Exercise 1.1.25 (Equivalence of Jordan measure and Riemann-Darboux integral)

First we look at the case where $f$ is non-negative $f\ge 0$, which is then easily extendable to the general case. Notice that the theorem assertion is true for the following cases:

(i)

constant functions and boxes
Let $f:[a,b]\to ℝ$ be a constant function $c>0$. Then it is Riemann integrable, and we have $$\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}f=c\cdot m([a,b])=c(b-a)$$

This is equivalent to the fact that the area under the graph of $f$ is a box, which is satisfies:

$$E=\left\{(x,t):x\in [a,b],t\in [0,f(x)]\right\}=[a,b]\times [0,c]$$

with the Jordan measure of $m(E)=c(b-a)$.

(ii)

piecewise constant functions and elementary sets
Let $f:[a,b]\to ℝ$ be a piecewise constant function w.r.t. partition $I_1,\dots ,I_n$ and values $c_1,\dots ,c_n$. This function is again automatically Riemann integrable, and the piecewise constant integral is given by $$\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}f=c_1\cdot m(I_1)+\dots +c_n\cdot m(I_n)$$

This is equivalent to the graph of $f$ being an elementary set which satisfies:

$$\begin{array}{llll}\hfill E& =\left\{(x,t):x\in [a,b],t\in [0,f(x)]\right\}& \hfill & \\ \hfill & =\left\{(x,t):x\in I_1,t\in [0,f(x)]\right\}\cup \dots \cup \left\{(x,t):x\in I_n,t\in [0,f(x)]\right\}& \hfill & \\ \hfill & =I_1\times [0,c_1]\cup \dots \cup I_n\times [0,c_n].& \hfill & \end{array}$$

Since this is a disjoint box partition of $E$, by properties of elementary measure we conclude that $m(E)=c_1\cdot m(I_1)+\dots +c_n\cdot m(I_n)$.

Now we prove the theorem assertion.

Let $g$ be an arbitrary piecewise constant function majorizing $f$. Then the area $A(g)$ under the graph of $g$ is an elementary set that outer covers the area $E=A(f)$ under the graph of $f$, and we have $\int g=m(G(g))\in \{m(B):B\supset E,B\text{ is elementary}\}$. Similarly, let $B$ be an elementary set that outer covers the area $E=A(f)$, i.e., $B=I_1\times [0,c_1]\cup \dots \cup I_n\times [0,c_n]$. Then, this value can be represent as the integral of a piecewise constant function $g$ w.r.t. $I_1,\dots ,I_n$ with values $c_1,\dots ,c_n$. In other words, the set of measures of elementary outer covers of $E$ is the same set as the piecewise constant integrals of functions that majorize $f$. Similar line of thought proves the assertion about elementary inner covers. The conclusion is thus

In case when $f$ has negative parts, we separate its piecewise constant outer and inner covers into positive and negative parts.