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,


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)

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


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

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


\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}

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


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

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.


\begin{aligned}
\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) \\
&= \sup_{x\in [x_{i-1}, x_i]} f(x)-\sup_{t\in [x_{i-1}, x_i]} -f(t) \\
&=\sup_{x\in [x_{i-1}, x_i]} f(x)-\inf_{t\in [x_{i-1}, x_i]} f(t) \\
&= M_i - m_i
\end{aligned}