1. Differentiability of Monotonic Function
Definition 1.1 Suppose \(E \subset \mathbb{R}\), \(\mathcal{V}=\{I_\alpha\}\) is a family of intervals with all positive lengths. If \[\forall x \in E, \forall \varepsilon>0, \exists I_\alpha \in \mathcal{V}\ \text{s.t.}\ x \in I_\alpha, m(I_\alpha)<\varepsilon,\] then \(\mathcal{V}\) is a Vitali cover of \(E\).
Theorem 1.1 \(\mathcal{V}\) is a Vitali cover if and only if \[\forall x \in E, \exists \{I_n\} \subset \mathcal{V}\ \text{s.t.}\ x \in I_n, m(I_n) \to 0.\]
Theorem 1.2 (Vitali Covering Theorem) Suppose \(E \subset \mathbb{R}\) and \(m^*(E)<\infty\). If \(\mathcal{V}\) is a Vitali cover of \(E\), then there exists a sequence of intervals \(\{I_n\} \subset \mathcal{V}\) such that \(I_i \cap I_j=\varnothing\) for all \(i \neq j\), and \[m^*\left(E-\bigcup_{i=1}^\infty I_i\right)=0.\]
Lemma 1.2.1 For \(x \in E \cap U\), where \(U\) is open, and a Vitali cover \(\mathcal{V}\) of \(E\), there exists an \(I_x \in \mathcal{V}\) such that \(x \in I_x\) and \(I_x \subset U\).
Proof. Since \(x \in U\), where \(U\) is open, then \[\exists \varepsilon>0\ \text{s.t.}\ (x-\varepsilon, x+\varepsilon) \subset U.\] Since \(\mathcal{V}\) is a Vitali cover of \(E\), then \[\exists I_x \in \mathcal{V}\ \text{s.t.}\ x \in I_x, m(I_x)<\frac{\varepsilon}{4}.\] Hence, \(I_x \subset (x-\varepsilon, x+\varepsilon) \subset U\).\(\square\)
Proof. Since \[m^*\left(E-\bigcup_{i=1}^\infty I_i\right)=m^*\left(E-\bigcup_{i=1}^\infty \overline{I_i}\right),\] then we can assume without loss of generality that \(I_i\) is closed for all \(i\).
In addition, since \(m^*(E)<\infty\), then there exists an open set \(U \supset E\) such that \(m(U)<\infty\). Let \[\mathcal{W}=\{I \in \mathcal{V}: I \subset U\}.\] By lemma 1.2.1, for all \(x \in E\), there exists an \(I_x\) such that \(x \in I_x \subset U\), and thus \(I_x \in \mathcal{W}\), i.e., \(\mathcal{W} \neq \varnothing\). Take an arbitrary \(\delta>0\), \[\exists \widetilde{I}_x \in \mathcal{V}\ \text{s.t.}\ x \in \widetilde{I}_x, m(\widetilde{I}_x)<\min\left\{\frac{\varepsilon}{4}, \delta\right\}.\] Hence, \(\widetilde{I}_x \in \mathcal{W}\), and thus \(\mathcal{W}\) is a Vitali cover of \(E\). Therefore, we can assume without loss of generality that the intervals in \(\mathcal{V}\) are all contained in an open set \(U\) with a finite measure.
Take an arbitrary \(I_1 \in \mathcal{V}\). If \(E-I_1=\varnothing\), then the theorem holds. If \(E-I_1 \neq \varnothing\), then for \(x \in E-I_1\), \(x \in I_1^c\), where \(I_1^c\) is open. By lemma 1.2.1, there exists an \(I_x \in \mathcal{V}\) such that \(x \in I_x\) and \(I_x \subset I_1^c\), i.e., \(I_x \cap I_1=\varnothing\). Hence, \(\{I \in \mathcal{V}: I \cap I_1=\varnothing\} \neq \varnothing\). Let \(r_1=\sup\{m(I): I \in \mathcal{V}, I \cap I_1=\varnothing\} \leq m(U)<\infty\). We now can find \(I_2 \in \mathcal{V}\), \(I_2 \cap I_1=\varnothing\) such that \(\displaystyle m(I_2)>\frac{r_1}{2}\). Similarly, we can find \(r_3, r_4, \ldots\). Therefore, we can find a sequence of intervals \(\{I_n\} \subset \mathcal{V}\) such that \(I_i \cap I_j=\varnothing\), where \[r_n=\sup\{m(I): I \in \mathcal{V}, I \cap I_1=\varnothing, \ldots, I \cap I_n=\varnothing\}\] and \[m(I_{n+1})>\frac{r_n}{2}.\]
Since \(\displaystyle \bigcup_{i=1}^\infty I_i \subset U\), then \[\sum_{i=1}^\infty m(I_i)=m\left(\bigcup_{i=1}^\infty I_i\right) \leq m(U)<\infty.\] Hence, \(m(I_i) \to 0\) as \(i \to \infty\), and \(\displaystyle \sum_{i=k+1}^\infty m(I_i) \to 0\) as \(k \to \infty\).
Take an arbitrary \(\displaystyle y \in E-\bigcup_{i=1}^\infty I_i\), and fix a \(k \in \mathbb{N}\) such that \(\displaystyle y \notin \bigcap_{i=1}^k I_i\), i.e., \(\displaystyle y \in \left(\bigcup_{i=1}^k I_i\right)^c\). By lemma 1.2.1, there exists an \(I_y \in \mathcal{V}\) such that \(y \in I_y\) and \(\displaystyle I_y \subset \left(\bigcup_{i=1}^k I_i\right)^c\).
We can find a \(n>k\) such that \(I_y \cap I_n \neq \varnothing\); otherwise, \(m(I_y) \leq r_n \to 0\), i.e., \(m(I_y)=0\), which is a contradiction. Take an \(n(y) \in \mathbb{N}\) such that \(I_y \cap I_{n(y)} \neq \varnothing\) and \(I_y \cap I_{n(y)-1}=\varnothing, \ldots, I_y \cap I_1=\varnothing\). Note that \(n(y)>k\). Let \(x_{n(y)}\) be the midpoint of \(I_{n(y)}\), then \[\begin{aligned} |y-x_{n(y)}| &\leq m(I_y)+\frac{1}{2}m(I_{n(y)}) \\&\leq r_{n(y)-1}+\frac{1}{2}m(I_{n(y)}) \\&<2m(I_{n(y)})+\frac{1}{2}m(I_{n(y)}) \\&=\frac{5}{2}m(I_{n(y)}). \end{aligned}\]
Let \(J_n\) be a closed interval with the midpoint of \(I_n\) as the midpoint, and a length that becomes five times that of \(I_n\), then \(y \in J_{n(y)}\). Since \(n(y)>k\), then \(\displaystyle y \in \bigcup_{i=k+1}^\infty J_i\). Hence, \(\displaystyle E-\bigcup_{i=1}^\infty I_i \subset \bigcup_{i=k+1}^\infty J_i\).
Therefore, \[m^*\left(E-\bigcup_{i=1}^\infty I_i\right) \leq m\left(\bigcup_{i=k+1}^\infty J_i\right) \leq \sum_{i=k+1}^\infty m(J_i)=5\sum_{i=k+1}^\infty m(I_i).\] Since \(\displaystyle \sum_{i=k+1}^\infty m(I_i) \to 0\) as \(k \to \infty\), then \[m^*\left(E-\bigcup_{i=1}^\infty I_i\right)=0.\]\(\square\)
Corollary 1.3 Suppose \(E \subset \mathbb{R}\) and \(m^*(E)<\infty\). If \(\mathcal{V}\) is a Vitali cover of \(E\), then for all \(\varepsilon>0\), there exist finite disjoint intervals \(\{I_i\} \subset \mathcal{V}\) such that \[m^*\left(E-\bigcup_{i=1}^n I_i\right)<\varepsilon.\]
Definition 1.2 The upper right-hand derivative, lower right-hand derivative, upper left-hand derivative, and lower left-hand derivative (collectively called Dini derivative) are defined as \[\begin{aligned} D^+f(x_0)&=\varlimsup_{h \to 0^+} \frac{f(x_0+h)-f(x_0)}{h}, \\D_+f(x_0)&=\varliminf_{h \to 0^+} \frac{f(x_0+h)-f(x_0)}{h}, \\D^-f(x_0)&=\varlimsup_{h \to 0^-} \frac{f(x_0+h)-f(x_0)}{h}, \end{aligned}\] and \[D_-f(x_0)=\varliminf_{h \to 0^-} \frac{f(x_0+h)-f(x_0)}{h}.\] Note that \(\displaystyle \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}\) exists if and only if \(D^+f(x_0)=D_+f(x_0)=D^-f(x_0)=D_-f(x_0)\). Moreover, \(f\) is differentiable at \(x_0\) if and only if \[-\infty<f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}=D^+f(x_0)=D_+f(x_0)=D^-f(x_0)=D_-f(x_0)<\infty.\]
Definition 1.3 Let \[Df(x_0)=\lim_{n \to \infty} \frac{f(x_0+h_n)-f(x_0)}{h_n},\] where \(h_n \to 0\) and \(Df(x_0)\) exists.
Theorem 1.4 Suppose \(f\) is a strictly increasing function on \([a, b]\), and \(E \subset [a, b]\). If for all \(x \in E\), there exists a sequence \(\{h_n\}\), where \(h_n \to 0\) such that \(p \geq Df(x) \geq q\), where \(p, q \geq 0\), then \[p \cdot m^*(E) \geq m^*(f(E)) \geq q \cdot m^*(E).\]
Proof. Take an arbitrary \(\varepsilon>0\), there exists an open set \(G \supset E\) such that \(m^*(E)>m^*(G)-\varepsilon\). Let \(p_0>p\), for \(x_0 \in E\), there exists a sequence \(\{h_n\}\), where \(h_n \to 0\), such that \[\lim_{n \to \infty} \frac{f(x_0+h_n)-f(x_0)}{h_n}<p_0.\]
Let \[I_n(x_0)=\begin{cases} [x_0, x_0+h_n], &h_n>0 \\ [x_0+h_n, x_0], &h_n<0 \end{cases},\] and \[\Delta_n(x_0)=\begin{cases} [f(x_0), f(x_0+h_n)], &h_n>0 \\ [f(x_0+h_n), f(x_0)], &h_n<0 \end{cases}.\] Since \(f\) is a strictly increasing function, then \(f(I_n(x_0)) \subset \Delta_n(x_0)\).
When \(n\) is large enough, \(I_n(x_0) \subset G\), and \[\frac{f(x_0+h_n)-f(x_0)}{h_n}<p_0.\] By inequality above, \[m^*(f(I_n(x_0))) \leq m^*(\Delta_n(x_0))<p_0 \cdot m(I_n(x_0)).\]
Since \(m^*(I_n(x_0)) \to 0\), then \(m^*(\Delta_n(x_0)) \to 0\), i.e., \(\{\Delta_n(x_0)\}\) is a Vitali cover of \(f(E)\). By Vitali covering theorem, there exist countable disjoint intervals \(\{\Delta_{n_j}(x_j)\}\) such that \[m\left(f(E)-\bigcup_{j=1}^\infty \Delta_{n_j}(x_j)\right)=0.\] Besides, \(\{I_{n_j}(x_j)\}\) are disjoint. Hence, \[m^*(f(E)) \leq \sum_{j=1}^\infty m^*(\Delta_{n_j}(x_j))<p_0\sum_{j=1}^\infty m^*(I_{n_j}(x_j))=p_0 \cdot m^*\left(\bigcup_{j=1}^\infty I_{n_j}(x_j)\right).\] Since \(\displaystyle \bigcup_{j=1}^\infty I_{n_j}(x_j) \subset G\), then \[m^*(f(E))<p_0 \cdot m^*(G)<p_0(m^*(E)+\varepsilon).\] Let \(\varepsilon \to 0\) and \(p_0 \to p\), we have \(m^*(f(E)) \leq p \cdot m^*(E)\).
Assume without loss of generality that \(f\) is continuous and \(q>0\). Since \(f\) is a strictly increasing function on \([a, b]\), then there is a strictly increasing function \(f^{-1}\) on \(f([a, b])\). Take \(x_0 \in E\) and \(y_0=f(x_0)\), we have \[\lim_{n \to \infty} \frac{f^{-1}(y_0+k_n)-f^{-1}(y_0)}{k_n}=\lim_{n \to \infty} \frac{(x_0+h_n)-x_0}{f(x_0+h_n)-f(x_0)}=\frac{1}{Df(x_0)} \leq \frac{1}{q},\] where \(k_n=f(x_0+h_n)-f(x_0) \to 0\). Hence, \[m^*(f^{-1}(f(E))) \leq \frac{1}{q} \cdot m^*(f(E)),\] i.e., \[q \cdot m^*(E) \leq m^*(f(E)).\]\(\square\)
Theorem 1.5 (Lebesgue's Theorem) Suppose \(f\) is an increasing function on \([a, b]\). Then \(f\) is differentiable almost everywhere on \([a, b]\), \(f'\) is Lebesgue integrable on \([a, b]\), and \[\int_a^b f'(x)\text{d}x \leq f(b)-f(a).\]
Proof. Let \(g(x)=f(x)+x\), which is strictly increasing. Let \[E=\{x \in [a, b]: \text{$g'(x)$ does not exist}\}.\] For all \(x \in E\), there exist \(D_1g(x)\) and \(D_2g(x)\) such that \(D_1g(x)<D_2g(x)\). There exist \(p, q \in \mathbb{Q}\) such that \(D_1g(x)<p<q<D_2g(x)\). Let \(E_{pq}=\{x \in [a, b]: 0<D_1g(x)<p<q<D_2g(x)\}\), then \[E=\bigcup_{p, q \in \mathbb{Q}} E_{pq}.\]
By theorem 1.4, \[q \cdot m^*(E_{pq}) \leq m^*(g(E_{pq})) \leq p \cdot m^*(E_{pq}).\] Since \(q>p\), then \(m^*(E_{pq})=0\), and thus \(m(E)=0\), i.e., \(g\) is differentiable almost everywhere on \([a, b]\), and \(f\) is differentiable almost everywhere on \([a, b]\).
Let \(\displaystyle g_n(x)=n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)\), and \(f(x)=f(b)\) when \(x \geq b\), then \(g_n(x) \to f'(x)\ \text{a.e.}\ x \in [a, b]\), where \(g_n(x) \geq 0\). Let \(f'(x)=0\) if the derivative does not exist at \(x\). We have \[\begin{aligned} \int_a^b f'(x)\text{d}x &\leq \varliminf_{n \to \infty} \int_a^b g_n(x)\text{d}x \\&=\varliminf_{n \to \infty} \int_a^b n\left[f\left(x+\frac{1}{n}\right)-f(x)\right]\text{d}x \\&=\varliminf_{n \to \infty} \left(n\int_a^b f\left(x+\frac{1}{n}\right)\text{d}x-n\int_a^b f(x)\text{d}x\right) \\&=\varliminf_{n \to \infty} \left(n\int_{a+1/n}^{b+1/n} f(x)\text{d}x-n\int_a^b f(x)\text{d}x\right) \\&=\varliminf_{n \to \infty} \left(n\int_b^{b+1/n} f(x)\text{d}x-n\int_a^{a+1/n} f(x)\text{d}x\right) \\&=\varliminf_{n \to \infty} \left(f(b)-n\int_a^{a+1/n} f(x)\text{d}x\right) \\&\leq \varliminf_{n \to \infty} (f(b)-f(a)) \\&=f(b)-f(a)<\infty. \end{aligned}\] Therefore, \(f'\) is Lebesgue integrable on \([a, b]\).\(\square\)
Example 1.1 Suppose \(E \subset [a, b]\) and \(m(E)=0\). Then there exists an increasing function \(f\) on \([a, b]\) such that for all \(x \in E\), \(f'(x)=\infty\).
Proof. There exist open sets \(G_1 \supset G_2 \supset \cdots\) such that \(E \subset G_n\) and \(\displaystyle m(G_n)<\frac{1}{2^n}\). Let \[f_n(x)=m([a, x] \cap G_n),\] where \(f_n\) is increasing on \([a, b]\), and \(\displaystyle 0 \leq f_n(x) \leq m(G_n)<\frac{1}{2^n}\). Besides, since \(f_n(x+h)-f_n(x) \leq |h|\), it is easy to show that \(f_n\) is continuous on \([a, b]\). Let \[f(x)=\sum_{n=1}^\infty f_n(x),\] where \(f\) is increasing and continuous on \([a, b]\).
For all \(x \in E\), when \(x \neq b\), take a \(k \in \mathbb{N}\), then \[\exists \varepsilon>0\ \text{s.t.}\ 0<h<\varepsilon, [x, x+h] \subset G_1, \ldots, G_k.\] We have \[\begin{aligned} \frac{f(x+h)-f(x)}{h}&=\frac{1}{h}\left(\sum_{i=1}^\infty f_i(x+h)-\sum_{i=1}^\infty f_i(x)\right) \\&=\sum_{i=1}^\infty \frac{f_i(x+h)-f_i(x)}{h} \\&\geq \sum_{i=1}^k \frac{f_i(x+h)-f_i(x)}{h} \\&=\sum_{i=1}^k \frac{m([a, x+h] \cap G_i)-m([a, x] \cap G_i)}{h} \\&=\sum_{i=1}^k \frac{h}{h}=k. \end{aligned}\] Hence, \(f'_+(x)=\infty.\) Similarly, for all \(x \in E\), when \(x \neq a\), we can show \(f'_-(x)=\infty\). Therefore, \(f'(x)=\infty\) for all \(x \in E\).\(\square\)
Example 1.2 In Lebesgue's theorem, we can find a function \(\theta\) such that \[\int_a^b \theta'(x)\text{d}x<\theta(b)-\theta(a).\]
Proof. Suppose \(P_0\) is a Cantor ternary set. Let \(\displaystyle G_0=P_0^c\). Define a function \(\theta\) on \(G_0\): when \(\displaystyle x \in \left(\frac{1}{3}, \frac{2}{3}\right)\), \(\displaystyle \theta(x)=\frac{1}{2}\); when \(\displaystyle x \in \left(\frac{1}{9}, \frac{2}{9}\right)\), \(\displaystyle \theta(x)=\frac{1}{4}\); when \(\displaystyle x \in \left(\frac{7}{9}, \frac{8}{9}\right)\), \(\displaystyle \theta(x)=\frac{3}{4}\); \(\ldots\). For \(x_0 \in P_0\), let \[\theta(x_0)=\sup_{\substack{x \in G_0 \\ x<x_0}}\{\theta(x)\}.\] Hence, \(\theta\) is increasing. In addition, since \[\lim_{x \to x_0^-} \theta(x)=\lim_{x \to x_0^+} \theta(x)=\theta(x_0),\] then \(\theta\) is continuous.
For \(x \in G_0\), \(\theta'(x)=0\), i.e., \(\theta'(x)=0\ \text{a.e.}\ x \in [0, 1]\). Hence, \[0=\int_0^1 \theta'(x)\text{d}x<1=\theta(1)-\theta(0).\]\(\square\)
2. Bounded Variation Function
Definition 2.1 Suppose \(f\) is a real function on \([a, b]\). Take a partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\). The variation of \(f\) on \([a, b]\) is \[v_\Delta=\sum_{i=1}^n |f(x_i)-f(x_{i-1})|.\] The total variation of \(f\) on \([a, b]\) is \[\bigvee_a^b (f)=\sup\{v_\Delta: \text{$\Delta$ is an arbitrary partition of $[a, b]$}\}.\] If \[\bigvee_a^b (f)<\infty,\] then \(f\) is a bounded variation function on \([a, b]\). The collection of all bounded variation functions on \([a, b]\) is denoted \(\text{BV}([a, b])\).
Theorem 2.1 If \(f \in \text{BV}([a, b])\), then \(f\) is bounded on \([a, b]\).
Proof. Take an arbitrary \(x \in [a, b]\), and take a partition \(\Delta: a=x_0<x_1=x<x_2=b\). We have \[\begin{aligned} |f(x)| &\leq |f(x)-f(a)|+|f(a)| \\&\leq |f(x)-f(a)|+|f(b)-f(x)|+|f(a)| \\&=v_\Delta+|f(a)| \\&\leq \bigvee_a^b (f)+|f(a)|. \end{aligned}\] Since \(f \in \text{BV}([a, b])\), then \[\bigvee_a^b (f)+|f(a)|<\infty,\] i.e., \(f\) is bounded on \([a, b]\).\(\square\)
Theorem 2.2 \(\text{BV}([a, b])\) is a linear space.
Proof. Take an arbitrary partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\). Suppose \(c \in \mathbb{R}\), \(f, g \in \text{BV}([a, b])\), \[\sum_{i=1}^n |f(x_i)-f(x_{i-1})| \leq M_1,\] and \[\sum_{i=1}^n |g(x_i)-g(x_{i-1})| \leq M_2.\] Hence, \[\begin{aligned} \sum_{i=1}^n |(cf+g)(x_i)-(cf+g)(x_{i-1})|&=\sum_{i=1}^n |(cf(x_i)+g(x_i))-(cf(x_{i-1})+g(x_{i-1}))| \\&=\sum_{i=1}^n |c(f(x_i)-f(x_{i-1}))+(g(x_i)-g(x_{i-1}))| \\&\leq |c|\sum_{i=1}^n |f(x_i)-f(x_{i-1})|+\sum_{i=1}^n |g(x_i)-g(x_{i-1})| \\&\leq |c|M_1+M_2<\infty, \end{aligned}\] i.e., \(cf+g \in \text{BV}([a, b])\).\(\square\)
Theorem 2.3 If \(f, g \in \text{BV}([a, b])\), then \(fg \in \text{BV}([a, b])\).
Proof. Take an arbitrary partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\). Since \(f, g \in \text{BV}([a, b])\), then there is an \(M\) such that \(|f(x)| \leq M\), \(|g(x)| \leq M\), \(\displaystyle \sum_{i=1}^n |f(x_i)-f(x_{i-1})| \leq M\), and \(\displaystyle \sum_{i=1}^n |g(x_i)-g(x_{i-1})| \leq M\). Hence,
\[\begin{aligned} \sum_{i=1}^n |(fg)(x_i)-(fg)(x_{i-1})|&=\sum_{i=1}^n |f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})| \\&=\sum_{i=1}^n |f(x_i)g(x_i)-f(x_i)g(x_{i-1})+f(x_i)g(x_{i-1})-f(x_{i-1})g(x_{i-1})| \\&=\sum_{i=1}^n |f(x_i)(g(x_i)-g(x_{i-1}))+g(x_{i-1})(f(x_i)-f(x_{i-1}))| \\&\leq \sum_{i=1}^n |f(x_i)| \cdot |g(x_i)-g(x_{i-1})|+\sum_{i=1}^n |g(x_{i-1})| \cdot |f(x_i)-f(x_{i-1})| \\&\leq 2nM^2<\infty, \end{aligned}\] i.e., \(fg \in \text{BV}([a, b])\).\(\square\)
Theorem 2.4 If \(f \in \text{BV}([a, b])\) and \(a<c<b\), then \[\bigvee_a^b (f)=\bigvee_a^c (f)+\bigvee_c^b (f).\]
Proof. Take an arbitrary partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\). If \(c=x_k\) for some \(k\), then \[\sum_{i=1}^n |f(x_i)-f(x_{i-1})|=\sum_{i=1}^k |f(x_i)-f(x_{i-1})|+\sum_{i=k+1}^n |f(x_i)-f(x_{i-1})| \leq \bigvee_a^c (f)+\bigvee_c^b (f).\] If \(x_k<c<x_{k+1}\) for some \(k\), then \[\begin{aligned} \sum_{i=1}^n |f(x_i)-f(x_{i-1})|&=\sum_{i=1}^k |f(x_i)-f(x_{i-1})|+|f(x_{k+1})-f(x_k)|+\sum_{i=k+2}^n |f(x_i)-f(x_{i-1})| \\&\leq \sum_{i=1}^k |f(x_i)-f(x_{i-1})|+|f(c)-f(x_k)| \\&\ \ \ \ +|f(x_{k+1})-f(c)|+\sum_{i=k+2}^n |f(x_i)-f(x_{i-1})| \\&=\bigvee_a^c (f)+\bigvee_c^b (f). \end{aligned}\] Hence, \[\bigvee_a^b (f) \leq \bigvee_a^c (f)+\bigvee_c^b (f).\]
For all \(\varepsilon>0\), we can find two partitions \[\Delta_1: a=y_0<y_1<\cdots<y_m=c\] and \[\Delta_2: c=z_0<z_1<\cdots<z_n=b\] such that \[\bigvee_a^c (f)<\sum_{i=1}^m |f(y_i)-f(y_{i-1})|+\frac{\varepsilon}{2}\] and \[\bigvee_c^b (f)<\sum_{i=1}^n |f(z_i)-f(z_{i-1})|+\frac{\varepsilon}{2}.\] Let \(\Delta'=\Delta_1 \cup \Delta_2\), then \(\Delta'\) is a partition of \([a, b]\) with \[v_{\Delta'}=\sum_{i=1}^m |f(y_i)-f(y_{i-1})|+\sum_{i=1}^n |f(z_i)-f(z_{i-1})|>\bigvee_a^c (f)+\bigvee_c^b (f)-\varepsilon,\] i.e., \[\bigvee_a^b (f) \geq \bigvee_a^c (f)+\bigvee_c^b (f).\] Therefore, \[\bigvee_a^b (f)=\bigvee_a^c (f)+\bigvee_c^b (f).\]\(\square\)
Theorem 2.5 (Jordan Decomposition Theorem) \(f \in \text{BV}([a, b])\) if and only if \(f=g-h\), where \(g\) and \(h\) are increasing on \([a, b]\).
Proof. \((\Leftarrow)\) For any increasing function \(\theta\) on \([a, b]\) with an arbitrary partition \(\Delta\), we have \(v_\Delta=|\theta(b)-\theta(a)|\), and thus \[\bigvee_a^b (\theta)=|\theta(b)-\theta(a)|<\infty,\] i.e., \(\theta \in \text{BV}([a, b])\).
Suppose \(g\) and \(h\) are increasing on \([a, b]\), then \(g, h \in \text{BV}([a, b])\), and thus \(f=g-h \in \text{BV}([a, b])\).
\((\Rightarrow)\) Suppose \(f \in \text{BV}([a, b])\). Let \[g(x)=\bigvee_a^x (f),\] where \(g\) is increasing.
Let \(h(x)=g(x)-f(x)\). Take arbitrary \(x_1<x_2\) such that \(a \leq x_1<x_2 \leq b\). We have \[\begin{aligned} h(x_2)-h(x_1)&=\bigvee_a^{x_2} (f)-f(x_2)-\left(\bigvee_a^{x_1} (f)-f(x_1)\right) \\&=\bigvee_{x_1}^{x_2} (f)-(f(x_2)-f(x_1)) \\&\geq |f(x_2)-f(x_1)|-(f(x_2)-f(x_1)) \geq 0. \end{aligned}\] Hence, \(h\) is increasing, and \(f=g-h\).\(\square\)
If \(f \in \text{BV}([a, b])\), then \(f=g-h\), where \(g\) and \(h\) are increasing on \([a, b]\). Take an arbitrary increasing function \(a\), we know \(g+a\) and \(h+a\) are increasing, and \(f=(g+a)-(h+a)\). Hence, the increasing decomposition functions are not unique.
We define positive variation function as \[p(x)=\frac{1}{2}\left(\bigvee_a^x (f)+f(x)-f(a)\right),\] and negative variation function as \[n(x)=\frac{1}{2}\left(\bigvee_a^x (f)-f(x)+f(a)\right).\] Hence, \(p\) and \(n\) are increasing, \(\displaystyle \bigvee_a^x (f)=p(x)+n(x)\), and \(f(x)-f(a)=p(x)-n(x)\), which is called normal decomposition.
Theorem 2.6 Suppose \(f \in \text{BV}([a, b])\). Then \(f\) is differentiable almost everywhere on \([a, b]\), and \(f' \in L([a, b])\).
Proof. Since \(f \in \text{BV}([a, b])\), then \(f=g-h\), where \(g\) and \(h\) are increasing on \([a, b]\).
By Lebesgue's theorem, \(g\) and \(h\) are differentiable almost everywhere on \([a, b]\); \(g\) and \(h\) are Lebesgue integrable on \([a, b]\).
Hence, \(f=g-h\) is differentiable almost everywhere on \([a, b]\), and is Lebesgue integrable on \([a, b]\).\(\square\)
The bounded variation function is not necessarily a continuous function. For example, \[f(x)=\begin{cases}0, &x \leq 0 \\1, &x>0\end{cases}.\]
The continuous function is not necessarily a bounded variation function. For example, \[f(x)=\begin{cases}\displaystyle x\sin\frac{1}{x}, &0<x \leq 1 \\0, &x=0\end{cases}.\] Take a partition: \[x_0=0, x_1=\frac{1}{(n-2)\pi+\pi/2}, x_2=\frac{1}{(n-3)\pi+\pi/2}, \ldots, x_{n-1}=\frac{1}{\pi+\pi/2}, x_n=1.\] We have \[\begin{aligned}\sum_{i=1}^n |f(x_i)-f(x_{i-1})|&=\left|x_1\sin\frac{1}{x_1}-0\right|+\cdots+\left|1\sin1-x_{n-1}\sin\frac{1}{x_{n-1}}\right|\\&\geq \frac{4}{\pi}\sum_{k=2}^n \frac{1}{2k-1}+\sin1.\end{aligned}\] As \(n \to \infty\), \(\displaystyle \sum_{i=1}^n |f(x_i)-f(x_{i-1})=\infty\), and thus \(f \notin \text{BV}([a, b])\).
Theorem 2.7 If \(f \in \text{BV}([a, b])\), then \(f=g+h\), for some step function \(g\) and continuous function \(h\).
Proof. Let \(f\) be an increasing function. Let \[\theta_l(x)=\begin{cases} 0, &x<0 \\ 1, &x \geq 0 \end{cases}\] and \[\theta_r(x)=\begin{cases} 0, &x \leq 0 \\ 1, &x>0 \end{cases}.\] Let \(d_n\) be the discontinuous point of \(f\), and \[g(x)=\sum_{n=1}^\infty \mu_n\theta_l(x)+\sum_{n=1}^\infty \lambda_n\theta_r(x),\] where \(\mu_n=f(d_n)-f(d_n^-)\) and \(\lambda_n=f(d_n^+)-f(d_n)\). Note that \(g\) is a step function, with discontinuous points \(\{d_n\}\). Hence, \(f-g\) is continuous, i.e., \[\text{Increasing function}=\text{Step function}+\text{Continuous function}.\] Since a bounded variation function is difference of two increasing functions, then \[\text{Bounded variation function}=\text{Step function}+\text{Continuous function}.\]\(\square\)
3. Differentiation of Indefinite Integral
Theorem 3.1 Suppose \(f \in L([a, b])\), and \(f(x)=0\) if \(x \notin [a, b]\). Then \[\lim_{h \to 0} \int_\mathbb{R} |f(x+h)-f(x)|\text{d}x=0.\]
Proof. Take an arbitrary \(\varepsilon>0\), and let \(f=f_1+f_2\), where \(f_1\) is a continuous function with compact support set on \(\mathbb{R}\), and \(f_2\) satisfies \[\int_\mathbb{R} |f_2(x)|\text{d}x<\frac{\varepsilon}{4}.\] Since \(f_1\) has compact support set and is uniformly continuous, then there exists a \(\delta>0\) such that when \(|h|<\delta\) we have \[\int_\mathbb{R} |f_1(x+h)-f_1(x)|\text{d}x<\frac{\varepsilon}{2}.\]
Hence, \[\begin{aligned} \int_\mathbb{R} |f(x+h)-f(x)|\text{d}x &\leq \int_\mathbb{R} |f_1(x+h)-f_1(x)|\text{d}x+\int_\mathbb{R} |f_2(x+h)-f_2(x)|\text{d}x \\&<\frac{\varepsilon}{2}+\int_\mathbb{R} |f_2(x+h)|\text{d}x+\int_\mathbb{R} |f_2(x)|\text{d}x \\&=\frac{\varepsilon}{2}+2\int_\mathbb{R} |f_2(x)|\text{d}x<\varepsilon. \end{aligned}\]\(\square\)
Theorem 3.2 Suppose \(f \in L([a, b])\). Then \[\frac{\text{d}}{\text{d}x}\left(\int_a^x f(t)\text{d}t\right)=f(x)\ \text{a.e.}\ x \in [a, b].\]
Lemma 3.2.1 Suppose \(f \in L([a, b])\). Then \[\lim_{h \to 0}\frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t=0\ \text{a.e.}\ x \in [a, b].\]
Proof. Suppose \(h>0\), and assume \(f(x)=0\) if \(x \notin [a, b]\). We have \[\begin{aligned}\int_a^b \left(\lim_{h \to 0} \frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t\right)\text{d}x &\leq \int_\mathbb{R} \left(\lim_{h \to 0} \frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t\right)\text{d}x\\&\leq \varliminf_{h \to 0} \int_\mathbb{R} \left(\frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t\right)\text{d}x\\&=\varliminf_{h \to 0} \int_0^h \frac{1}{h}\left(\int_\mathbb{R} |f(x+t)-f(x)|\text{d}x\right)\text{d}t.\end{aligned}\] Take an arbitrary \(\varepsilon>0\), there exists a \(\delta>0\) such that when \(|t|<\delta\), \[\int_\mathbb{R} |f(x+t)-f(x)|\text{d}x<\varepsilon.\] When \(h<\delta\), we have \[\int_0^h \frac{1}{h}\left(\int_\mathbb{R} |f(x+t)-f(x)|\text{d}x\right)\text{d}t<\frac{\varepsilon}{h}\int_0^h \text{d}t=\varepsilon.\] Therefore, \[\lim_{h \to 0}\frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t=0\ \text{a.e.}\ x \in [a, b].\] We define \(x\) as a Lebesgue point if \(x\) satisfies \[\lim_{h \to 0}\frac{1}{h}\int_0^h |f(x+t)-f(x)|\text{d}t=0.\]
Proof. Suppose \(h>0\), and assume \(f(x)=0\) if \(x \notin [a, b]\). We have \[\begin{aligned} \int_a^b \left|\lim_{h \to 0} \frac{1}{h}\int_0^h (f(x+t)-f(x))\text{d}t\right|\text{d}x&=\int_a^b \left(\lim_{h \to 0} \frac{1}{h}\left|\int_0^h (f(x+t)-f(x))\text{d}t\right|\right)\text{d}x \\&\leq \int_a^b \left(\lim_{h \to 0} \frac{1}{h} \int_0^h |f(x+t)-f(x)|\text{d}t\right)\text{d}x \\&=0. \end{aligned}\] Hence, \[\int_a^b \left|\lim_{h \to 0} \frac{1}{h}\int_0^h (f(x+t)-f(x))\text{d}t\right|\text{d}x=0,\] i.e., \[\lim_{h \to 0} \frac{1}{h}\int_0^h (f(x+t)-f(x))\text{d}t=0\ \text{a.e.}\ x \in [a, b].\]
Similarly, if \(h<0\), we have \[\lim_{h \to 0} \frac{1}{h}\int_0^h (f(x+t)-f(x))\text{d}t=0\ \text{a.e.}\ x \in [a, b].\]
Therefore, \[\lim_{h \to 0} \frac{1}{h}\int_0^h (f(x+t)-f(x))\text{d}t=\lim_{h \to 0} \frac{1}{h}\int_0^h f(x+t)\text{d}t-f(x)=0\ \text{a.e.}\ x \in [a, b].\] Since \[\int_0^h f(x+t)\text{d}t=\int_x^{x+h} f(t)\text{d}t=\int_a^{x+h} f(t)\text{d}t-\int_a^x f(t)\text{d}t,\] then \[\lim_{h \to 0}\frac{1}{h}\left(\int_a^{x+h} f(t)\text{d}t-\int_a^x f(t)\text{d}t\right)=f(x)\ \text{a.e.}\ x \in [a, b],\] i.e., \[\frac{\text{d}}{\text{d}x}\left(\int_a^x f(t)\text{d}t\right)=f(x)\ \text{a.e.}\ x \in [a, b].\]\(\square\)
4. Fundamental Theorem of Calculus
Theorem 4.1 Suppose \(f\) is differentiable almost everywhere on \([a, b]\) and \(f'(x)=0\ \text{a.e.}\ x \in [a, b]\). If \(f\) is not a constant function on \([a, b]\), then there exists an \(\varepsilon>0\) such that for all \(\delta>0\), there exists a finite sequence of pairwise disjoint subintervals \((x_i, y_i)\) of \([a, b]\) with \(x_i<y_i \in [a, b]\) such that \[\sum_{i=1}^n |y_i-x_i|<\delta,\] and \[\sum_{i=1}^n |f(y_i)-f(x_i)|>\varepsilon.\] \(f\) is defined as a singularity function.
Proof. Since \(f\) is not a constant function, then there exists a \(c \in [a, b]\) such that \(f(a) \neq f(c)\). Since \(f'(x)=0\ \text{a.e.}\ x \in [a, b]\), then there exists an \(E \subset [a, c]\) such that \(m([a, c]-E)=0\), where \(f'(x)=0\) for \(x \in E\).
Let \(2\varepsilon=|f(c)-f(a)|>0\) and \(\displaystyle r=\frac{\varepsilon}{c-a}\). There exits a \(\delta_1>0\) such that when \(0<h<\delta_1\), \[|f(x+h)-f(x)|<rh.\] It is obvious that the collection of \([x, x+h]\) is a Vitali cover of \(E\). By Vitali covering theorem, for all \(\delta>0\), there exist finite disjoint intervals \([x_1, x_1+h_1], \ldots, [x_n, x_n+h_n]\) such that \[m\left(E-\bigcup_{i=1}^n [x_i, x_i+h_i]\right)<\delta.\] Assume \(x_1<x_1+h_1<\cdots<x_n<x_n+h_n\). Hence, \[m\left((a, x_1) \cup \bigcup_{i=1}^n (x_i+h_i, x_{i+1}) \cup (x_n+h_n, c)\right)<\delta.\]
Since \[\sum_{i=1}^n |f(x_i+h_i)-f(x_i)|<r\sum_{i=1}^n h_i<r(c-a)=\varepsilon,\] and \[\begin{aligned} &|f(x_1)-f(a)|+\sum_{i=1}^n |f(x_{i+1})-f(x_i+h_i)|+|f(c)-f(x_n+h_n)|+\sum_{i=1}^n |f(x_i+h_i)-f(x_i)| \\\geq &\left|f(x_1)-f(a)+\sum_{i=1}^n (f(x_{i+1})-f(x_i+h_i))+f(c)-f(x_n+h_n)+\sum_{i=1}^n (f(x_i+h_i)-f(x_i))\right| \\=&|f(c)-f(a)|=2\varepsilon, \end{aligned}\] then \[|f(x_1)-f(a)|+\sum_{i=1}^n |f(x_{i+1})-f(x_i+h_i)|+|f(c)-f(x_n+h_n)|>\varepsilon.\]\(\square\)
Definition 4.1 A function \(f\) is absolutely continuous on \([a, b]\) if for all \(\varepsilon>0\), there exists a \(\delta>0\) such that whenever a finite sequence of pairwise disjoint subintervals \((x_i, y_i)\) of \([a, b]\) with \(x_i<y_i \in [a, b]\) satisfies \[\sum_{i=1}^n |y_i-x_i|<\delta,\] then \[\sum_{i=1}^n |f(y_i)-f(x_i)|<\varepsilon.\] The collection of all absolutely continuous functions on \([a, b]\) is denoted \(\text{AC}([a, b])\).
Example 4.1 If \(f\) is Lipschitz continuous on \([a, b]\), i.e., \(|f(x)-f(y)| \leq M \cdot |x-y|\), then \(f \in \text{AC}([a, b])\).
Example 4.2 If \(f \in \text{AC}([a, b])\), then \(f\) is continuous on \([a, b]\).
Example 4.3 If \(f\) is a step function, then \(f\) is a singularity function.
Theorem 4.2 Suppose \(f \in L([a, b])\). Then \(\displaystyle \int_a^x f(t)\text{d}t\) is absolutely continuous on \([a, b]\).
Proof. Suppose \(f \in L([a, b])\). We have \[\forall \varepsilon>0, \exists \delta>0\ \text{s.t.}\ m([a, b])<\delta \Rightarrow \int_{[a, b]} |f(x)|\text{d}x<\varepsilon.\] Take an arbitrary finite sequence of pairwise disjoint subintervals \((x_i, y_i)\) of \([a, b]\) with \(x_i<y_i \in [a, b]\), where \[\sum_{i=1}^n |y_i-x_i|<m([a, b])<\delta.\] We have \[\begin{aligned} \sum_{i=1}^n \left|\int_a^{y_i} f(t)\text{d}t-\int_a^{x_i} f(t)\text{d}t\right|&=\sum_{i=1}^n \left|\int_{x_i}^{y_i} f(t)\text{d}t\right| \\&\leq \sum_{i=1}^n \int_{x_i}^{y_i} |f(t)|\text{d}t \\&=\int_{\bigcup_{i=1}^n (x_i, y_i)} |f(t)|\text{d}t<\varepsilon, \end{aligned}\] i.e., \(\displaystyle \int_a^x f(t)\text{d}t\) is absolutely continuous on \([a, b]\).\(\square\)
Theorem 4.3 \(\text{AC}([a, b])\) is a linear space.
Proof. Take an arbitrary finite sequence of pairwise disjoint subintervals \((x_i, y_i)\) of \([a, b]\) with \(x_i<y_i \in [a, b]\), where \[\sum_{i=1}^n |y_i-x_i|<\delta.\] Suppose \(c \in \mathbb{R}-\{0\}\), \(f, g \in \text{AC}([a, b])\), \[\sum_{i=1}^n |f(y_i)-f(x_i)|<\frac{\varepsilon}{2|c|},\] and \[\sum_{i=1}^n |g(y_i)-g(x_i)|<\frac{\varepsilon}{2}.\]
We have \[\begin{aligned} \sum_{i=1}^n |(cf+g)(y_i)-(cf+g)(x_i)|&=\sum_{i=1}^n |(cf(y_i)+g(y_i))-(cf(x_i)+g(x_i))| \\&=\sum_{i=1}^n |c(f(y_i)-f(x_i))+(g(y_i)-g(x_i))| \\&\leq |c|\sum_{i=1}^n |f(y_i)-f(x_i)|+\sum_{i=1}^n |g(y_i)-g(x_i)| \\&<|c|\frac{\varepsilon}{2|c|}+\frac{\varepsilon}{2}=\varepsilon, \end{aligned}\] i.e., \(cf+g \in \text{AC}([a, b])\).\(\square\)
Theorem 4.4 If \(f \in \text{AC}([a, b])\), then \(f \in \text{BV}([a, b])\).
Proof. Take \(\varepsilon=1\), and an arbitrary finite sequence of pairwise disjoint subintervals \((x_i, y_i)\) of \([a, b]\) with \(x_i<y_i \in [a, b]\), where \[\sum_{i=1}^n |y_i-x_i|<\delta.\] We have \[\sum_{i=1}^n |f(y_i)-f(x_i)|<1.\] Take a partition \(\Delta: a=c_0<c_1<\cdots<c_n=b\) such that \(c_{i+1}-c_i<\delta\), then \[\bigvee_{c_i}^{c_{i+1}} (f)<1,\] and thus \[\bigvee_a^b (f)=\sum_{i=0}^n \bigvee_{c_i}^{c_{i+1}} (f)<n<\infty,\] i.e., \(f \in \text{BV}([a, b])\).\(\square\)
Corollary 4.5 If \(f \in \text{AC}([a, b])\), then \(f\) is differentiable almost everywhere on \([a, b]\), and \(f' \in L([a, b])\).
Theorem 4.6 (Fundamental Theorem of Calculus) If \(f \in \text{AC}([a, b])\), then \[f(x)-f(a)=\int_a^x f'(t)\text{d}t.\]
Proof. We know \(\displaystyle \int_a^x f'(t)\text{d}t \in \text{AC}([a, b])\) and \(\displaystyle \frac{\text{d}}{\text{d}x}\left(\int_a^x f'(t)\text{d}t\right)=f'(x)\ \text{a.e.}\ x \in [a, b]\). Let \[h(x)=f(x)-\int_a^x f'(t)\text{d}t,\] then \(h \in \text{AC}([a, b])\), and \(h'(x)=f'(x)-f'(x)=0\ \text{a.e.}\ x \in [a, b]\). By theorem 4.1, \(h\) is constant, where \[h(a)=f(a)-\int_a^a f'(t)\text{d}t=f(a)-0=f(a).\] Hence, \[f(x)-h(a)=f(x)-f(a)=\int_a^x f'(t)\text{d}t.\]\(\square\)
Theorem 4.7 (Lebesgue Decomposition Theorem) Suppose \(f \in \text{BV}([a, c])\). We can find a step function \(\varphi\), an absolutely continuous function \(f_1\), and a continuous singularity function \(f_2\) such that \[f=\varphi+f_1+f_2.\]
Proof. By theorem 2.7, if \(f \in \text{BV}([a, c])\), we can find a step function \(\varphi\), and a continuous function \(g\) such that \(f=\varphi+g\). Let \[f_1(x)=\int_a^x g'(t)\text{d}t,\] then \(f_1\) is an absolutely continuous function. Let \(f_2=g-f_1\), then \(f_2\) is a continuous singularity function.\(\square\)