# Math4121 Lecture 10

## Recap

### Properties of Riemann-Stieltjes Integral

#### Linearity (Theorem 6.12 (a))

If $f,g\in \mathscr{R}(\alpha)$ on $[a, b]\subset \mathbb{R},c,d\in \mathbb{R}$, then $cf+dg\in \mathscr{R}(\alpha)$ on $[a, b]$ and

$$
\int_a^b (cf+dg)d\alpha = c\int_a^b f d\alpha + d\int_a^b g d\alpha
$$

#### Composition (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 $[a, b]$.

#### Monotonicity (Theorem 6.12 (b))

If $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, and $f(x)\leq g(x),\forall x\in [a, b]$, then $\int_a^b f d\alpha \leq \int_a^b g d\alpha$.

## Continue on Chapter 6

### Properties of Integrable Functions

#### Theorem 6.13

Suppose $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, and $c\in (a, b)$. Then

(a) $fg\in \mathscr{R}(\alpha)$ on $[a, b]$.

Proof:

By linearity, $f+g,f-g\in \mathscr{R}(\alpha)$ on $[a, b]$.

Moreover, let $\phi(x)=x^2$, which is continuous on $\mathbb{R}$.

By **Theorem 6.11**, $f^2,g^2\in \mathscr{R}(\alpha)$ on $[a, b]$.

By linearity, $fg=1/4((f+g)^2-(f-g)^2)\in \mathscr{R}(\alpha)$ on $[a, b]$.

QED

(b) $|f|\in \mathscr{R}(\alpha)$ on $[a, b]$, and $|\int_a^b f d\alpha|\leq \int_a^b |f| d\alpha$.

Proof:

Let $\phi(x)=|x|$, which is continuous on $\mathbb{R}$.

By **Theorem 6.11**, $|f|\in \mathscr{R}(\alpha)$ on $[a, b]$.

Let $c=-1$ or $c=1$. such that $c\int_a^b f d\alpha=| \int_a^b f d\alpha|$.

By linearity, $c\int_a^b f d\alpha=\int_a^b cfd\alpha$. Since $cf\leq |f|$, by monotonicity, $|\int_a^b cfd\alpha|=\int_a^b cfd\alpha\leq \int_a^b |f| d\alpha$.

QED

### Indicator Function

#### Definition 6.14

The unit step function is defined as

$$
I(x)=\begin{cases}
0, & x\le 0 \\
1, & x>0
\end{cases}
$$

#### Theorem 6.15

Let $a<s<b$. $f$ is bounded on $[a, b]$ and continuous at $s$. Define $\alpha(x)=I(x-s)$ on $[a, b]$. Then $f\in \mathscr{R}(\alpha)$ on $[a, b]$, and $\int_a^b f d\alpha=f(s)$.

Proof:

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,
$$
\begin{aligned}
U(P,f,\alpha)&=\sum_{i=1}^3 M_i(\alpha(x_i)-\alpha(x_{i-1}))\\
&=M_1(0-0)+M_2(1-0)+M_3(1-1)\\
&=\sup_{x\in [s,x_2]}f(x)(\alpha(x_2)-\alpha(s))\\
&=\sup_{x\in [s,x_2]}f(x)(1-0)\\
&=M_2 \\
\end{aligned}
$$

$$
\begin{aligned}
L(P,f,\alpha)&=\sum_{i=1}^3 m_i(\alpha(x_i)-\alpha(x_{i-1}))\\
&=m_1(0-0)+m_2(1-0)+m_3(1-1)\\
&=\inf_{x\in [s,x_2]}f(x)(\alpha(x_2)-\alpha(s))\\
&=\inf_{x\in [s,x_2]}f(x)(1-0)\\
&=m_2 \\
\end{aligned}
$$

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 $\int_a^b f d\alpha=f(s)$.

QED