Exercise 1.1.22 (Equivalence of Riemann and Darboux integral)

We start with the easy one.

$⟸$

Suppose that $f$ is Darboux integrable with the Darboux integral $D\in ℝ$, i.e., for an arbitrary $𝜖>0$ we have an upper piecewise integral with $p.c.{\int }_{[a,b]}\overline{g}-D\le 𝜖∕2$ and a lower piecewise integral with $D-p.c.{\int }_{[a,b]}\underline{g}\le 𝜖∕2$. Since $\overline{g},\underline{g}$ are piecewise constant, we can find a common partition $\mathit{P\#}{P}^{\prime }$ of $[a,b]$ which we can associate with the points $a=x_0<x_1<\dots <x_n=b$ and the piecewise constant values ${\overline{c}}_1,\dots ,{\overline{c}}_n<+\infty$ and ${\underline{c}}_1,\dots ,{\underline{c}}_n>-\infty$ in an obvious manner. For each $a=x_0<x_1<\dots <x_n=b$ choose an arbitrary tag $x_{i-1}\le x_i^{\text{∗}}\le x_i$. We then have $$D-\frac{𝜖}{2}<\sum _{i=1}^n{\underline{c}}_i(x_{i-1}-x_i)\le \sum _{i=1}^{n}f(x_i^{\text{∗}})(x_{i-1}-x_i)\le \sum _{i=1}^n{\overline{c}}_i(x_{i-1}-x_i)<D+\frac{𝜖}{2}$$

In other words, we have found a tagged partition $P$ such that $R(f,P)$ is $𝜖$-close to $D$. Notice that if we choose any other tagged partition $P^{\prime }$ with the size of ${\delta }^{\prime }\le \delta =\inf <!--\; FUNCTION\; APPLICATION\; -->x_i-x_{i-1}$, then it is easy to verify (by taking common refinement again) that the inequality above still holds; thus the convergence along the filter is ensured.

$⟹$

Now suppose that $f$ is Riemann-integrable with the Riemann integral $R\in ℝ$. Fix $𝜖>0$ and choose a tagged partition $P$ such that $|R(f,P)|-R\le 𝜖∕2$. Notice that we can establish the bounds (which also happen to be upper and lower piecewise integrals for $f$):

$$\sum _{i=1}^n\inf <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\cdot \delta (x_i)\le \sum _{i=1}^{n}f(x_i^{\text{∗}})\delta (x_i)\le \sum _{i=1}^n\sup <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\cdot \delta (x_i)$$

where $x_i^{\text{∗}}$ can be chosen arbitrarily by definition. In particular, we can set ${\overline{x}}_i^{\text{∗}}$ such that $\sup <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)-f({\overline{x}}_i^{\text{∗}})\le \frac{𝜖}{2n\mathrm{\Delta }}$ and ${\underline{x}}_i^{\text{∗}}$ such that $f({\underline{x}}_i^{\text{∗}})-\inf <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\le \frac{𝜖}{2n\mathrm{\Delta }}$ where $\mathrm{\Delta }=\sup <!--\; FUNCTION\; APPLICATION\; -->x_i-x_{i-1}$. For both cases we have:

$$\begin{array}{llll}\hfill \sum _{i=1}^{n}f({\underline{x}}_i^{\text{∗}})\delta (x_i)-\sum _{i=1}^n\inf <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\cdot \delta (x_i)& =\sum _{i=1}^n\left[f({\underline{x}}_i^{\text{∗}})-\inf <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\right]\delta (x_i)& \hfill & \\ \hfill & \le \sum _{i=1}^n\frac{𝜖}{n\mathrm{\Delta }}\delta (x_i)\le \frac{𝜖}{2}& \hfill & \end{array}$$

and

$$\begin{array}{llll}\hfill \sum _{i=1}^n\sup <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)\cdot \delta (x_i)-\sum _{i=1}^{n}f({\overline{x}}_i^{\text{∗}})\delta (x_i)& =\sum _{i=1}^n\left[\sup <!--\; FUNCTION\; APPLICATION\; -->f[x_{i-1},x_i)-f({\overline{x}}_i^{\text{∗}})\right]\delta (x_i)& \hfill & \\ \hfill & \le \sum _{i=1}^n\frac{𝜖}{n\mathrm{\Delta }}\delta (x_i)\le \frac{𝜖}{2}& \hfill & \end{array}$$

To summarize, we have found a Riemann sum which is $𝜖∕2$-close to the Riemann integral $R$, and we have constructed upper and lower piecewise integrals for $f$ which are $𝜖∕2$ close to the Riemann sum. Thus, by triangle inequality, we have found upper and lower piecewise integrals which are $𝜖$-close to $R$. Since our choice of $𝜖$ was arbitrary, by properties of supremum and infimum the lower Darboux integral must be equal to upper Darboux integral, both being equal to $R$.

Let $E\subseteq {ℝ}^d$ be a box and $f:E\to ℝ$ be a bounded function. Show that $f$ is Riemann integrable if and only if it is Darboux integrable, in which case the Riemann integral and the Darboux integral are equal.

Fix an arbitrary $𝜖>0$. Let $f$ be Riemann integrable on $E$. Then there exists a $\delta >0$ such that

for all tagged partitions $P$ with $\mathrm{\Delta }P\le \delta$.

Let $P^{\text{−}}$ be a tagged partition with $\mathrm{\Delta }{P}^{\text{−}}\le \delta$, such that for every $B_i$ we have $y_i^{\text{−}}\in \{f(y)=\inf <!--\; FUNCTION\; APPLICATION\; -->\{f(y):y\in B_i\}\}$.

Then define $g:E\to ℝ$ as $g(x)=y_i^{\text{−}}$ for $x\in B_i$. Now for all $x\in B_i$ we have $g(x)\le f(x)$ and thus $g\le f$ pointwise and $g$ is piecewise constant.

Similarly, let $P^{\text{+}}$ be a tagged partition with $\mathrm{\Delta }{P}^{\text{+}}\le \delta$, such that for every $B_i$ we have $y_i^{\text{+}}\in \{f(y)=\sup <!--\; FUNCTION\; APPLICATION\; -->\{f(y):y\in B_i\}\}$. Define $h:E\to ℝ$ as $h(x)=y_i^{\text{+}}$ for $x\in B_i$. Now for all $x\in B_i$ we have $h(x)\ge f(x)$ and thus $h\ge f$ pointwise and $h$ is piecewise constant.

Then

so that $f$ is Darboux integrable.

Given a $\delta >0$, denote $G_{\delta }$ as the set of piecewise continuous functions on $E$ with respect to an arbitrary partition $P^{\text{−}}$ such that $\mathrm{\Delta }{P}^{\text{−}}\le \delta$ and with the property for $g\in G_{\delta }$ that

for any $B_i\in P^{\text{−}}$.

Let $g_{\delta }\in \{g\in G_{\delta }:{\int }_{E}g=\inf <!--\; FUNCTION\; APPLICATION\; -->\{{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_{E}g:g\in G_{\delta }\}\}$.

So we have $g_{\delta }\le f$ and that $g_{\delta }$ gives the lowest possible Darboux integral over $E$ for any piecewise continuous function with respect to a partition $P^{\text{−}}$ with $\mathrm{\Delta }{P}^{\text{−}}\le \delta$, but that still has the property that it can serve as a tagged partition for $R(P^{\text{−}},f)$, since $c_i=f(x)$ for some $x\in B_i$ (or at least is $𝜖$-close to such an $f(x)$).

We have to prove that

Let $g^{\prime }$ be piecewise constant on $E$ and be such that $\underset{̲}{{\int }_{E}f}-\text{pc}{\int }_{E}g\le 𝜖$. This is guaranteed by $f$ being Darboux integrable. Let $P^{\prime }$ be the partition alongside which $g$ is piecewise constant with $\mathrm{\Delta }{P}^{\prime }={\delta }^{\prime }$. Then choose $g_{{\delta }^{\prime }}$ like above. It is possible that $\underset{̲}{{\int }_{E}f}-{\int }_E{g}_{{\delta }^{\prime }}>𝜖$ by the difference in the partitions chosen. Now consider the refinement partition $P^{\prime }\#P^{″}$ with $\mathrm{\Delta }(P^{\prime }\#P^{″})={\delta }^{\text{∗}}$ and consider $g_{{\delta }^{\text{∗}}}$.

Assume to the contrary that $\underset{̲}{{\int }_{E}f}-\text{pc}{\int }_E{g}_{{\delta }^{\text{∗}}}>𝜖$. Then for some $B_i^{\text{∗}}\in P^{\prime }\#P^{″}$ for which $B_i^{\text{∗}}\subseteq B_j$ for $B_j^{\prime }\in P^{\prime }$, we must have $c_i^{\text{∗}}<c_j^{\prime }$, because otherwise by monotonicity we would get

But we also know $c_i^{\text{∗}}=f(x)$ for some $x\in B_i^{\text{∗}}$ and then $g^{\prime }(x)=c_j^{\prime }>f(x)$, which is a contradiction to $g\le f$. Therefore,

.

The proof for $h\ge f$ goes analogously, with similar notation.

Let $P$ be an arbitrary tagged partition with $\mathrm{\Delta }(P)\le {\delta }^{\text{∗}}$. Then for any box $B_i\in P$ we must have $y_i=f(x)$ for some $x\in B_i$, that is

for all $x\in B_i$, from which we can deduce

that is

in other words

for all tagged partitions $P$ with $\mathrm{\Delta }(P)\le {\delta }^{\text{∗}}$.