Classification of Absolutely Continuous Functions

# Classification of Absolutely Continuous Functions

 Lemma 1: Let $f$ be absolutely continuous on $[a, b]$ and let $f'(x) = 0$ almost everywhere on $[a, b]$. Then $f$ is a constant function on $[a, b]$.
• Proof: Let $c \in (a, b]$. Since $f$ is absolutely continuous on $[a, b]$, from the Absolutely Continuous Functions are of Bounded Variation page we have that $f$ is of bounded variation on $[a, b]$ and so $f$ is differentiable almost everywhere on $[a, b]$. Since $f'(x) = 0$ almost everywhere on $[a, b]$ there exists a Lebesgue measurable subset $E$ of $(a, c)$ such that $m(E) = c - a$ and that $f$ is differentiable on $E$ with $f'(x) = 0$ for all $x \in E$.
• Let $\epsilon > 0$ and consider the following collection of closed intervals:
(1)
\begin{align} \quad \mathcal I = \{ [x, y] \subseteq [a, b] : |f(y) - f(x)| < \epsilon |y - x| \} \end{align}
• Then $\mathcal I$ is a Vitali cover of $[a, c]$.
• Since $f$ is absolutely continuous on $[a, b]$, for this $\epsilon > 0$ there exists a $\delta > 0$ such that if $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ is any finite collection of disjoint subintervals on $[a, b]$ such that if $\displaystyle{\sum_{i=1}^{n} (y_i - x_i) < \delta}$ then:
(2)
\begin{align} \quad \sum_{i=1}^{n} | f(y_i) - f(x_i) | < \epsilon \end{align}
• Also, by the Vitali covering lemma, since $m([a, c]) \leq m([a, b]) < \infty$ and $\mathcal I$ is a Vitali cover of $[a, c]$, for $\delta > 0$ there exists a finite collection of mutually disjoint intervals in $\mathcal I$, $\{ [x_1, y_1], [x_2, y_2], ..., [x_n, y_n] \}$ such that:
(3)
\begin{align} \quad m \left ( [a, c] \setminus \bigcup_{i=1}^{n} \right ) < \delta \end{align}
• Now:
(4)
\begin{align} \quad f(c) - f(a) &= f(c) - f(y_n) + f(y_n) - f(a) \\ &= f(c) - f(y_n) + f(y_n) - f(x_n) + f(x_n) - f(a) \\ &= \vdots \\ &= f(c) - f(y_n) + f(y_n) - f(x_n) + f(x_n) - f(y_{n-1}) + f(y_{n-1}) - f(x_{n-1}) + f(x_{n-1}) - ... - f(x_1) + f(x_1) - f(a) \end{align}
• Taking the absolute value of both sides of the equality above yields:
(5)
\begin{align} \quad | f(c) - f(a) | \leq \biggr \lvert \sum_{i=1}^{n} [f(x_i) - f(y_{i-1})] + \sum_{i=1}^{n} [f(y_i) - f(x_i)] \biggr \rvert \leq \sum_{i=1}^{n} |f(x_i) - f(y_{i-1})| + \sum_{i=1}^{n} |f(y_i) - f(x_i)| < \epsilon(1 + c - a) \end{align}
• Let $\epsilon \to 0$. Then $| f(c) - f(a) | = 0$ so $f(c) = f(a)$. But $c \in (a, b]$ was arbitrary. So $f$ is constant on $[a, b]$. $\blacksquare$

The next theorem is of large significance. It classifies the set of all absolutely continuous functions on $[a, b]$ to be exactly the integral functions $\displaystyle{\int_a^x f(t) \: dt + F(a)}$

 Theorem 2: Let $F$ be a real-valued function defined on $[a, b]$. Then $F$ is absolutely continuous on $[a, b]$ if and only if there exists a Lebesgue integrable function $f$ on $[a, b]$ such that $\displaystyle{F(x) = \int_a^x f(t) \: dt + F(a)}$.
• Proof: $\Rightarrow$ Let $F$ be absolutely continuous on $[a, b]$. Then $F$ is of bounded variation on $[a, b]$ and so $F = F_1 - F_2$ (where $F_1$ and $F_2$ are increasing functions on $[a, b]$) and $F$ is differentiable almost everywhere on $[a, b]$ and $F'$ is also integrable on $[a, b]$.
• Define a new function $G$ on $[a, b]$ for all $x \in [a, b]$ by:
(6)
\begin{align} \quad G(x) = \int_a^x F'(t) \: dt + F(a) \end{align}
(7)
\begin{align} \quad G'(x) = F'(x) \end{align}
• Define a new function $H$ on $[a, b]$ for all $x \in [a, b]$ by:
(8)
\begin{align} \quad H(x) = F(x) - G(x) \end{align}
• Then $H$ is also absolutely continuous on $[a, b]$ and moreover, $H'(x) = F'(x) + G'(x) = F'(x) + F'(x) = 0$ almost everywhere on $[a, b]$. So by the previous lemma, $H(x) = C$ for some constant $C$. Thus:
(9)
\begin{align} \quad F(x) - G(x) = C \end{align}
• Plugging in $x = a$ yields:
(10)
\begin{align} \quad F(a) - G(a) = C \\ \quad F(a) - \int_a^a F'(t) \: dt + F(a) = C \\ \quad 0 = C \end{align}
• Therefore $F(x) = G(x)$, that is:
(11)
\begin{align} \quad F(x) = \int_a^x F'(t) \: dt + F(a) \end{align}
• By letting $f = F'$ be out Lebesgue integrable function, we conclude that:
(12)
\begin{align} \quad F(x) = \int_a^x f(t) \: dt + F(a) \end{align}
• $\Leftarrow$ Let $f$ be a Lebesgue integrable function on $[a, b]$ and let:
(13)
\begin{align} \quad F(x) = \int_a^x f(t) \: dt \end{align}
• Let $\epsilon > 0$ be given. Then there exists a $\delta > 0$ such that if $E$ is a Lebesgue measurable subset of $[a, b]$ and $m(E) < \delta$ then:
(14)
\begin{align} \quad \int_E |f(t)| \: dt < \epsilon \end{align}
• Let $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ be a finite collection of disjoint open subintervals of $[a, b]$ such that $\displaystyle{\sum_{i=1}^{n} (y_i - x_i) < \delta}$. Then $\displaystyle{m \left ( \bigcup_{i=1}^{n} (x_i, y_i) \right ) < \delta}$ and so $\displaystyle{\int_{\bigcup_{i=1}^{n} (x_i, y_i)} | f(t) | \: dt < \epsilon}$. Now:
(15)
\begin{align} \quad \sum_{i=1}^{n} | F(y_i) - F(x_i) | = \sum_{i=1}^{n} \biggr \lvert \int_{x_i}^{y_i} f(t) \: dt \biggr \rvert \leq \sum_{i=1}^{n} \int_{x_i}^{y_i} | f(t) | \: dt = \int_{\bigcup_{i=1}^{n} (x_i, y_i)} | f(t) | \: dt < \epsilon \end{align}
• So $F$ is absolutely continuous on $[a, b]$. Thus for every Lebesgue integrable function $f$ we have that $\displaystyle{\int_a^x f(t) \: dt + F(a)}$ is an absolutely continuous function on $[a, b]$. $\blacksquare$