3.5 KiB
Math4121 Lecture 6
Chapter 6: Riemann-Stieltjes Integral
A nice point to restart your learning, LOL.
Differentiation and existence of the integral
Definition 6.1
Let [a,b]\subseteq \mathbb{R}. A partition of [a,b] is a finite sequence of points \{x_0,x_1,\cdots,x_n\}\subseteq [a,b] such that x_0<x_1<\cdots<x_n.
Let \alpha:[a,b]\to \mathbb{R} be monotone increasing. (\alpha(x)\leq \alpha(y) for x\leq y)
We will use \alpha for monotone increasing function in later sections.
Definition 6.2
For a partition P of [a,b], we define \Delta \alpha_i=\alpha(x_i)-\alpha(x_{i-1}) for i=1,2,\cdots,n.
Let f:[a,b]\to \mathbb{R} be bounded.
Then we define
m_i=\inf_{x\in [x_{i-1},x_i]}f(x),\quad M_i=\sup_{x\in [x_{i-1},x_i]}f(x).
Defined the lower and upper Riemann sum by
L(P,f,\alpha)=\sum_{i=1}^n m_i\Delta \alpha_i,\quad U(P,f,\alpha)=\sum_{i=1}^n M_i\Delta \alpha_i.
Defined the lower and upper integral by
\underline{\int_a^b}f(x)d\alpha=\sup_P L(P,f,\alpha),\quad \overline{\int_a^b}f(x)d\alpha=\inf_P U(P,f,\alpha).
If \underline{\int_a^b}f(x)d\alpha=\overline{\int_a^b}f(x)d\alpha, then we say f is Riemann-Stieltjes integrable with respect to \alpha on [a,b], written as f\in \mathscr{R}(\alpha), and the common value is called the Riemann-Stieltjes integral of f with respect to \alpha on [a,b], denoted by
\int_a^b f(x)d\alpha=\underline{\int_a^b}f(x)d\alpha=\overline{\int_a^b}f(x)d\alpha.
If \alpha(x)=x, then we write \int_a^b f(x)dx instead of \int_a^b f(x)d\alpha.
Damn, that's a really loooong definition.
Definition 6.3
A partition P^* is called a refinement of P if P\subseteq P^*.
Given two partitions P_1 and P_2, we define their common refinement P^*=P_1\cup P_2. we can merge two partitions by adding all points in both partitions.
Theorem 6.4
If P^* is a refinement of P, then
L(P^*,f,\alpha)\geq L(P,f,\alpha)
Refinement of a partition will never make the lower sum smaller.
U(P^*,f,\alpha)\leq U(P,f,\alpha)
Refinement of a partition will never make the upper sum larger.
Proof:
Main idea:
Let P=P_0\subset P_1\subset P_2\subset \cdots \subset P_K=P^*.
Where P_k has more points than P_{k-1}.
It suffices to show that L(P_k,f,\alpha)\geq L(P_{k-1},f,\alpha) for all k=1,2,\cdots,K.
Let P_{k-1}=\{y_0,y_1,\cdots,y_J\} and P_k=\{y_0,y_1,\cdots,y_{j-1},y^*,y_j,\cdots,y_J\}.
Then, since \alpha is monotone increasing, we have y_{j-1}\leq y^*\leq y_j.
\begin{aligned}
L(P_k,f,\alpha)-L(P_{k-1},f,\alpha)&=\sum_{i=1}^{j+1}\inf_{x\in [y_{i-1},y_i]}f(x)(\alpha(y_i)-\alpha(y_{i-1}))-\sum_{i=1}^j\inf_{x\in [y_{i-1},y_i]}f(x)(\alpha(y_i)-\alpha(y_{i-1}))\\
&=\inf_{x\in [y^*,y_j]}f(x)(\alpha(y_j)-\alpha(y^*))+\inf_{x\in [y_{j-1},y^*]}f(x)(\alpha(y^*)-\alpha(y_{j-1}))-\inf_{x\in [y_{j-1},y_j]}f(x)(\alpha(y_j)-\alpha(y_{j-1}))\\
&\geq m_j(\alpha(y_j)-\alpha(y^*))+m_j(\alpha(y^*)-\alpha(y_{j-1}))-m_{j-1}(\alpha(y_j)-\alpha(y_{j-1}))\\
&=0
\end{aligned}
Same for U(P_k,f,\alpha)\geq U(P_{k-1},f,\alpha).
QED
Theorem 6.5
\underline{\int_a^b}f(x)d\alpha\leq \overline{\int_a^b}f(x)d\alpha
Proof:
Let P^* be a common refinement of P_1 and P_2.
By Theorem 6.4, we have
L(P_1,f,\alpha)\leq L(P^*,f,\alpha)\leq U(P^*,f,\alpha)\leq U(P_2,f,\alpha)
Fixing P_1 and take the supremum over all P_2, we have
\underline{\int_a^b}f(x)d\alpha\leq \sup_{P_1}L(P_1,f,\alpha)\leq \inf_{P_2}U(P_2,f,\alpha)=\overline{\int_a^b}f(x)d\alpha
QED