From 59513fe4230e8d068034cf3dff2a446a931ac797 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 5 Feb 2025 13:50:35 -0600
Subject: [PATCH] Update Math4121_L10.md

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-# Lecture 10
\ No newline at end of file
+# Lecture 10
+
+## Recap
+
+### Properties of Riemann-Stieltjes Integral
+
+#### Linearity (Theorem 6.12 (a))
+
+If $f,g\in \mathscr{R}(\alpha)$ on $[a, b]\subset \mathbb{R},c,d\in \mathbb{R}$, then $cf+dg\in \mathscr{R}(\alpha)$ on $[a, b]$ and
+
+$$
+\int_a^b (cf+dg)d\alpha = c\int_a^b f d\alpha + d\int_a^b g d\alpha
+$$
+
+#### Composition (Theorem 6.11)
+
+Suppose $f\in \mathscr{R}(\alpha)$ on $[a, b]$, $m\leq f(x)\leq M$ for all $x\in [a, b]$, and $\phi$ is continuous on $[m, M]$, and let $h(x)=\phi(f(x))$ on $[a, b]$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+#### Monotonicity (Theorem 6.12 (b))
+
+If $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, and $f(x)\leq g(x),\forall x\in [a, b]$, then $\int_a^b f d\alpha \leq \int_a^b g d\alpha$.
+
+## Continue on Chapter 6
+
+### Properties of Integrable Functions
+
+#### Theorem 6.13
+
+Suppose $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, and $c\in (a, b)$. Then
+
+(a) $fg\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+Proof:
+
+By linearity, $f+g,f-g\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+Moreover, let $\phi(x)=x^2$, which is continuous on $\mathbb{R}$.
+
+By **Theorem 6.11**, $f^2,g^2\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+By linearity, $fg=1/4((f+g)^2-(f-g)^2)\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+EOP
+
+(b) $|f|\in \mathscr{R}(\alpha)$ on $[a, b]$, and $|\int_a^b f d\alpha|\leq \int_a^b |f| d\alpha$.
+
+Proof:
+
+Let $\phi(x)=|x|$, which is continuous on $\mathbb{R}$.
+
+By **Theorem 6.11**, $|f|\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+Let $c=-1$ or $c=1$. such that $c\int_a^b f d\alpha=| \int_a^b f d\alpha|$.
+
+By linearity, $c\int_a^b f d\alpha=\int_a^b cfd\alpha$. Since $cf\leq |f|$, by monotonicity, $|\int_a^b cfd\alpha|=\int_a^b cfd\alpha\leq \int_a^b |f| d\alpha$.
+
+EOP
+
+### Indicator Function
+
+#### Definition 6.14
+
+The unit step function is defined as
+
+$$
+I(x)=\begin{cases}
+0, & x\le 0 \\
+1, & x>0
+\end{cases}
+$$
+
+#### Theorem 6.15
+
+Let $a<s<b$. $f$ is bounded on $[a, b]$ and continuous at $s$. Define $\alpha(x)=I(x-s)$ on $[a, b]$. Then $f\in \mathscr{R}(\alpha)$ on $[a, b]$, and $\int_a^b f d\alpha=f(s)$.
+
+Proof:
+
+Under the hypothesis, $f$ is bounded on $[a, b]$ and continuous at $s$.
+
+We can choose partition $P=\{x_0,x_1,x_2,x_3\}$ such that $a=x_0<x_1=s<x_2<x_3=b$.
+
+Then $U(P,f,\alpha)=\sup_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s))$. $L(P,f,\alpha)=\inf_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s))$.
+
+Since $f$ is continuous at $s$, when $x_3\to s$, $U(P,f,\alpha)\to f(s)$ and $L(P,f,\alpha)\to f(s)$.
+
+Therefore, $U(P,f,\alpha)-L(P,f,\alpha)\to 0$, $f\in \mathscr{R}(\alpha)$ on $[a, b]$, and $\int_a^b f d\alpha=f(s)$.
+
+EOP
+
+