Math4121 Lecture 7

Continue on Chapter 6

Riemann integrable

Theorem 6.6

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 $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.

Proof:

$\impliedby$

For every $P$,

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

Thus $0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha < \epsilon,\forall \epsilon > 0$.

Then, $\overline{\int}_a^b f d\alpha = \underline{\int}_a^b f d\alpha$.

So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.

$\implies$

If $f\in \mathscr{R}(\alpha)$ on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.

Then by the definition of Riemann integrable, $\sup_{P} L(f, P, \alpha) =\int^b_a f d\alpha = \inf_{P} U(f, P, \alpha)$.

Given any $\epsilon > 0$, by definition of infimum and supremum, there exists a partition $P_1,P_2$ such that

Taking $P = P_1 \cup P_2$, by Theorem 6.4  we have

So $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.

QED

Theorem 6.8

If $f$ is continuous on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.

Proof:

Main idea:

$U(f, P, \alpha) - L(f, P, \alpha) = \sum_{i=1}^n \left( M_i - m_i \right) \Delta \alpha_i$

If we can make $M_i - m_i$ small enough, then $U(f, P, \alpha) - L(f, P, \alpha)$ can be made arbitrarily small.

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.

Suppose we can find a partition $P$ such that $M_i - m_i < \eta$. Then $U(f, P, \alpha) - L(f, P, \alpha) \leq\eta\sum_{i=1}^n \Delta \alpha_i = \eta (\alpha(b)-\alpha(a))$.

Let $\epsilon >0$ and choose $\eta = \frac{\epsilon}{\alpha(b)-\alpha(a)}$. Then there exists a partition $P$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.

Since $f$ is continuous on $[a, b]$ (a compact set), then $f$ is uniformly continuous on $[a, b]$. Theorem 4.19 

If $f$ is continuous on $x$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.

If $f$ is continuous on $[a, b]$, then $f$ is continuous at $x,\forall x\in [a, b]$.

So, there exists a $\delta > 0$ such that for all $x, t\in [a, b]$ with $|x-t| < \delta$, we have $|f(x)-f(t)| < \eta$.

Let $P=\{x_0, x_1, \cdots, x_n\}$ be a partition of $[a, b]$ such that $\Delta x_i < \delta$ for all $i$.

So, $\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| < \eta$ for all $i$.

So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.

QED