update

pages/Math4111/Math4111_L24.md

@@ -133,6 +133,28 @@ By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\i
 
 EOP
 
+---
+
+Supplemental materials:
+
+_I found this section is not covered in the lecture but is used in later chapters._
+
+#### Definition 4.18
+
+Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. $f$ is **uniformly continuous** on $X$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x, y\in X$, $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.
+
+#### Theorem 4.19
+
+If $f$ is a continuous mapping of a compact metric space $X$ into a metric space $Y$, then $f$ is uniformly continuous on $X$.
+
+Proof:
+
+See the textbook.
+
+EOP
+
+---
+
 ### Continuity and connectedness
 
 > **Definition 2.45**: Let $X$ be a metric space. $A,B\subset X$ are **separated** if $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.

pages/Math4121/Math4121_L7.md

@@ -1 +1,96 @@
 # Lecture 7
+
+## Continue on Chapter 6
+
+### Riemann integrable
+
+#### Theorem 6.6
+
+A function $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
+
+Proof:
+
+$\impliedby$
+
+For every $P$,
+
+$$
+L(f, P, \alpha) \leq \underline{\int}_a^b f d\alpha \leq \overline{\int}_a^b f d\alpha \leq U(f, P, \alpha)
+$$
+
+So if $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$, then for every $\epsilon > 0$, there exists a partition $P$ such that
+
+$$
+0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha \leq U(f, P, \alpha) - L(f, P, \alpha) < \epsilon
+$$
+
+Thus $0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha < \epsilon,\forall \epsilon > 0$.
+
+Then, $\overline{\int}_a^b f d\alpha = \underline{\int}_a^b f d\alpha$.
+
+So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+$\implies$
+
+If $f\in \mathscr{R}(\alpha)$ on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+Then by the definition of Riemann integrable, $\sup_{P} L(f, P, \alpha) =\int^b_a f d\alpha = \inf_{P} U(f, P, \alpha)$.
+
+Given any $\epsilon > 0$, by definition of infimum and supremum, there exists a partition $P_1,P_2$ such that
+
+$$
+\int^b_a f d\alpha - \frac{\epsilon}{2} < L(f, P_1, \alpha) \leq \sup_{P} L(f, P, \alpha) = \inf_{P} U(f, P, \alpha) < \int^b_a f d\alpha + \frac{\epsilon}{2}
+$$
+
+Taking $P = P_1 \cup P_2$, by [Theorem 6.4](https://notenextra.trance-0.com/Math4121/Math4121_L6#theorem-64) we have
+
+$$
+U(f, P, \alpha) - L(f, P, \alpha) \leq \left( \int^b_a f d\alpha + \frac{\epsilon}{2} \right) - \left( \int^b_a f d\alpha - \frac{\epsilon}{2} \right) = \epsilon
+$$
+
+So $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+EOP
+
+#### Theorem 6.8
+
+If $f$ is continuous on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+Proof:
+
+> Main idea:
+>
+> $$U(f, P, \alpha) - L(f, P, \alpha) = \sum_{i=1}^n \left( M_i - m_i \right) \Delta \alpha_i$$
+>
+> If we can make $M_i - m_i$ small enough, then $U(f, P, \alpha) - L(f, P, \alpha)$ can be made arbitrarily small.
+>
+> Since $M_i=\sup_{x\in [t_{i-1}, t_i]} f(x)$ and $m_i=\inf_{x\in [t_{i-1}, t_i]} f(x)$, we can make $M_i - m_i$ small enough by making the partition $P$ sufficiently fine.
+
+Suppose we can find a partition $P$ such that $M_i - m_i < \eta$. Then $U(f, P, \alpha) - L(f, P, \alpha) \leq\eta\sum_{i=1}^n \Delta \alpha_i = \eta (\alpha(b)-\alpha(a))$.
+
+> Let $\epsilon >0$ and choose $\eta = \frac{\epsilon}{\alpha(b)-\alpha(a)}$. Then there exists a partition $P$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
+
+Since $f$ is continuous on $[a, b]$ (a compact set), then $f$ is uniformly continuous on $[a, b]$. [Theorem 4.19](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-419)
+
+> If $f$ is continuous on $x$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.
+>
+> If $f$ is continuous on $[a, b]$, then $f$ is continuous at $x,\forall x\in [a, b]$.
+
+So, there exists a $\delta > 0$ such that for all $x, t\in [a, b]$ with $|x-t| < \delta$, we have $|f(x)-f(t)| < \eta$.
+
+Let $P=\{x_0, x_1, \cdots, x_n\}$ be a partition of $[a, b]$ such that $\Delta x_i < \delta$ for all $i$.
+
+So, $\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| < \eta$ for all $i$.
+
+$$
+\begin{aligned}
+\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| &= \sup_{x,t\in [x_{i-1}, x_i]} f(x)-f(t) \\
+&= \sup_{x\in [x_{i-1}, x_i]} f(x)-\sup_{t\in [x_{i-1}, x_i]} -f(t) \\
+&=\sup_{x\in [x_{i-1}, x_i]} f(x)-\inf_{t\in [x_{i-1}, x_i]} f(t) \\
+&= M_i - m_i
+\end{aligned}
+$$
+
+So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+EOP