final update on 4121

pages/Math4121/Math4121_L9.md

@@ -75,30 +75,9 @@ Let $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
 Proof:
 
-Property (aa), (b), (e) holds for Riemann Sums themselves.
+**Property (aa), (b), (e) holds for Riemann Sums themselves.**
 
-$$
-\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R}
-$$
-
-$$
-U(P,cf, \alpha) = cU(P,f,\alpha)
-$$
-
-For (b), notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$.
-
-For (e), notice that
-
-$$
-\begin{aligned}
-\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\
-&= \Delta \alpha_i + \Delta \beta_i
-\end{aligned}
-$$
-
-(c),(d) are left as homework.
-
-For (a), Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
+**For (a)**, Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
 
 Since $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, for any $\epsilon > 0$, there exists a partition $P_1,P_2$ of $[a, b]$ such that $U(f,P_1,\alpha)-L(f,P_1,\alpha) < \epsilon$ and $U(g,P_2,\alpha)-L(g,P_2,\alpha) < \epsilon$.
 
@@ -108,4 +87,76 @@ So $U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_
 
 Since $\epsilon$ is arbitrary, $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
 
+
+$$
+\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R}
+$$
+
+$$
+U(P,cf, \alpha) = cU(P,f,\alpha)
+$$
+
+**For (b)**, notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$.
+
+**For (c)**, if $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and if $a<c<b$, then $f\in \mathscr{R}$ on $[a,c]$ and $[c,b]$, and 
+
+$$
+\int_a^c f d\alpha + \int_c^b f d\alpha=\int_a^b f d\alpha
+$$
+
+For every partition $P=\{x_0,x_1,\cdots,x_n\}$ of $[a,b]$, we have a refinement $P^*=P\cup\{c\}$ of $[a,b]$. Let $P_1=\{x_0,x_1,\cdots,x_j,c\}$ and $P_2=\{c,x_{j+1},\cdots,x_n\}$ be the partitions of $[a,c]$ and $[c,b]$ respectively. So 
+$$
+\begin{aligned}
+U(P^*,f,\alpha)&=\sum_{i=0}^{n}M_i(x_i-x_{i+1})\\
+&=M_c(c-x_j)+\sum_{i=0}^{j-1}M_i(x_i-x_{i+1})+\sum_{i=j+1}^{n}M_i(x_i-x_{i+1})\\
+&=U(P_1,f,\alpha)+U(P_2,f,\alpha) 
+\end{aligned}
+$$
+and
+$$
+\begin{aligned}
+L(P^*,f,\alpha)&=\sum_{i=0}^{n}m_i(x_i-x_{i+1})\\
+&=m_c(x_j-c)+\sum_{i=0}^{j-1}m_i(x_i-x_{i+1})+\sum_{i=j+1}^{n-1}m_i(x_i-x_{i+1})\\
+&=L(P_1,f,\alpha)+L(P_2,f,\alpha)
+\end{aligned}
+$$
+
+Since $P^*$ is a refinement of $P$, by \textbf{Theorem 6.4}, we have $U(P^*,f,\alpha)\leq U(P,f,\alpha)$ and $L(P^*,f,\alpha)\geq L(P,f,\alpha)$.
+
+So $\int_a^c f d\alpha+\int_c^b f d\alpha\leq U(P^*,f,\alpha)\leq U(P,f,\alpha)=\int_a^b f d\alpha$.
+
+Similarly, we have $\int_a^c f d\alpha+\int_c^b f d\alpha\geq L(P^*,f,\alpha)\geq L(P,f,\alpha)=\int_a^b f d\alpha$.
+
+Therefore, $\int_a^c f d\alpha+\int_c^b f d\alpha=\int_a^b f d\alpha$.
+
+**For (d)**, if $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and if $|f(x)| \leq M$ on $[a,b]$, then 
+$$
+\left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a))
+$$
+
+Since $|f(x)|\leq M$ on $[a,b]$, $\forall x\in [a,b]$, we have $f(x)\in [-M,M]$ on $[a,b]$. So $\sup|f(x)|\leq M$ and $\inf|f(x)|\leq M$. Since $L(P,f,\alpha)\leq \int_a^b f d\alpha\leq U(P,f,\alpha)$, we have
+
+So
+
+$$
+\begin{aligned}
+    \left|\int_a^b f d\alpha\right|&\leq \max\left\{|L(P,f,\alpha)|,|U(P,f,\alpha)|\right\}\\
+    &=\max\left\{\sum_{i=0}^{n-1}|M_i|\Delta x_i,\sum_{i=0}^{n-1}|m_i|\Delta x_i\right\}\\
+    &\leq \sum_{i=0}^{n-1}\max\{|M_i|,|m_i|\}\Delta x_i\\
+    &\leq \sum_{i=0}^{n-1}M\Delta x_i\\
+    &=M(\alpha(b)-\alpha(a))
+\end{aligned}
+$$
+
+Therefore, $\left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a))$.
+
+**For (e)**, notice that
+
+$$
+\begin{aligned}
+\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\
+&= \Delta \alpha_i + \Delta \beta_i
+\end{aligned}
+$$
+
 QED