2.8 KiB
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