Chapter 6 The Riemann-Stieltjes Integral
Exercise 1
(By Matt Frito Lundy) Note: I should probably consider the cases where $x \pm d \notin [a,b]$ in the solutions to 1 and 2 below. First note that for any partition $P$ of $[a,b]$, $L(P,f,\alpha) = 0$, so we have $$\underline{\int_a^b} f \ d\alpha = 0$$ and $0 \leq U(P,f,\alpha)$.
Let $\varepsilon > 0$ be given. $\alpha$ is continuous at $x_0$ means there exists a $\delta > 0$ such that $\left| x – x_0 \right| < \delta$ and $a \leq x \leq b$ implies $\left| \alpha(x) – \alpha(x_0) \right| < \varepsilon$.
Let $P = \{a, x_0 – \delta, x_0 + \delta, b\}$. Then we have
$$U(P,f,\alpha) = \alpha(x_0 + \delta) – \alpha(x_0 – \delta) < 2\varepsilon$$ and
$$0 \leq U(P,f,\alpha) < 2\varepsilon.$$ Because $\varepsilon$ was arbitrary, we have
$$\overline{\int_a^b} f \ d\alpha = 0$$ so that $f \in \mathscr{R}(\alpha)$ and
$$\int_a^b f \ d\alpha = 0.$$
Exercise 2
(By Matt Frito Lundy) Suppose there exists an $x \in [a,b]$ such that $f(x) > 0$. $f$ is continuous at on $[a,b]$ means there exists a $\delta$ such that $|t – x| < \delta$ and $a \leq t \leq b$ implies $f(t) > 0$. Let $P = \{a, x – \delta, x+ \delta, b\}$, then $L(P,f) > 0$, and for any partition $P$ $$\int_a^b f(x) \ dx \geq L(P,x) > 0,$$ a contradiction.
Exercise 3
(By Matt Frito Lundy)
(a) First suppose that $f(0+) = f(0)$ and let $\varepsilon > 0$ be given. Then there exists a $\delta^* >0$ such that $0 < x < \delta^*$ implies $\left| f(x) – f(0) \right| < \epsilon$. Let $\delta = \min\{1,\delta^*/2\}$ and form a partition $P = \{-1,0,\delta,1\}$. Then we have
$$U(P,f,\beta_1) – L(P,f,\beta_1) = f(s) – f(t)$$ for some $s,t \in [0,\delta]$. But
$$f(s) – f(t) \leq \left| f(s) – f(0) \right| + \left| f(0) – f(t) \right| $$ so we have
$$U(P,f,\beta_1) – L(P,f,\beta_1) \leq \left| f(s) – f(0) \right| + \left| f(0) – f(t) \right| < 2\varepsilon,$$ which shows that $f \in \mathscr R (\beta_1)$.
Now suppose that $f \in \mathscr R (\beta_1)$ and let $\varepsilon >0$ be given. Then there exists a partition $P$ for which $$U(P,f,\beta_1) – L(P,f,\beta_1) < \varepsilon.$$ Let $P^*$ be a refinement of $P$ that includes $0$. Then we have:
$$U(P^*,f,\beta_1) – L(P^*,f,\beta_1) < \varepsilon.$$ But if $[0,\delta]$ is the subinterval of $P^*$ that contains $0$, then:
$$U(P^*,f,\beta_1) – L(P^*,f,\beta_1) = f(s) – f(t)$$ for some $s,t\in[0,\delta]$ where $f(s) – f(t) \geq \left| f(x_1) – f(x_2) \right|$ for any $x_1,x_2 \in [0,\delta]$. So we have for any $0 < x < \delta$
$$\left| f(0) – f(x)\right| \leq f(s) – f(t) < \varepsilon,$$ which means that $f(0+) = f(0)$.
As above, for any partition $P$ that contains $0$ we have:
$$U(P,f,\beta_1) = M_l \quad L(P,f,\beta_1) = m_l$$ in the interval $[0,x_l]$ of $P$. Because $f$ is right-continuous at $0$, both $M_l$ and $m_l$ converge to $f(0)$ as $x_l \to 0$, so $$\int f \ d\beta_1 = f(0).$$
(b) The statement is: $f\in \mathscr R (\beta_2)$ if and only if $f(0-) = f(0)$ and then
$$\int f \ d\beta_2 = f(0).$$ The proof is similar to part (a).
(c) Suppose that $f$ is continuous at $0$ and let $\varepsilon$ be given. Then there exists a $\delta^* > 0$ such that $\left| x \right| < 0$ implies $\left| f(x) – f(0) \right| < \varepsilon$. Let $\delta = \min\{1,\delta^*/2\}$, and $P = \{-1,-\delta,\delta,1\}$. Then
$$U(P,f,\beta_3) – L(P,f,\beta_3) = f(s) – f(t)$$ where $s,t \in [-\delta,\delta]$. But
$$f(s) – f(t) \leq \left| f(s) – f(0) \right| + \left| f(0) – f(t) \right| $$ so
$$U(P,f,\beta_3) – L(P,f,\beta_3) < 2\varepsilon$$ and $f\in \mathscr R (\beta_3)$.
Now suppose that $f\in \mathscr R (\beta_3)$ and let $\varepsilon > 0$ be given. There exists a partition $P$ such that
$$U(P,f,\beta_3) – L(P,f,\beta_3) < \varepsilon.$$ Let $P^*$ be a refinement of $P$ that contains $0$ so that the partitions around $0$ are $[x_{l-1},0]$ and $[0,x_l]$, let $\delta = \min\{\left| x_l \right| ,\left| x_{l-1} \right| \}$ and let $P^\#$ be a refinement of $P^*$ that contains $\pm \delta$. Then
$$U(P^\#,f,\beta_3) – L(P^\#,f,\beta_3) = \frac{1}{2} \left[ f(s) – f(t) + f(q) – f(r)\right] $$ where $s,t \in [0,\delta]$, $q,r \in [-\delta,0]$ $$f(s) – f(t) \geq \left| f(0) – f(x) \right|
\quad \text{for any} \quad x \in [0,\delta]$$ $$f(q) – f(q) \geq \left| f(0) – f(x) \right|
\quad \text{for any} \quad x \in [-\delta, 0].$$ So for any $x \in [-\delta,\delta]$ we have
$$\left| f(0) – f(x) \right| < \epsilon$$ which shows that $f$ is continuous at $0$.
(d) The result follows from parts (a) – (c) and the fact that if $f$ is continuous at $0$, then $f(0-) = f(0) = f(0+)$.
Exercise 4
(By Matt Frito Lundy) For any partition $P$, we have
$$U(P,f) = b-a > 0$$ $$L(P,f) = 0,$$ so $f \notin \mathscr R$.
Exercise 5
(By Matt Frito Lundy) To answer the question Does $f^2 \in \mathscr R$ imply that $f \in \mathscr R$? Consider the function
$$f(x) =
\begin{cases}
1 & x \text{ rational} \\
-1 & x \text{ irrational}.
\end{cases} $$Then $f^2 = 1$ and $f^2 \in \mathscr R$, but $f \notin \mathscr R$.
To answer the question Does $f^3 \in \mathscr R$ imply that $f \in \mathscr R$? Use the fact that $f$ is bounded on $[a,b]$ implies that $f^3$ is bounded on $[a,b]$, so there exists $m, M \in \mathbf R$ such that $m\leq f^3 \leq M$ for all $x \in [a,b]$. Let $\phi (x) = x^{1/3}$, then $\phi$ is continuous on $[m,M]$, and $f = \phi (f^3)$, so $f \in \mathscr R$ on $[a,b]$ by theorem 6.11.
Notice that $\phi(x) = x^{1/2}$ does not work for the first question because $\phi$ is not continuous on $[-1,1]$.
Exercise 6
(By analambanomenos) Following the hint, recall that $P=\cup E_n$ where $E_n$ is a set of $2^n$ disjoint close intervals of length $3^{-n}$ obtained by removing the middle thirds of the intervals in $E_{n-1}$. The total length of the intervals of $E_n$ is $(2/3)^n$ and so $\rightarrow 0$ as $n\rightarrow\infty$. We can replace the closed intervals of $E_n$, $[a_{i,n},b_{i,n}]$ with slightly larger open intervals $(a_{i,n}-\delta/2^n,b_{i,n}+\delta/2^n)$, with total length $(2/3)^n+\delta$, so that we can cover $P$ with a set of disjoint open intervals with total length as small as possible.
Now proceeding as in the proof of Theorem 6.10, let $M=\sup\big|f(x)\big|$, and cover $P$ with a collection of open intervals $(u_j,v_j)$ with total length less than $\varepsilon$. The complement $K$ of the union of the open intervals in $[a,b]$ is compact. Then $f$ is uniformly continuous on $K$ and there exists $\delta>0$ such that $\big|f(s)-f(t)\big|<
\varepsilon$ if $s,t\in K$, $|s-t|<\delta$.
Form a partition $Q=\{x_0,\ldots,x_n\}$ of $[a,b]$ such that each $u_j$ and $v_j$ occurs in $Q$, no point of any segment $(u_j,v_j)$ occurs in $Q$, and if $x_{i-1}$ is not one of the $u_j$, then $\Delta x_i<\delta$.
Note that $M_i-m_i\le 2M$ for every $i$, and that $M_i-m_i\le\varepsilon$ if $x_{i-1}$ is not one of the $u_j$. Hence
\begin{align*}
U(Q,f)-L(Q,f) &= \sum_i (M_i-m_i)\Delta x_i \\
&= \sum_{x_{i-1}\in\{u_j\}}(M_i-m_i)\Delta x_i+\sum_{x_{i-1}\notin\{u_j\}}(M_i-m_i)\Delta x_i \\
&\le 2M\varepsilon + \varepsilon(b-a)=(2M+b-a)\varepsilon.
\end{align*}Hence $f\in\mathscr R$ by Theorem 6.6.
Exercise 7
(By analambanomenos)
(a) By Theorems 6.12(c) and 6.20, $$\lim_{c\rightarrow0}\int_c^1f(x)\,dx=\int_0^1f(x)\,dx-\lim_{c\rightarrow0}\int_0^cf(x)\,dx=\int_0^1f(x)\,dx.$$
(b) Let
$$f(x) =
\begin{cases}
0 & x=0 \\
-\displaystyle\frac{2^{2n-1}}{2n-1} & 2^{-(2n-1)}<x\le2^{-(2n-2)},\;n=1,2,\ldots \\
\displaystyle\frac{2^{2n}}{2n} & 2^{-2n}<x\le2^{-(2n-1)},\;n=1,2,\ldots
\end{cases} $$Then for any positive integer $N$, $$\int_{2^{-N}}^1f(x)\,dx=\sum_{n=1}^N\frac{(-1)^n}{n}$$ which converges as $N\rightarrow\infty$ by Theorem 3.43. However,
$$\int_{2^{-N}}^1\big|f(x)\big|\,dx=\sum_{n=1}^N\frac{1}{n}$$ fails to converge as $N\rightarrow\infty$ by Theorem 3.28.
Exercise 8
(By analambanomenos) For any positive integer $n$ define $g_1(x)=f(n)$ and $g_2(x)=f(n+1)$ for $x\in(n,n+1]$. Since $f$ decreases monotonically, $g_1(x)\ge f(x)\ge g_2(x)$ for all $x\in[1,\infty)$. And since $$\int_n^{n+1}g_1(x)\,dx=f(n)\quad\quad\int_n^{n+1}g_2(x)\,dx=f(n+1)$$ we get by Theorem 6.12(b), for any positive integer $N$, $$\sum_1^Nf(n)=\int_1^Ng_1(x)\,dx\le
\int_1^Nf(x)\,dx\le\int_1^Ng_2(x)\,dx=\sum_2^{N+1}f(n).$$ Hence $\int_1^Nf(x)\,dx$ converges as $N\rightarrow\infty$ if and only if $\sum_1^Nf(n)$ converges. In that case, if $A=\sum_1^\infty f(n)$, then $\int_1^\infty f(x)\,dx$ lies between $A-f(1)$ and $A$.
Exercise 9
(By analambanomenos) If $F$ and $G$ are functions which are differentiable on $[c,1]$ for all $c>0$ and such that $F’=f\in\mathscr R$ and $G’=g\in\mathscr R$ on $[c,1]$ for all $c>0$, then by Theorem 6.22 we have $$\int_c^1F(x)g(x)\,dx=F(1)G(1)-F(c)G(c)-\int_c^1f(x)G(x)\,dx.$$ Suppose that two of the limits $$\lim_{c\rightarrow0}\int_c^1F(x)g(x)\,dx=\int_0^1F(x)g(x)\,dx,\;\;
\lim_{c\rightarrow0}\int_c^1f(x)G(x)\,dx=\int_0^1f(x)G(x)\,dx,\;\;\lim_{c\rightarrow0}F(c)G(c)$$ exist and are finite. Then the third limit exists and is finite, and we have
$$\int_0^1F(x)g(x)\,dx=F(1)G(1)-\lim_{c\rightarrow0}F(c)G(c)-\int_0^1F(x)g(x)\,dx.$$ Similarly, if $F$ and $G$ are functions which are differentiable on $[a,b]$ for all $b>a$ and such that $F’=f\in\mathscr R$ and $G’=g\in\mathscr R$ on $[a,b]$ for all $b>a$, then by Theorem 6.22 we have $$\int_a^bF(x)g(x)\,dx=F(b)G(b)-F(a)G(a)-\int_a^bf(x)G(x)\,dx.$$ Suppose that two of the limits $$\lim_{b\rightarrow\infty}\int_a^bF(x)g(x)\,dx=\int_a^\infty F(x)g(x)\,dx,\;\;\lim_{b\rightarrow\infty}\int_a^bf(x)G(x)\,dx=\int_a^\infty
f(x)G(x)\,dx,\;\;\lim_{b\rightarrow\infty}F(b)G(b)$$ exist and are finite. Then the third limit exists and is finite, and we have $$\int_a^\infty F(x)g(x)\,dx=
\lim_{b\rightarrow\infty}F(b)G(b)-F(a)G(a)-\int_a^\infty F(x)g(x)\,dx.$$
For the example, let $$F(x)=\frac{1}{1+x},\quad F’(x)=f(x)=-\frac{1}{(1+x)^2},\quad G(x)=\sin x,\quad G’(x)=g(x)=\cos x.$$ The functions $F$ and $G$ are differentiable on $[0,b)$, for all $b>0$, and $f\in\mathscr R$, $g\in\mathscr R$ on $[0,b]$ for all $b>0$. Also $$\lim_{b\rightarrow\infty}F(b)G(b)=\lim_{b\rightarrow\infty}\frac{\sin b}{1+b}=0$$ and
$$\lim_{b\rightarrow\infty}\bigg|\int_0^b\frac{\sin x}{(1+x)^2}\,dx\bigg|\le\lim_{b\rightarrow\infty}\int_0^b\frac{|\sin x|}{(1+x)^2}\,dx\le\lim_{b\rightarrow\infty}
\int_0^b\frac{1}{(1+x)^2}\,dx$$ which converges by Exercise 8 since $\sum_0^\infty1/(1+n)^2$ converges. Hence we can apply the results of the first part of this exercise and conclude that $$\int_0^\infty\frac{\cos x}{1+x}\,dx=\lim_{b\rightarrow\infty}\frac{\sin b}{1+b}-\frac{\sin 0}{1+0}+\int_0^\infty\frac{\sin x}{(1+x)^2}\,dx=\int_0^\infty
\frac{\sin x}{(1+x)^2}\,dx.$$We’ve seen above that $\sin x/(1+x)^2$ converges absolutely on $[0,\infty)$. To show that $\cos x/(1+x)$ diverges absolutely,
\begin{align*}
\int_0^\infty\frac{|\cos x|}{1+x}\,dx &= \sum_{k=0}^\infty\int_{2\pi k}^{2\pi(k+1)}\frac{|\cos x|}{1+x}\,dx \\
&\ge\sum_{k=0}^\infty\frac{1}{2\pi(k+1)+1}\int_{2\pi k}^{2\pi(k+1)}|\cos x|\,dx \\
&\ge\sum_{k=0}^\infty\frac{1}{2\pi(k+1)+2\pi}\int_0^{2\pi}|\cos x|\,dx \\
&=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+2}
\end{align*}a sum which diverges.
Exercise 10
(By analambanomenos) Note that since $q$ and $p=q/(q-1)$ are positive, then $q-1$ is positive.
(a) Fix $u$ and let $f(v)=(u^p/p)+(v^q/q)-uv$. Then $f’(v)=v^{q-1}-u$ and $f”(v)=(q-1)v^{q-2}$ is non-negative for non-negative $v$, so the critical point $v=u^{1/(q-1)}$ is a minimum. Hence
\begin{align*}
\frac{u^p}{p}+\frac{v^q}{q}-uv &\ge \frac{u^p}{p}+\frac{u^{q/(q-1)}}{q}-u^{1+1/(q-1)} \\
&=\biggl(\frac{1}{p}+\frac{1}{q}-1\biggr)u^p \\
&= 0
\end{align*}so that $uv\le(u^p/p)+(v^q/q)$. Equality holds at the critical value $v=u^{1/(q-1)}$, or $v^q=u^{q/(q-1)}=u^p$.
(b) From part (a) we have
\begin{align*}
\int_a^bfg\,d\alpha &\le \int_a^b\biggl(\frac{f^p}{p}+\frac{g^q}{q}\biggr)\,d\alpha \\
&= \frac{1}{p}\int_a^b f^p\,d\alpha+\frac{1}{q}\int_a^bg^q\,d\alpha \\
&= \frac{1}{p}+\frac{1}{q} \\
&= 1.
\end{align*}(c) Define $$A=\biggl(\int_a^b|f|^p\,d\alpha\biggr)^{1/p}\quad\quad B=\biggl(\int_a^b|g|^q\,d\alpha\biggr)^{1/q}.$$ Then
\begin{align*}
\int_a^b\bigg|\frac{f}{A}\bigg|^p\,d\alpha &= \frac{1}{A^p}\int_a^b|f|^p\,d\alpha=1 \\
\int_a^b\bigg|\frac{g}{B}\bigg|^q\,d\alpha &= \frac{1}{B^q}\int_a^b|g|^q\,d\alpha=1.
\end{align*}Applying part (b), we get $$\int_a^b\bigg|\frac{f}{A}\bigg|\cdot\bigg|\frac{g}{B}\bigg|\,d\alpha \le 1,$$ or $$\bigg|\int_a^bfg\,d\alpha\bigg|\le\int_a^b|fg|\,d\alpha\le AB=
\biggl(\int_a^b|f|^p\,d\alpha\biggr)^{1/p}\biggl(\int_a^b|g|^q\,d\alpha\biggr)^{1/q}.$$(d) Let $f\in\mathscr R$, $g\in\mathscr R$ on $[c,1]$ for all $c>0$ such that the improper integrals $\int_0^1|f|^p\,dx$ and $\int_0^1|g|^q\,dx$ exist. Then for all $c>0$ we have
$$\bigg|\int_c^1fg\,dx\bigg|\le\biggl(\int_c^1|f|^p\,dx\biggr)^{1/p}\biggl(\int_c^1|g|^q\,dx\biggr)^{1/q}.$$ Since the right side increases monotonically as $c\rightarrow0$, we can take the limit of both sides to get the desired result.
Similarly, let $f\in\mathscr R$, $g\in\mathscr R$ on $[a,b]$ for all $b>a$ such that the improper integrals $\int_a^\infty|f|^p\,dx$ and $\int_a^\infty|g|^q\,dx$ exist. Then for all $b>a$ we have $$\bigg|\int_a^bfg\,dx\bigg|\le\biggl(\int_a^b|f|^p\,dx\biggr)^{1/p}\biggl(\int_a^b|g|^q\,dx\biggr)^{1/q}.$$ Since the right side increases monotonically as $b\rightarrow\infty$, we can take the limit of both sides to get the desired result.
