Ch3. Differentiation and Integration (Ex.12 ~ Ex.20)

(i) \(F'(x)\) exists for every \(x\in\mathbb{R}\)

proof
\[ F'(x)=2x\sin\!\left(\frac{1}{x^2}\right)-\frac{2}{x}\cos\!\left(\frac{1}{x^2}\right)\quad(x\neq 0) \] \[ \begin{aligned} F'(0) &=\lim_{x\to 0}\frac{F(x)-F(0)}{x-0} =\lim_{x\to 0}\frac{1}{x}\left\{x^2\sin\!\left(\frac{1}{x^2}\right)\right\}\\ &=\lim_{x\to 0}x\sin\!\left(\frac{1}{x^2}\right)=0 \qquad(\because\ |\sin(1/x^2)|\le 1). \end{aligned} \]

(ii) \(F'(x)\) is not integrable on \([-1,1]\)

proof

\[ \begin{aligned} F'(x) &=2x\sin\!\left(\frac{1}{x^2}\right)+x^2\cos\!\left(\frac{1}{x^2}\right)(-2)x^{-3}\\ &=2x\sin\!\left(\frac{1}{x^2}\right)-\frac{2}{x}\cos\!\left(\frac{1}{x^2}\right). \end{aligned} \] \(\displaystyle \int_0^1\left|2x\sin\!\left(\frac{1}{x^2}\right)\right|dx\le \int_0^1 2x\,dx<\infty\), it remains to show \(\displaystyle \int_0^1 \frac{2}{x}\left|\cos\!\left(\frac{1}{x^2}\right)\right|dx=\infty\).

1. Let \(g(x)=\dfrac{2}{x}\left|\cos\!\left(\dfrac{1}{x^2}\right)\right|\), \(a_k=\left(2k\pi+\dfrac{\pi}{4}\right)^{-1/2}\) and \(b_k=\left(2k\pi-\dfrac{\pi}{4}\right)^{-1/2}\) \((k\in\mathbb{N},\ a_k<b_k)\). Note that \([a_k,b_k]:= I_k\) doesn’t intersect for all \(k\in\mathbb{N}\). \((b_{k+1}<a_k)\) When \(x\in I_k\), \(2k\pi-\dfrac{\pi}{4}\le \dfrac{1}{x^2}\le 2k\pi+\dfrac{\pi}{4}\) and \(\dfrac{1}{\sqrt{2}}\le \cos\!\left(\dfrac{1}{x^2}\right)\le 1\). Also, \(\dfrac{2}{x}\ge 2\cdot\dfrac{1}{b_k}\). So, \(g(x)\ge \dfrac{2}{b_k}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{b_k}\) for \(x\in I_k\) and \(g(x)\ge \sum_{k=1}^N \dfrac{\sqrt{2}}{b_k}\chi_{I_k}(x)\ (N\in\mathbb{N})\).

2. \(\displaystyle \int_0^1 g(x)\,dx=\infty\)
   Then, \[ \begin{aligned} \displaystyle \int_0^1 g(x)\,dx&\ge \sum_{k=1}^\infty \frac{\sqrt{2}}{b_k}(b_k-a_k) \\ &=\sum_{k=1}^\infty \sqrt{2}\left(1-\frac{a_k}{b_k}\right) \\ &=\sum_{k=1}^\infty \sqrt{2}\left(1-\sqrt{\frac{2k\pi-\pi/4}{2k\pi+\pi/4}}\right) \\ &=\sum_{k=1}^\infty \sqrt{2}\left(1-\sqrt{1-\frac{\pi/2}{2k\pi+\pi/4}}\right). \end{aligned} \] Since \(\sqrt{1-t}\le 1-\dfrac{1}{2}t\) for \(t\in[0,1]\), we have \(1-\sqrt{1-t}\ge \dfrac{1}{2}t\). Then, \(\displaystyle \int_0^1 g(x)\,dx\ge \sum_{k=1}^\infty \sqrt{2}\cdot\frac{1}{2}\cdot\frac{\pi/2}{2k\pi+\pi/4}=\frac{\pi\sqrt{2}}{4}\sum_{k=1}^\infty \frac{1}{2\pi(k+1/8)}>\frac{\sqrt{2}}{8}\sum_{k=1}^\infty \frac{1}{k+1}\). Since \(\displaystyle \sum_{k=1}^\infty \frac{1}{k+1}=\infty\), \(\displaystyle \int_0^1 g(x)\,dx=\infty\ \therefore\ \int_0^1 |F'(x)|\,dx=\infty\).

(a) \(\int_a^b |F'(x)|dx \le T_F(a,b)\)

proof
① \(F(x)-F(a)=P_F(a,x)-N_F(a,x)\)
\(F\) is of bounded variation on \([a,b]\), so \(F(x)-F(a)=P_F(a,x)-N_F(a,x)\), where \(P_F\) is a positive variation of \(F\) on \([a,x]\), \(N_F\) is a negative variation of \(F\) on \([a,x]\), and \(a\le x\le b\).

② \(f\overset{\mathrm{let}}{=}g+h\) and \(\exists f'\) a.e. Then \(g',h'\) exist a.e
\(G\overset{\mathrm{let}}{=}\{x:\ g'(x)\ \text{doesn't exist}\}\) and \(H\overset{\mathrm{let}}{=}\{x:\ h'(x)\ \text{doesn't exist}\}\). Then \(f'(x)\) doesn’t exist for \(x\in G\cup H\). So \(m(G\cup H)=0\).

If \(m(G)>0\) (wlog), \(m(G\cup H)\ge m(G)>0\), contradiction.
If \(m(G)\cdot m(H)>0\), \(m(G\cup H)\ge m(G)>0\).

Thus, \(m(G)=m(H)=0\) and \(g',h'\) exist a.e \(x\in[a,b]\).

③ \(|F'|\le T_F'\), conclusion
Then, \(F'(x)=P_F(a,x)'-N_F(a,x)'\) a.e \(x\in[a,b]\) (\(\because\) ②, thm 3.3.4)
Also, \(|F'(x)|=|P_F(a,x)'-N_F(a,x)'|\le P_F(a,x)'+N_F(a,x)'=T_F'(a,x)\) (lemma 3.3.2)
Note that \(T_F'(a,x)\) exists a.e

Thus, \(\int_a^b |F'(x)|dx \le \int_a^b T_F'(a,x)\,dx = T_F(a,b)\)

(b)-1 if \(\int_a^b |F'(x)|dx = T_F(a,b)\), then \(F\) is absolutely continuous

proof
\(F\) is of bounded variation, so \(T_F\) is finite. Then, \(F'\) is integrable on \([a,b]\). Thus, \(F(x)=\int_a^x F'(t)\,dt\) is absolutely continuous. (p.183)

(b)-2 \(F\) is absolutely continuous on \([a,b]\)

(c) For a rectifiable curve \(z\), \(L=\int_a^b |z'(t)|dt\) holds iff \(z\) is absolutely continuous.

proof
① \(\leftarrow\)
\(z(t)\overset{\mathrm{let}}{=}x(t)+iy(t)\in\mathbb{C}\ (x,y\in\mathbb{R}).\) Then \(x,y\) are absolutely continuous.

cf) Assume \(x\) is not absolutely continuous on \([a,b]\) (wlog). Then, there is \(\epsilon_0>0\) such that for every \(\delta>0\) with \(\sum_{i=1}^N|b_i-a_i|<\delta\) and each \((a_i,b_i)\subset[a,b]\) is disjoint, \(\sum_{i=1}^N|x(b_i)-x(a_i)|\ge \epsilon_0\) ~1)

\(z\) is absolutely continuous, so there is \(\delta_0>0\) such that \(\sum_{i=1}^N|b_i-a_i|<\delta_0\) implies \(\sum_{i=1}^N|z(b_i)-z(a_i)|<\epsilon_0\) ~2).

\[|z(b_i)-z(a_i)|=\left|(x(b_i)+iy(b_i))-(x(a_i)+iy(a_i))\right|\ge |x(b_i)-x(a_i)|,\] so \[\sum_{i=1}^N|z(b_i)-z(a_i)|\ge \sum_{i=1}^N|x(b_i)-x(a_i)|\ge \epsilon_0\ (\because\ \sim1),\] which is contradiction to \(\sim2)\).

The case when \(y\) is not absolutely continuous (only) leads to same contradiction. Thus, both \(x\) and \(y\) are absolutely continuous.

Then, using thm 3.4.1, \[ \begin{aligned} L &=\int_a^b \left(x'(t)^2+y'(t)^2\right)^{1/2}dt =\int_a^b \left(z'(t)\overline{z'(t)}\right)^{1/2}dt =\int_a^b |z'(t)|dt. \end{aligned} \]

② \(\rightarrow\)
\[ \int_a^b |z'(t)|dt=\int_a^b \left(x'(t)^2+y'(t)^2\right)^{1/2}dt=L<\infty. \] This means \(z'\) is integrable, so \(z(t)=\int_a^t z'(t)\,dt\) is absolutely continuous.

(a) \(f\) maps a set of measure \(0\) to a set of measure \(0\)

proof
\(f\) is absolutely continuous on \(\mathbb{R}\), so there is \(\delta_\epsilon>0\) such that \(\sum_{i=1}^N|f(x_i)-f(y_i)|<\epsilon\) whenever \(\sum_{i=1}^N|x_i-y_i|<\delta_\epsilon\) \((N\in\mathbb{N})\) for given \(\epsilon>0\).

for given \(\delta>0\) ✓
\(Z\) is measurable, so there is an open set \(O_\epsilon\subset\mathbb{R}\) such that \(Z\subset O_\epsilon\) and \(m(O_\epsilon-Z)=m(O_\epsilon)-m(Z)<\delta_\epsilon\) ; \(m(O_\epsilon)<\delta_\epsilon\).

Note that \(O_\epsilon\) can be written as a countable union of open & disjoint intervals in \(\mathbb{R}\): \(O_\epsilon=\bigsqcup_{k=1}^\infty (a_k,b_k)\ \overset{\mathrm{let}}{=}\ \bigsqcup_{k=1}^\infty I_k\).

\(F\) is continuous on \([a_k,b_k]\) which is compact, so there exist \(c_k,d_k\in[a_k,b_k]\) such that \(f(c_k)=\inf_{x\in I_k}f(x)\) and \(f(d_k)=\sup_{x\in I_k}f(x)\) and thus \(f(I_k)=[f(c_k),f(d_k)]\). (PMA 4.16)

Note that \(|c_k-d_k|\le (b_k-a_k)\) for all \(k\in\mathbb{N}\).
Then \(\sum_{k=1}^N|c_k-d_k|\le \sum_{k=1}^N|b_k-a_k|\le m(O_\epsilon)<\delta_\epsilon\).
This means \(\sum_{k=1}^N|f(c_k)-f(d_k)|<\epsilon\) for all \(N\in\mathbb{N}\), so \(\sum_{k=1}^\infty |f(c_k)-f(d_k)|\le \epsilon\).

Then, \[ \begin{aligned} m(f(O_\epsilon)) &=m\!\left(f\!\left(\bigsqcup_{k=1}^\infty I_k\right)\right) =m\!\left(\bigsqcup_{k=1}^\infty f(I_k)\right) \le \sum_{k=1}^\infty m(f(I_k)) =\sum_{k=1}^\infty |f(c_k)-f(d_k)| \le \epsilon. \end{aligned} \] \(Z\subset O_\epsilon\) leads to \(f(Z)\subset f(O_\epsilon)\), so \(m(f(Z))\le m(f(O_\epsilon))\le \epsilon\).

\(m(f(Z))\le \epsilon\) holds for all \(\epsilon>0\), so \(m(f(Z))=0\).

(b) \(f\) maps a measurable set to a measurable set \(E\).

① \(K_n=F\cap\overline{B_n}\) is compact

Let \(F\) be a closed set, and \(\overline{B_n}\) be a closed ball centered at the origin with radius \(n\in\mathbb{N}\). \(\overline{B_n}\) is also bounded, so it is compact. \(K_n=F\cap\overline{B_n}\) is compact as well (PMA 2.35).

② \(F=\bigcup_n K_n\), thus measurable

If \(x\in\bigcup_{n=1}^\infty K_n\), there is an integer \(N\) such that \(x\in F\cap\overline{B_N}\). So, \(x\in F\). For \(y\in F\), there is an integer \(M\) such that \(M-1<|y|\le M\). Since \(y\in\overline{B_M}\), \(y\in K_M\). So, \(y\in\bigcup_{n=1}^\infty K_n\). This leads to \(F=\bigcup_{n=1}^\infty K_n\), a countable union of measurable sets. Thus, \(F\) is measurable.

③ \(E=F_\sigma\cup(E-F_\sigma)\)

Suppose \(E\) is a measurable set in \(\mathbb{R}\). Then \(E=F_\sigma\cup(E-F_\sigma)\), where \(F_\sigma\) is a countable union of closed sets and \(E-F_\sigma\) is a null set: \(F_\sigma=\bigcup_{k=1}^\infty F_k\) & \(m(E-F_\sigma)=0\).

④ \(F_\sigma=\bigcup K_\ell\)

Every closed set \(F_k\) can be written as a countable union of compact sets: \(F_k=\bigcup_{m=1}^\infty K_{k,m}\). Then, \(F_\sigma=\bigcup_{k=1}^\infty\bigcup_{m=1}^\infty K_{k,m}\ \overset{\mathrm{let}}{=}\ \bigcup_{\ell=1}^\infty K_\ell\).

⑤ conclusion

Then, \(f(E)=f(E-F_\sigma)\cup\left(\bigcup_{\ell=1}^\infty f(K_\ell)\right)\). \(f(E-F_\sigma)\) has measure \(0\), which is measurable, and \(f(K_\ell)\) is compact, which is measurable. Since \(f(E)\) is a countable union of measurable sets, \(f(E)\) is a measurable set in \(\mathbb{R}\).

(c) \(E\) is a measurable subset of \([A,B]\), then show that \(F^{-1}(E)\cap\{x\in[a,b]:F'(x)>0\}\) is measurable (let \(H=\{x:F'>0\}\))

proof
① \(\int_{F^{-1}(O)}F'(x)\,dx=m(O)\) for open set \(O\subset[A,B]\)

\(O=\dot\bigcup_{i=1}^\infty (A_i,B_i)\) (\(\because\) \(O\) is open). \(F\) is continuous and increasing, so each \((A_i,B_i)\) can be written as \((a_i,b_i)=F^{-1}((A_i,B_i))\). Then \(F^{-1}(O)=\dot\bigcup_{i=1}^\infty F^{-1}((A_i,B_i))=\dot\bigcup_{i=1}^\infty (a_i,b_i)\).

Then, \[ \begin{aligned} \int_{F^{-1}(O)}F'(x)\,dx &=\int_{[a,b]}\chi_{\cup_i (a_i,b_i)}(x)\,F'(x)\,dx =\sum_{i=1}^\infty\int_{(a_i,b_i)}F'(x)\,dx \qquad(\mathrm{MCT})\\ &=\sum_{i=1}^\infty\big(F(b_i)-F(a_i)\big) =\sum_{i=1}^\infty|B_i-A_i| =m(O)\qquad(\because\ F\ \text{is A.C.}) \end{aligned} \]

② \(N\subset[A,B]\) with \(m(N)=0\), then \(m(F^{-1}(N)\cap H)=0\)

1. \(G_N=N\cup(G_N-N)\), \(m(G_N)=0\)
   \(N\) is measurable, so there is a \(G_\delta\) set \(G_N\subset[A,B]\) such that \(N\subset G_N\), \(m(G_N-N)=0\), and \(G_N=N\cup(G_N-N)\). Let \(G_N=\bigcap_{k=1}^\infty O_k\), where \(O_k\subset[A,B]\) is an open set, and \(E_n=\bigcap_{k=1}^n O_k\). Note that \(m(G_N)=0\).

2. \(\bigcap_{k=1}^\infty O_k=\bigcap_{n=1}^\infty E_n\)
   \(x\in\bigcap_{k=1}^\infty O_k\), then \(x\in O_k\) for all \(k\in\mathbb{N}\). This means \(x\in E_n=\bigcap_{k=1}^n O_k\) for all \(n\in\mathbb{N}\). So, \(x\in\bigcap_{n=1}^\infty E_n\).

\(x\in\bigcap_{n=1}^\infty E_n\), then \(x\in E_n=\bigcap_{k=1}^n O_k\) for all \(n\in\mathbb{N}\). \(x\in O_k\) for \(1\le k\le n\) and this holds for all \(n\in\mathbb{N}\), so \(x\in O_k\ \forall k\in\mathbb{N}\). Thus \(x\in\bigcap_{k=1}^\infty O_k\).

3. \(F^{-1}(G_N)\) is a \(G_\delta\)
   Then, \(F^{-1}(G_N)=F^{-1}\!\left(\bigcap_{n=1}^\infty E_n\right)=\bigcap_{n=1}^\infty F^{-1}(E_n)\). \(F\) is continuous, so \(F^{-1}(E_n)\) is open and \(F^{-1}(G_N)\) is a set of \(G_\delta\). Note that \(F^{-1}(E_n)\supset F^{-1}(E_{n+1})\) (\(\because\ E_n\supset E_{n+1}\)).

4. \(\int_{F^{-1}(G_N)} F' = m(G_N)=0\) \((\because\ F^{-1}(G_N)\supset F^{-1}(E_n))\) \(\{\chi_{F^{-1}(E_n)}(x)F'(x)\}\) is a sequence of measurable functions and \(\chi_{F^{-1}(E_n)}(x)F'(x)\downarrow \chi_{F^{-1}(G_N)}(x)F'(x)\). Using MCT, \[ \lim_{n\to\infty}\int_{F^{-1}(E_n)}F'(x)\,dx=\int_{F^{-1}(G_N)}F'(x)\,dx. \qquad(\because\ ①) \] Since \(\int_{F^{-1}(E_n)}F'(x)\,dx=m(E_n)\) and \(\lim_{n\to\infty}m(E_n)=m(G_N)\) (each \(m(E_n)<\infty\)), \(\int_{F^{-1}(G_N)}F'=m(G_N)=0\).

5. \(m(F^{-1}(N)\cap H)=0\) So, \(m_*(F^{-1}(N)\cap H)\le m(F^{-1}(G_N)\cap H)\) (\(\because\ N\subset G_N,\ F^{-1}(G_N)\ \&\ H\ \text{measurable?}\)) \[ \begin{aligned} &=m(\{x\in F^{-1}(G_N):F'(x)>0\}) =m\!\left(\bigcup_{n=1}^\infty\left\{x\in F^{-1}(G_N):F'(x)\ge \frac{1}{n}\right\}\right)\\ &=\lim_{n\to\infty}m\!\left(\left\{x\in F^{-1}(G_N):F'(x)\ge \frac{1}{n}\right\}\right) \le \lim_{n\to\infty}\left(\frac{1}{1/n}\right)\int_{F^{-1}(G_N)}F'(x)\,dx \qquad(\text{Chebyshev inequality})\\ &=\lim_{n\to\infty}0=0 \end{aligned} \] ( \(E_\alpha:=\{f>\alpha\}\Rightarrow m(E_\alpha)\le \frac{1}{\alpha}\int f\) )

Thus, \(m(F^{-1}(N)\cap H)=0\).

③ \(F^{-1}(E)\cap H\) is measurable

1. \(G_E=E\cup(G_E-E),\ G_H=H\cup(G_H-H)\)
   \(E\) is measurable, so there is a \(G_\delta\) set \(G_E=\bigcap_{n=1}^\infty E_n\) such that \(E_n\) is open, \(E\subset G_E\subset[A,B]\), \(G_E=E\cup(G_E-E)\), and \(m(G_E-E)=0\). \(H\) is measurable, so there is a \(G_\delta\) set \(G_H=\bigcap_{i=1}^\infty H_i\) such that \(H_i\) is an open set, \(H\subset G_H\subset[a,b]\), \(G_H=H\cup(G_H-H)\), and \(m(G_H-H)=0\).

2. \(F^{-1}(G_E)\cap G_H\) is a \(G_\delta\) \[ \begin{aligned} F^{-1}(G_E)\cap G_H &=F^{-1}\!\left(\bigcap_{n=1}^\infty E_n\right)\cap\left(\bigcap_{i=1}^\infty H_i\right) =\left(\bigcap_{n=1}^\infty F^{-1}(E_n)\right)\cap\left(\bigcap_{i=1}^\infty H_i\right)\\ &=\bigcap_{n,i}\left(F^{-1}(E_n)\cap H_i\right). \end{aligned} \] \(F^{-1}(E_n)\) is open (\(\because\) \(F\) is continuous), as well as \(H_i\), so \(F^{-1}(E_n)\cap H_i\) is open.
   So, \(F^{-1}(G_E)\cap G_H\) is a set of \(G_\delta\).

3. \((F^{-1}(G_E)\cap G_H)-(F^{-1}(E)\cap H)\subset (F^{-1}(G_E-E)\cap H)\cup(G_H-H)\) \[ \begin{aligned} &\{F^{-1}(G_E)\cap G_H\}-\{F^{-1}(E)\cap H\}\\ &=\{F^{-1}(G_E)\cap G_H\}\cap\{F^{-1}(E)\cup H\}^c\\ &=F^{-1}(G_E)\cap\left[G_H\cap\{F^{-1}(E)\cup H\}^c\right]\qquad(\because\ [F^{-1}(E)]^c=F^{-1}(E^c))\\ &=F^{-1}(G_E)\cap\left[\{G_H\cap F^{-1}(E^c)\}\cup\{G_H-H\}^c\right]\\ &=\{F^{-1}(G_E)\cap G_H\cap F^{-1}(E^c)\}\cup\{F^{-1}(G_E)\cap(G_H-H)\}\\ &=\{F^{-1}(G_E-E)\cap G_H\}\cup\{F^{-1}(G_E)\cap(G_H-H)\}\\ &=\{F^{-1}(G_E-E)\cap H\}\cup\{F^{-1}(G_E-E)\cap(G_H-H)\}\cup\{F^{-1}(G_E)\cap(G_H-H)\}\\ &=\{F^{-1}(G_E-E)\cap H\}\cup\{F^{-1}(G_E)\cap(G_H-H)\}\\ &\subset \{F^{-1}(G_E-E)\cap H\}\cup(G_H-H). \end{aligned} \]

4. \(m\big((F^{-1}(G_E)\cap G_H)-(F^{-1}(E)\cap H)\big)=0\) \(m(G_E-E)=0\), so \(m(\{F^{-1}(G_E-E)\cap H\})=0\) (\(\because\) ②). Since \(m(G_H-H)=0\), \[m\big(\{F^{-1}(G_E)\cap G_H\}-\{F^{-1}(E)\cap H\}\big) \le m(\{F^{-1}(G_E-E)\cap H\})+m(G_H-H)=0.\] So, \(m\big(\{F^{-1}(G_E)\cap G_H\}-\{F^{-1}(E)\cap H\}\big)=0\).

5. conclusion Since \(F^{-1}(G_E)\cap G_H=(F^{-1}(E)\cap H)\cup\{(F^{-1}(G_E)\cap G_H)-(F^{-1}(E)\cap H)\}\), \(F^{-1}(E)\cap H\) is measurable (Cor 1.3.5).