I am trying to understand Spivak's development of integration in Calculus, 4th ed. The integral is introduced in Chapter 13, and I find the proofs quite difficult and unfamiliar. The Appendix to the chapter discusses Riemann sums, which here are defined similarly to upper and lower sums, but involve function evaluations at arbitrary points. The first Problem in this appendix is the following:
- Suppose that $f$ and $g$ are continuous functions on $\left[a,b\right]$. For a partition $P=\left\{t_0,\ldots,t_n\right\}$ of $\left[a,b\right]$ choose a set of points $x_i$ in $\left[t_{i-1}, t_i\right]$ and another set of points $u_i$ in $\left[t_{i-1}, t_i\right]$. Consider the sum $$ \sum_{i=1}^n f\left(x_i\right)g\left(u_i\right)\left(t_i - t_{i-1}\right).$$ Notice that this is not a Riemann sum of $fg$ for $P$. Nevertheless, show that all such sums will be within $\epsilon$ of $\int_a^b fg $ provided that the partition $P$ has all lengths $t_i - t_{i-1}$ small enough. Hint: Estimate the difference between such a sum and a Riemann sum; you will need to use uniform continuity (Chapter 8, Appendix).
Below is my attempt. I think it is correct, but I would appreciate some critique, since I still find it pretty non-intuitive.
My Attempt
Let $\epsilon >0$ be arbitrary. Since $f,g$ are continuous on $\left[a,b\right]$, they are bounded, say $\left|f\right|,\left|g\right| \leq M$. Also by continuity on a closed interval, we know that they are uniformly continuous on every closed subinterval of $\left[a,b\right]$; thus, let $\delta>0$ such that $\left|x-y\right| < \delta$ implies $$ \left|f\left(x\right) - f\left(y\right)\right| < \frac{\epsilon}{2M\left(b-a\right)}$$ and $$ \left|g\left(x\right) - g\left(y\right)\right| < \frac{\epsilon}{2M\left(b-a\right)}.$$
Next, let $P=\left\{t_0,\ldots,t_n\right\}$ be a partition of $\left[a,b\right]$ such that $t_j - t_{j-1} < \delta$ for all $j=1,\ldots,n$. Let $x_j, u_j \in \left[t_{j-1},t_j\right]$ be arbitrary points in each of the intervals. We will write the sum of interest as $$ S = \sum_{i=1}^n f\left(x_i\right)g\left(u_i\right) \left(t_i - t_{i-1}\right) $$ for simplicity. Note also that since $f$ and $g$ are continuous on each $\left[t_{j-1},t_j\right]$, we know that $fg$ is also continuous and so $fg$ attains its maximum on $\left[t_{j-1}, t_j\right]$ at some $v_j$ for each $j=1,\ldots,n$. That is, $$ \left(fg\right)\left(v_i\right) = f\left(v_i\right)g\left(v_i\right) = \sup \left\{ \left(fg\right)\left(t\right) \, : \, t \in \left[t_{i-1}, t_i\right] \right\}.$$
Next we will check how far off our $S$ is from the upper and lower sums of $fg$ given this partition $P$. \begin{align} U\left(fg, P\right) - S & = \sum_{i=1}^n \left[f\left(v_i\right)g\left(v_i\right) - f\left(x_i\right)g\left(u_i\right)\right] \left(t_i - t_{i-1}\right) \\ & = \sum_{i=1}^n \left[f\left(v_i\right)g\left(v_i\right) - f\left(v_i\right)g\left(u_i\right) + f\left(v_i\right)g\left(u_i\right) - f\left(x_i\right)g\left(u_i\right) \right] \left(t_i - t_{i-1}\right) \\ & = \sum_{i=1}^n \left[f\left(v_i\right)\left[g\left(v_i\right) - g\left(u_i\right) \right] + g\left(u_i\right) \left[f\left(v_i\right) - f\left(x_i\right)\right] \right] \left(t_i - t_{i-1}\right) \\ & \leq \sum_{i=1}^n \left[M\frac{\epsilon}{2M\left(b-a\right)} + M\frac{\epsilon}{2M\left(b-a\right)} \right] \left(t_i - t_{i-1}\right) \\ & = \frac{\epsilon}{b-a}\sum_{i=1}^n \left(t_i - t_{i-1}\right) \\ & = \epsilon \end{align} so that we have $U\left(fg, P\right) - S < \epsilon$. It can be shown similarly that $S - L\left(fg, P\right) < \epsilon$. Taken together, we have $$ S - \epsilon < L\left(fg,P\right) \leq \int_a^b fg \leq U\left(fg, P\right) < S + \epsilon, $$ or $$ \left| S - \int_a^b fg\right| < \epsilon. $$
$$\tag*{$\blacksquare$}$$
