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pages/Math4121/Math4121_L36.md

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+# Math 4121 Lecture 36
+
+## Random visit for Lebesgue Integration
+
+### Convergence Theorem
+
+#### Theorem 6.14 Monotone Convergence Theorem
+
+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 $n$)
+
+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
+
+$$
+\int_E f \, dm = \lim_{n\to\infty} \int_E f_n \, dm
+$$
+
+Proof:
+
+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
+
+$$
+\frac{A}{\epsilon}m(E_n)=\int_E \frac{A}{\epsilon}\chi_{E_n}dm\leq \int_E f_n dm\leq A
+$$
+
+In particular, $m(E_n)<\epsilon$. Since $E_n\subset E_{n+1}$ for all $n$, $m(U)=\lim_{n\to\infty} m(E_n)<\epsilon$.
+
+$\lim_{n\to\infty} \int_E f_n dm\leq \int_E f dm\leq \lim_{n\to\infty}$. 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 $\bigcup_{i=1}^{k} S_i\subseteq E$.
+
+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 $x\in A_n$.
+
+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 $i$.
+
+$\int_{A_n} \phi dm=\sum_{i=1}^{k} a_i m(S_i\cap A_n)$.
+
+As $n\to\infty$, $m(A_n\cap S_i)\to m(S_i)$ and $\sum_{i=1}^{k} a_i m(S_i\cap A_n)\to \int_E \phi dm$.
+
+Let $\epsilon>0$. There exists $n_0$ large such that $\int_{A_n} \phi dm>\int_E \phi dm-\epsilon$ for all $n\geq n_0$.
+
+Then for such $n\geq n_0$,
+
+$$
+\int_E f_n dm\geq \int_{A_n} f_n dm\geq \int_{A_n} \alpha \phi dm>\alpha(\int_E \phi dm-\epsilon)
+$$
+
+So, $\lim_{n\to\infty} \int_E f_n dm\geq \alpha(\int_E \phi dm-\epsilon)$.
+
+Since $\alpha,\epsilon$ are arbitrary, set $\alpha\to 1$ and $\epsilon\to 0$ to get $\lim_{n\to\infty} \int_E f_n dm\geq \int_E \phi dm$.
+
+For any simple function $\phi\leq f$, taking sup over all simple functions $\phi\leq f$ gives $\lim_{n\to\infty} \int_E f_n dm\geq \int_E f dm$.
+
+QED
+
+#### Lemma Absolute Integrability
+
+$f$ is integrable on $E$ if and only if $|f|$ is integrable on $E$ and $\left|\int_E f \, dm\right|\leq \int_E |f| \, dm$.
+
+Proof:
+
+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 $E=E_1\cup E_2$.
+
+$$
+\int_E |f| dm=\int_{E_1} f^+ dm+\int_{E_2} f^- dm
+$$
+
+For the reverse inequality, note that
+
+$$
+\int_E f^+ dm\leq \int_E |f| dm
+$$
+
+and
+
+$$
+\int_E f^- dm\leq \int_E |f| dm
+$$
+
+QED
+
+#### Corollary Properties of Integrals
+
+Let $f$ and $g$ be integrable on $E$, and $c\in \mathbb{R}$.
+
+1. $\int_E (cf) dm=c\int_E f dm$
+2. $\int_E (f+g) dm=\int_E f dm+\int_E g dm$
+
+Proof:
+
+First we prove it for $f,g$ nonnegative and $c\geq 0$.
+
+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.
+
+By Monotone Convergence Theorem,
+
+$$
+\int_E cf dm=\lim_{n\to\infty} \int_E c\phi_n dm=c\lim_{n\to\infty} \int_E \phi_n dm=c\int_E f dm
+$$
+
+Second part leave as homework.
+
+QED
+
+### Theorem 6.8
+
+Riemann integrable functions are Lebesgue integrable and the values of the integrals are the same.
+
+Proof:
+
+Say $f$ is Riemann integrable on $[a,b]$. $m\leq f(x)\leq M$ for all $x\in [a,b]$.
+
+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 $U(P_n,f)\searrow \int_a^b f dx$.
+
+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 $P_n$.
+
+So $\int_a^b \phi_n dm=L(P_n,f)$ and $\int_a^b \psi_n dm=U(P_n,f)$.
+
+$m\leq \phi_n\leq f\leq \psi_n\leq M$ for all $n$. almost everywhere.
+
+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 $x\in [a,b]$.
+
+$g(x)\leq f(x)\leq h(x)$ almost everywhere.
+
+So
+
+$$
+\int_a^b g dm= \int_a^b f dm= \int_a^b h dm =\int_a^b f dx
+$$
+
+So $h(x)=f(x)=g(x)$ almost everywhere.
+
+QED
+
+
+

pages/Math4121/Math4121_L37.md

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+# Math4121 Lecture 37
+
+## Extended fundamental theorem of calculus with Lebesgue integration
+
+### Density of continuous functions
+
+#### Lemma: 
+
+Let $K\subseteq U$ be bounded sets in $\mathbb{R}$, $K$ is closed and $U$ is open. Then there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.
+
+Proof in homework.
+
+Hint: Consider the basic intervals cases.
+
+#### Theorem for continuous functions
+
+Let $f$ be integrable. For each $\epsilon>0$, there is a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $\int_{\mathbb{R}}|f-g|dm<\epsilon$.
+
+Proof:
+
+First where $f=\chi_S$ for some bounded means set $S$. then extended to all $f$ integrable.
+
+First, assume $f=\chi_S$. Let $\epsilon>c$, we can find $K\subseteq S\subseteq U$. and $K$ is closed and $U$ is open such that (by definition of Lebesgue outer measure)
+
+$$
+m(K)+\frac{\epsilon}{2}>m(S)>m(U)-\frac{\epsilon}{2}
+$$
+
+In particular, $m(U\setminus K)=m(U)-m(K)<\epsilon$.
+
+By lemma, there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.
+
+So
+
+$$
+\int_E |\chi_S -g|dm=\int_{U\setminus K} |\chi_S -g|dm\leq m(U\setminus K)<\epsilon
+$$
+
+For the general case,
+
+By the Monotone Convergence Theorem (use $|f|\chi_{[-N,N]}$ to approximate $|f|$), we can find $N$ large such that
+
+$$
+\int_{E_N^c}|f|dm<\frac{\epsilon}{2}
+$$
+
+where $E_N=E\cap [-N,N]$.
+
+Notice that by the definition of Lebesgue integral, $\int f^+ dm=\sup\{\int \phi^+ dm:\phi\text{ is simple and } \phi\leq f^+\}$ and $\int f^- dm=\sup\{\int \phi^- dm:\phi\text{ is simple and } \phi\leq f^-\}$.
+
+By considering $f^+$ and $f^-$ separately, we can find a simple function $\phi$ such that
+
+$$
+\int_{E_N} |f-\phi|dm<\frac{\epsilon}{3}
+$$
+
+
+
+
+QED
+