## Section5.1The Riemann integral

Note: 1.5 lectures

An integral is a way to “sum” the values of a function. There is often confusion among students of calculus between integral and antiderivative. The integral is (informally) the area under the curve, nothing else. That we can compute an antiderivative using the integral is a nontrivial result we have to prove. In this chapter we define the Riemann integral 1  using the Darboux integral 3 , which is technically simpler than (but equivalent to) the traditional definition of Riemann.

### Subsection5.1.1Partitions and lower and upper integrals

We want to integrate a bounded function defined on an interval $$[a,b]\text{.}$$ We first define two auxiliary integrals that are defined for all bounded functions. Only then can we talk about the Riemann integral and the Riemann integrable functions.

#### Definition5.1.1.

A partition $$P$$ of the interval $$[a,b]$$ is a finite set of numbers $$\{ x_0,x_1,x_2,\ldots,x_n \}$$ such that

\begin{equation*} a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b . \end{equation*}

We write

\begin{equation*} \Delta x_i := x_i - x_{i-1} . \end{equation*}

Let $$f \colon [a,b] \to \R$$ be a bounded function. Let $$P$$ be a partition of $$[a,b]\text{.}$$ Define

\begin{equation*} \begin{aligned} & m_i := \inf \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} , \\ & M_i := \sup \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} , \\ & L(P,f) := \sum_{i=1}^n m_i \Delta x_i , \\ & U(P,f) := \sum_{i=1}^n M_i \Delta x_i . \end{aligned} \end{equation*}

We call $$L(P,f)$$ the lower Darboux sum and $$U(P,f)$$ the upper Darboux sum.

The geometric idea of Darboux sums is indicated in Figure 5.1. The lower sum is the area of the shaded rectangles, and the upper sum is the area of the entire rectangles, shaded plus unshaded parts. The width of the $$i$$th rectangle is $$\Delta x_i\text{,}$$ the height of the shaded rectangle is $$m_i\text{,}$$ and the height of the entire rectangle is $$M_i\text{.}$$

#### Proof.

Let $$P$$ be a partition. Note that $$m \leq m_i$$ for all $$i$$ and $$M_i \leq M$$ for all $$i\text{.}$$ Also $$m_i \leq M_i$$ for all $$i\text{.}$$ Finally, $$\sum_{i=1}^n \Delta x_i = (b-a)\text{.}$$ Therefore,

\begin{multline*} m(b-a) = m \left( \sum_{i=1}^n \Delta x_i \right) = \sum_{i=1}^n m \Delta x_i \leq \sum_{i=1}^n m_i \Delta x_i \leq \\ \leq \sum_{i=1}^n M_i \Delta x_i \leq \sum_{i=1}^n M \Delta x_i = M \left( \sum_{i=1}^n \Delta x_i \right) = M(b-a) . \end{multline*}

Hence we get (5.1). In particular, the set of lower and upper sums are bounded sets.

#### Definition5.1.3.

As the sets of lower and upper Darboux sums are bounded, we define

\begin{equation*} \begin{aligned} & \underline{\int_a^b} f(x)\,dx := \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\} , \\ & \overline{\int_a^b} f(x)\,dx := \inf \, \bigl\{ U(P,f) : P \text{ a partition of } [a,b] \bigr\} . \end{aligned} \end{equation*}

We call $$\underline{\int}$$ the lower Darboux integral and $$\overline{\int}$$ the upper Darboux integral. To avoid worrying about the variable of integration, we often simply write

\begin{equation*} \underline{\int_a^b} f := \underline{\int_a^b} f(x)\,dx \qquad \text{and} \qquad \overline{\int_a^b} f := \overline{\int_a^b} f(x)\,dx . \end{equation*}

If integration is to make sense, then the lower and upper Darboux integrals should be the same number, as we want a single number to call the integral. However, these two integrals may differ for some functions.

#### Example5.1.4.

Take the Dirichlet function $$f \colon [0,1] \to \R\text{,}$$ where $$f(x) := 1$$ if $$x \in \Q$$ and $$f(x) := 0$$ if $$x \notin \Q\text{.}$$ Then

\begin{equation*} \underline{\int_0^1} f = 0 \qquad \text{and} \qquad \overline{\int_0^1} f = 1 . \end{equation*}

The reason is that for every $$i\text{,}$$ we have $$m_i = \inf \{ f(x) : x \in [x_{i-1},x_i] \} = 0$$ and $$M_i = \sup \{ f(x) : x \in [x_{i-1},x_i] \} = 1\text{.}$$ Thus

\begin{equation*} \begin{aligned} & L(P,f) = \sum_{i=1}^n 0 \cdot \Delta x_i = 0 , \\ & U(P,f) = \sum_{i=1}^n 1 \cdot \Delta x_i = \sum_{i=1}^n \Delta x_i = 1 . \end{aligned} \end{equation*}

#### Remark5.1.5.

The same definition of $$\underline{\int_a^b} f$$ and $$\overline{\int_a^b} f$$ is used when $$f$$ is defined on a larger set $$S$$ such that $$[a,b] \subset S\text{.}$$ In that case, we use the restriction of $$f$$ to $$[a,b]$$ and we must ensure that the restriction is bounded on $$[a,b]\text{.}$$

To compute the integral, we often take a partition $$P$$ and make it finer. That is, we cut intervals in the partition into yet smaller pieces.

#### Definition5.1.6.

Let $$P = \{ x_0, x_1, \ldots, x_n \}$$ and $$\widetilde{P} = \{ \widetilde{x}_0, \widetilde{x}_1, \ldots, \widetilde{x}_{\ell} \}$$ be partitions of $$[a,b]\text{.}$$ We say $$\widetilde{P}$$ is a refinement of $$P$$ if as sets $$P \subset \widetilde{P}\text{.}$$

That is, $$\widetilde{P}$$ is a refinement of a partition if it contains all the numbers in $$P$$ and perhaps some other numbers in between. For example, $$\{ 0, 0.5, 1, 2 \}$$ is a partition of $$[0,2]$$ and $$\{ 0, 0.2, 0.5, 1, 1.5, 1.75, 2 \}$$ is a refinement. The main reason for introducing refinements is the following proposition.

#### Proof.

The tricky part of this proof is to get the notation correct. Let $$\widetilde{P} = \{ \widetilde{x}_0, \widetilde{x}_1, \ldots, \widetilde{x}_{\ell} \}$$ be a refinement of $$P = \{ x_0, x_1, \ldots, x_n \}\text{.}$$ Then $$x_0 = \widetilde{x}_0$$ and $$x_n = \widetilde{x}_{\ell}\text{.}$$ In fact, there are integers $$k_0 < k_1 < \cdots < k_n$$ such that $$x_j = \widetilde{x}_{k_j}$$ for $$j=0,1,2,\ldots,n\text{.}$$

Let $$\Delta \widetilde{x}_p := \widetilde{x}_p - \widetilde{x}_{p-1}\text{.}$$ See Figure 5.2. We get

\begin{equation*} \Delta x_j = x_j - x_{j-1} = \widetilde{x}_{k_j} - \widetilde{x}_{k_{j-1}} = \sum_{p=k_{j-1}+1}^{k_j} \widetilde{x}_{p} - \widetilde{x}_{p-1} = \sum_{p=k_{j-1}+1}^{k_j} \Delta \widetilde{x}_p . \end{equation*} Figure 5.2. Refinement of a subinterval. Notice $$\Delta x_j = \Delta \widetilde{x}_{p-2} + \Delta \widetilde{x}_{p-1} + \Delta \widetilde{x}_{p}\text{,}$$ and also $$k_{j-1}+1 = p-2$$ and $$k_{j} = p\text{.}$$

Let $$m_j$$ be as before and correspond to the partition $$P\text{.}$$ Let $$\widetilde{m}_j := \inf \{ f(x) : \widetilde{x}_{j-1} \leq x \leq \widetilde{x}_j \}\text{.}$$ Now, $$m_j \leq \widetilde{m}_p$$ for $$k_{j-1} < p \leq k_j\text{.}$$ Therefore,

\begin{equation*} m_j \Delta x_j = m_j \sum_{p=k_{j-1}+1}^{k_j} \Delta \widetilde{x}_p = \sum_{p=k_{j-1}+1}^{k_j} m_j \Delta \widetilde{x}_p \leq \sum_{p=k_{j-1}+1}^{k_j} \widetilde{m}_p \Delta \widetilde{x}_p . \end{equation*}

So

\begin{equation*} L(P,f) = \sum_{j=1}^n m_j \Delta x_j \leq \sum_{j=1}^n \, \sum_{p=k_{j-1}+1}^{k_j} \widetilde{m}_p \Delta \widetilde{x}_p = \sum_{j=1}^{\ell} \widetilde{m}_j \Delta \widetilde{x}_j = L(\widetilde{P},f). \end{equation*}

The proof of $$U(\widetilde{P},f) \leq U(P,f)$$ is left as an exercise.

Armed with refinements we prove the following. The key point of this next proposition is that the lower Darboux integral is less than or equal to the upper Darboux integral.

#### Proof.

By Proposition 5.1.2, for every partition $$P\text{,}$$

\begin{equation*} m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a). \end{equation*}

The inequality $$m(b-a) \leq L(P,f)$$ implies $$m(b-a) \leq \underline{\int_a^b} f\text{.}$$ The inequality $$U(P,f) \leq M(b-a)$$ implies $$\overline{\int_a^b} f \leq M(b-a)\text{.}$$

The middle inequality in (5.2) is the main point of this proposition. Let $$P_1, P_2$$ be partitions of $$[a,b]\text{.}$$ Define $$\widetilde{P} := P_1 \cup P_2\text{.}$$ The set $$\widetilde{P}$$ is a partition of $$[a,b]\text{,}$$ which is a refinement of $$P_1$$ and a refinement of $$P_2\text{.}$$ By Proposition 5.1.7, $$L(P_1,f) \leq L(\widetilde{P},f)$$ and $$U(\widetilde{P},f) \leq U(P_2,f)\text{.}$$ So

\begin{equation*} L(P_1,f) \leq L(\widetilde{P},f) \leq U(\widetilde{P},f) \leq U(P_2,f) . \end{equation*}

In other words, for two arbitrary partitions $$P_1$$ and $$P_2\text{,}$$ we have $$L(P_1,f) \leq U(P_2,f)\text{.}$$ Recall Proposition 1.2.7, and take the supremum and infimum over all partitions:

\begin{equation*} \underline{\int_a^b} f = \sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\} \leq \inf \, \bigl\{ U(P,f) : P \text{ a partition of } [a,b] \bigr\} = \overline{\int_a^b} f . \qedhere \end{equation*}

### Subsection5.1.2Riemann integral

We can finally define the Riemann integral. However, the Riemann integral is only defined on a certain class of functions, called the Riemann integrable functions.

#### Definition5.1.9.

Let $$f \colon [a,b] \to \R$$ be a bounded function such that

\begin{equation*} \underline{\int_a^b} f(x)\,dx = \overline{\int_a^b} f(x)\,dx . \end{equation*}

Then $$f$$ is said to be Riemann integrable. The set of Riemann integrable functions on $$[a,b]$$ is denoted by $$\sR[a,b]\text{.}$$ When $$f \in \sR[a,b]\text{,}$$ we define

\begin{equation*} \int_a^b f(x)\,dx := \underline{\int_a^b} f(x)\,dx = \overline{\int_a^b} f(x)\,dx . \end{equation*}

As before, we often write

\begin{equation*} \int_a^b f := \int_a^b f(x)\,dx. \end{equation*}

The number $$\int_a^b f$$ is called the Riemann integral of $$f\text{,}$$ or sometimes simply the integral of $$f\text{.}$$

By definition, a Riemann integrable function is bounded. Appealing to Proposition 5.1.8, we immediately obtain the following proposition. See also Figure 5.3. Figure 5.3. The area under the curve is bounded from above by the area of the entire rectangle, $$M(b-a)\text{,}$$ and from below by the area of the shaded part, $$m(b-a)\text{.}$$

Often we use a weaker form of this proposition. That is, if $$\abs{f(x)} \leq M$$ for all $$x \in [a,b]\text{,}$$ then

\begin{equation*} \abs{\int_a^b f} \leq M(b-a) . \end{equation*}

#### Example5.1.11.

We integrate constant functions using Proposition 5.1.8. If $$f(x) := c$$ for some constant $$c\text{,}$$ then we take $$m = M = c\text{.}$$ In inequality (5.2) all the inequalities must be equalities. Thus $$f$$ is integrable on $$[a,b]$$ and $$\int_a^b f = c(b-a)\text{.}$$

#### Example5.1.12.

Let $$f \colon [0,2] \to \R$$ be defined by

\begin{equation*} f(x) := \begin{cases} 1 & \text{if } x < 1,\\ \nicefrac{1}{2} & \text{if } x = 1,\\ 0 & \text{if } x > 1. \end{cases} \end{equation*}

We claim $$f$$ is Riemann integrable and $$\int_0^2 f = 1\text{.}$$

Proof: Let $$0 < \epsilon < 1$$ be arbitrary. Let $$P := \{0, 1-\epsilon, 1+\epsilon, 2\}$$ be a partition. We use the notation from the definition of the Darboux sums. Then

\begin{equation*} \begin{aligned} m_1 &= \inf \, \bigl\{ f(x) : x \in [0,1-\epsilon] \bigr\} = 1 , & M_1 &= \sup \, \bigl\{ f(x) : x \in [0,1-\epsilon] \bigr\} = 1 , \\ m_2 &= \inf \, \bigl\{ f(x) : x \in [1-\epsilon,1+\epsilon] \bigr\} = 0 , & M_2 &= \sup \, \bigl\{ f(x) : x \in [1-\epsilon,1+\epsilon] \bigr\} = 1 , \\ m_3 &= \inf \, \bigl\{ f(x) : x \in [1+\epsilon,2] \bigr\} = 0 , & M_3 &= \sup \, \bigl\{ f(x) : x \in [1+\epsilon,2] \bigr\} = 0 . \end{aligned} \end{equation*}

Furthermore, $$\Delta x_1 = 1-\epsilon\text{,}$$ $$\Delta x_2 = 2\epsilon$$ and $$\Delta x_3 = 1-\epsilon\text{.}$$ See Figure 5.4. Figure 5.4. Darboux sums for the step function. $$L(P,f)$$ is the area of the shaded rectangle, $$U(P,f)$$ is the area of both rectangles, and $$U(P,f)-L(P,f)$$ is the area of the unshaded rectangle.

We compute

\begin{equation*} \begin{aligned} & L(P,f) = \sum_{i=1}^3 m_i \Delta x_i = 1 \cdot (1-\epsilon) + 0 \cdot 2\epsilon + 0 \cdot (1-\epsilon) = 1-\epsilon , \\ & U(P,f) = \sum_{i=1}^3 M_i \Delta x_i = 1 \cdot (1-\epsilon) + 1 \cdot 2\epsilon + 0 \cdot (1-\epsilon) = 1+\epsilon . \end{aligned} \end{equation*}

Thus,

\begin{equation*} \overline{\int_0^2} f - \underline{\int_0^2} f \leq U(P,f) - L(P,f) = (1+\epsilon) - (1-\epsilon) = 2 \epsilon . \end{equation*}

By Proposition 5.1.8, we have $$\underline{\int_0^2} f \leq \overline{\int_0^2} f\text{.}$$ As $$\epsilon$$ was arbitrary, $$\overline{\int_0^2} f = \underline{\int_0^2} f\text{.}$$ So $$f$$ is Riemann integrable. Finally,

\begin{equation*} 1-\epsilon = L(P,f) \leq \int_0^2 f \leq U(P,f) = 1+\epsilon. \end{equation*}

Hence, $$\bigl\lvert \int_0^2 f - 1 \bigr\rvert \leq \epsilon\text{.}$$ As $$\epsilon$$ was arbitrary, we conclude $$\int_0^2 f = 1\text{.}$$

It may be worthwhile to extract part of the technique of the example into a proposition.

#### Proof.

If for every $$\epsilon > 0$$ such a $$P$$ exists, then

\begin{equation*} 0 \leq \overline{\int_a^b} f - \underline{\int_a^b} f \leq U(P,f) - L(P,f) < \epsilon . \end{equation*}

Therefore, $$\overline{\int_a^b} f = \underline{\int_a^b} f\text{,}$$ and $$f$$ is integrable.

#### Example5.1.14.

Let us show $$\frac{1}{1+x}$$ is integrable on $$[0,b]$$ for all $$b > 0\text{.}$$ We will see later that continuous functions are integrable, but let us demonstrate how we do it directly.

Let $$\epsilon > 0$$ be given. Take $$n \in \N$$ and pick $$x_j := \nicefrac{jb}{n}\text{,}$$ to form the partition $$P := \{ x_0,x_1,\ldots,x_n \}$$ of $$[0,b]\text{.}$$ We have $$\Delta x_j = \nicefrac{b}{n}$$ for all $$j\text{.}$$ As $$f$$ is decreasing, for every subinterval $$[x_{j-1},x_j]\text{,}$$ we obtain

\begin{equation*} m_j = \inf \left\{ \frac{1}{1+x} : x \in [x_{j-1},x_j] \right\} = \frac{1}{1+x_j} , \qquad M_j = \sup \left\{ \frac{1}{1+x} : x \in [x_{j-1},x_j] \right\} = \frac{1}{1+x_{j-1}} . \end{equation*}

Then

\begin{multline*} U(P,f)-L(P,f) = \sum_{j=1}^n \Delta x_j (M_j-m_j) = \frac{b}{n} \sum_{j=1}^n \left( \frac{1}{1+\nicefrac{(j-1)b}{n}} - \frac{1}{1+\nicefrac{jb}{n}} \right) = \\ = \frac{b}{n} \left( \frac{1}{1+\nicefrac{0b}{n}} - \frac{1}{1+\nicefrac{nb}{n}} \right) = \frac{b^2}{n(b+1)} . \end{multline*}

The sum telescopes, the terms successively cancel each other, something we have seen before. Picking $$n$$ to be such that $$\frac{b^2}{n(b+1)} < \epsilon\text{,}$$ the proposition is satisfied, and the function is integrable.

#### Remark5.1.15.

A way of thinking of the integral is that it adds up (integrates) lots of local information—it sums $$f(x)\,dx$$ over all $$x\text{.}$$ The integral sign was chosen by Leibniz to be the long S to mean summation. Unlike derivatives, which are “local,” integrals show up in applications when one wants a “global” answer: total distance travelled, average temperature, total charge, etc.

### Subsection5.1.3More notation

When $$f \colon S \to \R$$ is defined on a larger set $$S$$ and $$[a,b] \subset S\text{,}$$ we say $$f$$ is Riemann integrable on $$[a,b]$$ if the restriction of $$f$$ to $$[a,b]$$ is Riemann integrable. In this case, we say $$f \in \sR[a,b]\text{,}$$ and we write $$\int_a^b f$$ to mean the Riemann integral of the restriction of $$f$$ to $$[a,b]\text{.}$$

It is useful to define the integral $$\int_a^b f$$ even if $$a \not< b\text{.}$$ Suppose $$b < a$$ and $$f \in \sR[b,a]\text{,}$$ then define

\begin{equation*} \int_a^b f := - \int_b^a f . \end{equation*}

For any function $$f\text{,}$$ define

\begin{equation*} \int_a^a f := 0 . \end{equation*}

At times, the variable $$x$$ may already have some other meaning. When we need to write down the variable of integration, we may simply use a different letter. For example,

\begin{equation*} \int_a^b f(s)\,ds := \int_a^b f(x)\,dx . \end{equation*}

### Subsection5.1.4Exercises

#### Exercise5.1.1.

Define $$f \colon [0,1] \to \R$$ by $$f(x) := x^3$$ and let $$P := \{ 0, 0.1, 0.4, 1 \}\text{.}$$ Compute $$L(P,f)$$ and $$U(P,f)\text{.}$$

#### Exercise5.1.2.

Let $$f \colon [0,1] \to \R$$ be defined by $$f(x) := x\text{.}$$ Show that $$f \in \sR[0,1]$$ and compute $$\int_0^1 f$$ using the definition of the integral (but feel free to use the propositions of this section).

#### Exercise5.1.3.

Let $$f \colon [a,b] \to \R$$ be a bounded function. Suppose there exists a sequence of partitions $$\{ P_k \}$$ of $$[a,b]$$ such that

\begin{equation*} \lim_{k \to \infty} \bigl( U(P_k,f) - L(P_k,f) \bigr) = 0 . \end{equation*}

Show that $$f$$ is Riemann integrable and that

\begin{equation*} \int_a^b f = \lim_{k \to \infty} U(P_k,f) = \lim_{k \to \infty} L(P_k,f) . \end{equation*}

#### Exercise5.1.5.

Suppose $$f \colon [-1,1] \to \R$$ is defined as

\begin{equation*} f(x) := \begin{cases} 1 & \text{if } x > 0, \\ 0 & \text{if } x \leq 0. \end{cases} \end{equation*}

Prove that $$f \in \sR[-1,1]$$ and compute $$\int_{-1}^1 f$$ using the definition of the integral (but feel free to use the propositions of this section).

#### Exercise5.1.6.

Let $$c \in (a,b)$$ and let $$d \in \R\text{.}$$ Define $$f \colon [a,b] \to \R$$ as

\begin{equation*} f(x) := \begin{cases} d & \text{if } x = c, \\ 0 & \text{if } x \not= c. \end{cases} \end{equation*}

Prove that $$f \in \sR[a,b]$$ and compute $$\int_a^b f$$ using the definition of the integral (but feel free to use the propositions of this section).

#### Exercise5.1.7.

Suppose $$f \colon [a,b] \to \R$$ is Riemann integrable. Let $$\epsilon > 0$$ be given. Then show that there exists a partition $$P = \{ x_0, x_1, \ldots, x_n \}$$ such that for every set of numbers $$\{ c_1, c_2, \ldots, c_n \}$$ with $$c_k \in [x_{k-1},x_k]$$ for all $$k\text{,}$$ we have

\begin{equation*} \abs{\int_a^b f - \sum_{k=1}^n f(c_k) \Delta x_k} < \epsilon . \end{equation*}

#### Exercise5.1.8.

Let $$f \colon [a,b] \to \R$$ be a Riemann integrable function. Let $$\alpha > 0$$ and $$\beta \in \R\text{.}$$ Then define $$g(x) := f(\alpha x + \beta)$$ on the interval $$I = [\frac{a-\beta}{\alpha}, \frac{b-\beta}{\alpha}]\text{.}$$ Show that $$g$$ is Riemann integrable on $$I\text{.}$$

#### Exercise5.1.9.

Suppose $$f \colon [0,1] \to \R$$ and $$g \colon [0,1] \to \R$$ are such that for all $$x \in (0,1]\text{,}$$ we have $$f(x) = g(x)\text{.}$$ Suppose $$f$$ is Riemann integrable. Prove $$g$$ is Riemann integrable and $$\int_{0}^1 f = \int_{0}^1 g\text{.}$$

#### Exercise5.1.10.

Let $$f \colon [0,1] \to \R$$ be a bounded function. Let $$P_n = \{ x_0,x_1,\ldots,x_n \}$$ be a uniform partition of $$[0,1]\text{,}$$ that is, $$x_j = \nicefrac{j}{n}\text{.}$$ Is $$\{ L(P_n,f) \}_{n=1}^\infty$$ always monotone? Yes/No: Prove or find a counterexample.

#### Exercise5.1.11.

(Challenging)   For a bounded function $$f \colon [0,1] \to \R\text{,}$$ let $$R_n := (\nicefrac{1}{n})\sum_{j=1}^n f(\nicefrac{j}{n})$$ (the uniform right-hand rule).

1. If $$f$$ is Riemann integrable show $$\int_0^1 f = \lim \, R_n\text{.}$$

2. Find an $$f$$ that is not Riemann integrable, but $$\lim \, R_n$$ exists.

#### Exercise5.1.12.

(Challenging)   Generalize the previous exercise. Show that $$f \in \sR[a,b]$$ if and only if there exists an $$I \in \R\text{,}$$ such that for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that if $$P$$ is a partition with $$\Delta x_i < \delta$$ for all $$i\text{,}$$ then $$\abs{L(P,f) - I} < \epsilon$$ and $$\abs{U(P,f) - I} < \epsilon\text{.}$$ If $$f \in \sR[a,b]\text{,}$$ then $$I = \int_a^b f\text{.}$$

#### Exercise5.1.13.

Using Exercise 5.1.12 and the idea of the proof in Exercise 5.1.7, show that Darboux integral is the same as the standard definition of Riemann integral, which you have most likely seen in calculus. That is, show that $$f \in \sR[a,b]$$ if and only if there exists an $$I \in \R\text{,}$$ such that for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that if $$P = \{ x_0,x_1,\ldots,x_n \}$$ is a partition with $$\Delta x_i < \delta$$ for all $$i\text{,}$$ then $$\abs{\sum_{i=1}^n f(c_i) \Delta x_i - I} < \epsilon$$ for every set $$\{ c_1,c_2,\ldots,c_n \}$$ with $$c_i \in [x_{i-1},x_i]\text{.}$$ If $$f \in \sR[a,b]\text{,}$$ then $$I = \int_a^b f\text{.}$$

#### Exercise5.1.14.

(Challenging)   Construct functions $$f$$ and $$g\text{,}$$ where $$f \colon [0,1] \to \R$$ is Riemann integrable, $$g \colon [0,1] \to [0,1]$$ is one-to-one and onto, and such that the composition $$f \circ g$$ is not Riemann integrable.

Named after the German mathematician Georg Friedrich Bernhard Riemann 2  (1826–1866).
https://en.wikipedia.org/wiki/Riemann
Named after the French mathematician Jean-Gaston Darboux 4  (1842–1917).
https://en.wikipedia.org/wiki/Darboux
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