# Math4121 Lecture 8

## Continue on Riemann-Stieltjes Integral

### Integrable Functions

#### Theorem 6.9

If $f$ is monotonic (increasing) on $[a, b]$ and $\alpha$ is continuous on $[a, b]$, then $f\in \mathscr{R}(\alpha)$ on $[a, b]$.

Proof:

Given a partition $P = \{a = x_0, x_1, \cdots, x_n = b\}$, we have

$$
M_i = \sup_{x\in [x_{i-i}, x_i]} f(x)\leq f(x_{i})
$$

$$
m_i = \inf_{x\in [x_{i-1}, x_i]} f(x)\geq f(x_{i-1})
$$

So,

$$
\begin{aligned}
U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right] \left[ \alpha(x_i) - \alpha(x_{i-1}) \right] \\

&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right](\sup_{x\in [x_{i-1}, x_i]} \alpha(x) - \inf_{x\in [x_{i-1}, x_i]} \alpha(x)) \\
&=U(P,\alpha,f) - L(P,\alpha,f)
\end{aligned}
$$

By Theorem 6.8, $\alpha\in \mathscr{R}(f)$, so for any $\epsilon > 0$, there exists a partition $P$ such that

$$
U(P,\alpha,f) - L(P,\alpha,f) < \epsilon
$$

Therefore, $U(P,f,\alpha) - L(P,f,\alpha)<U(P,\alpha,f) - L(P,\alpha,f) < \epsilon$, so $f\in \mathscr{R}(\alpha)$ on $[a, b]$.

QED

#### Theorem 6.10

Suppose $f$ is bounded on $[a, b]$ and has finitely many points $\{y_1, y_2, \cdots, y_J\}$ of discontinuity, and $\alpha$ is continuous on $[a, b]$. Then $f\in \mathscr{R}(\alpha)$ on $[a, b]$.

Proof:

Since $f$ is bounded, there exists a $M>0$ such that $|f(x)|\leq M$ for all $x\in [a, b]$.

Let $\epsilon > 0$. Since $\alpha$ is continuous on $[a, b]$, we can find some intervals $[u_j,v_j]\subset (a,b)$ and $y_j\in [u_j,v_j]$ and $|\alpha(u_j) - \alpha(v_j)| < \epsilon$ for all $j=1,2,\cdots,J$.

Set $K=[a,b]\setminus \bigcup_{j=1}^{J}(u_j,v_j)$. Since $K$ is compact, $f$ is uniformly continuous on $K$. Hence, there exists a $\delta > 0$ such that for any $s,t\in K$ and $|s-t|<\delta$, we have $|f(s)-f(t)|<\epsilon$.

Let $P=\{x_0,x_1,\cdots,x_n=b\}$ containing all the points $u_j,v_j,\forall j=1,2,\cdots,J$ and $\Delta x_i<\delta,\forall x_i\notin \{u_j,v_j,\forall j=1,2,\cdots,J\}$.

Then,

If $x_i=u_j$ for some $j=1,2,\cdots,J$, then $M_i-m_i\leq M:=2\sup|f_x|$. But $\Delta \alpha_i\leq \epsilon$ for all $i=1,2,\cdots,n$.

If $x_i\neq u_j$ for all $j=1,2,\cdots,J$, then by uniform continuity of $f$ on $K$, we have $M_i-m_i\leq \epsilon$.

In either case, we have

$$
\begin{aligned}
U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
&\leq J M\epsilon + \sum_{i=1}^{n} \epsilon \Delta \alpha_i \\
&= \epsilon(J M + \sum_{i=1}^{n} \Delta \alpha_i)
\end{aligned}
$$

Since $\epsilon$ is arbitrary, we have $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.

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

QED

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