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