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Define
\[\varphi_{k}(x)=\begin{cases}\frac{j-1}{2^{k}}&\text{if }\frac{j-1}{2^{k}} \leq f(x)<\frac{j}{2^{k}}\ j=1,2,\ldots,k2^{k}\\ k&\text{if }f(x)\geq k\end{cases}.\]
Observe that each \(\varphi_{k}\) is a simple measurable function defined everywhere in the domain of \(f\). Also \(\varphi_{k}\leq\varphi_{k+1}\) because passing from \(\varphi_{k}\) to \(\varphi_{k+1}\) each subinterval \(\left[\frac{j-1}{2^{k}},\frac{j}{2^{k}}\right]\) is divided by half. The definition of \(\varphi_{k}\) implies that
\[|\varphi_{k}(x)-f(x)|\leq\frac{1}{2^{k}}\quad\text{for all}\quad x\in D\quad \text{such that}\quad f(x)\in[0,k].\qed\]
**Remark 62**.: One can prove the above theorem for an arbitrary measurable function \(f\), since we have \(f=f^{+}-f^{-}\) where both \(f^{+}\) and \(f^{-}\) are non-negative. Applying the above result to each non-negative function we obtain sequences say \((\varphi_{k}^{{}^{\prime}})\) and \((\varphi_{k}^{{}^{\prime\prime}})\) such that
\[\varphi_{k}^{{}^{\prime}}\to f^{+}\quad\text{and}\quad\varphi_{k}^{{}^{ \prime\prime}}\to f^{-}.\]
Then \(\varphi_{k}^{{}^{\prime}}-\varphi_{k}^{{}^{\prime\prime}}\) is also simple function and \(\varphi_{k}^{{}^{\prime}}-\varphi_{k}^{{}^{\prime\prime}}\to f^{+}-f^{-}=f\).
### Littlewood’s Three Principles
J. E. Littlewood stated the following principles in his 1944 “Lectures on the Theory of Functions.” These principles address the essential issues in measure theory and help us to understand the relationship between the new concepts of measurable sets or measurable functions and convergence of sequence of measurable functions with the older concepts they replaced. These principles are:
* Every measurable set is “nearly” a finite union of intervals.
* Every measurable function is “nearly” continuous.
* Every convergent sequence of measurable functions is “nearly” uniformly convergent.
The first of these principles is about the approximation property of measures. The second item is the well known Lusin’s Theorem and the third one is described in a theorem that carries the name Egorov. Of course the key word in the above principles is “nearly,” which we explain below when we give the statement of the theorems. First we need a notation. Given two sets \(E\) and \(F\), the _symmetric difference_ between these sets denoted by \(E\triangle F\) is defined as
\[E\triangle F=(E\setminus F)\cup(F\setminus E),\]
which consists of points that belong to exactly one of the sets \(E\) or \(F\).
**Theorem 113** (First Principle).: _If \(m(E)\) is finite, there exists a finite union \(F=\bigcup_{j=1}^{N}Q_{j}\) of closed intervals such that \(m(E\triangle F)\leq\epsilon\)._
**Theorem 114** (Second Principle-Lusin’s Theorem).: _Suppose \(f\) is measurable and finite valued on a set \(E\) with the measure of \(E\) being finite. Then for every \(\epsilon>0\) there is a closed set \(F_{\epsilon}\) with_
\[F_{\epsilon}\subset E\quad\text{and}\quad m(E\setminus F_{\epsilon})\leq\epsilon\]
_and \(f\) restricted to the set \(F_{\epsilon}\), that is \(f|_{F_{\epsilon}}\), is continuous._
**Theorem 115** (Third Principle-Egorov’s Theorem).: _Suppose \((f_{k})\) is a sequence of measurable functions defined on a measurable set \(E\) with \(m(E)<\infty\). Assume that \(f_{k}\to f\) a.e. on \(E\). Given \(\epsilon>0\), we can find a closed set \(F_{\epsilon}\subset E\) such that_
\[m(E\setminus F_{\epsilon})\leq\epsilon\quad\text{and}\quad f_{k}\to f\quad \text{uniformly on}\,\,\,F_{\epsilon}.\]
For the proofs of the above three theorems we refer to an excellent text [50], pp. 33-34.
In the following we first return to the Cantor set, after proving that the Cantor set has measure zero, using the construction of the Cantor set, we define Cantor-Lebesgue function. This function is a continuous function \(F:[0,1]\to[0,1]\) which is increasing with \(F(0)=0\) and \(F(1)=1\). However,we will see that \(F^{\prime}(x)=0\) almost everywhere. Hence
\[\int_{a}^{b}F^{\prime}(x)\,dx\neq F(b)-F(a).\]
### Cantor-Lebesgue Function
Recall the Cantor set from Chapter 1. The _Cantor set_, sometimes called the _Cantor ternary set_, is an intriguing subset of \(\mathbb{R}\) which will extend our understanding of the nature of subsets of \(\mathbb{R}\) and it is a valuable source of counterexamples in Analysis. For example, the Cantor set is the basis of the construction of a function called the _Cantor-Lebesgue function_, sometimes called the “Devil’s staircase” or the ” Cantor-Scheeffer function,” which is a continuous,non-decreasing function that is not constant, yet has zero derivative at almost every point.
The Cantor set is obtained by successively removing “middle thirds” from the interval \([0,1]\) as shown in Figure 2.23.
The Cantor set \(C\) is defined as:
\[C=\bigcap_{n=0}^{\infty}C_{n}.\]
The Cantor set is in some ways large (uncountable) and in other ways small (measure zero).
**Definition 95**.: A set \(A\subseteq\mathbb{R}\) is said to be _measure zero_ if, for \(\epsilon>0\), there exists a countable collection of open intervals \(I_{n}\) such that
\[A\subseteq\bigcup_{n=1}^{\infty}I_{n}\quad\text{and}\quad\sum_{n=1}^{\infty} |I_{n}|\leq\epsilon,\]
where by \(|I_{n}|\) we mean the length of the interval \(I_{n}\).
**Proposition 36**.: _The Cantor set \(C\) has measure zero._
Proof
: We start by observing the length of the intervals removed from \([0,1]\) to form the \(C\). To form \(C_{1}\) we removed \(I_{1}=(\frac{1}{3},\frac{2}{3})\) and the length of \(I_{1}=1/3\). In the second step we removed \(I_{2}=(\frac{1}{9},\frac{2}{9})\cup(\frac{7}{9},\frac{8}{9})\) and the length of \(I_{2}=2\left(1/9\right)\), and to construct \(C_{n}\) we removed \(2^{n-1}\) middle thirds of length \(\frac{1}{3^{n}}\), and so the total length of \([0,1]\setminus C\) must be:
\[\frac{1}{3}+2\left(\frac{1}{9}\right)+4\left(\frac{1}{27}\right) +\dots+2^{n-1}\left(\frac{1}{3^{n}}\right)+\dots =\sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}}\] \[=\frac{1}{3}\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n-1}\] \[=\frac{1}{3}\cdot\frac{1}{1-2/3}=1.\]
Figure 2.23: Cantor set construction
Since \(|\bigcup_{n=1}^{\infty}I_{n}\mid=1\), the Cantor set, which is equal to \([0,1]\setminus\bigcup_{n=1}^{\infty}I_{n}\) has Lebesgue measure zero. Here we used \(m(E\setminus F)=m(E)-m(F)\) provided that \(m(F)<\infty\).
Using the Cantor set one can define the following simple construction to describe the so called _Cantor-Lebesgue function_. As above we define the Cantor set
\[C=\bigcap_{k=0}^{\infty}C_{k},\]
where each \(C_{k}\) is a disjoint union of \(2^{k}\) intervals, such as
\[C_{1}=[0,1/3]\cup[2/3,1],\]
\[C_{2}=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1].\]
**Definition 96**.: The Cantor-Lebesgue function \(F:[0,1]\to[0,1]\) is a function with \(F(0)=0\) and \(F(1)=1\) defined inductively as follows:
* \(F_{1}(x)\) is a continuous increasing function such that \(F_{1}(0)=0\), \(F_{1}(1)=1\), \(F_{1}\) is linear on \(C_{1}\) and \(F_{1}(x)=\frac{1}{2}\) when \(x\in[1/3,2/3]\).
* \(F_{2}(x)\) is a continuous increasing function such that \(F_{2}(0)=0\), \(F_{2}(1)=1\), \(F_{2}\) is linear on \(C_{2}\) and \[F_{2}(x)=\begin{cases}1/4&\text{if }\,1/9\leq x\leq 2/9,\\ 1/2&\text{if }\,1/3\leq x\leq 2/3\\ 3/4&\text{if }\,7/9\leq x\leq 8/9.\end{cases}\] See Figure 2.24 for \(F_{1}\) and \(F_{2}\):
This process yields a sequence of continuous increasing functions \((F_{n})\) such that \(|\)\(F_{n+1}-F_{n}\mid\leq\frac{1}{2^{n+1}}\) and \(F_{n}\to F\) uniformly; thus the limit function \(F\) is also continuous and called the _Cantor-Lebesgue function_ (Figure 2.25).
Figure 2.24: Graphs of \(F_{1}\) and \(F_{2}\)
Note that this function can also be defined by converting base 3 representation into base 2. The following are some well-known consequences:
* \(F\) is constant on each interval of the complement of the Cantor set. Since \(m(C)=0\), we find that \(F^{\prime}(x)=0\) a.e.. This is particularly interesting in the context of the Fundamental Theorem of Calculus. We see that the Cantor-Lebesgue function does not satisfy \[F(b)-F(a)=\int_{a}^{b}F^{\prime}(x)\,dx,\] since \(F(1)-F(0)=1\neq 0\).
* A continuous image of a measurable set can be a non-measurable set. One can see this using the Cantor-Lebesgue function. To understand this, first observe that \(F(C)=[0,1]\) and recall that not every subset of \(\mathbb{R}\) is measurable (see the section on the Banach-Tarski paradox). If \(E\) is a subset of \([0,1]\) that is not Lebesgue measurable, and if we let \[B=C\cap F^{-1}(E),\] then \(m(B)=0\) because \(B\subset C\) and \(m(C)=0\). But every subset of \(\mathbb{R}\) with Lebesgue measure zero is measurable, thus \(B\) is measurable and \(F(B)=E\).
### Lebesgue Integration
For the Riemann integral, we divide the area under the graph of \(f\) into vertical rectangles to find upper and lower sums. How about dividing this area into horizontal rectangles? This leads to a more complicated mathematics. Recalling Remark 30 in Section 1.8, suppose \(f:[a,b]\to\mathbb{R}\) to be a non-negative bounded function defined on \([a,b]\). Let \(R=\{y_{0} Figure 2.25: The Cantor-Lebesgue function
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