We want to find a broad condition on functions \(F:[a, b]\to \mathbb R\) that guarantees the identity \[ \qquad\qquad\int_a^bF'(x)dx=F(b)=F(a) \qquad\qquad {\color{purple}(*)}.\] There are two phenomena that make this question problematic.
First, the identity (*) might not be meaningful if we merely assumed \(F\) was continuous due to the existence of non-differentiable functions.
Second, even if \(F'(x)\) existed for every \(x\), the function \(F'(x)\) would not necessarily be Lebesgue integrable. Consider the function \[ F(x)= \begin{cases} x^2\sin(x^{-2}) & \text{ if } x\neq0,\\ 0 & \text{ if } x=0. \end{cases}\] One can show that \(F'(x)\) exists for \(x\in[-1, 1]\), but \(F'\not\in L^1([-1, 1])\).
Let \(I\subseteq\mathbb R\) be an interval and let \(f:I\to\mathbb R\) be a monotonic function. Then the set of points of \(I\) of which \(f\) is discontinuous is at most countable.
Proof. Wlog we may assume that \(f\) is increasing and \(I=\mathbb R\).
Let \(E\) be the set of points at which \(f\) is discontinuous.
With every point \(x \in E\) we associate a rational number \(r(x) \in \mathbb{Q}\) such that \[f(x-) <r(x)<f(x+),\] so \(r:E \to \mathbb{Q}\). Here \(f(x+)=\lim_{y\searrow x}f(y)\) and \(f(x+)=\lim_{y\nearrow x}f(y)\).
Since \(x_1<x_2\) implies \(f(x_1+) \leq f(x_2-)\) we see that \(r(x_1) \neq r(x_2)\) if \(x_1 \neq x_2\).
We have established that the function \(r:E \to \mathbb{Q}\) is injective, thus \[\qquad\qquad {\rm card}(E) \leq {\rm card}(\mathbb{Q}). \qquad\qquad \tag*{$\blacksquare$}\]
If \(F:\mathbb{R} \to \mathbb{R}\) is an increasing, right-continuous function on \(\mathbb{R}\), then \(\mu_F\) is the Borel measure given by the relation \[{\color{purple}\mu_F((a,b])=F(b)-F(a)}.\] The term almost everywhere will always refer to Lebesgue measure.
A Borel measure \(\nu\) on \(\mathbb{R}^d\) is called regular if
\(\nu(K)<\infty\) for every compact \(K \subseteq \mathbb{R}^d\),
\(\nu(E)=\inf\{\nu(U): E \subseteq U, \text{ and } U \text{ is open}\}\) for every \(E \in {\rm Bor}(\mathbb{R}^d)\).
Remark. In fact, (ii) is implied by (i), and this definition of regular measures coincides with the regularity discussed at the beginning of the course.
Let \(F:\mathbb{R} \to \mathbb{R}\) be increasing and let \[G(x)=F(x+)=\lim_{t \searrow x}F(t).\] Then \(F\) and \(G\) are differentiable a.e. and \(F'=G'\) a.e., and \(F\) and \(G\) are measurable on \(\mathbb R\).
Proof. Observe that \(G\) is increasing and right-continuous, and \(G=F\) except perhaps where \(F\) is discontinuous. Moreover, \[G(x+h)-G(x)= \begin{cases} \mu_G((x,x+h])&\text{ if }h>0,\\ -\mu_G((x+h,x])& \text{ if }h<0. \end{cases}\] The families \((x-r,x]\) and \((x,x+r]\) shrink nicely to \(x\) as \(r=|h| \to 0\).
Last time we proved the following theorem:
Let \(\nu\) be a regular signed or complex Borel measure on \(\mathbb{R}^d\) and let \(d\nu=d\mu+f\,d\lambda_d\) be the Lebesgue–Radon–Nikodym decomposition. Then for \(\lambda_d\)-a.e. \(x \in \mathbb{R}^d\) we have \[\lim_{r \to 0}\frac{\nu(E_r)}{\lambda_d(E_r)}=f(x), \quad \text{ and } \quad \lim_{r \to 0}\frac{\mu(E_r)}{\lambda_d(E_r)}=0,\] for every family \((E_r)_{r >0} \subseteq \mathbb{R}^d\) that shrinks nicely to \(x\in \mathbb R^d\).
Applying this theorem to \(\mu_G\) we see that \(G'(x)\) exists for a.e. \(x\in \mathbb R\).
It remains to show that if \(H=G-F\), then \(H'=0\) exists a.e. on \(\mathbb R\).
Let \(E=\{x_j: j \in \mathbb{N}\}\) be an enumeration of points for which \(H \neq 0\).
Then \(H(x_j)>0\) and for \(A\subseteq\mathbb R\) define a Borel measure \[{\color{blue}\mu(A)=\sum_{j \in \mathbb N}H(x_j)\delta_{x_j}(A)}, \quad \text{ where }\quad \delta_{x_j}(x)= \begin{cases} 1 &\text{ if } x=x_j,\\ 0 &\text{ otherwise}. \end{cases}\]
Observe that \(\mu((-N, N))<\infty\) for any \(N \in \mathbb{N}\). In particular, \(\mu\) is finite on compact subsets of \(\mathbb R\), thus regular.
Note also that \(\mu \perp \lambda_1\) since \(\lambda_1(E)=\lambda_1(E^c)=0\). But then again by the Lebesgue differentiation theorem for Borel measures we may conclude \[\left|\frac{H(x+h)-H(x)}{h}\right| \leq \frac{\mu((x-2|h|,x+2|h|))}{|h|} \ _{\overrightarrow{h\to 0}} \ 0\] for a.e. \(x \in \mathbb{R}\). Thus \(H'(x)=0\) for a.e. \(x \in \mathbb{R}\). $$\tag*{$\blacksquare$}$$
Suppose that \(a\le b\) in \(\mathbb R\), and that \(F:[a,b]\to\mathbb R\) is a non-decreasing function. Then \(\int_a^bF'(x)dx\) exists and \[\qquad\qquad \int_a^bF'(x)dx\le F(b)-F(a). \qquad\qquad {\color{purple}(*)}\]
Remark.
In general, equality in (*) might not hold, even if we assume that \(F\) is a continuous function.
In order to see this, let \(F:[0, 1]\to [0, 1]\) be the Cantor–Lebesgue function. This is continuous increasing function such that \(F(1)=1\) and \(F(0)=0\). Moreover, \(F\) is constant on each interval of the complement of the Cantor set \(\mathcal C\). Since \(\lambda_1(\mathcal C) = 0\), we find that \(F'(x) = 0\) a.e. on \([0, 1]\), yielding that equality in (*) cannot hold.
Proof. The result is trivial if \(a=b\); let us suppose that \(a<b\). Let us extend the function \(F\) from \([a, b]\) to \(\mathbb R\) by setting \[F(x)= \begin{cases} F(a) & \text{ if } x\in (-\infty, a),\\ F(x) & \text{ if } x\in [a, b],\\ F(b) & \text{ if } x\in (b, \infty). \end{cases}\] By Theorem (A) we see that \(F'\) is measurable and defined a.e. on \(\mathbb R\). Let \[F_n(x)=\frac{F(x+1/n)-F(x)}{1/n} \quad \text{ for } \quad x\in \mathbb R, \ n\in \mathbb N.\] Then \(\lim_{n\to\infty}F_n(x)=F'(x)\) for a.e. \(x\in \mathbb R\). Since \(F_n(x)\ge0\) by Fatou’s lemma we obtain that \[\int_a^bF'(x)dx\le \liminf_{n\to \infty}\int_a^bF_n(x)dx.\]
By translation invariance of Lebesgue measure we note that \[\begin{aligned} \int_a^bF_n(x)dx&=n\int_a^bF(x+1/n)dx-n\int_a^bF(x)dx\\ &=n\int_{a+1/n}^{b+1/n}F(x)dx-n\int_a^bF(x)dx\\ &=n\int_{b}^{b+1/n}F(x)dx-n\int_{a}^{a+1/n}F(x)dx\\ &\le n\big(\lambda_1((b, b+1/2))F(b)-\lambda_1((a, a+1/2))F(a)\big)\\ &=F(b)-F(a). \end{aligned}\] To justify the inequality it suffices to note that \(F(a)\le F(x)\le F(b)\) for all \(x\in \mathbb R\). This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
If \(F:\mathbb{R} \to \mathbb{C}\) and \(x \in \mathbb{R}\) we define the total variation function of \(F\) by \[T_F(x)=\sup\Big\{\sum_{j=1}^n|F(x_j)-F(x_{j-1})|: n \in \mathbb{N}, \ \ -\infty <x_0<\ldots<x_n=x\Big\}.\]
Remarks.
The sums in \(T_F\) are made bigger if we add more subdivision points.
If \(a<b\) the definition of \(T_F(b)\) is unaffected if we assume that \(a\) is always one of the subdivision point. Then \[\begin{aligned} T_F(b)&-T_F(a)\\ &=\sup\Big\{\sum_{j=1}^n |F(x_j)-F(x_{j-1})|: n \in \mathbb{N}, \ \ a=x_0<\ldots<x_n=b\Big\}. \end{aligned}\]
Thus \(T_F\) is an increasing function with values in \([0,\infty]\).
If \(T_F(\infty)=\lim_{x \to \infty}T_F(x)<\infty\) we say that \(F\) is of bounded variation on \(\mathbb{R}\) and define
\[{\rm BV}=\{F:\mathbb{R} \to \mathbb{C}:T_F(\infty)<\infty\}.\] The total variation of \(F\) on \([a,b]\) is defined by \[\begin{aligned} T_F(b)-T_F(a) =\sup\Big\{\sum_{j=1}^n |F(x_j)-F(x_{j-1})|: n \in \mathbb{N}, \ \ a=x_0<\ldots<x_n=b\Big\}. \end{aligned}\] It depends only on the values of \(F\) on \([a,b]\) and we may set \[{\rm BV}([a,b])=\{F:[a,b] \to \mathbb{C}:T_F(b)-T_F(a)<\infty\}.\] If \(F \in {\rm BV}\), then \(F \in {\rm BV}([a,b])\) for all \(a<b\). If \(F \in {\rm BV}([a,b])\), then setting \(F(x)=F(a)\) for \(x<a\) and \(F(x)=b\) for \(x>b\) we have \(F \in {\rm BV}\).
If \(F:\mathbb{R} \to \mathbb{R}\) is bounded and increasing then \(f \in {\rm BV}\) and \[T_F(x)=F(x)-F(-\infty).\]
If \(F,G \in {\rm BV}\), then \(aF+bG\in {\rm BV}\) for all \(a,b \in \mathbb{C}\).
If \(F\) is differentiable and \(F'\) is bounded then \(F \in {\rm BV}([a,b])\) for \(-\infty<a<b<\infty\) by the mean value theorem.
If \(F(x)=\sin(x)\), then \(F \in {\rm BV}([a,b])\) for \(-\infty<a<b<\infty\), but \(F \not\in {\rm BV}\).
Let \[F(x)= \begin{cases} x\sin(x^{-1}) &\text{ for }x \neq 0,\\ 0 &\text{ for }x = 0. \end{cases}\] Then \(F \not\in {\rm BV}([a,b])\) for \(a \leq 0 < b\) or \(a < 0 \leq b\).
If \(F \in {\rm BV}\) and \(F:\mathbb{R} \to \mathbb{R}\), then \(F=\frac{1}{2}\left((T_F+F)-(T_F-F)\right),\) and \(T_F+F\) and \(T_F-F\) are increasing.
Proof. If \(x<y\) and \(\varepsilon>0\) choose \(x_0<\ldots<x_n=x\) such that \[\sum_{j=1}^n |F(x_j)-F(x_{j-1})| \geq T_F(X)-\varepsilon.\] Then \(\sum_{j=1}^n|F(x_j)-F(x_{j-1})|+|F(y)-F(x)|\) is an approximation for \(T_F(y)\) and \(F(y)=(F(y)-F(x))+F(x)\). Therefore, \[\begin{aligned} T_F(y) \pm F(y) &\geq \sum_{j=1}^n |F(x_j)-F(x_{j-1})| +|F(y)-F(x)|\\ &\pm (F(y)-F(x)) \pm F(x) \geq T_F(x)-\varepsilon\pm F(x). \end{aligned}\] Since \(\varepsilon>0\) is arbitrary, we obtain \(T_F(y) \pm F(y) \geq T_F(x) \pm F(x).\) $$\tag*{$\blacksquare$}$$
\(F \in {\rm }BV\) iff \({\rm Re }(F) \in {\rm BV}\) and \({\rm Im}(F) \in {\rm BV}\).
If \(F:\mathbb{R} \to \mathbb{R}\) then \(F \in {\rm BV}\) iff \(F\) is the difference of two bounded increasing functions. We have the Jordan decomposition of \(F \in {\rm BV}\) \[ F={\color{red}\frac{1}{2}(T_F+F)}-{\color{blue}\frac{1}{2}(T_F-F)},\] into positive \(\frac{1}{2}(T_F+F)\) and negative \(\frac{1}{2}(T_F-F)\) variations.
If \(F \in {\rm BV}\), then \[ F(x+)=\lim_{y \searrow x}F(y)\quad \text{ and }\quad F(x-)=\lim_{y \nearrow x}F(y)\] exist for all \(x \in \mathbb{R}\), as do \(F(\pm \infty)=\lim_{y \to \pm\infty}F(y)\).
If \(F \in {\rm BV}\) the set of points for which \(F\) is discontinuous if countable.
If \(F \in {\rm BV}\) and \(G(x)=F(x+)\) then \(F'\) and \(G'\) exist and are equal a.e.
Proof of (a). It is obvious since \(|{\rm Re}(z)| \leq |z|\) and \(|{\rm Im}(z)| \leq |z|\).$$\tag*{$\blacksquare$}$$
Proof of (b). If \(F=F_1-F_2\) and \(F_1,F_2\) are increasing and bounded, then \(F_1,F_2 \in {\rm BV}\) and so \(F=F_1-F_2 \in {\rm BV}\). By the previous lemma we have \(F=F_1-F_2\), where \(F_1=\frac{1}{2}(T_F+F)\) and \(F_2=\frac{1}{2}(T_F-F)\), and \(F_1, F_2\) are both increasing.
Since \(T_F(y) \pm F(y) \geq T_F(x) \pm F(x)\) for \(y>x\), then \[\mp F(y) \pm F(x) \leq T_F(y)-T_F(x) \leq T_F(\infty)-T_F(-\infty)<\infty,\] so that \(F\), and hence \(T_F\pm F\) is bounded.$$\tag*{$\blacksquare$}$$
Proof of (c). If \(F \in {\rm BV}\), then \(F\) is a difference of two bounded increasing functions, thus the desired limits exist.$$\tag*{$\blacksquare$}$$
Proof of (d). If \(F \in {\rm BV}\), then \(F\) is a difference of two bounded increasing functions, thus the set of discontinuities of \(F\) is countable.$$\tag*{$\blacksquare$}$$
Proof of (e). It follows from Theorem (A) and items (a) and (b).$$\tag*{$\blacksquare$}$$
Since \({\color{blue}x^+=\max\{x, 0\}=\frac{1}{2}(|x|+x)}\) and \({\color{red}x^-=\max\{-x, 0\}=\frac{1}{2}(|x|-x)}\), for \(x \in \mathbb{R}\) we have \[\begin{aligned} \frac{1}{2}(T_F \pm F)(x)=&\sup\Big\{\sum_{j=1}^n(F(x_j)-F(x_{j-1}))^{\pm}:x_0<\ldots<x_N=x\Big\}\\ &\pm\frac{1}{2}F(-\infty). \end{aligned}\] Items (a) and (b) from Theorem (B) lead to the connection between \({\rm BV}\) and the space of complex Borel measures on \(\mathbb{R}\). To make this precise we introduce the space \[{\rm NBV}=\{F \in {\rm BV}: F\text{ is right-continuous and } F(-\infty)=0\}\] of normalize \(\rm BV\) functions.
If \(F \in {\rm BV}\) then \[G(x)=F(x+)-F(-\infty) \in {\rm NBV}\] and \(G'=F'\) a.e. on \(\mathbb R\). It only suffices to show that \(G \in {\rm BV}\). Using the Jordan decomposition for \(F\), we may write \(F=F_1-F_2\), where \(F_1,F_2\) are increasing and bounded functions. Then \[G(x)=F_1(x+)-(F_2(x+)-F(-\infty)),\] which is the difference of two bounded increasing functions, thus \(G \in {\rm BV}\).
If \(F \in {\rm BV}\) then \(T_F(-\infty)=0\). If \(F\) is also right-continuous, then so is \(T_F\).
Proof. If \(\varepsilon>0\) and \(x \in \mathbb{R}\) choose \(x_0<\ldots<x_n=x\) so that \[\sum_{j=1}^{n}|F(x_j)-F(x_{j-1})| \geq T_F(x)-\varepsilon.\] Hence \(T_F(x)-T_F(x_0) \geq T_F(x)-\varepsilon\), which ensures that \(T_F(-\infty)=0\), since \(T_F(y) \leq T_F(x_0) \leq \varepsilon\) for all \(y \leq x_0\).
Suppose that \(F\) is right–continuous. Given \(x \in \mathbb{R}\) and \(\varepsilon>0\) let us define \(\alpha=T_F(x+)-T_F(x)\), and choose \(\delta>0\) so that \(|F(x+h)-F(x)|<\varepsilon\) and \(T_F(x+h)-T_F(x+)<\varepsilon\) when \(0<h<\delta\).
For any such \(h\) there exist \(x=x_0<\ldots<x_n=x+h\) such that \[\sum_{j=1}^n|F(x_j)-F(x_{j-1})| \geq \frac{3}{4}(T_F(x+h)-T_F(x)) \geq \frac{3}{4}\alpha,\] since \(\alpha=T_F(x+)-T_F(x)\).
Hence \[\sum_{j=1}^n|F(x_j)-F(x_{j-1})| \geq \frac{3}{4}\alpha-|F(x_1)-F(x_0)| \geq \frac{3}{4}\alpha-\varepsilon.\] Likewise, there exist \(x=t_0<\ldots<t_m=x_1\) such that \[\sum_{j=1}^m |F(t_j)-F(t_{j-1})| \geq \frac{3}{4}\alpha\] and hence \[\begin{aligned} \alpha+\varepsilon &>T_F(x+h)-T_F(x) \\ &\geq \sum_{j=1}^m |F(y)-F(t_{j-1})|+\sum_{j=1}^n |F(x_j)-F(x_{j-1})|\\ &\geq \frac{3}{2}\alpha-\varepsilon, \end{aligned}\] since \(\alpha=T_F(x+)-T_F(x)\). Thus \(\alpha<4\varepsilon\) which gives \(\alpha=0\).$$\tag*{$\blacksquare$}$$
If \(\mu\) is a complex Borel measure on \(\mathbb{R}\) and \(F(x)=\mu((-\infty,x])\), then \(F \in {\rm NBV}\). Conversely, if \(F \in {\rm NBV}\), there exists a unique complex Borel measure \(\mu_F\) such that \(F(x)=\mu_F((-\infty,x])\).
Proof. If \(\mu\) is a complex measure we have \[\mu=\mu_1^+-\mu_1^-+i(\mu_2^+-\mu_2^-),\] where \(\mu_j^{\pm}\) for \(j\in \{1, 2\}\) are finite measures.
If \({\color{blue}F_j^{\pm}(x)=\mu_j^{\pm}((-\infty,x])}\), then \(F_j^{\pm}\) is increasing and right continuous.
\(F_j^{\pm}(-\infty)=0\), and \(F_j^{\pm}(+\infty)=\mu_j^{\pm}(\mathbb{R})<\infty\), thus \(F_j^{\pm}\) are bounded.
By Theorem (B), see item (a) and (b), the function \[F=F_1^+-F_1^-+i(F_2^+-F_2^-)\in {\rm NBV}.\]
Conversely, suppose that \(F \in {\rm NBV}\).
By item (a) of Theorem (B) we have \(F=F_1+iF_2\) for some \(F_1, F_2 \in {\rm NBV}\).
By item (b) of Theorem (B) we have \[{\color{purple} F=F_1^+-F_1^-+i(F_2^+-F_2^-),\quad \text{ where } \quad F_j^{\pm}=F_j\pm T_{F_j}.}\]
By Lemma (C) we conclude that \(F_j^{\pm} \in {\rm NBV}\).
Now each \(F_j^{\pm}\) gives rise to a measure \(\mu_j^{\pm}\) so that \[{\color{purple}F(x)=\mu_F((-\infty,x])},\] where \[\qquad \qquad \mu_{F}=\mu_1^+-\mu_1^-+i(\mu_2^+-\mu_2^-).\qquad \qquad \tag*{$\blacksquare$}\]
Proposition (E). If \(F \in {\rm NBV}\), then \(F' \in L^1(\mathbb R)\). Moreover, \(\mu_F \perp \lambda\) iff \(F'=0\) a.e. on \(\mathbb R\), and \(\mu_F \ll \lambda\) iff \(F(x)=\int_{\infty}^{x}F'(x)dx.\)
Proof. Since \(F \in {\rm NBV}\), then by Theorem (D) there exists a regular complex Borel measure \(\mu_F\) such that \(F(x)=\mu_F((-\infty,x])\).
By the Lebesgue–Radon–Nikodym theorem: \(\mu_F={\color{red}(\mu_F)_s}+{\color{blue}(\mu_F)_a}\).
By the Lebesgue differentiation theorem for Borel measures we have \[{\color{blue}F'(x)=\lim_{r \to 0}\frac{\mu_F(E_r)}{\lambda(E_r)}=\lim_{r \to 0}\frac{(\mu_F)_a(E_r)}{\lambda(E_r)}=\frac{d\mu_F}{d\lambda}(x),} \ \text{ and } \ {\color{red}\lim_{r \to 0}\frac{(\mu_F)_s(E_r)}{\lambda(E_r)}=0,}\]
a.e. on \(\mathbb R\), where \(E_r=(x,x+r]\) or \((x-r,x]\). Hence \[{\color{red}\mu_F \perp \lambda \ \text{ iff } \ F'=0 \text{ a.e.},}\quad \text{ and } \quad {\color{blue}\mu_F \ll \lambda \ \text{ iff } \ F(x)=\int_{-\infty}^x F'(y)dy,}\] since \(\mu_F\) is finite measure and \(F'=\frac{d\mu_F}{d\lambda}\) is the Radon–Nikodym derivative we conclude that \(F' \in L^1(\mathbb R)\).$$\tag*{$\blacksquare$}$$
A function \(F:\mathbb{R} \to \mathbb{C}\) is called absolutely continuous (a.c.) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that for any finite set of disjoint intervals \((a_1,b_1),\ldots,(a_N,b_N)\), we have \[\sum_{j=1}^N (b_j-a_j)<\delta \quad \implies \quad \sum_{j=1}^N|F(b_j)-F(a_j)|<\varepsilon.\]
Remarks.
\(F\) is said to be (a.c.) on \([a,b]\) if this condition is satisfies when the intervals \((a_j,b_j) \subseteq [a,b]\).
Clearly, if \(F\) is (a.c.) then \(F\) is uniformly continuous, (take \(N=1\)).
If \(F\) is differentiable and \(|F'| \leq M\), then \(F\) is (a.c.), and more generally any function satisfying Lipschitz condition is (a.c.).
The concepts of absolutely continuous measures and functions are very close to each other. Recall:
Let \(\nu\) be a finite signed measure and \(\mu\) a positive measure on \((X,\mathcal M)\), then \(\nu \ll \mu\) iff for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[\begin{aligned} {\color{red}\mu(E)<\delta \quad \implies \quad |\nu|(E)<\varepsilon.} \end{aligned}\]
Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^1(X, \mu)\), then for every \(\varepsilon>0\) there is \(\delta>0\) such that \[\mu(E)<\delta \quad \implies \quad \Big|\int_E fd\mu\Big| <\varepsilon.\] We will write \({\color{red}d\nu=fd\mu}\) to express the relationship \({\color{purple}\nu(E)=\int_E fd\mu}.\)
Proposition (F). If \(F \in {\rm NBV}\), then \(F\) is (a.c.) iff \(\mu_F \ll \lambda\).
Proof (\(\Longleftarrow\)). Since \(F \in {\rm NBV}\), then by Theorem (D) there exsist a regular complex Borel measure \(\mu_F\) such that \(F(x)=\mu_F((-\infty,x])\). If \(\mu_F \ll \lambda\), by the \(\varepsilon\)-\(\delta\) criterion for absolute continuity, for every \(\varepsilon>0\) there is \(\delta>0\) such that \(|\mu_F(E)|<\varepsilon\) if \(\lambda(E)<\delta\) where \(E=\bigcup_{j=1}^N (a_j,b_j]\), thus \(F\) is (a.c.). $$\tag*{$\blacksquare$}$$
Proof (\(\Longrightarrow\)). To prove the converse let \(E\) be a Borel set with \(\lambda(E)=0\). If \(\varepsilon,\delta\) are as in the definition of (a.c.) of \(F\) then we can find open sets \(U_1 \supseteq U_2 \supseteq \ldots \supseteq E\) such that \(\lambda(U_j)<\delta\) and \(\mu_F(U_j) \ _{\overrightarrow{j\to \infty}} \ \mu_F(E)\). Each \(U_j\) is a disjoint union of open intervals \((a_j^k,b_j^k)\subseteq U_j\) and \[ \sum_{k=1}^N |\mu_F((a_j^k,b_j^k))| \leq \sum_{k=1}^N|F(b_j^k)-F(a_j^k)|<\varepsilon \quad \text{ for all } \quad N \in \mathbb{N}.\] Letting \(N \to \infty\) we obtain \(|\mu_F(U_1)|<\varepsilon\) and hence \(|\mu_F(E)| \leq \varepsilon\), thus \(\mu_F(E)=0\), since \(\varepsilon\) was arbitrary. This proves that \(\mu_F \ll \lambda\) as desired.$$\tag*{$\blacksquare$}$$
If \(f \in L^1(\mathbb R)\) then \(F(x)=\int_{-\infty}^{x}f(t)dt \in{\rm NBV}\) and \(F\) is (a.c.), and \(f=F'\) a.e. on \(\mathbb R\). Conversely, if \(F \in {\rm NBV}\) and \(F\) is (a.c.), then \(F' \in L^1(\mathbb R)\) and \(F(x)=\int_{-\infty}^xF'(t)dt\).
Proof. If \(f \in L^1(\mathbb R)\) then \(F(-\infty)=0\) and \(f=f_1^+-f_1^-+i(f_2^+-f_2^-)\), where \(f_j^{\pm}\ge0\). Then \(F_j^{\pm}(x)=\int_{-\infty}^{x}f_j^{\pm}(t)dt\) is increasing, thus \(F\in{\rm BV}\) by item (a) of Theorem (B). Moreover, \(F\) is right-continuous, so \(F\in {\rm NBV}\).
Since \(F \in {\rm NBV}\), then by Theorem (D) there exsist a regular complex Borel measure \(\mu_F\) such that \(F(x)=\mu_F((-\infty,x])\).
Since \(F(x)=\int_{-\infty}^{x}f(t)dt\) then \(\mu_F\ll \lambda\), which shows that \(F\) is (a.c.) by Proposition (F).
By Proposition (E) we deduce \(F(x)=\int_{-\infty}^{x}F'(t)dt\), since \(\mu_F\ll \lambda\) and \(F \in {\rm NBV}\), and consequently \(F'=f\) a.e. on \(\mathbb R\).
Conversely suppose that \(F \in {\rm NBV}\) and \(F\) is (a.c.), then we have \(\mu_F\ll \lambda\) by Proposition (F). By Proposition (E) we readily see that \(F' \in L^1(\mathbb R)\) and \(F(x)=\int_{-\infty}^xF'(t)dt\), and the proof is completed.$$\tag*{$\blacksquare$}$$
If \(F\) is (a.c.) on \([a,b]\) then \(F \in {\rm BV}([a,b])\).
Proof. Let \(\delta\) be as in the definition of (a.c.) of the function \(F\) with \(\varepsilon=1\) and \(N=\lfloor \delta^{-1}(b-a)+1\rfloor\). If \(a=x_0<\ldots<x_n=b\). By inserting more subdivision points if necessary we can collect the intervals \((x_{j-1},x_j)\) into at most \(N\) groups of consecutive intervals such that the sum of the lengths in each groups is less that \(\delta\). Then \[\sum_{j=1}^n|F(x_j)-F(x_{j-1})|\le N,\] since the sum over each group is at most \(1\) and hence the total variation of \(F\) on \([a,b]\) is at most \(N\).$$\tag*{$\blacksquare$}$$
If \(-\infty<a<b<\infty\) and \(F:[a,b] \to \mathbb{C}\) the following are equivalent:
\(F\) is absolutely continuous on \([a,b]\).
There is \(f \in L^1([a,b])\) such that \[F(x)-F(a)=\int_a^x f(t)dt.\]
\(F\) is differentiable a.e. on \([a,b]\), and \(F' \in L^1([a,b])\) and \[F(x)-F(a)=\int_a^x F'(t)dt.\]
Proof (a) \(\Longrightarrow\) (c). Let \(G(x)=F(x)-F(a)\) then \(G(a)=0\). We extend \(G(x)\) to the entire \(\mathbb R\) by setting \(G(x)=0\) for all \(x<a\) and \(G(x)=G(b)\) for all \(x>b\). Hence \(G\) is absolutely continuous on \(\mathbb R\).
Since \(F\) is absolutely continuous on \([a,b]\), then \(f\in {\rm BV([a, b])}\) by Lemma (H) and consequently \(G\in {\rm NBV}\) and \(G\) is (a.c.).
By the previous corollary we deduce that \(G'=F'\in L^1(\mathbb R)\) and \[F(x)-F(a)=G(x)=\int_{-\infty}^xG'(t)dt=\int_{a}^xF'(t)dt. \qquad \tag*{$\blacksquare$}\]
Proof (c) \(\Longrightarrow\) (b). We now see that this implication is obvious. $$\tag*{$\blacksquare$}$$
Proof (b) \(\Longrightarrow\) (a). We set \(f(x)=0\) for \(t \in [a,b]^c\) and apply the previous corollary. Then \[F(x)-F(a)=G(x)=\int_{-\infty}^xf(t)dt\in {\rm NBV},\] and \(G\) is (a.c.) on \(\mathbb R\), thus \(F\) must be (a.c.) on \([a, b]\).$$\tag*{$\blacksquare$}$$
Let \(f\) and \(g\) be two real-valued absolutely continuous functions on \([a, b]\). Then \(fg'\) and \(f'g\) are in \(L^1([a, b])\) and for every \(x \in [a, b]\) we have \[\int_a^xf'(t)g(t)dt=f(x)g(x)-f(a)g(a)-\int_a^xf(t)g'(t)dt \quad \text{ for } \quad x\in [a, b]\]
Proof. The absolute continuity of \(f\) and \(g\) implies that of \(fg\). By the previous theorem \(f', g'\) and \((fg)'\) exist a.e. on \([a, b]\) and all belong to \(L^1([a, b])\). Moreover, \(fg', f'g\in L^1([a, b])\). Thus for \(x\in [a, b]\) we have \[\begin{aligned} f(x)g(x)-f(a)g(a)&=\int_a^x(f(t)g(t))'dt\\ &=\int_a^xf(t)'g(t)dt+\int_a^xf(t)g(t)'dt. \end{aligned}\] This gives the desired result. $$\tag*{$\blacksquare$}$$
Suppose that \(\varphi: [a, b]\to [\alpha, \beta]\) is an increasing (a.c.) function such that \(\varphi(a)=\alpha\) and \(\varphi(b)=\beta\). Then for every \(f\in L^1([\alpha, \beta])\) one has \[\qquad \qquad \int_{\alpha}^{\beta}f(t)dt=\int_a^bf(\varphi(t))\varphi'(t)dt. \qquad \qquad {\color{purple}(*)}\]
Proof. We first verify (*) for \(f=\mathbf{1}_{{(\alpha_0, \beta_0)}}\), where \((\alpha_0, \beta_0)\subset[\alpha, \beta]\). Since \(\varphi\) is continuous and increasing function thus \(\varphi^{-1}[(\alpha_0, \beta_0)]=(a_0, b_0)\) for some \((a_0, b_0)\subset [a, b]\) and \(\varphi(a_0)=\alpha_0\) and \(\varphi(b_0)=\beta_0\). Then we see that \[\begin{aligned} \int_{\alpha}^{\beta}f(t)dt=\beta_0-\alpha_0&=\varphi(b_0)-\varphi(a_0)=\int_{a_0}^{b_0}\varphi'(t)dt = \int_{a}^{b}f(\varphi(t))\varphi'(t)dt, \end{aligned}\] which is precisely (*) with \(f=\mathbf{1}_{{(\alpha_0, \beta_0)}}\). Now we can quite easily pass from characteristic functions of open intervals to characteristic functions of Borel and then Lebesgue measurable sets, and then by standard approximation arguments to all \(f\in L^1([\alpha, \beta])\). Justify this!$$\tag*{$\blacksquare$}$$