E c\phi_n dm=c\lim_{n\to\infty} \int_E \phi_n dm=c\int_E f dm
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$$
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Second part leave as homework.
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QED
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### Theorem 6.8
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Riemann integrable functions are Lebesgue integrable and the values of the integrals are the same.
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Proof:
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Say $f$ is Riemann integrable on $[a,b]$. $m\leq f(x)\leq M$ for all $x\in [a,b]$.
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We can find a partition $P_n\subseteq P_{n+1}$ of $[a,b]$ such that $L(P_n,f)\nearrow \int_a^b f dx$ and $U(P_n,f)\searrow \int_a^b f dx$.
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Let $\phi_n=\sum_{i=1}^{k} m_i \chi_{I_i}$ and $\psi_n=\sum_{i=1}^{k} M_i \chi_{I_i}$ where $I_i$ is an interval in $P_n$.
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So $\int_a^b \phi_n dm=L(P_n,f)$ and $\int_a^b \psi_n dm=U(P_n,f)$.
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$m\leq \phi_n\leq f\leq \psi_n\leq M$ for all $n$. almost everywhere.
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By Monotone Convergence Theorem, to $\phi_{n-m}$ we have $g(x)=\lim_{n\to\infty} \phi_n(x)$, $h(x)=\lim_{n\to\infty} \psi_n(x)$ exists for almost every $x\in [a,b]$.
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$g(x)\leq f(x)\leq h(x)$ almost everywhere.
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So
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$$
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\int_a^b g dm= \int_a^b f dm= \int_a^b h dm =\int_a^b f dx
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$$
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So $h(x)=f(x)=g(x)$ almost everywhere.
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QED
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