Chapter 07 The Riemann Integral

7.2 The Definition of the Riemann Integral

7.2.1.

Let $f$ be abounded function on $[a,b]$, and let $P$ be an arbitrary partition of $[a,b]$. First, explain why $U(f) ≥ L(f,P)$. Now, prove Lemma 7.2.6.

Proof: Given any partition $P'$,

$$U(f, P') \geq U(f, P \cup P') \geq L(f, P \cup P') \geq L(f, P)$$

That means $L(f,P)$ is a lower bound of $\{U(f,P) : P \in \mathbb{P}\}$.

So $U(f) \geq L(f,P)$.

This means $U(f)$ is also an upper bound of $\{L(f,P) : P \in \mathbb{P}\}$.

So $U(f) \geq L(f)$.

$\square$

7.2.2.

Consider $f(x) = 1/x$ over the interval $[1,4]$. Let $P$ be the partition consisting of the points $\{1,3/2,2,4\}$.

(a) Compute $L(f,P)$, $U(f,P)$, and $U(f,P)−L(f,P)$.

Solution:

$L(f, P) = \frac{3}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot 2 = \frac{3}{2}$

$$U(f, P) = 1 \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 2 = \frac{9}{4}$$

The difference is $\frac{3}{4}$.

7.2.3 (Sequential Criterion for Integrability).

(a) Prove that a bounded function $f$ is integrable on $[a,b]$ if and only if there exists a sequence of partitions $(P_n)^{\infty}_{n=1}$ satisfying

$$\lim_{n \to \infty} [U(f, P_n) - L(f, P_n)] = 0,$$

and in this case $\int_{a}^{b}f = \lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} L(f, P_n)$.

Proof: $\Rightarrow$ given $1/n$, we can find $P_n$, such that

$$0 < U(f, P_n) - L(f, P_n) < \frac{1}{n}$$

Then we have

$$\lim_{n \to \infty} [U(f, P_n) - L(f, P_n)] = 0.$$

$\Leftarrow$

$U(f) \leq U(f, P_n)$ and $L(F) \geq L(f, P_n)$, so $U(f) - L(f) \leq U(f, P_n) - L(f, P_n)$.

Since $\lim_{n \to \infty} [U(f, P_n) - L(f, P_n)] = 0$ then $U(f) - L(f) \leq 0$.

On the other hand, exercise 7.2.1 proves $U(f) - L(f) \geq 0$. So $U(f) = L(f)$. So $f$ is integrable.

$\square$

7.2.4.

Let $g$ be bounded on $[a,b]$ and assume there exists a partition $P$ with $L(g,P) = U(g,P)$. Describe g. Is it integrable? If so, what is the value of $\int_{a}^{b}g$?

Solution: In each segment of $P$, it's flat.

$\square$

7.2.5.

Assume that, for each $n$, $f_n$ is an integrable function on $[a,b]$.

If $(f_n) → f$ uniformly on $[a,b]$, prove that $f$ is also integrable on this set. (We will see that this conclusion does not necessarily follow if the convergence is pointwise.)

Proof: Given $\epsilon$, we can find $N$ such that $n > N$, $|f_n(x) - f(x)| < \frac{\epsilon}{2(b-a)}$.

Fix this $n$, we can find a partition such that $U(f_n, P) - L(f_n, P) < \epsilon / 2$

$$U(f_n, P) - L(f_n, P) = \sum_{k = 1}^{n} (M_{n, k} - m_{n, k}) \Delta x_{n, k} \geq \sum_{k = 1}^{n} (M)$$

7.2.6.

A tagged partition $(P,{c_k})$ is one where in addition to a partition $P$ we choose a sampling point $c_k$ in each of the subintervals $[x_{k−1},x_k]$. The corresponding Riemann sum,

$$R(f, P) = \sum_{k = 1}^{n} f(c_k) \Delta x_k,$$

is discussed in Section 7.1, where the following definition is alluded to.

Riemann’s Original Definition of the Integral: A bounded function f is integrable on $[a,b]$ with $\int_{a}^{b} f = A$, if for all $ϵ > 0$ there exists a $δ > 0$ such that for any tagged partition $(P,{c_k})$ satisfying $\Delta x_k < \delta$ for all $k$, it follows that

$$\left| R(f, P) - A \right| < \epsilon$$

Show that if $f$ satisfies Riemann’s definition above, then $f$ is integrable in the sense of Definition 7.2.7. (The full equivalence of these two characterizations of integrability is proved in Section 8.1.)

Proof:

$$M_k - m_k < f(c_{M, k}) + \frac{\epsilon}{4(b-a)} - (f(c_{m, k}) - \frac{\epsilon}{4(b-a)}) = f(c_{M, k}) - f(c_{m, k}) + \frac{\epsilon}{2(b-a)}$$

So

$$\sum_{k=1}^{n} (M_k - m_k) \Delta x_k \\ < \sum_{k=1}^{n} (f(c_{M, k}) - f(c_{m, k}) + \frac{\epsilon}{2(b-a)}) \Delta x_k \\ = R_1(f, P) - R_2(f, P) + \epsilon / 2 \\ < \epsilon / 2 + \epsilon / 2 = \epsilon.$$

$\square$

7.3 Integrating Functions with Discontinuities

7.3.1.

Consider the function

$$h(x) = \begin{cases} 1 &\text{for } 0 \leq x < 1\\ 2 &\text{for } x = 1\\ \end{cases}$$

over the interval $[0,1]$

(a) show that $L(f,P) =1$ for every partition $P$ of $[0,1]$.

Proof: Assume $P$ is $0 = x_0 < x_1 < x_2 < \cdots x_{n-1} < x_n = 1$. In the first $n-1$ segments, $h(x)$ is constant $1$, so the $m_k = 1$. In the nth segment, the $m_n$ is still $1$. So $L(f, P) = 1$.

(b) Construct a partition $P$ for which $U(f,P) < 1+1/10$.

Solution: Let $P = \{0, \frac{19}{20}, 1\}$,

$$U(f, P) = \frac{19}{20} + \frac{1}{10} < 1 + \frac{1}{10}$$ .

(c) Given $ϵ > 0$, construct a partition $P_ϵ$ for which $U(f,P_ϵ) < 1+ϵ$.

Proof: Let $P = \{0, 1 - \frac{\epsilon}{2}, 1\}$.

$$U(f, P) = 1 - \frac{\epsilon}{2} + \epsilon = 1 + \frac{\epsilon}{2}.$$

7.3.2.

Recall that Thomae’s function

$$t(x) = \begin{cases} 1 &\text{if } x = 0\\ 1/n &\text{if } x = m/n \\ 0 &\text{if } x \not\in \mathbf{Q} \\ \end{cases}$$

has a countable set of discontinuities occurring at precisely every rational number. Follow these steps to prove t(x) is integrable on $[0,1]$ with $\int_{0}^{1} t = 0$.

(a) First argue that $L(t,P) = 0$ for any partition $P$ of $[0,1]$.

Proof: Given any partition $0 = x_0 < x_1 < x_2 < \cdots x_{n-1} < x_n = 1$. For any segment, the $m_k$ is always $0$. So $L(f,P) = 0$.

(b) Let $ϵ > 0$, and consider the set of points $D_{ϵ/2} = \{x ∈ [0,1] : t(x) ≥ ϵ/2\}$. How big is $D_{ϵ/2}$?

Solution: Assume $n < 2/\epsilon$, the the size is at most $n(n+1)/2$.

(c) To complete the argument, explain how to construct a partition $P_ϵ$ of $[0,1]$ so that $U(t,P_ϵ) < ϵ$.

Solution: For each of $x \in [0, 1]$ such that $t(x) > \epsilon /2$, we make it a partition $[x, x + \frac{\epsilon}{n(n+1)}]$. So the upper sum is less than $\frac{\epsilon}{n(n+1)}$. Since there are only $n(n+1)/2$ of them, then total upper sum is $\epsilon / 2$. For other segments, each upper sum is $\epsilon / 2 \Delta x_k$, so the total upper sum is less than $\epsilon / 2$. Then the total sum is $\epsilon$.

$\square$

7.3.3.

Let

$$f(x) = \begin{cases} 1 &\text{if } x = 1/n \text{ for some } n \in N\\ 0 &\text{otherwise }\\ \end{cases}$$

Show that $f$ is integrable on $[0,1]$ and compute $\int_{0}^{1}f$.

Proof: Let

$$P_n = \{ 0, \frac{1}{n}, \frac{1}{n} + \frac{1}{n^2}, \frac{1}{n-1}, \frac{1}{n-1} + \frac{1}{n^2}, \frac{1}{n-2}, \frac{1}{n-2} + \frac{1}{n^2}, \}$$

So

$$U(f, P_n) = \frac{1}{n} + \frac{1}{n^2} \times (n-1) \\ < \frac{2}{n} \rightarrow 0$$

So $\int_{0}^{1} f = 0$.

$\square$

7.3.4.

Let $f$ and $g$ be functions defined on (possibly different) closed intervals, and assume the range of $f$ is contained in the domain of $g$ so that the composition $g ◦f$ is properly defined.

(a) Show, by example, that it is not the case that if $f$ and $g$ are integrable, then $g ◦f$ is integrable.

Solution: We use the example in the previous 2 exercises. Consider $f(x)$ is the Thomae function, and $g(x)$ is the function in exercise 7.3.3. Then $g◦f$ is the Dirichlet function, which is not differentiable.

(b) If $f$ is increasing and $g$ is integrable, then $g ◦f$ is integrable.

Proof: The following is not correct.

Assume $f$ is defined on $[a,b]$, and $g$ is defined on $[c,d]$, and $[f(a), f(b)] \subseteq [c, d]$. Since $g$ is integrable, and given $\epsilon$. We can find a $P = \{x_0, x_1, \cdots, x_n\}$ for $[c,d]$ such that $U(g,P) - L(g,P) < \epsilon.$

Assume $x_{k-1} \leq f(a) < x_k$ for some $k$. And let $c$ be the supremum of

$$S_k = \{x : f(x) \leq x_k\}$$

If $f(c) \leq x_k$, then $(M_{g ◦f, [a,c]} - m_{g ◦f, [a,c]}) (c-a) \leq (M_k - m_k)\Delta _k$.

If $f(c) > x_k$, then assume $f(c) \in (x_{l-1}, x_l]$ for some $l > k$. Then we can divide $[a,c]$ into $[a,d]$ and $[d,c]$ so that

$$(M_{g ◦f, [a,d]} - m_{g ◦f, [a,d]}) (d-a) + (M_{g ◦f, [d,c]} - m_{g ◦f, [d,c]}) (c-d) \leq \\ (M_k - m_k)\Delta _k + (M_{g, [x_{l-1}, f(c)]} - m_{g, [x_{l-1}, f(c)]})(f(c) - x_{l-1})$$

In this way, we can construct a partition of $[a,b]$ such that

$U(g ◦f,P') - L(g ◦f,P') < \epsilon.$

$\square$

(c) If $f$ is integrable and $g$ is increasing, then $g ◦f$ is integrable.

Solution: This is not true. Consider $f$ is the Thomae function.

$$g(x) = \begin{cases} 0 &\text{if } x = 0\\ 1 &\text{if } x > 0\\ \end{cases}$$

Then $g ◦f$ is again the Dirichlet function.

$\square$

7.3.5.

Provide an example or give a reason why the request is impossible.

(a) A sequence $(f_n) → f$ pointwise, where each $f_n$ has at most a finite number of discontinuities but $f$ is not integrable.

Solution:

Consider

$$f_n(x) = \begin{cases} 1 &\text{if } x = p/q, q \leq n \\ 0 &\text{otherwise}\\ \end{cases}$$

Then $f$ is the Dirichlet function which is not integrable.

(b) A sequence $(g_n) → g$ uniformly where each gn has at most a finite number of discontinuities and $g$ is not integrable.

Proof: Impossible. Each $g_n$ is integrable, and $(g_n) → g$ uniformly. So from exercise 7.2.5, $g$ is integrable.

(c) A sequence $(h_n) → h$ uniformly where each $h_n$ is not integrable but $h$ is integrable.

Solution: This is possible. Consdier

$$h_n(x) = \begin{cases} 1/n &\text{if } x \in \mathbf{Q}\\ 0 &\text{otherwise }\\ \end{cases}$$

$\square$

7.3.6.

Let $\{r_1,r_2,r_3,...\}$ be an enumeration of all the rationals in $[0,1]$, and define

$$g_n(x) = \begin{cases} 1 &\text{if } x = r_n \\ 0 &\text{otherwise.}\\ \end{cases}$$

Anwser:

(a) Is $G(x) = \sum_{n = 1}^{\infty} g_n(x)$ integrable on $[0, 1]$?

Solution: $G(x)$ is the Dirichlet function. So it's not integrable.

(b) $F(x) = \sum_{n = 1}^{\infty} g_n(x)/n$ integrable on $[0, 1]$?

Proof: Yes. Given $\epsilon$, we can find $n$ such that $1/n < \epsilon/2$, for $r_1,r_2,r_3,...r_n$, we create a segment $[r_i, r_i + 1/n^2]$. Then upper sum of those segments will be less than $1/n < \epsilon / 2$. For the rest of them, the upper sum will be less than $1/n \times 1 = 1/n < \epsilon / 2$. So overall the upper sum is less than $\epsilon$.

So it's integrable and the value is 0.

$\square$

7.3.7.

Assume $f : [a,b] → R$ is integrable.

(a) Show that if $g$ satisfies $g(x) = f(x)$ for all but a finite number of points in $[a,b]$, then $g$ is integrable as well.

Proof: Consider $h(x) = f(x) - g(x)$, then $h(x)$ is $0$ but a finite number of points. So it's integrable.

(b) Find an example to show that $g$ may fail to be integrable if it differs from $f$ at a countable number of points.

Solution: Consider $f = 0$ and $g$ is the Thomae function.

$\square$

7.3.8.

As in Exercise 7.3.6, let $\{r_1,r_2,r_3,...\}$ be an enumeration of all the rationals in $[0,1]$, but this time define

$$h_n(x) = \begin{cases} 1 &\text{if } r_n < x \leq 1 \\ 0 &\text{if } 0 \leq x \leq r_n \\ \end{cases}$$

Show $H(x) = \sum_{n = 1}^{\infty}h_n(x) / 2^n$ is integrable on $[0,1]$ even though it has discontinuities at every rational point.

Proof:

Simply notice that the $H(x)$ is a monotoneous increasing function will prove it.

We can also find the partition.

Given $\epsilon$, we can find $n$ such that $\frac{1}{2^n} < \epsilon / (2n)$. For partition $P$ which is $x_0 = 0, x_1,x_2,x_3,...,x_n = 1$ which is the set $\{0, 1, r_1,r_2,r_3,...,r_n\}$.

Between $(x_{k-1}, x_k]$, $M_k - m_k \leq \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \cdots < \frac{1}{2^n}$.

Also consider the interval $[x_{k-1}, x_{k-1} + \frac{1}{2^n}]$, for this interval, $m = H(x_{k-1})$,

$$M < m + \frac{1}{2} + \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \cdots < m + \frac{1}{2} + \frac{1}{2^n}$$

So

$$(M - m) \Delta < (\frac{1}{2} + \frac{1}{2^n}) \frac{1}{2^n} < \frac{1}{2^n}$$

since there are $n$ of them, then the sum is less than $\frac{n}{2^n} < \epsilon / 2$, for other part $[x_{k-1} + \frac{1}{2^n}, x_k]$, then sum is less than $\epsilon / 2$.

$\square$

7.3.9 (Content Zero).

A set $A ⊆ [a,b]$ has content zero if for every $ϵ > 0$ there exists a finite collection of open intervals $\{O_1,O_2,...,O_N\}$ that contain $A$ in their union and whose lengths sum to $ϵ$ or less.

Using $|O_n|$ to refer to the length of each interval, we have

$$A \subseteq \bigcup_{n=1}^{N} O_n \text{ and } \sum_{n=1}^{N} |O_n| \leq \epsilon.$$

Anwser

(a) Let $f$ be bounded on $[a,b]$. Show that if the set of discontinuous points of $f$ has content zero, then $f$ is integrable.

Proof: Assume $|f(x)| < M$, and given $\epsilon$ we can find a set of open intervals $\{O_1,O_2,...,O_N\}$ such that

$$A \subseteq \bigcup_{n=1}^{N} O_n \text{ and } \sum_{n=1}^{N} |O_n| \leq \epsilon / (4M).$$

Let $O_n = (a_n, b_n)$, let $P = \{a_1, b_1, \cdots, a_n, b_n\}$. In these intervals, the difference of the upper sum and lower sum is less than $2M \times \epsilon / (4M) = \epsilon / 2$.

In other interval, $f(x)$ is continuous, so we can find a partition such that the upper sum and lower sum is less than $\epsilon /2$.

$\square$

(b) Show that any finite set has content zero.

Proof: Assume there are $n$ points in $A$, $x_1, x_2, x_3, \dots x_{n}$, let

$$O_n = (x_n - \epsilon / 2n, x_n + \epsilon / 2n)$$

$\square$

(c) Content zero sets do not have to be finite. They do not have to be countable. Show that the Cantor set $C$ defined in Section 3.1 has content zero.

Proof:

Note $|C_1| = \frac{2}{3}$, we can use 2 open sets with length of $\frac{3}{8}$ to cover it.

Similarly, $|C_2|$ can be coverd by $4$ intervals of length $\frac{1}{8}$.

$|C_n|$ can be coverd by $2^n$ intervals of length $\frac{1}{8 \times 3^{n-2}}$.

So

$$\sum_{n=1}^{N} |O_n| = \frac{1}{2} \times \left( \frac{2}{3} \right) ^{n-2}.$$

In addition,

$$C \subset C_n \subset \bigcup_{k=1}^{n} O_n$$

$\square$

(d) Prove that

$$h(x) = \begin{cases} 1 &\text{if } x \in C\\ 0 &\text{if } x \not\in C\\ \end{cases}$$

is integrable, and find the value of the integral.

Proof: if $x \not\in C$, $x$ is in an open set. So $h(x)$ is continuous at $x$. Then we can apply (a). So $h(x)$ is integrable. and the value is $0$.

7.4 Properties of the Integral

7.4.1.

Let $f$ be a bounded function on a set $A$, and set

$$M = \sup \{f(x) : x \in A\} \\ m = \inf \{f(x) : x \in A\} \\ M' = \sup \{|f(x)| : x \in A\} \\ m' = \inf \{|f(x)| : x \in A\} \\ $$

Anwser:

(a) Show that $M - m \geq M' - m'$.

Proof:

• case 1: $M > 0, m > 0$ then $M' = M, m' = m$
• case 2: $M > 0, m = 0$ then $M' = M, m' = m$
• case 3: $M > 0, m < 0$ then $M' = M, m' > 0 > m$
• case 4: $M = 0, m = 0$ then $M' = M, m' = m$
• case 5: $M = 0, m < 0$ then $M' = -m, m' = -M$
• case 6: $M < 0, m < 0$ then $M' = -m, m' = -M$

In all 6 cases, we have $M - m \geq M' - m'$.

$\square$

(b) Show that if $f$ is integrable on the interval $[a,b]$, then $|f|$ is also integrable on this interval.

Proof: Given a partition, $P: a = x_0 < x_1 < x_2 < x_3 < \cdots < x_n = b$.

For segment $k$, the difference between upper sum and lower sum for $f$ is

$$(M_k - m_k) \Delta _k \geq (M_k' - m_k') \Delta _k$$

So we have $U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f, P) < \epsilon$.

$\square$

(c) Provide the details for the argument that in this case we have $\left| \int_{a}^{b} f \right| \leq \int_{a}^{b} |f|$.

Proof: $|f| - f \geq 0$, so $\int_{a}^{b}(|f| - f) \geq 0$, so $\int_{a}^{b}|f| - \int_{a}^{b} f \geq 0$, so $\int_{a}^{b} f \leq \int_{a}^{b}|f|$.

On the other hand, $f + |f| \geq 0$, so so $\int_{a}^{b}(f + |f|) \geq 0$, so $\int_{a}^{b} f \geq -\int_{a}^{b}|f|$.

Together, we have $\left| \int_{a}^{b} f \right| \leq \int_{a}^{b} |f|$.

$\square$

7.4.2.

(a) Let $g(x) = x^3$, and classify each of the following as positive, negative, or zero.

(i) $\int_{0}^{-1}g + \int_{0}^{1} g$

Solution:

$\int_{0}^{-1}g$ = $-\int_{-1}^{0}g > 0$, so overall it's positive.

(ii) $\int_{1}^{0}g + \int_{0}^{1} g$

Solution:

$\int_{1}^{0}g = - \int_{0}^{1} g$, so it's 0.

(iii) $\int_{1}^{-2}g + \int_{0}^{1} g$

Solution:

$$\int_{1}^{-2}g + \int_{0}^{1} g = - (\int_{-2}^{1}g - \int_{0}^{1} g) = \\ - \int_{-2}^{0} g > 0$$

$\square$

(b) Show that if $b ≤ a ≤ c$ and f is integrable on the interval $[b,c]$, then it is the case

$$\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$$

Proof:

We have

$$\int_{b}^{c} f = \int_{b}^{a} f + \int_{a}^{c} f$$

So

$$- \int_{c}^{b} f = - \int_{a}^{b} f + \int_{a}^{c} f$$

So

$$\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$$

$\square$

7.4.3.

Decide which of the following conjectures is true and supply a short proof. For those that are not true, give a counterexample.

(a) If $|f|$ is integrable on $[a,b]$, then $f$ is also integrable on this set.

Solution: False. Consdier $f$ defined on $[0,1]$

$$f(x) = \begin{cases} -1 &\text{if } x \in \mathbf{Q}\\ 1 &\text{if } x \in \mathbf{I}\\ \end{cases}$$

$\square$

(b) Assume $g$ is integrable and $g(x) ≥ 0$ on $[a,b]$. If $g(x) > 0$ for an infinite number of points $x ∈ [a,b]$, then $\int_{a}^{b}f > 0$.

Solution: False. Consider Thomae function.

(c) If $g$ is continuous on $[a,b]$ and $g(x) ≥ 0$ with $g(y_0) > 0$ for at least one point $y_0 ∈ [a,b]$, then $\int_{a}^{b}f > 0$.

Proof: Since $g$ is continuous, we can find a closed interval $y_0 \in [c, d] \subseteq [a,b]$ such that $f(x) > \epsilon$ for $[c, d]$. Then

$$\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{d} f + \int_{d}^{b} f \geq 0 + \epsilon (d-c) + 0 > 0.$$

$\square$

7.4.4.

Show that if $f(x) > 0$ for all $x ∈ [a,b]$ and $f$ is integrable, then $\int_{a}^{b} f > 0$.

Proof: No idea for now. Seems like we need to use section 7.5.

7.4.5.

Let $f$ and $g$ be integrable functions on $[a,b]$.

(a) Show that if $P$ is any partition of $[a,b]$, then

$$U(f+g, P) \leq U(f, P) + U(g, P)$$

Provide a specific example where the inequality is strict. What does the corresponding inequality for lower sums look like?

Proof: For any segment $[x_k, x_{k+1}]$, assume $M_{f+g, k}$ is the supremum of $f+g$ on this segment. For any $c \in [x_k, x_{k+1}]$, $(f+g)(c) = f(c) + g(c)\leq M_{f,k} + M_{g, k}$. So $M_{f+g, k} \leq M_{f,k} + M_{g, k}$.

Therefore $U(f+g, P) \leq U(f, P) + U(g, P)$.

Let $f(x) = x, g(x) = -x, [a,b] = [0,1], P = \{0, 1\}$.

So $f+g = 0$, $U(f+g, P) = 0$. On the other hand, $U(f,P) = 1, U(g,P) = 0$, so $U(f+g, P) = U(f,P) + U(g,P)$.

The lower sums look like

$$L(f+g, P) \geq L(f, P) + L(g, P)$$

$\square$

(b) Review the proof of Theorem 7.4.2 (ii), and provide an argument for part (i) of this theorem.

Proof: Since $f, g$ are all integrable, we can find $P_{f, n}, Q_{g, n}$, such that

$$\lim_{n \to \infty} U(f, P_{f, n}) - L(f, P_{f, n}) = 0 \\ \lim_{n \to \infty} U(g, P_{g, n}) - L(g, P_{g, n}) = 0 \\ $$

Then let $P_n = P_{f, n} \cup P_{g, n}$, then

$$\lim_{n \to \infty} U(f, P_{n}) - L(f, P_{n}) = 0 \\ \lim_{n \to \infty} U(g, P_{n}) - L(g, P_{n}) = 0 \\ $$

Adding them together

$$\lim_{n \to \infty} (U(f, P_{n}) + U(g, P_{n})) - (L(f, P_{n}) + L(g, P_{n})) = 0$$

Furthermore, We apply (a), then we know

$$\lim_{n \to \infty} U(f+g, P_n) - L(f+g, P_n) \leq 0$$

On the other hand, $U(f+g, P_n) - L(f+g, P_n) \geq 0$. So

$$\lim_{n \to \infty} U(f+g, P_n) - L(f+g, P_n) = 0$$

So $f+g$ is integrable. Since $U(f+g, P) \leq U(f, P) + U(g, P)$, then $U(f+g) \leq U(f) + U(g)$. Similarly, $L(f+g) \geq L(f) + L(g)$.

So $\int_{a}^{b} (f+g) = \int_{a}^{b}f + \int_{a}^{b} g$.

$\square$

7.4.6.

Although not part of Theorem 7.4.2, it is true that the product of integrable functions is integrable. Provide the details for each step in the following proof of this fact:

(a) If $f$ satisfies $|f(x)| ≤ M$ on $[a,b]$, show

$$\left| (f(x))^2 - (f(y))^2 \right| \leq 2M \left| f(x) - f(y) \right|$$

Proof:

$$\left| (f(x))^2 - (f(y))^2 \right| = \\ \left| f(x) + f(y) \right| \left| f(x) - f(y) \right| \leq (|f(x)| + |f(y)|) \left| f(x) - f(y) \right| \\ \leq 2M \left| f(x) - f(y) \right|$$

$\square$

(b) Prove that if $f$ is integrable on $[a,b]$, then so is $f^2$.

Proof:

Consider the partial sum $M_k - m_k$ for $f$. For any $x, y \in [x_k, x_{k+1}]$

$$(f(x) - f(y)) ^ 2 \leq 2M \left| f(x) - f(y) \right| < 2M (M_{f,k} - m_{f,k})$$

So $U(f^2, P) - L(f^2, P) \leq 2M U(f, P) - L(f, P)$. Then $f^2$ is also integrable.

$\square$

(c) Now show that if $f$ and $g$ are integrable, then $fg$ is integrable. (Consider $(f+g)^2$.)

Proof:

$$fg = \frac{1}{2} ((f+g)^2 - f^2 - g^2)$$

All functions on the right side are integrable. The sum of them is integrable.

$\square$

7.4.7.

Review the discussion immediately preceding Theorem 7.4.4.

(a) Produce an example of a sequence $(f_n) \rightarrow 0$ pointwise on [0,1] where $\lim_{n \to \infty} \int_{0}^{1}f_n$ does not exist.

Solution:

Consider

$$f_n(x) = \begin{cases} n^2 &\text{if } 0 \leq x \leq \frac{1}{n}\\ 0 &\text{if } x > \frac{1}{n^2} \\ \end{cases}$$

So we have

$$\int_{0}^{1} f_n = n$$

$\square$

(b) Produce an example of a sequence $g_n$ with $\int_{0}^{1}g_n \rightarrow 0$ but $g_n$ does not converge to zero for any $x \in [0,1]$. To make it more interesting, let’s insist that $g_n(x) ≥ 0$ for all $x$ and $n$.

Solution:

I borrowed the idea from exercise 6.4.2 (c) Consdier

$$f_1(x) = 1 \\ f_2(x) = \begin{cases} 1 &\text{if } x < \frac{1}{2}\\ 0 &\text{if } x \geq \frac{1}{2}\\ \end{cases} \\ f_3(x) = \begin{cases} 0 &\text{if } x < \frac{1}{2}\\ 1 &\text{if } x \geq \frac{1}{2}\\ \end{cases} \\ f_4(x) = \begin{cases} 1 &\text{if } \frac{0}{4} < x < \frac{1}{4}\\ 0 &\text{otherwise} \\ \end{cases} \\ f_5(x) = \begin{cases} 1 &\text{if } \frac{1}{4} < x < \frac{2}{4}\\ 0 &\text{otherwise} \\ \end{cases} \\ f_6(x) = \begin{cases} 1 &\text{if } \frac{2}{4} < x < \frac{3}{4}\\ 0 &\text{otherwise} \\ \end{cases} \\ f_7(x) = \begin{cases} 1 &\text{if } \frac{3}{4} < x < \frac{4}{4}\\ 0 &\text{otherwise} \\ \end{cases} \\ \cdots$$

$\square$

7.4.8.

For each $n \in \mathbf{N}$, let

$$h_n(x) = \begin{cases} 1/2^n &\text{if } 1/2^n < x \leq 1\\ 0 &\text{if } 0 \leq x \leq 1/2^n\\ \end{cases}$$

And set $H(x) = \sum_{n = 1}^{\infty} h_n(x)$. Show $H$ is integrable and compute $\int_{0}^{1} H$.

Solution: Because each $h_n(x)$ is increasing, $H(x)$ is also increasing. $H(x)$ is bounded since $H(x) \leq 1$. So $H$ is integrable.

Consider

$$H_k(x) = \sum_{n = 1}^{k} h_n(x)$$

Note that $H_k(x)$ converges uniformly to $H(x)$ and

$$\int_{0}^{1} H_k(x) = \frac{1}{2} (1 - \frac{1}{2}) + \frac{1}{4} (1 - \frac{1}{4}) + \cdots + \frac{1}{2^k} (1 - \frac{1}{2^k}) \\ = \sum_{n = 1}^{k} \frac{1}{2^k} - \sum_{n = 1}^{k} \frac{1}{2^{2k}}$$

So $\lim_{n \to \infty} H_k(x) = 1 - \frac{1/4}{1 - 1/4} = 2/3$.

$\square$

7.4.9.

Let $g_n$ and $g$ be uniformly bounded on $[0,1]$, meaning that there exists a single $M > 0$ satisfying $|g(x)| ≤ M$ and $|g_n(x)| ≤ M$ for all $n ∈ \mathbf{N}$ and $x ∈ [0,1]$.

Assume $g_n → g$ pointwise on $[0,1]$ and uniformly on any set of the form $[0,α]$, where $0 < α < 1$.

If all the functions are integrable, show that

$$\lim_{n \to \infty} \int_{0}^{1} g_n = \int_{0}^{1} g$$

Proof:

Given $\epsilon$, we want to find $N$ such that if $n > N$ then

$$\left| \int_{0}^{1} g_n - \int_{0}^{1} g \right| < \epsilon$$

Since $g_n$ and $g$ be uniformly bounded on $[0,1]$, we can have a $M$.

Given this $M$, we can find $\alpha$, such that $(1 - \alpha ) M < \epsilon / 3$.

Then

$$\left| \int_{0}^{1} g_n - \int_{0}^{1} g \right| < \left| \int_{0}^{1} g_n - \int_{0}^{\alpha} g_n \right| + \left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right| + \left| \int_{0}^{1} g_n - \int_{0}^{\alpha} g_n \right| \\ = \left| \int_{\alpha}^{1} g_n \right| + \left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right| + \left| \int_{\alpha}^{1} g \right| \\ \int_{\alpha}^{1} |g_n| + \left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right| + \int_{\alpha}^{1} |g| \leq (1 - \alpha ) M + \left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right| + (1 - \alpha ) M \\ 2 / 3 \epsilon + \left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right|$$

Since $g_n → g$ uniformly on any set of the form $[0,α]$, we can find $N$ such that if $n > N$ then

$$\left| \int_{0}^{\alpha} g_n - \int_{0}^{\alpha} g \right| < \epsilon / 3$$

So

$$\left| \int_{0}^{1} g_n - \int_{0}^{1} g \right| < \epsilon$$

$\square$

7.4.10.

Assume $g$ is integrable on $[0,1]$ and continuous at $0$. Show

$$\lim_{n \to \infty} \int_{0}^{1} g(x^n) dx = g(0)$$

Proof: We want to find $N$ such that if $n > N$ then

$$\left| \int_{0}^{1} g(x^n) dx - g(0) \right| < \epsilon$$

We use the technique in theorem 7.5.1

$$\left| \int_{0}^{1} g(x^n) dx - g(0) \right| = \\ \left| \int_{0}^{1} g(x^n) dx - \int_{0}^{1} g(0) dx \right| = \left| \int_{0}^{1} (g(x^n) - g(0))dx \right|$$

Since $g$ is integrable, $g$ is bounded, i.e. $\left| g(x) \right| \leq M$. Then we can find $c \in [0, 1]$, such that $2M(1-c) < \epsilon /2$.

Fix this $c$. Next, since $g$ is continuous at $0$, we can find $U_{\delta}(0)$ such that $x \in U_{\delta}(0)$, $|g(x) - g(0)| < \epsilon / 2$.

Then we can find $N$ such that if $n > N, x \in [0,c]$, $x^n \in U_{\delta}(0)$.

So

$$\left| \int_{0}^{1} (g(x^n) - g(0))dx \right| \leq \int_{0}^{1} \left| g(x^n) - g(0) \right| dx \\ = \int_{0}^{c} \left| g(x^n) - g(0) \right| dx + \int_{c}^{1} \left| g(x^n) - g(0) \right| dx \\ < \epsilon / 2 c + 2M(1-c) < \epsilon / 2 + \epsilon / 2= \epsilon$$

$\square$

7.4.11.

Review the original definition of integrability in Section 7.2, and in particular the definition of the upper integral $U(f)$. One reasonable sug- gestion might be to bypass the complications introduced in Definition 7.2.7 and simply define the integral to be the value of $U(f)$. Then every bounded function is integrable! Although tempting, proceeding in this way has some significant drawbacks. Show by example that several of the properties in Theorem 7.4.2 no longer hold if we replace our current definition of integrability with the proposal that $\int_{a}^{b} f= U(f)$ for every bounded function $f$.

Solution:

(i) Let $f(x)$ is Thomae function and $g(x)$ be the negative Thomae function.

$$U(f) = 1, U(g) = 0$$

So $U(f) + U(g) = 1$, but $U(f+g) = 0$.

$\square$

I think (ii), (iii), (iv), (v) holds.

$\square$

7.5 The Fundamental Theorem of Calculus

7.5.1.

(a) Let $f(x) = |x|$ and define $F(x) = \int_{-1}^{x} f$. Find a piecewise algebraic formula for $F(x)$ for $F(x)$ for all $x$. Where is $F$ continuous? Where is $F$ differentiable? Where does $F'(x) = f(x)$?

Solution:

$$F(x) = \begin{cases} -\frac{1}{2} x^2 + \frac{1}{2} &\text{if } -1 \leq x \leq 0 \\ \frac{1}{2} x^2 + \frac{1}{2} &\text{if } x > 0 \\ \end{cases}$$

$F(x)$ is continuous for $[-1, +\infty )$. It is differentiable on this interval. Also $F'(x) = f(x)$.

(b) Repeat part (a) for the function

$$f(x) = \begin{cases} 1 &\text{if } x < 0\\ 2 &\text{if } x \geq 0\\ \end{cases}$$

Solution:

$$F(x) = \begin{cases} x &\text{if } -1 \leq x \leq 0\\ 2x &\text{if } x > 0\\ \end{cases}$$

$F(x)$ is continuous for $[-1, +\infty )$. It's not differentiable at $0$. Also $F'(x) = f(x)$ when $x \not = 0$.

$\square$

7.5.2.

Decide whether each statement is true or false, providing a short justification for each conclusion.

(a) If $g= h'$ for some $h$ on $[a,b]$, then $g$ is continuous on $[a,b]$.

Solution:

Consider

$$h(x) = \begin{cases} 0 &\text{if } x = 0\\ x^{1.5}\sin \frac{1}{x} &\text{if } 0 < x \leq 1\\ \end{cases}$$

$$g(0) = \lim_{x \to 0} \frac{x^a \sin (1/x)}{x} = \lim_{x \to 0} x^{a-1} \sin (1/x) = 0 \\ g(c) = ac^{a-1} \sin(1/c) - c^{a-2} \cos (1/c)$$

$g$ is not continuous at $0$.

(b) If $g$ is continuous on $[a,b]$, then $g= h'$ for some $h$ on $[a,b]$.

Proof: This is true.

$g$ is continuous, so it's integrable. Then consider $G(x) = \int_{a}^{x} g(x)$. Since $g$ is continuous, $G'(x) = g(x)$.

$\square$

(c) If $H(x) = \int_{a}^{x} h$ is differentiable at $c ∈ [a,b]$, then $h$ is continuous at $c$.

Solution: false.

Consider $h(x)$ defined on $[0,2]$.

$$h(x) = \begin{cases} 0 &\text{if } x \not = 1\\ 1 &\text{if } x = 1\\ \end{cases}$$

Then $H(x) = 0$ is differentiable at $c = 0$, but $h(x)$ is not continuous at $0$.

$\square$

7.5.3.

The hypothesis in Theorem 7.5.1 (i) that $F'(x) = f(x)$ for all $x ∈ [a,b]$ is slightly stronger than it needs to be. Carefully read the proof and state exactly what needs to be assumed with regard to the relationship between $f$ and $F$ for the proof to be valid.

Solution: $F'(x) = f(x)$ for all $(a, b)$ is enough.

$\square$

7.5.4.

Show that if $f : [a,b] → \mathbf{R}$ is continuous and $\int_{a}^{x} f = 0$ for all $x ∈ [a,b]$, then $f(x) = 0$ everywhere on $[a,b]$. Provide an example to show that this conclusion does not follow if f is not continuous.

Proof: For theorem 7.5.1, $f$ is integrable and continuous, then $F(x) = \int_{a}^{x}f$ and $F'(x) = f(x)$. Since $F(x) = 0$, so $F'(x) = 0$ for $x \in [a, b]$, then $f(x) = 0$ for $x \in [a, b]$.

$\square$

7.5.5.

The Fundamental Theorem of Calculus can be used to supply a shorter argument for Theorem 6.3.1 under the additional assumption that the sequence of derivatives is continuous

Assume $f_n → f$ pointwise and $f'_n → g$ uniformly on $[a,b]$. Assuming each $f'_n$ is continuous, we can apply Theorem 7.5.1 (i) to get

$$\int_{a}^{x} f_n' = f_n(x) - f_n(a)$$

for all $x ∈ [a,b]$. Show that $g(x) = f'(x)$.

Proof:

Since $f'_n → g$ and $f'_n$ is continuous, then $g$ is continuous. So $g$ is integrable. Furthermore

$$\int_{a}^{x} g(x) = \lim_{n \to \infty} \int_{a}^{x} f_n'(x)$$

On the left side

$$G(x) = \int_{a}^{x} g(x) \text{ and } G'(x) = g(x)$$

On the right side

$$\lim_{n \to \infty} \int_{a}^{x} f_n'(x) = \lim_{n \to \infty} f_n(x) - f_n(a)\\ = f(x) - f(a)$$

So $G(x) = f(x) - f(a)$, so $g(x) = f'(x)$.

$\square$

7.5.6 (Integration-by-parts).

(a) Assume $h(x)$ and $k(x)$ have continuous derivatives on $[a,b]$ and derive the familiar integration-by-parts formula

$$\int_{a}^{b} h(t) k'(t) dt = h(b) k(b) - h(a) k (a) - \int_{a}^{b} h'(t) k(t) dt .$$

Proof:

Since $h(x)$ and $k(x)$ have continuous derivatives on $[a,b]$, so does $h(x)k(x)$. And its derivative at $t \in [a, b]$ is

$$h(t)k'(t) + h'(t)k(t)$$

In addition, since $h(x), k(x), h'(x), k'(x)$ are all continuous, then $h(t)k'(t), h'(t)k(t)$ are both integrable.

So we can directly apply Theorem 7.5.1 to get

$$\int_{a}^{b} (h(t)k'(t) + h'(t)k(t)) dt = h(b) k(b) - h(a) k (a)$$

Using theorem 7.4.2 (i) and rearrange it, we can prove it.

(b) Explain how the result in Exercise 7.4.6 can be used to slightly weaken the hypothesis in part (a).

Solution: exercise 7.4.6 states it is true that the product of integrable functions is integrable.

So we just need to assume $h'(x), k'(x)$ are both integrable. Then the argument in (a) still holds.

$\square$

7.5.7.

Use part (ii) of Theorem 7.5.1 to construct another proof of part (i) of Theorem 7.5.1 under the stronger hypothesis that $f$ is continuous. (To get started, set $G(x) = \int_{a}^{x}f$.)

Proof: $f$ is continuous, then $G'(x) = f(x)$. If $F'(x) = f(x)$, then $(F(x) - G(x))' = 0$ for $x \in [a, b]$. That means $F(x) = G(x) + C$.

$F(b) - F(a) = G(b) - G(a) = \int_{a}^{b}f(x) - \int_{a}^{a}f(x)$.

$\square$

7.5.8 (Natural Logarithm and Euler’s Constant).

Let

$$L(x) = \int_{1}^{x} \frac{1}{t} dt$$

where we consider only $x > 0$.

(a) What is $L(1)$? Explain why $L$ is differentiable and find $L'(x)$.

Solution: $L(1) = 0$. This is because $\frac{1}{t}$ is continuous.

$\square$

(b) Show that $L(xy) = L(x)+L(y)$. (Think of $y$ as a constant and differentiate $g(x) = L(xy).$)

Proof:

$$L(xy) = \int_{1}^{xy} \frac{1}{t} dt \\ L'(xy) = yL'(xy) = y \frac{1}{xy} = \frac{1}{x} = L'(x)$$

So $L(xy) = L(x) + C$ and plug $x = 1$, so $L(y) = C$. So $L(xy) = L(x) + L(y)$.

$\square$

(c) Show $L(x/y) = L(x)−L(y)$.

Proof:

$$L'(1/x) = x \cdot (- \frac{1}{x^2}) = - \frac{1}{x} = -L'(x) \\ L(x) + L(1/x) = C$$

plug $x = 1$, so $C = L(1) + L(1) = 0$. So $L(1/x) = -L(x)$.

Then with (b), we proved.

(d) Let

$$\gamma _n = (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}) - L(n)$$

Prove that $(γ_n)$ converges. The constant $γ = \lim γ_n$ is called Euler’s constant.

Proof:

For $n > m$, consdier

$$γ_n - γ_m = \frac{1}{m+1} + \cdots + \frac{1}{n} - \int_{m}^{n} \frac{1}{t} dt$$

Then note

$$\int_{m}^{n} \frac{1}{t} dt \geq \frac{1}{m+1} \cdot 1 + \cdots + \frac{1}{n} \cdot 1 \\ \int_{m}^{n} \frac{1}{t} dt \leq \frac{1}{m} \cdot 1 + \cdots + \frac{1}{n-1} \cdot 1 \\ $$

So

$$|γ_n - γ_m| \leq \frac{1}{m} - \frac{1}{n} < \frac{1}{m}$$

So it's Cauchy sequence.

$\square$

(d) Show how consideration of the sequence $γ_{2n}−γ_n$ leads to the interesting identity

$$L(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$$

Proof:

$$γ_{2n}−γ_n = 1 + \frac{1}{2} + \cdots + \frac{1}{2n} - L(2n) - (1 + \frac{1}{2} + \cdots + \frac{1}{n} - L(n)) \\ = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{2n} - L(2)$$

Since $(γ_n)$ converges, then $\lim_{n \to \infty} γ_{2n} - γ_n = 0$,

Then we proved.

$\square$

7.5.9.

Given a function $f$ on $[a,b]$, define the total variation of $f$ to be

$$Vf = \sup \{ \sum_{k = 1}^{n} \left| f(x_k) - f(x_{k-1}) \right| \}$$

where the supremum is taken over all partitions $P$ of $[a,b]$.

(a) If $f$ is continuously differentiable ($f'$ exists as a continuous function), use Fundamental Theorem of Calculus to show $Vf \leq \int_{a}^{b} |f'|$.

Proof: $f'$ is integrable. So from theorem 7.5.1 (i), we know $f(x_k) - f(x_{k-1}) = \int_{x_{k-1}}^{x_k} f'(x)$.

So

$$\sum_{k = 1}^{n} \left| f(x_k) - f(x_{k-1}) \right| = \sum_{k = 1}^{n} \left| \int_{x_{k-1}}^{x_k} f'(x) \right| \\ \leq \sum_{k = 1}^{n} \int_{x_{k-1}}^{x_k} |f'(x)| \\ = \int_{a}^{b} |f'(x)|$$

So $Vf \leq \int_{a}^{b} |f'|$

$\square$

(b) Use the Mean Value Theorem to establish the reverse inequality and conclude that $Vf = \int_{a}^{b} |f'|$.

Proof:

$$\sum_{k = 1}^{n} \left| f(x_k) - f(x_{k-1}) \right| = \sum_{k = 1}^{n} |f'(x_{k, c})(x_k - x_{k-1})| \geq L(|f|, P)$$

So $Vf \geq L(|f|) = \int_{a}^{b} |f'(x)|$.

$\square$

7.5.10 (Change-of-variable Formula).

Let $g : [a,b] → \mathbf{R}$ be differentiable and assume $g'$ is continuous. Let $f : [c,d] → \mathbf{R}$ be continuous, and assume that the range of $g$ is contained in $[c,d]$ so that the composition $f◦g$ is properly defined.

(a) Why are we sure $f$ is the derivative of some function? How about $(f◦g)g'$?

Proof: $f$ is continuous, so from Theorem 7.5.1 (ii) we know if $F(x) = \int_{c}^{x}f(x)$, then $f(x) = F'(x)$.

So consider the composite function $F◦g$,

$$(F◦g)'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$$

So $(f◦g)g'$ is the derivative of $F◦g$.

$\square$

(b) Prove the change-of-variable formula

$$\int_{a}^{b} f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(t) dt$$

Proof:

Since $f$ is continuous, then $f$ is integrable. Furthermore $F'(x) = f(x)$. Then

$$\int_{g(a)}^{g(b)} f(t) dt = F(g(b)) - F(g(a))$$

For the left side, $f(g(x))$ is continuous and $g'(x)$ is continuous, so $f(g(x)) g'(x)$ is integrable, and $(F◦g)' = (f◦g)g'$. So

$$\int_{a}^{b} f(g(x)) g'(x) dx = F(g(b)) - F(g(a))$$

$\square$

7.5.11.

Assume $f$ is integrable on [a,b] and has a “jump discontinu- ity” at $c ∈ (a,b)$. This means that both one-sided limits exist as $x$ approaches $c$ from the left and from the right, but that

$$\lim_{x \to c^-} f(x) \not = \lim_{x \to c^+} f(x)$$

(This phenomenon is discussed in more detail in Section 4.6.)

(a) Show that in this case, $F(x) = \int_{a}^{x}f(x)$ is not differentiable at $x = c$.

Proof: Assume

$$\lim_{x \to c^-} f(x) = a < b = \lim_{x \to c^+} f(x)$$

We following the proof of Theorem 7.5.1 (ii), assume $x < c$ and given $\epsilon$, we can find $(c-\delta, c)$ if $x \in (c-\delta, c)$, then $$\left| \frac{ F(c) - F(x) }{c - x} - a \right| = \frac{1}{c - x} |\int_{x}^{c} f(x) - a| \\ \leq \frac{1}{c - x} \int_{x}^{c} |f(x) - a| <\epsilon$$

That means

$$\lim_{x \to c^-} \frac{ F(c) - F(x) }{c - x} = a$$

$\square$

(b) The discussionin Section 5.5 mentions the existence of a continuous monotone function that fails to be differentiable on a dense subset of R. Combine the results of part (a) with Exercise 6.4.10 to show how to construct such a function.

Solution: In 6.4.10, we define this function.

Let $\{r_1,r_2,r_3,...\}$ be an enumeration of the set of rational numbers. For each $r_n ∈ \mathbf{Q}$, define

$$u_n(x) = \begin{cases} 1/2^n &\text{for } x > r_n \\ 0 &\text{for } x \leq r_n\\ \end{cases}$$

Now let $h(x) = \sum_{n = 1}^{\infty}u_n(x)$.

Since $h(x)$ is bounded and monotone increasing, then it is integrable. Let

$$H(x) = \int_{0}^{x} h(x)$$

Since $h(x) \geq 0$, then $H(x)$ is monotone increasing. From theorem 7.5.1 (ii), we know $H(x)$ is continuous.

$\lim_{x \to r_n^-} h(x)$ and $\lim_{x \to r_n^-} h(x)$ exist.

Since from any $r_n$, $\lim_{x \to r_n^-} h(x) < \lim_{x \to r_n^+} h(x)$, then based on (a), $H(x)$ is not differentiable at all $r_n$ which is $\mathbf{Q}$, which is dense in $\mathbf{R}$.

$\square$

7.6 Lebesgue’s Criterion for Riemann Integrability

7.6.14.

(a) Find $g'(0)$.

Solution:

$$\lim_{x \to 0} \frac{ x^2 \sin \frac{1}{x} - 0 }{x - 0} = \lim_{x \to 0} x \sin \frac{1}{x} = 0$$

$\square$

(b) Use the standard rules of differentiation to compute $g'(x)$ for $x \not = 0$.

$$g'(x) = 2x \sin \frac{1}{x} - x^2 \cos \frac{1}{x} \frac{1}{x^2}\\ = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$$

$\square$

(c) Explain why, for every $δ > 0$, $g'(x)$ attains every value between $1$ and $−1$ as $x$ ranges over the set $(−δ,δ)$. Conclude that $g'$ is not continuous at $x = 0$.

Proof: Given $(−δ,δ)$, we can find $k$ such that

$$-δ < -\frac{1}{2k \pi} < -\frac{1}{2 (k+1) \pi} < 0 < \frac{1}{2 (k+1) \pi} < \frac{1}{2k \pi} < δ$$

Note

$$g'(-\frac{1}{2k \pi}) = g'(\frac{1}{2k \pi}) = -1 \\ g'(-\frac{1}{2(k+1) \pi}) = g'(\frac{1}{2(k+1) \pi}) = 1 \\ $$

Since $g'(x)$ is continuous on $[\frac{1}{2 (k+1) \pi}, \frac{1}{2k \pi}]$, then because of the intermediate value property, $g'(x)$ attains every value between $1$ and $−1$.

Since $g'(x) = 0$, $g'$ is not continuous at $x = 0$.

$\square$

7.6.15.

(a) If $c ∈ C$, what is $\lim_{n→∞} f_n(c)$?

Solution: In each $f_n$, $f_n(c) = 0$. So $\lim_{n→∞} f_n(c) = 0$.

(b) Why does $\lim_{n→∞} f_n(x)$ exist for $x \not ∈ C$?

Solution: If $x \not \in C$, then there exists $N$, if $n \geq N$, $x \not \in C_n$. Let $N$ be the smallest integer such that $x \not \in C_N$.

It will be replaced by the $x^2 \sin \frac{1}{x}$ function and won't be changed any more.

So $\lim_{n→∞} f_n(x)$ exists.

7.6.16.

(a) Explain why $f'(x)$ exists for all $x \not \in C$.

Proof: if $x \not \in C$, then we can assue it was taken out during the interval $(a,b)$. On this interval, it is either $g(x-a)$ or $g(-x + b)$ or in the splice part. Wherever it is, it's differentiable.

$\square$

(b) If $c ∈ C$, argue that $|f(x)| ≤ (x− c)^2$ for all $x ∈ [0,1]$. Show how this implies $f'(c) = 0$.

Proof: If $x \in C$, $f(x) = 0$. So it holds.

If $x \not \in C$, assume it is taken out with $(a, b)$. Note that $a, b \in C$. So $(x-c)^2 \geq \max \{ (x-a)^2, (x-b)^2\}$. By our construction of $f(x)$, we know

$$|f(x)|^2 \leq (x - a)^2 \text{ and } |f(x)|^2 \leq (-x + b)^2$$

So the inequality also holds.

$$\lim_{x \to c} \left| \frac{f(x) - f(c)}{x - c} - 0 \right| =\\ \lim_{x \to c} \left| \frac{f(x)}{x-c} \right| \leq \lim_{x \to c} \left| \frac{(x-c)^2}{x-c} \right| = 0$$

So $f'(c) = 0$.

$\square$

(c) Give a careful argument for why $f'(x)$ fails to be continuous on $C$. Remember that $C$ contains many points besides the endpoints of the intervals that make up $C1,C2,C3,...$.

Proof: Let $c ∈ C$. And given $U_{\delta}(c)$, it is possible to find an open interval $(a, b) \subset U_{\delta}(c)$ and $(a, b) \cap C = \emptyset$. Based on our construction of $f$, $f'(x)$ can achieve any value in $[-1, 1]$ when $x \in (a, b)$. Since $f'(c) = 0$, $f'(x)$ cannot be continuous at $c$.

$\square$

7.6.17.

Why is $f'(x)$ Riemann-integrable on [0,1]?

Proof: This is because $f'(x)$ is continuous for $x \not \in C$, and Cantor set is measure $0$.

$\square$

7.6.18.

Show that, under these circumstances, the sum of the lengths of the intervals making up each $C_n$ no longer tends to zero as $n → ∞$. What is this limit?

Proof: The segment we take out is

$$\sum_{n=1}^{\infty} 2^{n-1} \left( \frac{1}{3^{n+1}} \right) = \frac{1}{9} (1 + \frac{2}{3} + \frac{4}{9} + \cdots) = \frac{1}{3}$$

$\square$

7.6.19.

As a final gesture, provide the example advertised in Exercise 7.6.13 of an integrable function $f$ and a continuous function $g$ where the composition $f ◦ g$ is properly defined but not integrable. Exercise 4.3.12 may be useful.

Solution: Consider $A$ is the set defined in 7.6.18, $g$ is the function used in Exercise 4.3.12. Since $A$ is a close set, $g$ is continuous. Let $f$ be this function defined in $[0,1]$ :

$$f(x) = \begin{cases} 0 &\text{if } x = 0 \\ \sin \frac{1}{x} &\text{if } x \not = 0 \\ \end{cases}$$

$f(x)$ is integrable because $f(x)$ is only discontinueous at $0$.

So let $h = f ◦ g$, $h(x) = 0$ if $x \in A$, because the distance between $x$ and $A$ is $0$.

Furthermore, $h(x)$ is not continuous at $x \in A$. Given $c \in A$ it is possible to find an open interval $(a, b) \subset U_{\delta}(c)$ and $(a, b) \cap C = \emptyset$, $a,b \in A$. $h(x)$ can achieve any value in $[-1, 1]$ when $x \in (a, b)$. Since $h(c) = 0$, $h(x)$ cannot be continuous at $c$.

$\square$