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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Zheng-Chao Han

Section 1.4 Extension of the Fundamental Theorem of Calculus; Absolutely Continuous Functions

Introduction.

Corollary 1.1.15 extends the Fundamental Theorem of Calculus to those monotone functions \(\alpha\) such that \(\alpha'\) exists and is Riemann integrable. Here we discuss its further generalization and related issues. Questions that need to be addressed include
  1. Identify conditions that guarantee that \(\alpha'\) exists and is Riemann integrable, or \(a.e.\) exists and is in \(L^{1}[a, b]\text{.}\)
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  2. Does (1.1.10) hold for any function \(\alpha\) such that \(\alpha'\) \(a.e.\) exists and is in \(L^{1}[a, b]\text{?}\) More generally, under what conditions on a function \(f\) on \([a, b]\) there would exist a function \(g\in L^{1}[a, b]\) such that
    \begin{equation} f(t)-f(a) =\int_{a}^{t} g(x)\, dx\tag{1.4.1} \end{equation}
    holds?
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The last question has two parts: (a). The necessary conditions for (1.4.1) to hold, in particular, does it hold that \(f'(x)=g(x)\) for \(a.e.\) \(x\in [a, b]\text{?}\) (b). The sufficient conditions for (1.4.1) to hold.
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Definition 1.4.1. Absolute Continuous Function.

A function \(f\) defined on \([a, b]\) is said to be absolutely continuous on \([a, b]\) if for any \(\epsilon > 0\text{,}\) there exists a \(\delta > 0\) such that for any finite collection of disjoint intervals \(\{[a_{i}, b_{i}]: 1\le i \le k\}\) of \([a, b]\) with its total length \(\sum_{i=1}^{k}(b_{i}-a_{i}) < \delta\text{,}\)
\begin{equation} \sum_{i=1}^{k}|f(b_{i})-f(a_{i})| < \epsilon.\tag{1.4.2} \end{equation}
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By Exercise 1.2.3, if (1.4.1) holds, then \(f\) is absolutely continuous on \([a, b]\text{.}\)
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If \(f\) is Lipschitz continuous on \([a, b]\text{,}\) namely, there exists some \(L > 0\) such that \(|f(x)-f(y)|\le L |y-x|\) for any \(x, y\in [a, b]\text{,}\) then \(f\) is absolutely continuous on \([a, b]\text{.}\)
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If \(f\in C[a, b]\) is differentiable everywhere in \((a, b)\) and \(f'\) is bounded on \((a, b)\text{,}\) then \(f\) is Lipschitz continuous on \([a, b]\text{,}\) therefore, is absolutely continuous on \([a, b]\text{.}\)
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Cantor’s function is differentiable \(a.e.\) in \((a, b)\text{,}\) yet is not absolutely continuous on \([a, b]\text{.}\)
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Exercise 1.4.2.

Suppose that \(f\in C[a, b]\) is differentiable \(a.e.\) in \((a, b)\) and \(|f'|\le M < \infty\) \(a.e.\) on \((a, b)\text{.}\) Is \(f\) necessarily Lipschitz continuous on \([a, b]\text{?}\) Is it necessarily absolutely continuous on \([a, b]\text{?}\)
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We will prove below that any absolutely continuous function on \([a, b]\) is differentiable \(a.e.\) on \([a, b]\) with its derivative a function in \(L^{1}[a, b]\text{.}\) Assuming this for now, we prove
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Proof.

According to our discussion above, \(f(x) :=\int_{a}^{x} g(t)\, dt\) is differentiable \(a.e.\) on \([a, b]\) and \(f'(x)\in L^{1}[a, b]\text{.}\) This implies that
\begin{equation*} g_{h}(x) := h^{-1}\int_{x}^{x+h} g(t)\, dt =h^{-1}(f(x+h)-f(x)) \to f'(x) \end{equation*}
\(a.e.\) on \([a, b]\) as \(|h|\to 0\text{.}\)
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On the other hand, according to Exercise 1.2.12, \(\int_{a}^{b}|g_{h}(x)-g(x)|\, dx\to 0\) as \(h\to 0\text{.}\) Then by (i) of Proposition 1.2.8, there exists a sequence \(h_{k}\to 0\) such that \(g_{h_{k}}(x)\to g(x)\) \(a.e.\) on \([a, b]\) as \(k\to \infty\text{.}\) This then identifies \(f'(x)=g(x)\) \(a.e.\) on \([a, b]\text{.}\)
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We could also complete the last part by using the Fatou Theorem:
\begin{align*} \int_{a}^{b} |f'(x)-g(x)|\, dx \amp= \int_{a}^{b} \lim_{k\to \infty }|g_{h_{k}}(x)-g(x)|\, dx \\ \amp \le \liminf{k\to \infty} \int_{a}^{b} |g_{h_{k}}(x)-g(x)|\, dx=0, \end{align*}
from which we conclude that \(f'(x)=g(x)\) \(a.e.\) on \([a, b]\) via (ii) of Proposition 1.2.8.
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For the remaining part, we will state and use Lebesgue’s differentiability theorem of a monotone function but will skip its proof.
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Proof.

According to Theorem 1.4.4, \(h^{-1}(f(x+h)-f(x))\to f'(x)\) \(a.e.\) on \([a, b]\text{.}\) For each \(h > 0\text{,}\) \(h^{-1}(f(x+h)-f(x)) \) is a non-negative function in \(\cR[a, b]\) (extend \(f(x)=f(b)\) for \(x > b\)), and
\begin{align*} \int_{a}^{b} h^{-1}(f(x+h)-f(x)) \, dx \amp =h^{-1}\left( \int_{b}^{b+h} f(x)\,dx - \int_{a}^{a+h} f(x)\, dx\right)\\ \amp \le f(b)- f(a), \end{align*}
using \(\int_{a}^{a+h} f(x)\, dx\ge f(a) h\text{.}\) Thus by Fatou Theorem,
\begin{align*} \int_{a}^{b} f'(x)\, dx \amp = \int_{a}^{b} \liminf_{h\to 0+} h^{-1}(f(x+h)-f(x)) \, dx\\ \amp \le \liminf_{h\to 0+} \int_{a}^{b} h^{-1}(f(x+h)-f(x)) \, dx \le f(b)-f(a). \end{align*}
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Definition 1.4.6. Function of Bounded Variation.

A function \(f\) on \([a, b]\) is said to have bounded variation on \([a, b]\text{,}\) if there exists some \(M > 0\) such that for any partition \(\cP=\{a=a_{0} < a_{1} < \cdots < a_{k}=b\}\text{,}\)
\begin{equation*} \sum_{i=1}^{k} |f(a_{i})-f(a_{i-1})| \le M. \end{equation*}
\(\sup_{\cP} \sum_{i=1}^{k} |f(a_{i})-f(a_{i-1})|\) is called the total variation of \(f\) on \([a, b]\) and is denoted as \(V(f, [a, b])\text{.}\)
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If \(f\) is a monotone increasing function on \([a, b]\text{,}\) then it has bounded variation on \([a, b]\) and \(V(f, [a, b])=f(b)-f(a)\text{.}\)
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If \(f=g-h\) is the difference of two monotone increasing functions on \([a, b]\text{,}\) then it has bounded variation on \([a, b]\) and \(V(f, [a, b])\le V(g, [a, b])+V(h, [a, b])\text{.}\)
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Proof.

Let \(f\) be a function of bounded variation on \([a, b]\text{.}\) Define \(g(x) :=V(f, [a, x])\text{.}\) Then \(g(x)\) is a monotone increasing function on \([a, b]\text{.}\) We now prove that \(h(x):=g(x)-f(x)\) is a monotone increasing function on \([a, b]\text{.}\)
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Take \(x < y\) in \([a, b]\text{,}\) then
\begin{equation*} h(y)-h(x)=g(y)-g(x)-[f(y)-f(x)]=V(f, [x, y]) -[f(y)-f(x)]. \end{equation*}
\(V(f, [x, y])\) is the supremum of \(\sum_{i=1}^{k} |f(a_{i})-f(a_{i-1})|\) for any partition \(\{x= a_{0} < a_{1} < \cdots < a_{k}=y\}\text{,}\) so \(V(f, [x, y])\ge |f(y)-f(x)|\text{.}\)
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An absolutely continuous function on \([a, b]\) has bounded variation on \([a, b]\text{,}\) therefore, according to Proposition 1.4.7, Theorem 1.4.4 and Proposition 1.4.5, has its derivative in \(L^{1}[a, b]\text{.}\)
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Proof.

We already discussed the necessary part. For the sufficient part, suppose that \(f\) is absolutely continuous on \([a, b]\text{.}\) Then, according to our discussion above, \(f'(x)\) exists \(a.e.\) on \([a, b]\) and is an element in \(L^{1}[a, b]\text{.}\) Set \(F(x)=\int_{a}^{x} f'(t)\, dt\text{.}\) Then, according to Theorem 1.4.3, \(F(x)\) is is absolutely continuous on \([a, b]\) and \(F'(x)=f'(x)\) \(a.e.\) on \([a, b]\text{.}\)
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Now \(f(x)-F(x)\) is absolutely continuous on \([a, b]\) and \((f(x)-F(x))'=0\) \(a.e.\) on \([a, b]\text{.}\) We can draw our conclusion based on the following Lemma.
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We will use the concept and properties of Vitali covering in proving this.
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Definition 1.4.10. Vitali Covering.

Suppose that \(\Gamma =\{I_{\alpha}\}\) is a family of intervals covering a set \(E\subset (a,b)\text{.}\) Suppose that for any \(\epsilon >0\) and any \(x\in E\text{,}\) there exists an interval \(I\in \Gamma\) such that \(x\in I\) and \(|I| < \epsilon\text{.}\) Then \(\Gamma\) is said for form a Vitali covering of \(E\)
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Proof.

We may take the \(I_{\alpha}\) to be closed intervals in \((a, b)\text{.}\) We choose \(I_{i}\) inductively. Choose \(I_{1}\) from \(\Gamma\) such that \(2|I_{1}| > \sup |I_{\alpha}|\text{.}\) After \(\{I_{i}, 1\le i \le l\}\) are chosen, if \(E\setminus \cup_{i=1}^{l}I_{i} \ne \emptyset\text{,}\) define \(\delta_{l+1}=\sup\{|I_{\alpha}|: I_{\alpha} \cap I_{i} =\emptyset, i=1,\cdots, l\}\text{.}\) Then \(\delta_{l+1} > 0\) and we choose some \(I_{l+1}\in \Gamma\) such that \(I_{l+1} \cap I_{i} =\emptyset, i=1,\cdots, l\) and \(2|I_{l+1}| > \delta_{l+1}\text{.}\)
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Since this collection of disjoint intervals are all contained in \((a, b)\text{,}\) \(\sum_{l} |I_{l}| < \infty\text{.}\) Thus, for a given \(\epsilon > 0\text{,}\) there exists \(k\) such that \(\sum_{l> k} |I_{l}| < \epsilon/5\text{.}\)
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Any \(x\in E\setminus \cup_{i=1}^{l}I_{i}\) must be contained in some \(I\in \Gamma\) such that \(I \cap I_{i}=\emptyset, i=1,\cdots, k\text{.}\) Since \(|I| > 0\) and \(\delta_{l}\to 0\) as \(l\to\infty\text{,}\) we claim that \(I\cap I_{l}\ne \emptyset\) for some \(l > k\text{,}\) for, whenever \(\delta_{l'} < |I|\text{,}\) \(I\cap I_{l}\ne \emptyset\) for some \(l < l'\text{.}\) Let \(I_{l}'\) be the interval with the same center point as \(I_{l}\) but \(|I_{l}'|=5 |I_{l}|\text{,}\) then \(I\subset I_{l}'\) whenever \(I\cap I_{l}\ne \emptyset\text{.}\) Thus \(\cup_{l > k} I_{l}'\) covers \(E\setminus \cup_{i=1}^{l}I_{i}\) with \(\sum_{l > k} |I_{l}'| =5 \sum_{l > k} |I_{l}|< \epsilon\text{.}\)
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Proof of Lemma 1.4.9.

Suppose \(g\) is not a constant function on \([a, b]\text{.}\) Then there exists some \(c\in (a, b)\) such that \(|g(c)-g(a)| > 2\epsilon\) for some \(\epsilon > 0\text{.}\) Consider \(E_{c} :=\{x\in (a, c): g'(x)=0\}\text{.}\) Then \([a, c]\setminus E_{c}\) is negligible. For any \(x\in E_{c}\) and any \(r > 0\text{,}\) for all sufficiently small \(h > 0\text{,}\) we have \([x, x+h]\subset (a, c)\) and
\begin{equation*} |g(x+h)-g(x)|\le r h. \end{equation*}
Thus for any fixed such \(r > 0\text{,}\) the family \(\{[x, x+h]: x\in E_{c}\}\) forms a Vitali cover of \(E_{c}\text{.}\) Thus, for any \(\delta > 0\text{,}\) there exists disjoint intervals
\begin{equation*} [x_{1}, x_{1}+h_{1}], \cdots, [x_{k}, x_{k}+h_{k}] \end{equation*}
such that
\begin{equation*} [a, c] \setminus \cup_{i=1}^{k} [x_{i}, x_{i}+h_{i}] \subset \left([a, c]\setminus E_{c}\right) \cup \left( E_{c}\setminus \cup_{i=1}^{k} [x_{i}, x_{i}+h_{i}]\right) \end{equation*}
can be covered by an open set \(G\) with \(|G| < \delta\text{.}\)
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We may assume that
\begin{equation*} a < x_{1} < x_{1}+h_{1} < x_{2} < x_{2}+h_{2} < \cdots < x_{k} < x_{k}+h_{k} < c. \end{equation*}
These points, together with \(a \) and \(c\text{,}\) forms a partition of \([a, c]\text{.}\) The sum of the lengths of those intervals of this partition not-overlapping with any of \((x_{i}, x_{i}+h_{i})\) is \(< \delta\text{.}\) Then
\begin{equation*} 2\epsilon < |g(c)-g(a)| \le \sum_{i=1}^{k} |g(x_{i})-g(x_{i}+h_{i})| + \sum_{i=0}^{k} |g(x_{i}+h_{i})-g(x_{i+1})| \end{equation*}
where we set \(x_{0}+h_{0}=a\) and \(x_{k+1}=c\text{.}\)
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We fix \(r > 0\) such that \(r(b-a) < \epsilon\text{.}\) Using the absolute continuity of \(f\) on \([a, b]\text{,}\) there exists some \(\delta > 0\) such that for any collection of disjoint intervals \([a_{j}, b_{j}]\) with \(\sum_{j} (b_{j}-a_{j}) < \delta\text{,}\) we have \(\sum_{j}|g(b_{j})-g(a_{j})| < \epsilon\text{.}\) For this reason, \(\sum_{i=0}^{k} |g(x_{i}+h_{i})-g(x_{i+1})| < \epsilon\text{.}\) But
\begin{equation*} \sum_{i=1}^{k} |g(x_{i})-g(x_{i}+h_{i})| \le \sum_{i=1}^{k}r h_{i} \le r(b-a) < \epsilon. \end{equation*}
We now see a contradiction and must conclude that \(g\) is a constant function on \([a, b]\)
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Exercise 1.4.12.

Suppose that \(f\) is absolutely continuous on \((a, b)\) and \(|f'|\le M < \infty\) \(a.e.\) on \((a, b)\text{.}\) Is \(f\) necessarily Lipschitz continuous on \([a, b]\text{?}\)
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Exercise 1.4.13.

Suppose that \(f\) has bounded variation on \([0, 1]\) and is absolutely continuous on \([\epsilon ,1]\) for any \(0 < \epsilon < 1\text{.}\) Furthermore, suppose that \(f(x)\) is continuous at \(x=0\text{.}\) Prove that \(f\) is absolutely continuous on \([0 ,1]\text{.}\)
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