5.9 KiB
Math4121 Lecture 9
Exam next week.
Transition to new book.
Continue on Chapter 6
Integrable Functions
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].
Proof:
Since \phi is uniformly continuous on [m, M], for any \epsilon > 0, there exists a \delta > 0 such that if s, t\in [m, M] and |s-t| < \delta, then |\phi(s)-\phi(t)| < \epsilon.
Since f\in \mathscr{R}(\alpha) on [a, b], we can find a partition P=\{x_0, x_1, \cdots, x_n\} of [a, b] such that U(f, P, \alpha)-L(f, P, \alpha) < \epsilon \delta.
Set M_i=\sup \{f(x): x\in [x_{i-1}, x_i]\} and m_i=\inf \{f(x): x\in [x_{i-1}, x_i]\}. M_i^*=\sup \{h(x): x\in [x_{i-1}, x_i]\} and m_i^*=\inf \{h(x): x\in [x_{i-1}, x_i]\}.
We call a index i good if M_i-m_i < \delta.
If i is good, then \forall x, y\in [x_{i-1}, x_i], |f(x)-f(y)| < \delta and by uniform continuity of \phi, |\phi(f(x))-\phi(f(y))| < \epsilon.
Therefore, |M_i^*-m_i^*| < \epsilon.
If i is bad, then M_i-m_i\geq \delta.
Notice that
\begin{aligned}
\delta\sum_{i\in\text{bad}} \Delta \alpha_i &\leq \sum_{i\in\text{bad}} (M_i-m_i) \Delta \alpha_i \\
&\leq \sum_{i=1}^n (M_i-m_i) \Delta \alpha_i \\
&\leq U(f, P, \alpha)-L(f, P, \alpha) \\
&< \epsilon\delta
\end{aligned}
Therefore, \sum_{i=1}^n (M_i^*-m_i^*) \Delta \alpha_i < \epsilon.
So,
\begin{aligned}
U(P,h,\alpha)-L(P,h,\alpha) &= \sum_{i=1}^n (M_i^*-m_i^*) \Delta \alpha_i \\
&\leq \sum_{i\in\text{good}} \epsilon \Delta \alpha_i + \sum_{i\in\text{bad}}2 \sup \{|h(x)-h(y)|: x, y\in [x_{i-1}, x_i]\} \Delta \alpha_i \\
&\leq \epsilon [\alpha(b)-\alpha(a)] + 2\epsilon \sup \{|h(x)-h(y)|: x, y\in [a, b]\}\\
\end{aligned}
Since \epsilon is arbitrary, h\in \mathscr{R}(\alpha) on [a, b].
QED
Properties of Integrable Functions
Theorem 6.12
Let f,g\in \mathscr{R}(\alpha) on [a, b].
(a) f+g\in \mathscr{R}(\alpha) on [a, b], \int_a^b (f+g)d\alpha = \int_a^b f d\alpha + \int_a^b g d\alpha. (Linearity of the integral)
(aa) If c\in \mathbb{R}, then cf\in \mathscr{R}(\alpha) on [a, b], and \int_a^b cf d\alpha = c\int_a^b f d\alpha.
(b) If f(x)\leq g(x),\forall x\in [a, b], then \int_a^b f d\alpha \leq \int_a^b g d\alpha.
(c) c\in [a, b], then \int_a^c f d\alpha + \int_c^b f d\alpha = \int_a^b f d\alpha.
(d) If |f(x)|\leq M, then |\int_a^b f d\alpha| \leq M(\alpha(b)-\alpha(a)).
(e) If f\in \mathscr{R}(\beta) then f\in \mathscr{R}(\alpha+\beta) and \int_a^b f d(\alpha+\beta) = \int_a^b f d\alpha + \int_a^b f d\beta.
Proof:
Property (aa), (b), (e) holds for Riemann Sums themselves.
For (a), Set h(x)=f(x)+g(x). Then h\in \mathscr{R}(\alpha) on [a, b] and we will show \int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha.
Since f,g\in \mathscr{R}(\alpha) on [a, b], for any \epsilon > 0, there exists a partition P_1,P_2 of [a, b] such that U(f,P_1,\alpha)-L(f,P_1,\alpha) < \epsilon and U(g,P_2,\alpha)-L(g,P_2,\alpha) < \epsilon.
Let P=P_1\cup P_2. Then U(P,f,\alpha)\leq U(P_1,f,\alpha)< \int_a^b f d\alpha + \epsilon and U(P,g,\alpha)\leq U(P_2,g,\alpha)< \int_a^b g d\alpha + \epsilon.
So U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_a^b g d\alpha + 2\epsilon.
Since \epsilon is arbitrary, \int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha.
\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R}
U(P,cf, \alpha) = cU(P,f,\alpha)
For (b), notice that if f(x)\leq g(x), then \sup f(x)\leq \sup g(x), U(P,f,\alpha)\leq U(P,g,\alpha). and L(P,f,\alpha)\leq L(P,g,\alpha).
For (c), if f\in \mathscr{R}(\alpha) on [a,b], and if a<c<b, then f\in \mathscr{R} on [a,c] and [c,b], and
\int_a^c f d\alpha + \int_c^b f d\alpha=\int_a^b f d\alpha
For every partition P=\{x_0,x_1,\cdots,x_n\} of [a,b], we have a refinement P^*=P\cup\{c\} of [a,b]. Let P_1=\{x_0,x_1,\cdots,x_j,c\} and P_2=\{c,x_{j+1},\cdots,x_n\} be the partitions of [a,c] and [c,b] respectively. So
\begin{aligned}
U(P^*,f,\alpha)&=\sum_{i=0}^{n}M_i(x_i-x_{i+1})\\
&=M_c(c-x_j)+\sum_{i=0}^{j-1}M_i(x_i-x_{i+1})+\sum_{i=j+1}^{n}M_i(x_i-x_{i+1})\\
&=U(P_1,f,\alpha)+U(P_2,f,\alpha)
\end{aligned}
and
\begin{aligned}
L(P^*,f,\alpha)&=\sum_{i=0}^{n}m_i(x_i-x_{i+1})\\
&=m_c(x_j-c)+\sum_{i=0}^{j-1}m_i(x_i-x_{i+1})+\sum_{i=j+1}^{n-1}m_i(x_i-x_{i+1})\\
&=L(P_1,f,\alpha)+L(P_2,f,\alpha)
\end{aligned}
Since P^* is a refinement of P, by \textbf{Theorem 6.4}, we have U(P^*,f,\alpha)\leq U(P,f,\alpha) and L(P^*,f,\alpha)\geq L(P,f,\alpha).
So \int_a^c f d\alpha+\int_c^b f d\alpha\leq U(P^*,f,\alpha)\leq U(P,f,\alpha)=\int_a^b f d\alpha.
Similarly, we have \int_a^c f d\alpha+\int_c^b f d\alpha\geq L(P^*,f,\alpha)\geq L(P,f,\alpha)=\int_a^b f d\alpha.
Therefore, \int_a^c f d\alpha+\int_c^b f d\alpha=\int_a^b f d\alpha.
For (d), if f\in \mathscr{R}(\alpha) on [a,b], and if |f(x)| \leq M on [a,b], then
\left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a))
Since |f(x)|\leq M on [a,b], \forall x\in [a,b], we have f(x)\in [-M,M] on [a,b]. So \sup|f(x)|\leq M and \inf|f(x)|\leq M. Since L(P,f,\alpha)\leq \int_a^b f d\alpha\leq U(P,f,\alpha), we have
So
\begin{aligned}
\left|\int_a^b f d\alpha\right|&\leq \max\left\{|L(P,f,\alpha)|,|U(P,f,\alpha)|\right\}\\
&=\max\left\{\sum_{i=0}^{n-1}|M_i|\Delta x_i,\sum_{i=0}^{n-1}|m_i|\Delta x_i\right\}\\
&\leq \sum_{i=0}^{n-1}\max\{|M_i|,|m_i|\}\Delta x_i\\
&\leq \sum_{i=0}^{n-1}M\Delta x_i\\
&=M(\alpha(b)-\alpha(a))
\end{aligned}
Therefore, \left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a)).
For (e), notice that
\begin{aligned}
\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\
&= \Delta \alpha_i + \Delta \beta_i
\end{aligned}
QED