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\[\int_{0}^{2\pi}\left(f(x)\right)^{2}\,dx\leq\int_{0}^{2\pi}\left(f^{\prime}(x) \right)^{2}\,dx.\]
Hint: Consider Fourier series for \(f\) and use Parseval’s Identity (see [26], p. 577).
### 2.7 Lebesgue Measure and Integration
The treatment of integration developed by French mathematician Henri Lebesgue (1875-1941) has become indispensable in many areas of mathematics because it not only allows us to integrate many more functions but also enables us to understand when the integral of the limit functions is equal to the limit of the integrals. In his thesis in 1902, Lebesgue presented an extension of the Riemann integral. The Lebesgue integral is defined for what are called _measurable_ functions, a class that includes the Riemann integrable functions. Moreover, Lebesgue’s ideas gave new insights on the differentiability of monotone functions and led to an extension of the fundamental theorem of calculus.
The Riemann integral is surely one of the gems of calculus and works for many functions. Unfortunately it has some shortcomings. Before we list few of these difficulties below, let us recall how we defined the Riemann integral.
**Definition 87**.: Let \(f\) be a function defined on a bounded interval \([a,b]\). A _partition_\(P\) of \([a,b]\) is a finite set of points in \([a,b]\) such that
\[a=x_{0} For \(k=1,2,\ldots,n\), let \[M_{k}=\sup\{f(x):\ x_{k-1}\leq x\leq x_{k}\}\quad\text{and}\quad m_{k}=\inf\{f( x):\ x_{k-1}\leq x\leq x_{k}\};\] then by the _upper sum_, denoted by \(U(f,P)\), and the _lower sum_, denoted by \(L(f,P)\) we mean: \[U(f,P)=\sum_{k=1}^{n}M_{k}(x_{k}-x_{k-1})\quad\text{and}\quad L(f,P)=\sum_{k=1 }^{n}m_{k}(x_{k}-x_{k-1}).\] Clearly \(L(f,P)\leq U(f,P)\) for a particular partition \(P\), and it helps to think of upper sums as an overestimate and lower sums as an underestimate for the value of the integral. If \(P\) and \(Q\) are two partitions of \([a,b]\) with \(P\subset Q\), then \(Q\) is called a _refinement_ of \(P\). It can be seen that if \(P\subset Q\) then \[L(f,P)\leq L(f,Q)\quad\text{ and}\quad U(f,P)\geq U(f,Q).\]Let \(\mathcal{P}\) be the collection of all possible partitions of the interval \([a,b]\), and define (Figure 2.20) \[U(f)=\inf\{U(f,P):\ P\in\mathcal{P}\}\quad\text{and}\quad L(f)=\sup\{L(f,P):\ P\in \mathcal{P}\}.\] Let \(f\) be a bounded function on \([a,b]\) and \(P\) be any partition of \([a,b]\). First, observe that \(U(f,P)\geq L(f,P)\); then one can prove that for any bounded function \(f\) on \([a,b]\) we always have \(L(f)\leq U(f)\). **Definition 88**.: A bounded function \(f\) on \([a,b]\) is _Riemann integrable_ over \([a,b]\) if \(U(f)=L(f)\) and its integral is defined to be the common value, written as \[\int_{a}^{b}f(x)\,dx=U(f)=L(f).\] We can also say that \(f\) is Riemann integrable if and only if for each \(\epsilon>0\), there exists a partition \(P\) such that \(U(f,P)-L(f,P)<\epsilon\), that is, its integrability is equivalent to the existence of partitions whose upper and lower sums are arbitrarily close together. More precisely: **Proposition 33** (Riemann’s Criterion).: _A bounded real-valued function \(f\) on \([a,b]\) is Riemann integrable if and only if for each \(\epsilon>0\) there is a partition \(P_{\epsilon}\) such that_ \[U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon.\] Now we give some reasons why the Riemann integral is not good enough. First of all the Riemann integral does not handle functions with many discontinuities as illustrated in the following example. **Example 89**.: Consider Dirichlet’s function \(f:[0,1]\to\mathbb{R}\) given by \[f(x)=\left\{\begin{array}{ll}1&\text{if }x\in\mathbb{Q}\\ 0&\text{if }x\notin\mathbb{Q}\end{array}.\right.\] We have seen that this function is discontinuous at every point of \([0,1]\). Now we turn our attention to integrability. If \(P\) is some partition of \([a,b]\), then the density of rationals in \(\mathbb{R}\) implies that every subinterval of \(P\) contains points for which \(f(x)=1\), and thus \(U(f,P)=1\). Similarly, \(L(f,P)=0\) because irrational Figure 2.20: Upper and lower sums numbers are also dense in \(\mathbb{R}\). Since this is the case for every possible partition, \(U(f)=1\) and \(L(f)=0\); the two are not equal; therefore, Dirichlet’s function is not Riemann integrable. This example illustrates that not every bounded function is Riemann integrable. The following is another example showing that Riemann integral does not handle unbounded functions. **Example 90**.: Consider \[f(x)=\left\{\begin{array}{ll}\frac{1}{\sqrt{x}}&\quad\text{if}\;\;0 If \(x_{0} \[\int_{0}^{1}f(x)\,dx=\lim_{t\downarrow 0}\int_{t}^{1}\frac{1}{\sqrt{x}}\,dx= \lim_{t\downarrow 0}(2\sqrt{x})\,|_{t}^{1}=2;\] thus the area under \(f\) is 2 not \(\infty\). The Riemann integral does not work well with pointwise limits either. When a sequence of Riemann integrable functions \(f_{n}(x)\) converge pointwise to \(f(x)\) on an interval \([a,b]\), the limit function \(f\) need not be Riemann integrable. It is also possible that \[\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx\neq\int_{a}^{b}\lim_{n\to\infty}f_{n }(x)\,dx,\] in other words, we cannot interchange the limit and integral. **Example 91**.: Suppose we have a sequence of functions \(\{f_{n}\}\) where the graph of \(f_{n}(x)\) looks like Figure 2.21: Figure 2.21: Graph of \(f_{n}\) Then, it is easy to see that the sequence \(\{f_{n}(x)\}\) converges pointwise to \(f(x)=0\), yet \[\int_{0}^{1}f_{n}(x)\,dx=1,\ \ \text{for every n, whereas}\ \int_{0}^{1}f(x)\,dx=0.\] The Lebesgue approach of integration reduces the problem of integration to that of the integral of the characteristic function \(\chi_{E}\) of a set \(E\), where by the characteristic function of \(E\) we mean: \[\chi_{E}(x)=\left\{\begin{array}{ll}1&\text{if}\ \,x\in E\\ 0&\text{if}\ \,x\notin E\end{array}\right..\] Lebesgue finds a suitable way to define \[m(E)=\int_{a}^{b}\chi_{E}\,dx.\] By \(m(E)\) we mean the measure of the set \(E\) which we will define later. The point here is that the problem of integration connects with the problem of measure. That is why we first define measure of a set, then introduce measurable functions and then present Lebesgue integral and its fantastic convergence theorems. ### Lebesgue Outer Measure Let \(E\subset\mathbb{R}\), and suppose we want to assign a non-negative number \(m(E)\), called the _measure of \(E\)_, in such a way that the following properties are satisfied: * \(m([0,1])=1\). The above three conditions together will imply that the measure of an interval is simply its length. These three conditions are not only independent but if one assumes the Axiom of Choice they became inconsistent (see the Section 2.8 about the _Banach-Tarski Paradox_). If we look at the last condition and think of the geometric notions of length, area, and volume which only require for those measures to be additive across finitely many disjoint objects, we will see why Lebesgue departed from the last condition above and considered _countably additive_ measures. The notion of _outer measure_ is the first step toward an understanding the notion of length. Let \(I\) be an open interval of \(\mathbb{R}\); we write \(|I|\) to denote the length of \(I\) which is defined as \[|I|=\begin{cases}b-a&\text{if }\,I=(a,b)\\ 0&\text{if }\,I=\emptyset\\ \infty&\text{if }\,I=(-\infty,a)\text{ or }I=(a,\infty)\\ \infty&\text{if }\,I=(-\infty,\infty).\end{cases}\]
* \(m(E+h)=m(E)=m(-E)\), where \(E+h=\{x+h:x\in E\}\) and \(-E=\{-x:x\in E\}\). That is, congruent sets have the same measure.
* If \((E_{n})\) is any sequence, finite or infinite, of pairwise disjoint sets, then \[m\left(\bigcup_{n\geq 1}E_{n}\right)=\sum_{n\geq 1}m(E_{n}).\]
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