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-# Lecture 9
\ No newline at end of file
+# Lecture 9
+
+Exam next week.
+
+Transition to new book.
+
+## Continue on Chapter 6
+
+### Integrable Functions
+
+#### 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 $[a, b]$.
+
+Proof:
+
+Since $\phi$ is uniformly continuous on $[m, M]$, for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $s, t\in [m, M]$ and $|s-t| < \delta$, then $|\phi(s)-\phi(t)| < \epsilon$.
+
+Since $f\in \mathscr{R}(\alpha)$ on $[a, b]$, we can find a partition $P=\{x_0, x_1, \cdots, x_n\}$ of $[a, b]$ such that $U(f, P, \alpha)-L(f, P, \alpha) < \epsilon \delta$.
+
+Set $M_i=\sup \{f(x): x\in [x_{i-1}, x_i]\}$ and $m_i=\inf \{f(x): x\in [x_{i-1}, x_i]\}$. $M_i^*=\sup \{h(x): x\in [x_{i-1}, x_i]\}$ and $m_i^*=\inf \{h(x): x\in [x_{i-1}, x_i]\}$.
+
+We call a index $i$ good if $M_i-m_i < \delta$.
+
+If $i$ is good, then $\forall x, y\in [x_{i-1}, x_i]$, $|f(x)-f(y)| < \delta$ and by uniform continuity of $\phi$, $|\phi(f(x))-\phi(f(y))| < \epsilon$.
+
+Therefore, $|M_i^*-m_i^*| < \epsilon$.
+
+If $i$ is bad, then $M_i-m_i\geq \delta$.
+
+Notice that
+
+$$
+\begin{aligned}
+\delta\sum_{i\in\text{bad}} \Delta \alpha_i &\leq \sum_{i\in\text{bad}} (M_i-m_i) \Delta \alpha_i \\
+&\leq \sum_{i=1}^n (M_i-m_i) \Delta \alpha_i \\
+&\leq U(f, P, \alpha)-L(f, P, \alpha) \\
+&< \epsilon\delta
+\end{aligned}
+$$
+
+Therefore, $\sum_{i=1}^n (M_i^*-m_i^*) \Delta \alpha_i < \epsilon$.
+
+So,
+
+$$
+\begin{aligned}
+U(P,h,\alpha)-L(P,h,\alpha) &= \sum_{i=1}^n (M_i^*-m_i^*) \Delta \alpha_i \\
+&\leq \sum_{i\in\text{good}} \epsilon \Delta \alpha_i + \sum_{i\in\text{bad}}2 \sup \{|h(x)-h(y)|: x, y\in [x_{i-1}, x_i]\} \Delta \alpha_i \\
+&\leq \epsilon [\alpha(b)-\alpha(a)] + 2\epsilon \sup \{|h(x)-h(y)|: x, y\in [a, b]\}\\
+\end{aligned}
+$$
+
+Since $\epsilon$ is arbitrary, $h\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+EOP
+
+### Properties of Integrable Functions
+
+#### Theorem 6.12
+
+Let $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+(a) $f+g\in \mathscr{R}(\alpha)$ on $[a, b]$, $\int_a^b (f+g)d\alpha = \int_a^b f d\alpha + \int_a^b g d\alpha$. (Linearity of the integral)
+
+If $c\in \mathbb{R}$, then $cf\in \mathscr{R}(\alpha)$ on $[a, b]$, and $\int_a^b cf d\alpha = c\int_a^b f d\alpha$.
+
+(b) If $f(x)\leq g(x),\forall x\in [a, b]$, then $\int_a^b f d\alpha \leq \int_a^b g d\alpha$.
+
+(c) $c\in [a, b]$, then $\int_a^c f d\alpha + \int_c^b f d\alpha = \int_a^b f d\alpha$.
+
+(d) If $|f(x)|\leq M$, then $|\int_a^b f d\alpha| \leq M(\alpha(b)-\alpha(a))$.
+
+(e) If $f\in \mathscr{R}(\beta)$ then $f\in \mathscr{R}(\alpha+\beta)$ and $\int_a^b f d(\alpha+\beta) = \int_a^b f d\alpha + \int_a^b f d\beta$.
+
+
+
+