Exercise 1.1.24 (Basic properties of Riemann-Darboux integral)

We make the following conventions. If $f$ is a function, we denote by $\underline{f}$ a piecewise constant function which minorizes $f$, and by $\underline{F}$ the set of all such functions. Similarly we write $\overline{f}$ for a piecewise constant function that majorizes $f$, and denote the set of all such functions by $\overline{F}$. Also, we use the obvious notation $\mathit{cA}=\{\mathit{ca}:a\in A\}$ and $A+B=\{a+b:a\in A,b\in B\}$ for set multiplication and addition. We will use the following properties of $\inf <!--\; FUNCTION\; APPLICATION\; -->$ and $\sup <!--\; FUNCTION\; APPLICATION\; -->$

• $\inf <!--\; FUNCTION\; APPLICATION\; -->\mathit{cA}=c\inf <!--\; FUNCTION\; APPLICATION\; -->A$ and $\sup <!--\; FUNCTION\; APPLICATION\; -->\mathit{cA}=c\sup <!--\; FUNCTION\; APPLICATION\; -->A$
• $\inf <!--\; FUNCTION\; APPLICATION\; -->A+B=\inf <!--\; FUNCTION\; APPLICATION\; -->A+\inf <!--\; FUNCTION\; APPLICATION\; -->B$ and $\sup <!--\; FUNCTION\; APPLICATION\; -->A+B=\sup <!--\; FUNCTION\; APPLICATION\; -->A+\sup <!--\; FUNCTION\; APPLICATION\; -->B$
• $A\subseteq B⟹\sup <!--\; FUNCTION\; APPLICATION\; -->A\le \sup <!--\; FUNCTION\; APPLICATION\; -->B$ and $\inf <!--\; FUNCTION\; APPLICATION\; -->A\ge \inf <!--\; FUNCTION\; APPLICATION\; -->B$

(i)

By definition of Darboux, we need to demonstrate that the inequality $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{\mathit{cf}}:\begin{array}{c}\underline{\mathit{cf}}\text{ is p.c.}\hfill \\ \underline{\mathit{cf}}\le \mathit{cf}\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{\mathit{cf}}:\begin{array}{c}\overline{\mathit{cf}}\text{ is p.c.}\hfill \\ \overline{\mathit{cf}}\ge \mathit{cf}\hfill \end{array}\right\}$$

can be turned into an equality. Consider two cases:

• Notice that if $\underline{f}$ is a piecewise function minorizing $f$, then $c\cdot \underline{f}$ is a piecewise function minorizing $\mathit{cf}$. In other words, $c\underline{F}\subseteq \underline{\mathit{cF}}$. Since supremums are monotone, from $\sup <!--\; FUNCTION\; APPLICATION\; -->\mathit{cA}=c\sup <!--\; FUNCTION\; APPLICATION\; -->A$ follows that

  $$c{\int \; <!--\; nolimits\; -->}_{[a,b]}f\le c\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{\mathit{cf}}:\begin{array}{c}\underline{\mathit{cf}}\text{ is p.c.}\hfill \\ \underline{\mathit{cf}}\le \mathit{cf}\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{\mathit{cf}}:\begin{array}{c}\overline{\mathit{cf}}\text{ is p.c.}\hfill \\ \overline{\mathit{cf}}\ge \mathit{cf}\hfill \end{array}\right\}$$

  Similarly, from $c\overline{f}\ge \mathit{cf}$ follows that $c\overline{F}\subseteq \overline{\mathit{cF}}$; thus, $c\inf <!--\; FUNCTION\; APPLICATION\; -->\{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->\overline{f}\}\ge \inf <!--\; FUNCTION\; APPLICATION\; -->\{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->\overline{\mathit{cf}}\}$, i.e.,

  $$c{\int \; <!--\; nolimits\; -->}_{[a,b]}f\le c\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{\mathit{cf}}:\begin{array}{c}\underline{\mathit{cf}}\text{ is p.c.}\hfill \\ \underline{\mathit{cf}}\le \mathit{cf}\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{\mathit{cf}}:\begin{array}{c}\overline{\mathit{cf}}\text{ is p.c.}\hfill \\ \overline{\mathit{cf}}\ge \mathit{cf}\hfill \end{array}\right\}\le c{\int \; <!--\; nolimits\; -->}_{[a,b]}f$$

  Thus, we have squeezed both terms between the integral of $f$ multiplied by $c$, as desired.

• This follows by a symmetric argument (one needs to flip the right- and the left-hand sides of the above sandwich equation).

Now we prove the additivity property. We need to demonstrate that the following inequality can be turned into an equality.

$$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f+g}:\begin{array}{c}\underline{f+g}\text{ is p.c.}\hfill \\ \underline{f+g}\le f+g\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{f+g}:\begin{array}{c}\overline{f+g}\text{ is p.c.}\hfill \\ \overline{f+g}\ge f+g\hfill \end{array}\right\}$$

We use the sandwhich strategy again. Notice that any piecewise constant functions $\underline{f},\underline{g}$ that minorize $f,g$ also minorize their sum $\underline{f}+\underline{g}\le f+g$. Thus, we have $\underline{F}+\underline{G}\subseteq \underline{F+G}$, from which we conclude by properties of supremum that

$${\int \; <!--\; nolimits\; -->}_{[a,b]}f+{\int \; <!--\; nolimits\; -->}_{[a,b]}g\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f+g}:\begin{array}{c}\underline{f+g}\text{ is p.c.}\hfill \\ \underline{f+g}\le f+g\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{f+g}:\begin{array}{c}\overline{f+g}\text{ is p.c.}\hfill \\ \overline{f+g}\ge f+g\hfill \end{array}\right\}$$

Similar line of thought gives us

$$\begin{array}{llll}\hfill {\int \; <!--\; nolimits\; -->}_{[a,b]}f+{\int \; <!--\; nolimits\; -->}_{[a,b]}g\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f+g}:\begin{array}{c}\underline{f+g}\text{ is p.c.}\hfill \\ \underline{f+g}\le f+g\hfill \end{array}\right\}& \le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{f+g}:\begin{array}{c}\overline{f+g}\text{ is p.c.}\hfill \\ \overline{f+g}\ge f+g\hfill \end{array}\right\}& \hfill & \\ \hfill & \le {\int \; <!--\; nolimits\; -->}_{[a,b]}f+{\int \; <!--\; nolimits\; -->}_{[a,b]}g& \hfill & \end{array}$$

as desired.

(ii)

Let $\overline{G}$ be the set of all piecewise constant functions $\overline{g}$ that majorize $g$, and let $\underline{F}$ be the set of all piecewise constant functions $\underline{f}$ that minorize $f$ (to be very pedantic, we have used the power set and the specification axiom). By monotonicity of piecewise constant integral [ Exercise 1.1.21] we have $$\forall <!--\; FUNCTION\; APPLICATION\; -->\underline{f}\in \underline{F},\forall <!--\; FUNCTION\; APPLICATION\; -->\overline{g}\in \overline{G}:\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f}\le \text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{g}$$

In other words, any piecewise constant integral of $f$ is a lower bound for $\text{p.c.}\int \overline{g}$. Since the infimum is “the” lower bound, we can simply take infimum with respect to $\overline{g}$, obtaining

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\underline{f}\in \underline{F}:\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{g}:\overline{g}\text{ p.c. },g\le \overline{g}\right\}=\int \; <!--\; nolimits\; -->g$$

Similarly, we can take supremums with respect to $\underline{f}$, obtaining

$$\int \; <!--\; nolimits\; -->f=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\underline{f}:\underline{f}\text{ p.c. },\underline{f}\le f\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}\overline{g}:\overline{g}\text{ p.c. },g\le \overline{g}\right\}=\int \; <!--\; nolimits\; -->g$$

Thus, arriving at

$${\int \; <!--\; nolimits\; -->}_{[a,b]}f\le {\int \; <!--\; nolimits\; -->}_{[a,b]}g$$

as desired.

(iii)

Obviously, the indicator function is piecewice constant with respect to the partition $P=\{E,[a,b]∕E\}$ with $c_E=1$ and $c_{[a,b]∕E}=0$. By the properties of piecewise constant integral we have $${\int \; <!--\; nolimits\; -->}_{[a,b]}{1}_E=c_E\cdot m(E)+c_{[a,b]∕E}\cdot m([a,b]∕E)=1\cdot m(E)+0\cdot m([a,b]∕E)=m(E).$$

(iv)

Now suppose that aside our usual $\mathit{int}(f)={\int }_{[a,b]}f$ there exists another functional $\mathit{in}{t}^{\prime }(f)=c\in ℝ$ obeying the above properties. Notice that both must necessarily agree on piecewise constant functions: Let $f$ be any piecewise constant function with respect to the partition $I_1,\dots ,I_n$ and values $c_1,\dots ,c_n$. Then $f=1_{I_1}\cdot c_1+\dots +1_{I_n}\cdot c_n$. By property a) and c) we have $$\begin{array}{llll}\hfill \mathit{in}{t}^{\prime }(f)& =\mathit{in}{t}^{\prime }(1_{I_1}\cdot c_1+\dots +1_{I_n}\cdot c_n)& \hfill & \\ \hfill & \stackrel{a)}{\text{=}}{c}_1\mathit{in}{t}^{\prime }(1_{I_1})+\dots +c_n\mathit{in}{t}^{\prime }(1_{I_n})& \hfill & \\ \hfill & \stackrel{b)}{\text{=}}{c}_{1}m(I_1)+\dots +c_{n}m(I_n)& \hfill & \\ \hfill & \stackrel{b)}{\text{=}}{c}_1{\int \; <!--\; nolimits\; -->}_{[a,b]}{1}_{I_1}+\dots +c_n{\int \; <!--\; nolimits\; -->}_{[a,b]}{1}_{I_n}& \hfill & \\ \hfill & \stackrel{a)}{\text{=}}{\int \; <!--\; nolimits\; -->}_{[a,b]}{1}_{I_1}\cdot c_1+\dots +1_{I_n}\cdot c_n={\int \; <!--\; nolimits\; -->}_{[a,b]}f& \hfill & \\ \hfill & & \hfill \end{array}$$

But if int and int’ agree on all piecewise constant integrals, then they also have to agree on Riemann integrals, since we are taking supremums and infimums of the same sets.