Ch3. Differentiation and Integration (Ex.21 ~ Ex.32)

(a) Show that \(f(F(x))F'(x)\) is measurable on \([a,b]\)

proof
① \(\chi_I(F(x))F'(x)\), where \(I=[P,Q]\subset[A,B]\)

Let \(P=F(p)\) and \(Q=F(q)\) \((p,q\in[a,b])\). \(F\) is continuous and increasing, so \([p,q]=F^{-1}([P,Q])\).

Then, \(\chi_I(F(x))=1\) when \(P\le F(x)\le Q\), which is \(p\le x\le q\). So, \(\chi_I(F(x))=\chi_{[p,q]}(x)\), which is measurable.

\(F'\) is measurable (\(\because\) integrable), so \(\chi_I(F(x))F'(x)\) is measurable.

Note that \[ \int_{[a,b]} \chi_I(F(x))F'(x)\,dx =\int_{[a,b]}\chi_{[p,q]}(x)F'(x)\,dx =\int_p^q F'(x)\,dx =Q-P =|I| \qquad(\because\ F\ \text{is A.C.}) \]

② \(\chi_O(F(x))F'(x)\), \(O\) is an open set in \([A,B]\)

\(\chi_O(F(x))=1\Leftrightarrow F(x)\in O \Leftrightarrow x\in F^{-1}(O)\Leftrightarrow \chi_{F^{-1}(O)}(x)=1\).
Since \(F\) is continuous, \(F^{-1}(O)\) is open (in \([a,b]\)) \(\to\) measurable.
Then, \(\chi_O(F(x))=\chi_{F^{-1}(O)}(x)\) is measurable.

Thus, \(\chi_O(F(x))F'(x)\) is measurable.

③ \(\chi_G(F(x))F'(x)\), where \(G=\bigcap_{i=1}^\infty O_i\), \((O_i\ \text{is open})\subset[A,B]\)

\(\chi_G(F(x))=1\Leftrightarrow F(x)\in G \Leftrightarrow x\in F^{-1}\!\left(\bigcap_i O_i\right)=\bigcap_i F^{-1}(O_i)\Leftrightarrow \chi_{F^{-1}(G)}(x)=1\).
Then, \(\chi_G(F(x))=\chi_{F^{-1}(G)}(x)\) is measurable.
(\(\because\ \bigcap_i F^{-1}(O_i)\) is a countable union of open (\(\to\) measurable) sets)

\(\therefore\ \chi_G(F(x))F'(x)\) is measurable.

④ \(\chi_Z(F(x))F'(x)\), where \(m(Z)=0\) & \(Z\subset[A,B]\)

There is a covering of \(Z\subset\bigcup_i I_i\), such that \(\sum_i|I_i|\le m(Z)+\epsilon\) for given \(\epsilon>0\).

\(Z\subset\bigcup_i I_i\) leads to \(\chi_Z\le \chi_{\cup_i I_i}\le \sum_i\chi_{I_i}\), so \(\chi_Z(F(x))F'(x)\le \sum_i\chi_{I_i}(F(x))F'(x)\).

Then, \[ \begin{aligned} \int_a^b \chi_Z(F(x))F'(x)\,dx &\le \int_a^b \sum_i\chi_{I_i}(F(x))F'(x)\,dx\\ &=\sum_i\int_a^b \chi_{I_i}(F(x))F'(x)\,dx \qquad(\mathrm{MCT}\ \text{Cor }2.1.10)\\ &=\sum_i |I_i|\qquad(\because\ ①)\\ &<\epsilon \ \Rightarrow\ \text{integrable} \end{aligned} \]

\(\chi_Z\) is a characteristic function (\(\ge 0\)), and \(F\) is increasing (\(F'\ge 0\) a.e) so \(\chi_Z(F(x))F'(x)\ge 0\) a.e \(x\in[a,b]\) and \(\int_a^b \chi_Z(F(x))F'(x)\,dx\ge 0\).

\(0\le \int_a^b \chi_Z(F(x))F'(x)\,dx<\epsilon\) holds for all \(\epsilon>0\), so \(\int_a^b \chi_Z(F(x))F'(x)\,dx=0\) and \(\chi_Z(F(x))F'(x)=0\) a.e \(x\in[a,b]\).
Thus, \(\chi_Z(F(x))F'(x)\) is measurable.

⑤ \(\chi_E(F(x))F'(x)\), \(E\) is measurable

\(G=E\cup(G-E)\), where \(G=\bigcap_i O_i\), \(E\subset G\), and \(m(G-E)=0\).
Then, \(\chi_E(F(x))=\chi_G(F(x))-\chi_{G-E}(F(x))\).

Since \(\chi_G(F(x))F'(x)\), \(\chi_{G-E}(F(x))F'(x)\) is measurable (\(\because\) ③, ④)
\(\chi_E(F(x))F'(x)=\chi_G(F(x))F'(x)-\chi_{G-E}(F(x))F'(x)\) is measurable.

⑥ \(f(F(x))F'(x)\), conclusion

\(f\) is measurable, so there is a sequence of simple functions \(\{\psi_k\}\) such that \(|\psi_k|\le |\psi_{k+1}|\) and \(\lim_{k\to\infty}\psi_k(t)=f(t)\) (thm 1.4.2). Let \(\psi_k(t)=\sum_{i=1}^{N_k} a_{k,i}\chi_{E_{k,i}}(t)\), where \(a_{k,i}\in\mathbb{R}\) and \(E_{k,i}\) is a measurable set.

Let \(\phi_k(x)=\psi_k(F(x))F'(x)=\sum_{i=1}^{N_k} a_{k,i}\chi_{E_{k,i}}(F(x))F'(x)\).
Each \(\chi_{E_{k,i}}(F(x))F'(x)\) is measurable (\(\because\) ⑤), so their finite combination \(\phi_k\) is measurable.

Since \(\lim_{k\to\infty}\phi_k(x)=f(F(x))F'(x)\), \(f(F(x))F'(x)\) is a limit of measurable functions. Thus, \(f(F(x))F'(x)\) is measurable.

(b) If \(f\) is integrable on \([A,B]\), show that \(f(F(x))F'(x)\) is integrable and \(\int_A^B f(y)\,dy=\int_a^b f(F(x))F'(x)\,dx\)

proof
① \(\chi_I(F(x))F'(x)\), where \(I=[P,Q]\subset[A,B]\)

We’ve shown that \(\int_a^b \chi_I(F(x))F'(x)\,dx=|I|=\int_A^B \chi_I(y)\,dy\).

② \(\chi_O(F(x))F'(x)\), \(O\) is an open set in \([A,B]\)

\(O=\dot\bigcup_{i=1}^\infty (A_i,B_i)\) (\(\because\) \(O\) 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, \[ \begin{aligned} \int_a^b \chi_O(F(x))F'(x)\,dx &=\int_{F^{-1}(O)}F'(x)\,dx =\int_{[a,b]}\chi_{\cup_i (a_i,b_i)}(x)F'(x)\,dx\\ &=\sum_{i=1}^\infty\int_{(a_i,b_i)}F'(x)\,dx \qquad(\mathrm{MCT}) =\sum_{i=1}^\infty\big(F(b_i)-F(a_i)\big)\\ &=\sum_{i=1}^\infty|B_i-A_i| =m(O) =\int_A^B \chi_O(y)\,dy. \end{aligned} \] Thus, \(\int_a^b \chi_O(F(x))F'(x)\,dx=\int_A^B \chi_O(y)\,dy\).

③ \(\chi_G(F(x))F'(x)\), where \(G=\bigcap_i O_i\), \((O_i\ \text{is open})\subset[A,B]\)

Let \(S_n=\bigcap_{i=1}^n O_i\), which is open. Then \(\bigcap_{n=1}^\infty S_n=\bigcap_{i=1}^\infty O_i\). Note that \(S_n\supset S_{n+1}\).

(\(\because\) \(x\in\bigcap_{n=1}^\infty S_n\), then \(x\in S_n\ \forall n\in\mathbb{N}\Rightarrow x\in O_i\) for \(1\le i\le n\ \forall n\). So \(x\in O_i\ \forall i\in\mathbb{N}\) and \(x\in\bigcap_i O_i\).
\(x\in\bigcap_i O_i\), then \(x\in O_i\ \forall i\in\mathbb{N}\Rightarrow x\in\bigcap_{i=1}^n O_i=S_n\ \forall n\). So \(x\in\bigcap_n S_n\).)

\(\{\chi_{F^{-1}(S_n)}(x)F'(x)\}\) is a sequence of measurable functions s.t. \(\chi_{F^{-1}(S_n)}(x)F'(x)\downarrow \chi_{F^{-1}(G)}(x)F'(x)\) (\(\because\ S_n\downarrow G\)). Using MCT, \[ \lim_{n\to\infty}\int_{F^{-1}(S_n)}F'(x)\,dx=\int_{F^{-1}(G)}F'(x)\,dx. \] \(S_n\) is open, so \(\int_{F^{-1}(S_n)}F'(x)\,dx=m(S_n)\). Since \(\lim_{n\to\infty}m(S_n)=m(G)\), \(\int_{F^{-1}(G)}F'=m(G)\).

So, \[ \int_a^b \chi_G(F(x))F'(x)\,dx =\int_{F^{-1}(G)}F'(x)\,dx =m(G) =\int_A^B \chi_G(y)\,dy. \]

④ \(X_Z(F(x))F'(x)\), where \(m(Z)=0\) & \(Z\subset [A,B]\)

We’ve shown \[\begin{aligned} \int_a^b X_Z(F(x))F'(x)\,dx = m(Z) = \int_A^B X_Z(y)\,dy \end{aligned}\]

⑤ \(X_E(F(x))F'(x)\), \(E\) is measurable

\(G=E\cup(G-E)\), then \(X_E(F(x))F'(x)=X_G(F(x))F'(x)-X_{G-E}(F(x))F'(x)\)

Then, \[\begin{aligned} \int_a^b X_E(F(x))F'(x)\,dx = m(G)-m(G-E)=m(E)=\int_A^B X_E(y)\,dy \end{aligned}\]

cf) When \(\psi(y)=\sum_{i=1}^N a_i X_{E_i}(y)\), \(\psi(F(x))=\sum_{i=1}^N a_i X_{E_i}(F(x))\), so

\[\begin{aligned} \int_a^b \psi(F(x))F'(x)\,dx &= \sum_{i=1}^N a_i \int_a^b X_{E_i}(F(x))F'(x)\,dx \\ &= \sum_{i=1}^N a_i \int_A^B X_{E_i}(y)\,dy = \int_A^B \psi(y)\,dy \end{aligned}\]

⑥

Note that \(|f(F(x))|F'(x)=(|f|\circ F(x))F'(x)\) \((\because F'\ge 0\ \text{a.e})\)

\(Hf\) is measurable, so there is a sequence of simple functions \(\{\phi_n\}\) such that \(0\le \phi_n\le \phi_{n+1}\) and \(\lim_{n\to\infty}\phi_n=|Hf|\) (thm 1.4.1). \(\phi_n\uparrow |Hf|\), so \(\lim_{n\to\infty}\int_{[A,B]}\phi_n=\int_{[A,B]}|Hf|\) (MCT).

\(\phi_n\) is measurable, so \(\int_{[A,B]}\phi_n(y)\,dy=\int_{[a,b]}\phi_n(F(x))F'(x)\,dx\)

\(\phi_n(F(x))F'(x)\uparrow (|Hf|\circ F)(x)\cdot F'(x)\), so \(\lim_{n\to\infty}\int_{[a,b]}(\phi_n\circ F)(x)F'(x)\,dx=\int_{[a,b]}(|Hf|\circ F)(x)F'(x)\,dx\) (MCT)

The limit is same, so \(\int_{[a,b]}(Hf\circ g)(x)g'(x)\,dx=\int_{[A,B]}Hf(y)<\infty\) \((\because f\ \text{is integrable})\)

Thus, \((f\circ g)(x)g'(x)\) is integrable

Using the similar logic above with the existence of a sequence of simple functions \(\{\psi_k\}\) s.t. \(|\psi_k|\le |\psi_{k+1}|\) & \(\lim_{k\to\infty}\psi_k=f\), we can obtain the following result : thm 1.4.2

\[\int_{[a,b]} f(g(x))g'(x)\,dx=\int_A^B f(y)\,dy\]

(b) If \(f\) is integrable on \([A,B]\), show that \(f(F(x))F'(x)\) is integrable and

\[\int_A^B f(y)\,dy=\int_a^b f(F(x))F'(x)\,dx\]

proof Consider the case \(f\ge 0\) first. Let \(\tilde{F}(x)=\int_A^x f(t)\,dt\) and \(\widetilde{F}(x)=\int_a^x f(F(t))F'(t)\,dt\).

For \(n\in\mathbb{N}\), define \(f_n(x)\) as follows; \[ f_n(x)= \begin{cases} f(x) & (0\le f(x)\le n)\\ n & (f(x)>n) \end{cases} \]

Let \(\tilde{F}_n(x)=\int_A^x f_n(t)\,dt\) and \(\widetilde{F}_n(x)=\int_a^x f_n(F(t))F'(t)\,dt\) accordingly.

① \(f_n(F(x))F'(x)\) is integrable \(\ \forall n\in\mathbb{N}\)

\[\begin{aligned} \int_a^b f_n(F(x))F'(x)\,dx &= \int_A^B f_n(y)\,dy \quad (F(x)=y) \\ &\le \int_A^B f(y)\,dy < \infty \quad (\because f\in L^1(\mathbb{R})) \end{aligned}\]

② \(\tilde{F}_n\) follows Lipschitz conditions

For \(x\le y\in[A,B]\), \[\begin{aligned} |\tilde{F}_n(y)-\tilde{F}_n(x)| &= \left|\int_x^y f_n(t)\,dt\right| \\ &\le \int_x^y |f_n(t)|\,dt \\ &\le \int_x^y n\,dt = n(y-x) \end{aligned}\]

\(y\le x\in[A,B]\) leads to \(|\tilde{F}_n(y)-\tilde{F}_n(x)|\le n(x-y)\), so \(|\tilde{F}_n(y)-\tilde{F}_n(x)|\le n|y-x|\)

③ \(h\) is Lipschitz and \(g\) is absolutely continuous on \(\mathbb{R}\), then \(h\circ g\) is absolutely continuous on \(\mathbb{R}\)

There is \(C\ge 0\) such that \(|h(y)-h(x)|\le C|y-x|\) for all \(x,y\in\mathbb{R}\). For given \(\varepsilon>0\), there is \(\delta_\varepsilon>0\) such that whenever \(\sum_{i=1}^N|b_i-a_i|<\delta_\varepsilon\), \(\sum_{i=1}^N|g(b_i)-g(a_i)|<\varepsilon/C\)

\(g(a_i),g(b_i)\in\mathbb{R}\), so \(|h(g(b_i))-h(g(a_i))|\le C|g(b_i)-g(a_i)|\).

Then, \(\sum_{i=1}^N|h\circ g(b_i)-h\circ g(a_i)|\le \sum_{i=1}^N C|g(b_i)-g(a_i)|\le C\cdot \varepsilon/C=\varepsilon\) whenever \(\sum_{i=1}^N|b_i-a_i|<\delta_\varepsilon\).

This holds for all \(\varepsilon>0\), so \(h\circ g\) is absolutely continuous on \(\mathbb{R}\).

④ \((\tilde{F}_n\circ F)'(x)=\widetilde{F}_n'(x)\ \ a.e\)

We’ve shown \((\tilde{F}_n\circ F)(x)\) is absolutely continuous. Also, \(\widetilde{F}_n\) is absolutely continuous (\(\because\) ①), so \(\tilde{F}_n\circ F-\widetilde{F}_n\) is absolutely continuous. This means \((\tilde{F}_n\circ F-\widetilde{F}_n)'\) exists a.e \([a,b]\).

\[\begin{aligned} \big((\tilde{F}_n\circ F)(x)-\widetilde{F}_n(x)\big)' &= \tilde{F}_n'(F(x))F'(x)-f_n(F(x))F'(x) \\ &= 0\ \ a.e, \end{aligned}\] so \(\tilde{F}_n\circ F-\widetilde{F}_n\) is constant.

Then, \(\tilde{F}_n\circ F(x)-\widetilde{F}_n(x)=\tilde{F}_n\circ F(a)-\widetilde{F}_n(a)=\tilde{F}_n(A)-\widetilde{F}_n(a)=0\ \ \forall x\in[a,b]\).

So, \[\int_A^B f_n(x)\,dx=\int_a^b f_n(F(x))F'(x)\,dx\]

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cf) \(FG\) is absolutely continuous on \([a,b]\)

proof ① bounded
\(F\) and \(G\) are continuous on a compact set \([a,b]\), so they are bounded.
Then, there is \(M>0\) such that \(|F|\le M\) and \(|G|\le M\)

②
For given \(\varepsilon>0\), there is \(\delta_\varepsilon>0\) such that whenever \(\sum_{i=1}^N|b_i-a_i|<\delta_\varepsilon\) holds, \(\sum_{i=1}^N|F(b_i)-F(a_i)|<\varepsilon/(2M)\) and \(\sum_{i=1}^N|G(b_i)-G(a_i)|<\varepsilon/(2M)\).

③
Then, whenever \(\sum_{i=1}^N|b_i-a_i|<\delta_\varepsilon\),

\[\begin{aligned} \sum_{i=1}^N |F(b_i)G(b_i)-F(a_i)G(a_i)| &= \sum_{i=1}^N |F(b_i)G(b_i)-F(a_i)G(b_i)+F(a_i)G(b_i)-F(a_i)G(a_i)| \\ &\le \sum_{i=1}^N |G(b_i)|\,|F(b_i)-F(a_i)|+\sum_{i=1}^N |F(a_i)|\,|G(b_i)-G(a_i)| \\ &< \sum_{i=1}^N M|F(b_i)-F(a_i)|+\sum_{i=1}^N M|G(b_i)-G(a_i)| \\ &< M\cdot \frac{\varepsilon}{2M}+M\cdot \frac{\varepsilon}{2M}=\varepsilon \end{aligned}\]

This holds for all \(\varepsilon>0\), so \(FG\) is absolutely continuous.

(a) \([F(x)G(x)]_{x=a}^{x=b}=\int_a^b F(x)G'(x)\,dx+\int_a^b F'(x)G(x)\,dx\)

proof \(F\) and \(G\) are absolutely continuous, so \(F'\) and \(G'\) exist a.e \([a,b]\).
(\(\because\) (o)) \(FG\) is absolutely continuous, so \((FG)'\) exists a.e \([a,b]\).

\((FG)'=F'G+FG'\) (a.e), and \(\int_a^b (FG)'=\int_a^b F'G+\int_a^b FG'\).

Thus, \([F(x)G(x)]_{x=a}^{x=b}=\int_a^b F(x)G'(x)\,dx+\int_a^b F'(x)G(x)\,dx\)

(b) Let \(F\) be absolutely continuous on \([-\pi,\pi]\) with \(F(-\pi)=F(\pi)\), \(a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}F(x)e^{-inx}\,dx\), and \(F(x)\sim \sum_n a_ne^{inx}\). Show that \(F'(x)\sim \sum_n ina_ne^{inx}\)

proof Let \(b_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}F'(x)e^{-inx}\,dx\) be a Fourier coefficient of \(F'\).

Using the result of (a),

\[\begin{aligned} b_n &=\frac{1}{2\pi}\int_{-\pi}^{\pi}F'(x)e^{-inx}\,dx \\ &=-\frac{1}{2\pi}\int_{-\pi}^{\pi}F(x)(-in)e^{-inx}\,dx+\Big[F(x)e^{-inx}\Big]_{x=-\pi}^{x=\pi} \\ &=in\cdot \frac{1}{2\pi}\int_{-\pi}^{\pi}F(x)e^{-inx}\,dx + F(\pi)e^{-in\pi}-F(-\pi)e^{in\pi} \\ &=in\cdot a_n+0\ \ (\because F(\pi)=F(-\pi),\ e^{-in\pi}=e^{in\pi+i(2n)\pi}=e^{in\pi}) \\ &=in\cdot a_n \end{aligned}\]

Thus, \(F'(x)\sim \sum_n (ina_n)e^{inx}\)

(c) What happens if \(F(\pi)\ne F(-\pi)\)? (Consider \(F(x)=x\), and \(G(x)=e^{-inx}\))

proof \(F\) is absolutely continuous on \([-\pi,\pi]\).

\[\begin{aligned} a_n &=\frac{1}{2\pi}\int_{-\pi}^{\pi}xe^{-inx}\,dx = \frac{1}{2\pi}\Big[x\cdot\Big(\frac{1}{-in}\Big)e^{-inx}\Big]_{x=-\pi}^{x=\pi}-\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{-in}e^{-inx}\,dx \\ &=\frac{1}{2\pi}\Big[\frac{1}{-in}xe^{-inx}\Big]_{x=-\pi}^{x=\pi}-\frac{1}{2\pi}\Big[\frac{1}{(in)^2}e^{-inx}\Big]_{x=-\pi}^{x=\pi} \\ &=\frac{1}{2\pi}\Big[\frac{1}{-in}xe^{-inx}+\frac{1}{n^2}e^{-inx}\Big]_{x=-\pi}^{x=\pi} \\ &=\frac{1}{2\pi}\Big\{\Big(\frac{1}{-in}\pi e^{-in\pi}+\frac{1}{n^2}e^{-in\pi}\Big)-\Big(\frac{1}{-in}(-\pi)e^{in\pi}+\frac{1}{n^2}e^{in\pi}\Big)\Big\} \\ &=\frac{1}{2\pi}\Big(\frac{1}{-in}\pi(-1)^n+\frac{1}{in}\pi(-1)^n\Big) =\frac{i}{n}(-1)^n\quad (n\ne 0) \end{aligned}\]

and \(a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}x\,dx=0\).

\(F'(x)=1\), so \(b_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}1\,dx=1\), and

\[\begin{aligned} b_n &=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-inx}\,dx =\frac{1}{2\pi}\left[\frac{-1}{in}e^{-inx}\right]_{x=-\pi}^{x=\pi} \\ &=\frac{1}{2\pi}\left\{\left(\frac{-1}{in}(-1)^n\right)-\left(\frac{-1}{in}(-1)^n\right)\right\} =0 \quad (n\ne 0) \end{aligned}\]

For all \(n\in\mathbb{Z}\), \(b_n\ne ina_n\).

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(a) Suppose \((D^+F)(x)\ge 0\) for every \(x\in[a,b]\). Show that \(F\) is increasing on \([a,b]\)

proof ① \(D^+G(x)>0\) for continuous \(G\) on \([a,b]\), \(G\) is increasing

Assume there is \(c,d\in(a,b)\) such that \(c<d\) and \(G(x_1)>G(x_2)\) for all \(x_1,x_2\in[c,d]\) with \(x_1<x_2\). Intermediate Value Thm guarantees that for \(\mu\in(G(c),G(d))\), there is \(\xi\in(c,d)\) such that \(G(\xi)=\mu\) \((\because G\ \text{is continuous})\).

For each \(t\in(\xi,d)\), \(\frac{G(t)-G(\xi)}{t-\xi}<0\). So \(D^+G(\xi)=\limsup_{t\to\xi+}\frac{G(t)-G(\xi)}{t-\xi}\le 0\), contradiction. Thus, \(G\) is increasing

② \(F\) is increasing

Let \(G_\varepsilon(x)=F(x)+\varepsilon x\), which is continuous, for given \(\varepsilon>0\).

\[\begin{aligned} D^+G_\varepsilon(x) &=\limsup_{h\to 0+}\frac{1}{h}\Big(F(x+h)+\varepsilon(x+h)-F(x)-\varepsilon x\Big) \\ &=\limsup_{h\to 0+}\left\{\frac{1}{h}(F(x+h)-F(x))+\varepsilon\right\} \\ &=\limsup_{h\to 0+}\frac{1}{h}(F(x+h)-F(x))+\varepsilon = D^+F(x)+\varepsilon>0, \end{aligned}\]

cf) \(\limsup(f(x)+g(x))\le \limsup f(x)+\limsup g(x)\)
\(\to\) if one of them have lim, \(=\)

So \(G_\varepsilon\) is increasing \((\because\) ①\()\)

Then, for \(x<y\in[a,b]\), \[F(y)-F(x)=G_\varepsilon(y)-G_\varepsilon(x)-\varepsilon(y-x)\ge -\varepsilon(y-x).\] So \(\frac{F(y)-F(x)}{y-x}\ge -\varepsilon\). This holds for all \(\varepsilon>0\), so \(\frac{F(y)-F(x)}{y-x}\ge 0\) and \(F(y)\ge F(x)\). This holds for all \(x,y\in[a,b]\) with \(x<y\), so \(F\) is increasing.

(b) \(F'(x)\) exists for every \(x\in(a,b)\) and \(|F'(x)|\le M\). Show that \(|F(x)-F(y)|\le M|x-y|\) and \(F\) is absolutely continuous.

proof ① \(|F(x)-F(y)|\le M|x-y|\)

\(F\) is continuous on \((a,b)\) and differentiable on \((a,b)\), so there is \(c\) btw \(x\) and \(y\) such that \(F(x)-F(y)=F'(c)(x-y)\) (MVT). Since \(|F'(c)|\le M\), \(|F(x)-F(y)|\le M|x-y|\).

② \(F\) is absolutely continuous

For a finite collection \(\{(a_i,b_i)\}_{i=1}^N\) of disjoint intervals in \(\mathbb{R}\), \[\sum_{i=1}^N |F(b_i)-F(a_i)|\le M\sum_{i=1}^N (b_i-a_i)\ \ (\because ①)\]

For given \(\varepsilon>0\), choose \(\delta_\varepsilon=\varepsilon/M\) and \(\{(a_i,b_i)\}\) such that \(\sum_{i=1}^N(b_i-a_i)<\delta_\varepsilon\). Then, \(\sum_{i=1}^N|F(b_i)-F(a_i)|<\varepsilon\).

This holds for all \(\varepsilon>0\), so \(F\) is absolutely continuous.

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(a) Show that \(F=F_A+F_C+F_J\), where \(F_A\) is absolutely continuous, \(F_C\) is continuous and \(F_C'(x)=0\) a.e, and \(F_J\) is a jump function.

proof ① \(F-F_J\) is increasing & continuous

\(F\) is increasing on \([a,b]\), so \(F(a)\le F(x)\le F(b)\) which means \(F\) is bounded.
Let \(F_J\) be a jumpfunction of \(F\), then \(F-F_J\) is increasing & continuous (lem 3.3.13)
let \(f=F-F_J\). Note that \(F\) is increasing, so as \(F_J\)

② \(F_A(x)=\int_a^x f'(t)\,dt\)

\(f\) is increasing and continuous, so \(f'\) exists a.e. Also, \(f\) is bounded (\(::\) increasing on \([a,b]\)), so \(f'\) is integrable.

Let \(F_A(x)=\int_a^x f'(t)\,dt\). Since \(f'\) is integrable, \(F_A(x)\) is absolutely continuous.

For \(c<d\in[a,b]\), \(F_A(d)-F_A(c)=\int_{[c,d]} f'(t)\,dt\ge 0\) (\(::\) \(f\) is increasing, \(f'\ge 0\) a.e)

③ \(F_C(x)=f(x)-F_A(x)\)

1. \(F_C(x)\) is continuous and \(F_C'(x)=0\) a.e Let \(F_C(x)=f(x)-F_A(x)\). \(f\) and \(F_A\) are continuous, so \(F_C\) is continuous.
   \(f'\) exists a.e and \(F_A'=f'\) a.e (\(::\) \(F_A\) is absolutely continuous), so \(F_C'=0\) a.e.

2. \(F_C\) is increasing For \(c<d\in[a,b]\), \(F_C(d)-F_C(c)=(f(d)-f(c))-(F_A(d)-F_A(c))=f(d)-f(c)-\int_{[c,d]} f'\) Let \(g_n(x)=\frac{f(x+1/n)-f(x)}{1/n}\) which is measurable function and \(g_n\ge 0\) (\(::\) \(f\) is increasing)
   Since \(f'\) exist a.e, \(\lim_{n\to\infty} g_n(x)=f'(x)\) a.e. Then, \[\begin{aligned} \int_{[c,d]} f' &\le \liminf_{n\to\infty}\int_{[c,d]} g_n \quad (\because \text{Fatou, lemma 2.1.7})\\ &= \liminf_{n\to\infty}\frac{1}{1/n}\int_c^d\big(f(x+1/n)-f(x)\big)\,dx \quad (\because f\ \text{is continuous, so Riemann integrable})\\ &= \liminf_{n\to\infty}\frac{1}{1/n}\Big\{\big(\tilde{F}(d+1/n)-\tilde{F}(d)\big)-\big(\tilde{F}(c+1/n)-\tilde{F}(c)\big)\Big\} \quad (\text{let }\int f=\tilde{F})\\ &= \lim_{n\to\infty}\frac{\tilde{F}(d+1/n)-\tilde{F}(d)}{1/n}-\lim_{n\to\infty}\frac{\tilde{F}(c+1/n)-\tilde{F}(c)}{1/n} = f(d)-f(c). \quad (\because f\ \text{is cont, so }\tilde{F}\ \text{is differentiable})\\ &\hspace{13.3cm}\text{PMA 6.20} \end{aligned}\] So, \(f(d)-f(c)\ge \int_{[c,d]} f'\) and thus \(F_C(d)-F_C(c)\ge 0\). \(\therefore\ F=F_A+F_C+F_J\), where \(F_A,F_C,F_J\) are increasing.

(b) Each \(F_A,F_C,F_J\) is uniquely determined up to additive constant

proof Let \(F=F_{A1}+F_{C1}+F_{J1}=F_{A2}+F_{C2}+F_{J2}\).

① \(F_{J2}-F_{J1}\)
\(F_{J2}-F_{J1}=F_{A1}-F_{A2}+F_{C1}-F_{C2}\), which is continuous. Then, \(F_{J2}-F_{J1}\) is constant.

Let \(G=F_{J2}-F_{J1}\)

② \(F_{A1}-F_{A2}\)
\(F_{A1}-F_{A2}=F_{C2}-F_{C1}+G\), which is absolutely continuous. Then, \(F_{A1}'-F_{A2}'=F_{C2}'-F_{C1}'=0-0=0\) a.e \([a,b]\), and \(F_{A1}-F_{A2}=C_A\) (constant).

③ \(F_{C1}-F_{C2}\)
Then, \(F_{C1}-F_{C2}=(F_{J2}-F_{J1})+(F_{A2}-F_{A1})=G-C_A\) (constant)

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(a) there is a non-negative & integrable function \(f\) on \(\mathbb{R}^d\) such that

\[\liminf_{\substack{m(B)\to 0\\ x\in B}}\frac{1}{m(B)}\int_B f(y)\,dy=\infty \ \ \text{for }x\in E\]

proof ① open \(O_n\supset E\) with \(m(O_n)<\frac{1}{2^n}\)
\(m(E)=0\), so for every \(n\in\mathbb{N}\) there is an open set \(O_n\) such that \(E\subset O_n\) and \(m(O_n)<\frac{1}{2^n}\) (\(E\) is measurable).

② \(f=\sum_{n=1}^\infty \chi_{O_n}\) is integrable
Let \(f=\sum_{n=1}^\infty \chi_{O_n}\), then \[\begin{aligned} \int f &=\int \sum_{n=1}^\infty \chi_{O_n} = \sum_{n=1}^\infty \int \chi_{O_n}\ (\because \text{MCT}) \\ &= \sum_{n=1}^\infty m(O_n) < \sum_{n=1}^\infty \frac{1}{2^n}=1. \end{aligned}\] So \(f\) is integrable on \(\mathbb{R}^d\).

③ \(\frac{1}{m(B)}\int_B f(y)\,dy \ge \frac{1}{m(B)}\sum_{n=1}^\infty m(D_n\cap B)\).

Consider \(x\in E\). \(x\in O_n\) as well, so there is an open ball \(D_n\subset O_n\) centered at \(x\). Then, for a ball \(B\) containing \(x\), \[\begin{aligned} \int_B f(y)\,dy &=\sum_{n=1}^\infty \int_B \chi_{O_n}(y)\,dy =\sum_{n=1}^\infty \int_B \chi_{O_n\cap B}(y)\,dy \\ &=\sum_{n=1}^\infty m(O_n\cap B) \ge \sum_{n=1}^\infty m(D_n\cap B)\ \ (\because D_n\subset O_n). \end{aligned}\] So, \(\frac{1}{m(B)}\int_B f(y)\,dy \ge \frac{1}{m(B)}\sum_{n=1}^\infty m(D_n\cap B)\).

④ \(B_n=\bigcap_{k=1}^n D_k\) with \(\lim_{n\to\infty} m(B_n)=0\)

Let \(B_n=\bigcap_{k=1}^n D_k\), which is open, contains \(x\), and \(B_n\subset D_k\) for \(k=1,\cdots,n\). Note that \(m(B_n)\le m(D_n)\le m(O_n)<\frac{1}{2^n}\Rightarrow \lim_{n\to\infty} m(B_n)=0\). Note that \(x\in B_n,\ \forall n\in\mathbb{N}\).

⑤ \(\frac{1}{m(B_{N_\varepsilon})}\int_{B_{N_\varepsilon}} f(y)\,dy>\varepsilon\)

For given \(\varepsilon>0\), let \(N_\varepsilon\) be the smallest integer greater than \(\varepsilon\). Then \[\begin{aligned}\frac{1}{m(B_{N_\varepsilon})}\int_{B_{N_\varepsilon}} f(y)\,dy &\ge \sum_{k=1}^\infty \frac{m(D_k\cap B_{N_\varepsilon})}{m(B_{N_\varepsilon})}\ (\because ③) \\ &= N_\varepsilon+\sum_{k>N_\varepsilon}\frac{m(D_k\cap B_{N_\varepsilon})}{m(B_{N_\varepsilon})}\\ &\Bigl(\because B_{N_\varepsilon}\subset D_k\ \text{for }k=1,\cdots,N_\varepsilon\ \text{④} \to\ m(D_k\cap B_{N_\varepsilon})=m(B_{N_\varepsilon}) \Bigr) \\ &\ge N_\varepsilon>\varepsilon \end{aligned}\]

⑥ conclusion

To sum up, for every \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(N_\varepsilon>\varepsilon\) and \(n\ge N_\varepsilon\) implies \(\frac{1}{m(B_n)}\int_{B_n} f(y)\,dy = n>\varepsilon\). So \(\lim_{n\to\infty}\frac{1}{m(B_n)}\int_{B_n} f(y)\,dy=\infty\)

This holds for all \(\{B_n\}\) such that \(x\in B_n\ \forall n\) and \(\lim_{n\to\infty} m(B_n)=0\), so \[\liminf_{\substack{m(B)\to 0\\ x\in B}}\frac{1}{m(B)}\int_B f(y)\,dy=\infty.\]

Thus, \(\liminf_{\substack{m(B)\to 0\\ x\in B}}\frac{1}{m(B)}\int_B f(y)\,dy=\infty\) and this holds for all \(x\in E\).

(b) when \(d=1\), there is an increasing & absolutely continuous \(F\) such that \(D_+F(x)=D_-F(x)=\infty\) for each \(x\in E\).

proof ① \(F=\int_0^x f\)

Use the same \(f\) in (a), and let \(F(x)=\int_0^x f(t)\,dt\). Since \(f\) is integrable, \(F\) is absolutely continuous. Also, \(F\) is increasing, bcz \(f\) is non-negative.

② \(D_+F(x),\ D_-F(x)\)

\[\begin{aligned} D_+F(x) &=\liminf_{h\to 0+}\frac{F(x+h)-F(x)}{h} =\liminf_{h\to 0+}\frac{1}{h}\int_x^{x+h}f(t)\,dt =\infty \quad (\because m([x,x+h])=h) \\ D_-F(x) &=\liminf_{h\to 0-}\frac{F(x+h)-F(x)}{h} =\liminf_{h\to 0-}\frac{1}{h}\int_{x+h}^x f(t)\,dt =\infty \quad (\because m([x+h,x])=-h) \end{aligned}\]

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(a) \(\int_{\mathbb{R}}|F(x+h)-F(x)|\,dx \le A|h|\) for some constant \(A\) and all \(h\in\mathbb{R}\)

proof ① when \(F\) is bounded & increasing

It suffices to consider \(h>0\), because \(\int|F(x-h)-F(x)|dx=\int|F(x)-F(x-h)|dx=\int(F(x+h)-F(x))dx\) (\(::\) translation-invariance) \(h=0\) is trivial, \(\therefore \int 0\le 0\).

On an interval \([a,a+h]\) (\(a\in\mathbb{R}\)), \(F(x+h)-F(x)\le F(a+2h)-F(a)\), which leads to \(\int_a^{a+h}(F(x+h)-F(x))\,dx\le h(F(a+2h)-F(a))\).

Then, \[\int_{kh}^{(k+1)h}(F(x+h)-F(x))\,dx \le h\{F((k+2)h)-F(kh)\}\quad (k\in\mathbb{Z}).\]

Then, \[\begin{aligned} \int_{-Nh}^{Nh}(F(x+h)-F(x))\,dx &\le \sum_{k=-N}^{N-1} h\{F((k+2)h)-F(kh)\}\quad (N\in\mathbb{N}) \\ &= h\{F((N+1)h)+F(Nh)-F((-N+1)h)-F(-Nh)\} \\ &\le h\{2F(\infty)-2F(-\infty)\}=Ah \quad (\because F\ \text{is bounded}) \end{aligned}\]

This holds for all \(N\in\mathbb{N}\), so

\[\int_{\mathbb{R}}|F(x+h)-F(x)|\,dx = \lim_{N\to\infty}\int_{-Nh}^{Nh}(F(x+h)-F(x))\,dx \le Ah,\] and thus

\[\int_{\mathbb{R}}|F(x+h)-F(x)|\,dx \le A|h|\quad \text{for all }h\in\mathbb{R}.\]

② \(F(x)-F(-\infty)=P_F(x)-N_F(x)\)

\(P_F(x)\) and \(N_F(x)\) mean positive & negative variation of \(F\) on \((-\infty,x)\), and \(F(-\infty)\) is an abbreviation of \(\lim_{x\to-\infty}F(x)\).

\[P_F(x)=\sup\left\{\sum_{i=1}^k\big(F(t_i)-F(t_{i-1})\big): -\infty<t_1<\cdots<t_k=x\ \&\ F(t_i)\ge F(t_{i-1})\right\}\]

\[N_F(x)=\sup\left\{\sum_{i=1}^k-\big(F(t_i)-F(t_{i-1})\big): -\infty<t_1<\cdots<t_k=x\ \&\ F(t_i)<F(t_{i-1})\right\}\]

Then, there is a partition \(-\infty<t_1<\cdots<t_k=x\) for given \(\varepsilon>0\) that satisfies the following properties :

\[ \begin{aligned} &0\le P_F(x)-\sum_{i=1}^k(F(t_i)-F(t_{i-1}))<\varepsilon \text{ and } 0\le N_F(x)-\sum_{i=1}^k-\big(F(t_i)-F(t_{i-1})\big)<\varepsilon \\ &\Leftrightarrow -\varepsilon< P_F(x)-N_F(x)-\sum_{i=1}^k(F(t_i)-F(t_{i-1}))-\sum_{i=1}^k-\big(F(t_i)-F(t_{i-1})\big)<\varepsilon \\ &\Leftrightarrow \left|P_F(x)-N_F(x)-\sum_{i=1}^k(F(t_i)-F(t_{i-1}))\right|=\left|(P_F(x)-N_F(x))-(F(x)-F(-\infty))\right|<\varepsilon \end{aligned} \]

This holds for all \(\varepsilon>0\), so \(F(x)-F(-\infty)=P_F(x)-N_F(x)\). Each \(P_F\) and \(N_F\) is increasing and bounded (\(::\) \(P_F,N_F\le T_F\))
(\(::\) \(x\le y\to P_F(x)\le P_F(y)\))

Thus, \(F\) is the difference of two increasing and bounded function as well

③ \(F\) is of bounded variation

Go back to \(F\) of bounded variation: \(F=F_1-F_2\), where \(F_1,F_2\) are bounded and increasing function. There exist \(A_1,A_2>0\) such that \(\int_{\mathbb{R}}|F_1(x+h)-F_1(x)|dx\le A_1|h|\) and \(\int_{\mathbb{R}}|F_2(x+h)-F_2(x)|dx\le A_2|h|\) for all \(h\in\mathbb{R}\).

\[\begin{aligned} |F(x+h)-F(x)| &=|F_1(x+h)-F_2(x+h)-F(x)+F_2(x)| \\ &=\left|(F_1(x+h)-F_1(x))-(F_2(x+h)-F_2(x))\right| \\ &\le |F_1(x+h)-F_1(x)|+|F_2(x+h)-F_2(x)| \quad (\forall h\in\mathbb{R}) \end{aligned}\]

Then, \[\begin{aligned} \int_{\mathbb{R}}|F(x+h)-F(x)|dx &\le \int_{\mathbb{R}}|F_1(x+h)-F_1(x)|dx+\int_{\mathbb{R}}|F_2(x+h)-F_2(x)|dx \\ &\le (A_1+A_2)|h| \overset{\text{let}}{=} A|h| \end{aligned}\]

(b) \(\left|\int_{\mathbb{R}} F(x)\psi'(x)\,dx\right|\le A\), where \(\psi\) ranges over all \(C^1\) functions of bounded support with \(\sup_{x\in\mathbb{R}}|\psi(x)|\le 1\) (1st derivative is continuous)

proof ① \(\left|\int_{\mathbb{R}} F(x)\cdot \frac{\psi(x+h)-\psi(x)}{h}\,dx\right|\le A\)

\[\begin{aligned} F(x+h)\psi(x+h)-F(x)\psi(x) &= F(x+h)\psi(x+h)-F(x)\psi(x+h)+F(x)\psi(x+h)-F(x)\psi(x) \\ &= \{F(x+h)-F(x)\}\psi(x+h)+F(x)\{\psi(x+h)-\psi(x)\} \end{aligned}\]

By translation invariance, \(\int_{\mathbb{R}}F(x+h)\psi(x+h)\,dx=\int_{\mathbb{R}}F(x)\psi(x)\,dx\).

So, \(\int_{\mathbb{R}}F(x)\{\psi(x+h)-\psi(x)\}\,dx=-\int_{\mathbb{R}}\{F(x+h)-F(x)\}\psi(x+h)\,dx\), and

\[\begin{aligned} \left|\int_{\mathbb{R}}F(x)\{\psi(x+h)-\psi(x)\}\,dx\right| &=\left|\int_{\mathbb{R}}\{F(x+h)-F(x)\}\psi(x+h)\,dx\right| \\ &\le \int_{\mathbb{R}}|F(x+h)-F(x)|\,|\psi(x+h)|\,dx \\ &\le \int_{\mathbb{R}}|F(x+h)-F(x)|\,dx \quad (\because \sup|\psi(x)|\le 1) \\ &\le A|h| \quad \forall h\in\mathbb{R}\ (\because (a)) \end{aligned}\]

Then, \(\left|\int_{\mathbb{R}}F(x)\cdot \frac{\psi(x+h)-\psi(x)}{h}\,dx\right|\le A\)

② \(\psi,F\) are bounded

\(\psi\) is supported on a bounded set in \(\mathbb{R}\) (say, \([a,b]\)), so \(\psi\) is supported on a compact set \(K\subset\mathbb{R}\). \(\psi\) is continuous on \(K\), so there is \(M>0\) such that \(|\psi|\le M\).

\(F\) is bounded (\(::\) \(P_F,N_F\) are bounded), so there is \(L>0\) such that \(|F|\le L\).

③ \(\lim_{h\to 0}\int_{\mathbb{R}}F(x)\cdot \frac{1}{h}(\psi(x+h)-\psi(x))\,dx=\int_{\mathbb{R}}F(x)\psi'(x)\,dx\), DCT

Consider a sequence \(\{h_n\}\) such that \(h_n\ne 0\) and \(\lim_{n\to\infty}h_n=0\). Let \(f_n(x)=F(x)\cdot \frac{1}{h_n}\{\psi(x+h_n)-\psi(x)\}\).

1. \(f_n\) is measurable Consider the increasing \(F_1\) in (a) and let \(S_c=\{x\in\mathbb{R}: F_1(x)>c\}\) for \(c\in\mathbb{R}\). Let \(p\in S_c\): \(F_1(p)>c\). When \(q>p\), \(F_1(q)\ge F_1(p)>c\) so \(q\in S_c\). This is true for all \(q>p\), so \(S_c\) can be \((\inf S,\infty)\), \((\inf S,-\infty)\) as well as \(\mathbb{R}\) and \(\varnothing\). These sets are all measurable, so \(S_c\) is measurable. This holds for all \(c\in\mathbb{R}\), so \(F_1\) is measurable. \(F_2\) is increasing as well, so \(F=F_1-F_2\) is measurable. Since \(\psi\) is measurable (\(::\) continuous), \(f_n\) is measurable.

2. \(\lim_{n\to\infty} f_n = F(x)\cdot \lim_{n\to\infty}\frac{\psi(x+h_n)-\psi(x)}{h_n}=F(x)\psi'(x)\) (\(::\) \(\psi\in C^1\), differentiable)

3. \(|f_n|\le L\cdot U\,\chi_{\tilde{K}}(x)\) \(\frac{1}{h_n}\{\psi(x+h_n)-\psi(x)\}\) converges, so there is \(U>0\) such that \(\left|\frac{1}{h_n}\{\psi(x+h_n)-\psi(x)\}\right|\le U\) for all \(n\in\mathbb{N}\). Let \(\tilde{K}=[\inf K-\sup_n|h_n|,\ \sup K+\sup_n|h_n|]\). Then, \(\left|\frac{1}{h_n}\{\psi(x+h_n)-\psi(x)\}\right|\le U\cdot \chi_{\tilde{K}}(x)\). So, \(|f_n|\le L\cdot U\,\chi_{\tilde{K}}\) which is integrable ; \(L\cdot U\int \chi_{\tilde{K}}=L\cdot U\cdot m(\tilde{K})<\infty\).

4. By DCT, \(\lim_{n\to\infty}\int f_n=\int F(x)\psi'(x)\,dx\). This holds for all \(\{h_n\}\) with \(h_n\to 0\), so \(\lim_{h\to 0}\int F(x)\cdot \frac{1}{h}(\psi(x+h)-\psi(x))\,dx=\int F(x)\psi'(x)\,dx\)

④ Conclusion

Thus, \(\left|\int_{\mathbb{R}}F(x)\psi'(x)\,dx\right| =\lim_{h\to 0}\left|\int_{\mathbb{R}}F(x)\cdot \frac{\psi(x+h)-\psi(x)}{h}\,dx\right|\le A\ \ (\because ①)\)

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(a) \(\longleftarrow\)

proof ① \(f\) is absolutely continuous

For a finite collection \(\{(a_i,b_i)\}_{i=1}^N\) of disjoint intervals in \(\mathbb{R}\), \[\sum_{i=1}^N |f(b_i)-f(a_i)|\le M\sum_{i=1}^N(b_i-a_i).\]

For given \(\varepsilon>0\), choose \(\delta_\varepsilon=\varepsilon/M\) and \(\{(a_i,b_i)\}\) such that \(\sum_{i=1}^N(b_i-a_i)<\delta_\varepsilon\). Then, \(\sum_{i=1}^N|f(b_i)-f(a_i)|<\varepsilon\).

This holds for all \(\varepsilon>0\), so \(f\) is absolutely continuous.

② \(|f'(x)|\le M\) a.e \(x\)

Due to ①, \(f'\) exists a.e \(x\). For such \(x\), \(|f(x+h)-f(x)|\le M|h|\). Then

\[\frac{1}{|h|}|f(x+h)-f(x)|\le M\ \ \text{and thus}\ \ \lim_{h\to 0}\left|\frac{f(x+h)-f(x)}{h}\right|=|f'(x)|\le M\ \ (\text{a.e }x).\]

(b) \(\longrightarrow\)

proof

\(f\) is absolutely continuous, so (\(x\ge y\) wlog) \(f(x)-f(y)=\int_y^x f'(t)\,dt\). Let \(\mathcal{N}=\{t\in[y,x]: |f'(t)|>M\}\), then \[f(x)-f(y)=\int_{\mathcal{N}^c}f'(t)\,dt+\int_{\mathcal{N}}f'(t)\,dt=\int_{\mathcal{N}^c}f'(t)\,dt\] (\(\mathcal{N}^c=[y,x]-\mathcal{N}\) for simplicity, \(m(\mathcal{N})=0\))

\[|f(x)-f(y)|\le \int_{\mathcal{N}^c}|f'(t)|\,dt \le \int_{\mathcal{N}^c}M\,dt \le \int_{[y,x]}M\,dt=M(x-y).\]

The case \(y\ge x\) yields a same conclusion : \(|f(y)-f(x)|\le M(y-x)\)

This holds for all \(x,y\in\mathbb{R}\), so \(|f(x)-f(y)|\le M|x-y|\)