# Math4121 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 $[a, b]$.

QED

### 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)

(aa) 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 $\int_a^b f d(\alpha+\beta) = \int_a^b f d\alpha + \int_a^b f d\beta$.

Proof:

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 $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.

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 $U(g,P_2,\alpha)-L(g,P_2,\alpha) < \epsilon$.

Let $P=P_1\cup P_2$. Then $U(P,f,\alpha)\leq U(P_1,f,\alpha)< \int_a^b f d\alpha + \epsilon$ and $U(P,g,\alpha)\leq U(P_2,g,\alpha)< \int_a^b g d\alpha + \epsilon$.

So $U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_a^b g d\alpha + 2\epsilon$.

Since $\epsilon$ is arbitrary, $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.

QED