Note. This note is based on the seventh chapter of Zygmund's well-known textbook, Measure and integral. Some proofs will be skipped or left only ideas. You may find the proofs in Zygmund's book. Please tell me if there is any mistake in my note :).
0. Foreword
In the first part of this note, we will derive the relationship between Lebesgue integral and differentiation, which is the Lebesgue differentiation theorem. In the proof of this theorem, we introduce the simple Vitali lemma, which will be refined in the next. Then we will use this refined lemma to prove the differentiation of a monotone increasing function exists almost everywhere, and this fact is immediately generalized to function of bounded variation. Also, in the same theorem, we note that
\[0\leq \int_{a}^{b} f' \leq f(b-)-f(a+).\]
The last inequality cannot be replaced by equality, even if $f$ is continuous on $[a,b]$ (Consider the Cantor-Lebesgue function). To deal with this question, we introduce the concepts of absolute continuity and Singularity, and then we know that the equality holds if we added that $f$ is absolutely continuous. On the other hand, we will find out that a function of BV can split into a absolutely continuous function plus a singular function. Also, we will prove the Lebesgue version of integration by parts.
I skipped the part of convex function in the last part of chapter 7 in Zygmund's book.
1. The Indefinite Integral
Def. Given $f\in L(A)$, where $A$ is a measurable subset in $\mathbb{R}^n$. Define the indefinite integral of $f$ to be the function
\[F(E) = \int_E f,\]
where $E\subseteq A$ measurable.
Def [Set function]. A set function $F:\Sigma \rightarrow \mathbb{R}$, where $\Sigma$ is a $\sigma$-algebra, satisfies:
(i) $F(E)$ is finite for every $E\in \Sigma$
(ii) $F(E)$ is countably additive, i.e., if $E=\bigcup_k E_k$ is a union of disjoint $E_k \in \Sigma$, then
\[F(E)=\sum_k F(E_k).\]
Rmk. the indefinite integral of $f\in L(A)$ is a set function.
Recall that the diameter of a set $E$ is the number $\sup\{|x-y|:x,y\in E\}$.
Def. A set function $F$ is said to be continuous if $F(E)$ tends to zero as $E$'s diameter tends to zero.
Def. A set function $F$ is said to be absolutely continuous if $F(E)$ goes to zero as its measure goes to zero.
Rmk. There exists a continuous set function that is not absolutely continous.
Thm 1. If $f\in L(A)$, its indefinite integral is absolutely continous.
Idea. Assume f is nonnegative and chops the ``top'' of f.
Rmk. The converse is true, called the Radon-Nikodym theorem.
2. Lebesgue's Differentiation Theorem
Thm 2 [Lebesgue's Differentiation Theorem]. Let $f\in L(\mathbb{R})$, its indefinite integral is differentiable with derivative $f(x)$ at almost every $x\in \mathbb{R}$.
Proof strategy.
If f is continuous, the result follows immediately. Otherwise,
(i) Approximate $f$ with a continous functoin $C_k$ (global approx.).
(ii) Control the local behavior, the``average'', of $f-C_k$. To do this, we need a crucial lemma (Simple Vitali).
Lemma 3. If $f\in L(\mathbb{R})$, there exists a sequence $\{C_k\}$ of continuous functions with compact support such that
\[\int_{\mathbb{R}}|f-C_k|dx \rightarrow 0 \]
as $k \rightarrow \infty$.
Sketch of proof. Call a integrable f for which the statement hold has property $\scr{A}$. We may prove that
(i) a linear combination of functions with property $\scr{A}$ has property $\scr{A}$.
(ii) a sequence of function $f_k$ with property $\scr{A}$. Then its limit f in $L_1$ also has property $\mathscr{A}$.
So prove a characteristic function $\chi_I$ of an interval $I$ has property $\scr{A}$ first.
Lemma 4 [Simple Vitali]. Let $E$ be a subset of $\mathbb{R}^n$ with $|E|_e < +\infty$, and let $K$ be a collection of cubes $Q$ covering $E$. Then there exists a positive constant $\beta$, depending only on $n$ (dimension), and a finite number of disjoint cubes $Q_1$,...,$Q_N$ in $K$ such that
\[\sum_{j=1}^N |Q_j| \geq \beta |E|_e.\]
Proof. Define $K_1=K$ and
\[t^*_1=\sup\{t:Q(t)\in K_1\}.\]
If $t_1^*=+\infty$, then for any $\beta$ there always exists a cube $Q\in K$ such that $|Q| \geq \beta|E|_e$. Otherwise, we choose a cube $Q_1=Q(t)$ with $t>\frac{t_1^*}{2}$. Then we may split $K_1$ into $K_2$ and $K_2'$, where $K_2$ is of cubes disjoint with $Q_1$ and $K_2'$ is of cubes intersecting with $Q_1$. Define $Q_1^*$ concentric $Q_1$ with edge length $5t_1$. Note that $|Q_1^*|=5^n|Q_1|$ and every cubes in $K_{2}'$ ($5$ is the smallest natural number with this property ).
Recursively, we may construct a decreasing sequence ${t_j^*}$, and disjoint cubes ${Q_j}$ with the same method. Also, if any $K_{i}$ is empty, we define the corresponding $t_i^*=0$ and the process ends. Define $Q_j^*$ concentric $Q_j$ with edge length $5t_j$. Note that $|Q_j^*|=5^n|Q_j|$ and every cubes in $K_{j+1}'$ ($5$ is the smallest natural number with this property). Now,
(i) If K_{N+1} is empty, then
\[K=K_2\cup K_2'=K_{N+1}\cup K_{N+1}'\cup ... \cup K_2'.\]
So $E\subset \bigcup_{j=1}^N Q_j^*$. This means
\[|E|_e \geq \sum_{j=1}^N |Q_j^*| = 5^n \sum_{j=1}^N |Q_j|.\]
Pick $\beta=5^{-n}$.
(ii) If $t_j^* > \delta > 0$ for all $j\in\mathbb{N}$, $t_j > \frac{1}{2}t_j^* > \frac{1}{2}\delta$. This means $\sum_{j=1}^N |Q_j| \rightarrow \infty$ as $N \rightarrow \infty$. So for any $\delta$ there is a integer $N$ sufficiently large such that
\[\sum_{j=1}^N |Q_j| \geq \beta |E|_e.\]
(iii) If $t_j \rightarrow 0$, we claim that every cube in $K_1$ is contained in $\bigcup_{j=1}^\infty Q_j^*$. Otherwise, there exist a $Q(t)$ not intersecting any $Q_j$. So $Q(t)\in K_j$ for all $j\in \mathbb{N}$, $t\geq t_j^* \rightarrow 0$, which is a contradiction. So
\[|E|_e \leq \sum_{j=1}^\infty |Q_j^*| \leq 5^n \sum_{j=1}^\infty |Q_j|. \]
If we choose $0 < \beta < 5^{-n}$, then there exists $N$ sufficently large such that
\[\beta|E|_e \leq \sum_{j=1}^N |Q_j|.\]
$\blacksquare$
Def [Hardy-Littlewood maximal function of f]. Suppose f is defined on $\mathbb{R^n}$ is integrable over every cube $Q$ with edges parallel to the coordinate axes and center $x$, let
\[f^*(x)=\sup\frac{1}{|Q|} \int_Q |f(y)|dy,\]
is a gauge of the size of averages of $|f|$ around $x$.
Prop.
(i) $0 \leq f^*(x) \leq +\infty$
(ii) $(f+g)^*(x) \leq f^*(x)+g^*(x)$
(iii) $(cf)^*(x)=|c|f^*(x)$
(iv) $f^*$ is low semi-continuous.
(v) $f^*$ is not integrable unless f=0 a.e.
proof. Exercise!
Def [Weak $L(\mathbb{R}^n)$]. Given $f$ measurable, f is of weak $L(\mathbb{R})$ iff $|\{x\in \mathbb{R}^n\}| \leq \frac{c}{\alpha}$, for $\alpha < 0$.
Lemma 5 [Hardy-Littlewood]. If $f\in L(\mathbb{R^n})$, then $f^*$ belongs to weak $L(\mathbb{R}^n)$. Moreover, there exists a constant c independent of $f$ and $\alpha$ such that
\[|\{x\in\mathbb{R}^n:f^* > \alpha\}|\geq \frac{c}{\alpha} \int_\mathbb{R} |f|,\]
for $\alpha > 0$.
Proof. Suppose $f$ has compact support. Then there is a $c_1$ depending on $f$ such that $f^*(x)\leq c_1|x|^{-n}$ for sufficiently large |x|. Fix $\alpha > 0$ and let $E = \{f^* > \alpha\}$. If $x\in E$, there is a cube $Q_x$ such that
\[\frac{1}{|Q_x|}\int_{Q_x} |f| > \alpha \Rightarrow |Q_x| < \frac{1}{\alpha}\int_{Q_x} |f|.\]
Such collection of $Q_x$ covers $E$, so there exists $\beta > 0$ and $x_1,...,x_N$ in $E$ such that $Q_{x_1},...,Q_{x_N}$ are disjoint and
\[|E| < \beta^{-1}\sum_{j=1}^N\frac{1}{\alpha}\int_{Q_{x_j}}|f| \leq \frac{1}{\alpha\beta} \int_{\mathbb{R}^n}|f|.\]
Let $c=\beta^{-1}$.
For the general case, first note that we may assume $f \geq 0$, because changing $f$ to $|f|$ do no effect on $f^*$. Given $f_k$ be a nonnegative sequence increases to $f$. Then there is a $c$ independent of $k$ and $\alpha > 0$ such that
\[|\{ x \in \mathbb{R}^n : f_k^*(x) > \alpha\}| \geq \frac{c}{\alpha} \int_{\mathbb{R}^n} f_k \geq \frac{c}{\alpha} \int_{\mathbb{R}^n} f.\]
Since f_k^* \geq f^*, we have
\[|\{x\in \mathbb{R}^n : f^*(x) > \alpha \}| \geq \frac{c}{\alpha} \int_{\mathbb{R}^n} f.\]
$\blacksquare$
Proof of Lebesgue's theorem.
Given $f\in L(\mathbb{R})$, there exists $C_k$ continuous such that $\int_\mathbb{R} |f-C_k| \rightarrow 0$ when $k \rightarrow 0$. Let $F(Q)=\int_Q f$ and $F_k(Q)=\int_Q C_k$.
Then for all $k\in \mathbb{N}$,
\[\limsup_{Q\searrow x} |\frac{F(Q)}{|Q|} - f(x)| \leq \limsup_{Q\searrow x} |\frac{F(Q)}{|Q|}-\frac{F_k(Q)}{|Q|}|+\limsup_{Q\searrow x}|\frac{F_k(Q)}{|Q|}-C_k(x)|+|C_k(x)-f(x)|.\]
The second term is zero. Also,
\[\left|\frac{F(Q)}{|Q|}-\frac{F_k(Q)}{|Q|}\right| \leq \frac{1}{|Q|}\int_Q |f-C_k| \leq (f-C_k)^*(x).\]
Hence, for every $k$,
\begin{align}\limsup_{Q\searrow x} |\frac{F(Q)}{|Q|} - f(x)| \leq (f-C_k)^*(x)+|C_k(x)-f(x)|.\end{align}
Given $\epsilon > 0$, let $E_\epsilon$ be the set that the LHS of this equation exceeds $\epsilon$.
Note that
\begin{align}E_\epsilon \subseteq \{x:(f-C_k)^*(x)>\frac{\epsilon}{2}\}\cup\{x:|f(x)-C_k(x)| > \frac{\epsilon}{2}\}.\end{align}
Therefore,
\begin{align*} |E_\epsilon|_e &\leq |\{(f-C_k)^*(x)>\frac{\epsilon}{2}\}|+|\{|f(x)-C_k(x)|>\frac{\epsilon}{2}\}|\ \\ &\leq c\left(\frac{\epsilon}{2}\right)^{-1}\int_\mathbb{R} |f-C_k|+\left(\frac{\epsilon}{2}\right)^{-1}\int_\mathbb{R} |f-C_k| \end{align*}
tends to zero as $k\rightarrow \infty$. The last inequality is by Hardy-Littlewood and Chebyshev's inequality. Therefore, we may choose $\epsilon_k \rightarrow 0$ and let $E$ be the set that the LHS exceeds zero. Then $E=\bigcup_k E_k$. So $|E|=0$.
$\blacksquare$
Corollaries of Lebesgue's theorem.
(I) A measurable $f$ defined on $\mathbb{R}^n$ is said to be locally integrable on $\mathbb{R}^n$ if it is integrable over every bounded measurable subset on $\mathbb{R}^n$. The conclusion of Lebesgue's in valid even if $f$ is locally integrable on $\mathbb{R}^n$. To see this, consider a open set $G$ and let $g=f\chi_G$. Then g's integral is differentiable over $\mathbb{R}$ a.e., so is on $G$.
(II) For any measurable $E$, note that
\[\frac{1}{|Q|}\int_Q \chi_E = \frac{|E\cap Q|}{|Q|}\]
and
\[lim_{Q\searrow x} \frac{|E\cap Q|}{|Q|}=\chi_E(x) \]
almost everywhere. A point for which this limit is 1 is called a point of density of $E$, and a point for which it is zero is called a point of dispersion of $E$. Since
\[\frac{|Q\cap E|}{|Q|}+\frac{|Q\cap E^c|}{|Q|}=1,\]
every point of density of E is a point of dispersion of $E^c$, and vice versa.
Corollary. Let $E$ be a measurable set. Then almost every point of $E$ is a point of density of $E$.
(III) A point $x$ at which
\begin{align}\label{**}\lim_{Q \searrow x}\int_Q|f(y)-f(x)|dy=0\end{align}
is called a Lebesgue point of $f$, and the collection of all such $x$ is called a Lebesgue point.
Corollary. Let $f$ be locally integrable in $\mathbb{R}^n$. Then almost every point in $\mathbb{R}^n$ is a Lebesgue point of $f$; that is, there exists a set $Z$(depending on $f$) of measure zero such that ($\ref{**}$) holds for $x\notin Z$.
Proof. Skipped.
(IV) We can use other set to contract to $x$. A family $\{S\}$ of measurable set is said to shrink regularly to $x$ provided
(i) The diameter of $S$ tends to zero.
(ii) If $Q$ is the smallest cube with center $x$ containing $S$, there is a constant $k$ independent of $S$ such that
\[|Q|\leq k|S|.\]
The sets S need not contain $x$.
Corollary. Let $f$ be locally integrable in $\mathbb{R^n}$. Then at every point $x$ of the Lebesgue set of $f$ (a.e.),
\[\frac{1}{|S|}\int_S |f(y)-f(x)| \rightarrow 0,\]
for any $\{S\}$ which shrink regularly to x. Thus also
\[\frac{1}{|S|}\int_S f(y) \rightarrow f(x).\]
proof. Exercise!
3. The Vitali Covering Lemma
If we cover each point of a set not just by a single cube, but by a sequence of cubes with diameters tending to zero. In this case, it turns out that we can cover almost all points of the set by a sequence of disjoint cubes. This is a refined result of simple Vitali lemma.
Def. A family $K$ of cubes is said to cover a set $E$ in the Vitali sense if for every $x\in E$ and $\eta > 0$, there is a cube in $K$ containing $x$ whose diameter is less than $\eta$.
Thm 3 [Vitali Covering Lemma]. Suppose that $E$ is covered in the Vitali sense by a family $K$ of cubes, i.e., for any $x\in E$ and $\eta > 0$, there is a cube in $K$ containing $x$ whose diameter is less than $\eta$. Suppose $0<|E|_e<+\infty$. Then $\epsilon > 0$, there exists $\{Q_j\}$ of disjoint cubes in $K$ such that
\[|E-\bigcup_j Q_j|=0 ,\qquad \sum_j|Q_j| < (1+\epsilon)|E|_e.\]
Proof. The second relation is automatically satisfied if we choose an open set $G$ containing $E$ with $|G|<(1+\epsilon)|E|_e$ and consider only those $Q$ in $K$ which lies in $G$.
By Simple Vitali lemma, there exists a positive $\beta < 1$ depends only on dimension and disjoint $Q_1,...,Q_N$ in $K$ such that $\sum_{j=1}^N |Q_j| > \beta |E|_e.$ So
\[|E-\bigcup_{j=1}^{N_1} Q_j| \leq |G- \bigcup_{j=1}^{N_1}Q_j|=|G|-\sum_{j=1}^{N_1}|Q_j|<|E|_e(1+\epsilon-\beta).\]
If we consider from the start only those $\epsilon$ with $0 < \epsilon < \frac{\beta}{2}$, we have
\[|E-\bigcup_{j=1}^{N_1}Q_j|_e < |E|_e(1-\frac{\beta}{2}).\]
Repeat the process, we obtain at the m-th stage
\[|E-\bigcup_{j=1}^{N_m}Q_j|_e < |E|_e(1-\frac{\beta}{2})^m.\]
Since $(1-\frac{\beta}{2})^m \rightarrow 0$ as $m \rightarrow \infty$, there is a sequence of disjoint cubes in $K$ with the desired properties.
$\blacksquare$
Corollary. With the same assumption with the theorem, for all $\epsilon > 0$, there exists $Q_1$,...,$Q_N$ disjoint cubes in $K$ such that
\begin{align}\label{-}|E - \bigcup_{j=1}^N Q_j|_e < \epsilon ,\qquad \sum_{j=1}^N |Q_j| < (1+\epsilon)|E|_e.\end{align}
This is part of the proof of the previous theorem.
Note that by Carathéodary's theorem that
\begin{align*}|E|_e=|E-\bigcup_{j=1}^N Q_j|_e + |E\cap \bigcup_{j=1}^N Q_j|_e, \end{align*}
so
\begin{align}\label{***}|E|_e-\epsilon < |E\cap \bigcup_{j=1}^N Q_j|_e.\end{align}
In particular,
\begin{align}\label{****}|E|_e-\epsilon < \sum_{j=1}^N |Q_j|.\end{align}
4. Differentiation of Monotone Function.
Consider the four Dini numbers for an real-valued function $f$ at $x_0$
\begin{align*} &D_1 f(x_0) = \limsup_{h\rightarrow 0+}\frac{f(x_0+h)-f(x_0)}{h},\\ &D_2 f(x_0) = \liminf_{h\rightarrow 0+}\frac{f(x_0+h)-f(x_0)}{h},\\ &D_3 f(x_0) = \limsup_{h\rightarrow 0-}\frac{f(x_0+h)-f(x_0)}{h},\\ &D_4 f(x_0) = \liminf_{h\rightarrow 0-}\frac{f(x_0+h)-f(x_0)}{h}. \end{align*}
The following theorem is an application of Vitali's covering lemma.
Thm 4. Let $f(x)$ be a real-valued, monotone increasing function finite on an interval $(a,b)$. Then $f$ has a measurable, nonnegative derivative $f'$ almost everywhere in (a,b). Moreover,
\begin{align}\label{--} 0 \leq \int_a^b f' \leq f(b-)-f(a+). \end{align}
Proof. We may assume $(a,b)$ is finite; the general case is achieved by passage to the limit. We will show the set $\{x\in (a,b):D_1 f(x) > D_4 f(x)\}$ is of measure zero. It is enough to show for every positive rational number $r, s > 0$, the set
\[A_{r,s} = \{x \in (a,b) : D_1 f(x) > r > s > D_4 f(x)\}\]
is of measure zero. Fix $r > s > 0$. Write $A=A_{r,s}$ and suppose $|A|_e > 0$. If $x \in A$,
\[D_4 f(x) = \lim_{h \rightarrow 0+}\frac{f(x-h)-f)(x)}{-h} < s.\]
So there is $0 < h \ll 1$ such that
\[\frac{f(x-h)-f(x)}{-h} < s.\]
So by ($\ref{-}$) and ($\ref{***}$) for all $\epsilon > 0$ there exists disjoint $[x_j-h_j, x_j]$, $j=1,...,N$ such that
(i) $f(x_j)-f(x_j-h_j) < sh_j$
(ii) $|A\cap \bigcup_{j=1}^N [x_j-h_j,x_j]|_e > |A|_e - \epsilon$
(iii) $\sum_{j=1}^N h_j < s(1+\epsilon)|A|_e$.
Combining (i) and (iii), we obtain
(iv) $\sum_{j=1}^N [f(x_j)-f(x_j-h_j)] < s(1+\epsilon)|A|_e$
Let $B = A \cap \bigcup_{j=1}^N[x_j-h_j,x-j]$. For any $y\in B$ which is not an endpoint of some $[x_j-h_j,x_j]$. The fact that $D_1 f(y) > r$ implies $\exists\; 0 < k \ll 1$ such that $[y,y+k]\subset[x_j-h_j,x_j]$ for some $j$. So there exist disjoint $[y_i,y_i+k_i]$, $i=1,...,M$ such that
(v) Each $[y_i,y_i+k_i]$ lies in some $[x_j-h_j,x_j]$,
(vi) $f(y_i+k_i)-f(y_i) > rk_i$,
(vii) $\sum_i=1^M k_i > |B|_e-\epsilon > |A|_e - 2\epsilon$.
So we have
(viii) $\sum_{i=1}^M f(y_i+k_i)-f(y_i) > r(|A|_e - 2\epsilon)$.
The increase of f implies that
\[\sum_{i=1}^M[f(y_i+k_i)-f(y_i)] \leq \sum_{j=1}^N [f(x_j)-f(x_j-h_j)].\]
So $s(1+\epsilon)|A|_e > r(|A|_e - 2\epsilon).$ Take $\epsilon \rightarrow 0$. We deduce that $s \geq r$, a contradiction. Therefore, $|A|_e = 0$.
Since an analogous argument applies to any two Dini number, it follows that $f'(x)$ exists a.e. on $(a,b)$. Extend $f$ by setting $f(x)=f(b-)$ for $x \geq b$, and let
\[f_k(x)=\frac{f(x+h)-f(x)}{h},\]
where $h = \frac{1}{k}$.
Then for all $x\in (a,b)$ and natural number $k$, since $f_k$ nonnegative and measurable, by Fatou's lemma,
\[\int_a^b f' \leq \liminf \int_a^b f_k.\]
If $f(b-)$ is finite (otherwise, the fact is obvious), we have
\[\int_a^b f_k = f(b-)-\frac{1}{h}\int_a^{a+h}f.\]
Since $f(a+) \leq \frac{1}{h}\int_a^b f_k \leq f(a+h)$, we obtain
\[\int_a^b f' \leq \liminf \int_a^b f_k = lim_{k \rightarrow \infty} \int_a^b f_k = f(b-)-f(a+).\]
$\blacksquare$
Rmk. The ``$\leq$'' in ($\ref{--}$) cannot be replaced by ``='' (Consider Cantor-Lebesgue function on $[0,1]$).
Corollary. If $f$ is of bv on $[a,b]$, then $f'$ exists a.e. in $[a,b]$, and $f' \in L[a,b]$.
Thm 5. If $f$ is of bv on $[a,b]$ and $V(x)$ is the variation of $f$ on $[a,x]$, $a \leq x \leq b$, then $V'(x) = |f'(x)|$ a.e.
This theorem gives a relation between $V'(x)$ and $f'(x)$. Note that $V$ increases, so $V'(x)$ exists a.e.
Lemma [Fubini]. Let $f_k$ be a sequence of monotone increasing functions on $[a,b]$. If the series $s(x) = \sum f_k(x)$ converges on $[a,b]$, then $s'(x)=\sum f_k'(x)$ a.e. in $[a,b]$.
Proof. Let $s_m=\sum_{i=1}^m f_i$ and $r_m = \sum_{i=m+1}^\infty$. Then $s_m$ and $r_m$ are monotone increasing functions and $s=s_m+r_m$. So $f_1,...,f_m$ are differentiable and $s'=s_m'+r_m'$ with the exception of a set $Z_m$, $|Z_m|=0$. Since
\[s' \geq s_m'=\sum_{k=1}^m f_k\quad \forall x\notin Z_m,\]
so
\[s \geq \sum_{k=1}^\infty f_k \quad \forall x \notin Z\]
, where $Z=\bigcup Z_m$ of measure zero. Now we need to show $r_m'$ goes to zero a.e. for a sequence $\{m_j\}$ increases so rapidly such that $\sum r_{m_j}(x)$ converges at both $x=a$ and $x=b$. This implies the convergence of $\sum r_{m_j}(b)-r_{m_j}(a)$, and so the convergence of $\sum r_{m_j}(b-)-r_{m_j}(a+)$ in view of the monotonicity of $r_{m_j}$.
By theorem 4,
\[0 \leq \int_a^b \sum r_{m_j}' = \sum \int_a^b r_{m_j}' \leq \sum [r_{m_j}(b-)-r_{m_j}(a+)].\]
So $\sum r_{m_j}'$ integrable over $(a,b)$ $\Rightarrow$ finite a.e. on $(a,b)$ $\Rightarrow$ $r_{m_j}' \rightarrow 0$ a.e.
Proof of theorem. Choose $\{\Gamma_k:\Gamma_k=\{x_j^k\}\}$ a partition sequence of $[a,b]$ such that
\[0 \leq V-S_{\Gamma_k} \leq 2^{-k},\]
where $S_{\Gamma_k} = \sum |f(x_j^k)-f(x_{j-1}^k)$.
Define $f_k$ on $[a,b]$ as follows: if $x\in [x_{j-1}^k,x_j^k]$. let
\begin{equation*} f_k(x)=\begin{cases}f(x)+c_j^k &\text{if $f(x_j^k) \geq f(x_{j-1}^k)$,}\\[2ex]-f(x)+c_j^k &\text{if $f(x_j^k) < f(x_{j-1}^k)$,}\end{cases} \end{equation*}
where the $c_j^k$ are constants chosen so that $f_k(a)=0$ and $f_k$ is well-defined at $x_j$ for every $j$. Then for all $k$ and $j$,
\[f_k(x_{j}^k)-f_k(x_{j-1}^k)=|f(x_j^k)-f(x_{j-1}^k)|,\]
so that
\[S_{\Gamma_k} = \sum_j [f_k(x_j^k)-f_k(x_{j-1}^k)]=f_k(b).\]
Hence, for any $k$, we obtain $0 \leq V(b)-f_k(b) < 2^{-k}$.
It is left as exercise showing that each V(x)-f_k(x) is a increasing function of $x$. This will implies
\[\sum [V(x)-f_k(x)] < \sum 2^{-k} < \infty \quad \text{$\forall x\in [a,b]$}.\]
By Fubini's lemma, $\sum [V'(x)-f_k'(x)]$ converges a.e. in $[a,b]$. Hence, $f_k' \rightarrow V'$ a.e. However, $|f_k'|=|f'|$ a.e. Therefore, $|f'|=|V'|$ a.e. since $V' \geq 0$ whenever it exists.
where the $c_j^k$ are constants chosen so that $f_k(a)=0$ and $f_k$ is well-defined at $x_j$ for every $j$. Then for all $k$ and $j$,
\[f_k(x_{j}^k)-f_k(x_{j-1}^k)=|f(x_j^k)-f(x_{j-1}^k)|,\]
so that
\[S_{\Gamma_k} = \sum_j [f_k(x_j^k)-f_k(x_{j-1}^k)]=f_k(b).\]
Hence, for any $k$, we obtain $0 \leq V(b)-f_k(b) < 2^{-k}$.
It is left as exercise showing that each V(x)-f_k(x) is a increasing function of $x$. This will implies
\[\sum [V(x)-f_k(x)] < \sum 2^{-k} < \infty \quad \text{$\forall x\in [a,b]$}.\]
By Fubini's lemma, $\sum [V'(x)-f_k'(x)]$ converges a.e. in $[a,b]$. Hence, $f_k' \rightarrow V'$ a.e. However, $|f_k'|=|f'|$ a.e. Therefore, $|f'|=|V'|$ a.e. since $V' \geq 0$ whenever it exists.
$\blacksquare$
5. Absolutely Continuous and Singular Functions
Given $g$ is integrable on $[a,b]$ and $G(x) = \int_a^x g$ denotes its indefinite integral, then
\[\sum |G(b_i)-G(a_i)| \leq \int_{\bigcup [a_i,b_i]}|g|\]
for any non-overlapping $[a_i,b_i]$. Note that $\int_E |g|$ is an absolutely continuous set function, and therefore, $G$ is an absolutely continuous function. Thus, an indefinite integral is an absolutely continuous function. The converse is true and will proved below.
Prop. If f is absolutely continuous on a finite interval $[a,b]$, then it is of bv on $[a,b]$.
Proof. Exercise!
Def. A function $f$ is said to singular on $[a,b]$ iff $f'=0$ a.e. in $[a,b]$.
Rmk. C-L function is a non-constant singular function on $[0,1]$.
Thm 6. If $f$is both absolutely continuous and singular on $[a,b]$, then it is constant on $[a,b]$.
proof. It is enough to show that f(a)=f(b). Let $E$ be the set of those x in $(a,b)$ such that $f'(x)=0$. So $|E|=b-a$. Given $\epsilon > 0$ and $x \in E$, we have $[x,x+h] \subset (a,b)$ and $|f(x+h)-f(x)| < \epsilon h$ for all sufficiently small $h > 0$. For this $\epsilon$, there exists $\delta > 0$ in the definition of the absolute continuity of $f$. By corollary of Vitali's lemma and ($\ref(****)$), there are disjoint $Q_j=[x_j,x_j+h_j]$, $j=1,...,N$ lies in $(a,b)$ such that
(i) $|f(x_j+h_j)-f(x_j)| < \epsilon h_j$,
(ii) $\sum_{j=1}^N |Q_j| > (b-a) - \delta$.
By (i),
\[\sum_{j-1}^N |f(x_j+h_j)-f(x_j)| < \epsilon \sum_{j=1}^N |Q_j| \leq \epsilon (b-a).\]
Also, by (ii), the total length of the complementary intervals is less than $\delta$, the sum of the absolute values of the increments of $f$ over them is less than $\epsilon$. Thus, the sum of the absolute values of the increments of $(a,b)$ is no more than $\epsilon(b-a)+\epsilon$. This implies $|f(b)-f(a)| < \epsilon (b-a+1)$, so that $f(b)=f(a)$.
$\blacksquare$
Thm 7. A function $f$ is absolutely continuous on $[a,b]$ if and only if $f'$ exists a.e. in $[a,b]$, $f'$ is integrable on $[a,b]$ and
\[f(x)-f(a) = \int_a^x f' \quad\text{for $a \leq x \leq b$}.\]
Proof. Exercise!
Thm 8. If f is of bv on [a,b], then $f$ can be written as $f=g+h$, where $g$ is absolutely continuous and $h$ is singular on $[a,b]$, respectively. Moreover, $g$ and $h$ are unique up to additive constants.
Proof. Exercise!
There is a analogous version for $\mathscr{C}^1$ functions (see Zyg. 2.10).
Thm 8. Let f be absolutely continuous on $[a,b]$, and let $V(x)$, $P(x)$, and $N(x)$ denote its total, positive and negative variations on $[a,x]$, Then $V$, $P$, and $N$ are absolutely continuous on $[a,b]$, and
\[V(x)=\int_a^x |f'|,\qquad P(x)=\int_a^x (f')^+, \qquad N(x)= \int_a^x (f')^-. \]
Proof. Show $V$ is absolutely continuous.
Thm 9 [Integration by Parts].
(i) If $g$ is continuous on $[a,b]$ and $f$ is absolutely continuous on $[a,b]$, then
\[\int_a^b g df = \int_a^b gf' dx.\]
(ii) If both $f$ and $g$ are absolutely continuous on $[a,b]$, then
\[\int_a^b gf' dx = g(b)f(b)-g(a)f(a)-\int_a^b g'f dx.\]
Proof. To prove (i), recall that suppose $f$ continuous and $\phi$ of bv on $[a,b]$, respectively. Then $\int_a^b f d\phi$ exists. Moreover,
\[\left|\int_a^b fd\phi\right| \leq (sup_{[a,b]}|f|) V[\phi;a,b].\]
So the LHS exists. Since $f'\in L([a,b])$ and $g$ is bounded on $[a,b]$, then $f'g \in L([a,b])$ and thus the existence of RHS. Given $\Gamma={x_i}$ a partition of $[a,b]$ with norm $|\Gamma|$. Then
\begin{align*}\int_a^b gf' dx = \sum \int_{x_{i-1}}^{x_i} gf' dx=&\sum g(x_{i-1})\int_{x-1}^{x_i} f'(x)dx \\&+\sum \int_{x_{i-1}}^{x_i}[g(x)-g(x_{x-i})]f'(x)dx. \end{align*}
The first term on the right converges to $\int_a^bgdf$ as $|\Gamma|\rightarrow 0$. The second term converges to zero by the uniform continuity of $g$.
The proof of (ii) is left as exercise.
$\blacksquare$
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