97 lines
3.7 KiB
Markdown
97 lines
3.7 KiB
Markdown
# Math4121 Lecture 7
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## Continue on Chapter 6
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### Riemann integrable
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#### Theorem 6.6
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A function $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
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Proof:
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$\impliedby$
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For every $P$,
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$$
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L(f, P, \alpha) \leq \underline{\int}_a^b f d\alpha \leq \overline{\int}_a^b f d\alpha \leq U(f, P, \alpha)
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$$
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So if $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$, then for every $\epsilon > 0$, there exists a partition $P$ such that
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$$
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0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha \leq U(f, P, \alpha) - L(f, P, \alpha) < \epsilon
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$$
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Thus $0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha < \epsilon,\forall \epsilon > 0$.
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Then, $\overline{\int}_a^b f d\alpha = \underline{\int}_a^b f d\alpha$.
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So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
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$\implies$
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If $f\in \mathscr{R}(\alpha)$ on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
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Then by the definition of Riemann integrable, $\sup_{P} L(f, P, \alpha) =\int^b_a f d\alpha = \inf_{P} U(f, P, \alpha)$.
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Given any $\epsilon > 0$, by definition of infimum and supremum, there exists a partition $P_1,P_2$ such that
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$$
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\int^b_a f d\alpha - \frac{\epsilon}{2} < L(f, P_1, \alpha) \leq \sup_{P} L(f, P, \alpha) = \inf_{P} U(f, P, \alpha) < \int^b_a f d\alpha + \frac{\epsilon}{2}
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$$
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Taking $P = P_1 \cup P_2$, by [Theorem 6.4](https://notenextra.trance-0.com/Math4121/Math4121_L6#theorem-64) we have
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$$
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U(f, P, \alpha) - L(f, P, \alpha) \leq \left( \int^b_a f d\alpha + \frac{\epsilon}{2} \right) - \left( \int^b_a f d\alpha - \frac{\epsilon}{2} \right) = \epsilon
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$$
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So $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
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QED
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#### Theorem 6.8
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If $f$ is continuous on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
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Proof:
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> Main idea:
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>
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> $$U(f, P, \alpha) - L(f, P, \alpha) = \sum_{i=1}^n \left( M_i - m_i \right) \Delta \alpha_i$$
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>
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> If we can make $M_i - m_i$ small enough, then $U(f, P, \alpha) - L(f, P, \alpha)$ can be made arbitrarily small.
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>
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> Since $M_i=\sup_{x\in [t_{i-1}, t_i]} f(x)$ and $m_i=\inf_{x\in [t_{i-1}, t_i]} f(x)$, we can make $M_i - m_i$ small enough by making the partition $P$ sufficiently fine.
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Suppose we can find a partition $P$ such that $M_i - m_i < \eta$. Then $U(f, P, \alpha) - L(f, P, \alpha) \leq\eta\sum_{i=1}^n \Delta \alpha_i = \eta (\alpha(b)-\alpha(a))$.
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> Let $\epsilon >0$ and choose $\eta = \frac{\epsilon}{\alpha(b)-\alpha(a)}$. Then there exists a partition $P$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
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Since $f$ is continuous on $[a, b]$ (a compact set), then $f$ is uniformly continuous on $[a, b]$. [Theorem 4.19](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-419)
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> If $f$ is continuous on $x$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.
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>
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> If $f$ is continuous on $[a, b]$, then $f$ is continuous at $x,\forall x\in [a, b]$.
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So, there exists a $\delta > 0$ such that for all $x, t\in [a, b]$ with $|x-t| < \delta$, we have $|f(x)-f(t)| < \eta$.
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Let $P=\{x_0, x_1, \cdots, x_n\}$ be a partition of $[a, b]$ such that $\Delta x_i < \delta$ for all $i$.
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So, $\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| < \eta$ for all $i$.
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$$
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\begin{aligned}
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\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| &= \sup_{x,t\in [x_{i-1}, x_i]} f(x)-f(t) \\
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&= \sup_{x\in [x_{i-1}, x_i]} f(x)-\sup_{t\in [x_{i-1}, x_i]} -f(t) \\
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&=\sup_{x\in [x_{i-1}, x_i]} f(x)-\inf_{t\in [x_{i-1}, x_i]} f(t) \\
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&= M_i - m_i
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\end{aligned}
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$$
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So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
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QED
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