NoteNextra-origin/pages/Math4121/Math4121_L10.md

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].

EOP

(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.

EOP

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 U(P,f,\alpha)=\sup_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s)). L(P,f,\alpha)=\inf_{x\in [s,x_3]}f(x)(\alpha(x_3)-\alpha(s)).

Since f is continuous at s, when x_3\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).

EOP