Update Math4121_L10.md

pages/Math4121/Math4121_L10.md

@@ -79,9 +79,28 @@ 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 $U(P,f,\alpha)=\sup_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s))$. $L(P,f,\alpha)=\inf_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s))$.
+Then,
+$$
+\begin{aligned}
+U(P,f,\alpha)&=\sum_{i=1}^3 M_i(\alpha(x_i)-\alpha(x_{i-1}))\\
+&=M_1(0-0)+M_2(1-0)+M_3(1-1)\\
+&=\sup_{x\in [s,x_2]}f(x)(\alpha(x_2)-\alpha(s))\\
+&=\sup_{x\in [s,x_2]}f(x)(1-0)\\
+&=M_2 \\
+\end{aligned}
+$$
 
-Since $f$ is continuous at $s$, when $x_3\to s$, $U(P,f,\alpha)\to f(s)$ and $L(P,f,\alpha)\to f(s)$.
+$$
+\begin{aligned}
+L(P,f,\alpha)&=\sum_{i=1}^3 m_i(\alpha(x_i)-\alpha(x_{i-1}))\\
+&=m_1(0-0)+m_2(1-0)+m_3(1-1)\\
+&=\inf_{x\in [s,x_2]}f(x)(\alpha(x_2)-\alpha(s))\\
+&=\inf_{x\in [s,x_2]}f(x)(1-0)\\
+&=m_2 \\
+\end{aligned}
+$$
+
+Since $f$ is continuous at $s$, when $x\to s$, $U(P,f,\alpha)\to f(s)$ 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 $\int_a^b f d\alpha=f(s)$.