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provided that \(f(0)=2,\ g(1)=2,\ f^{\prime}(0)=\pi\), and \(g^{\prime}(1)=e\).
12. If \(f\) is a continuous one-to-one function defined on an interval and
\[f^{\prime}(f^{-1}(a))=0,\]
then show that \(f^{-1}\) is not differentiable at \(a\).
13. Let \(f_{n}:\mathbb{R}\to\mathbb{R}\) be differentiable for each \(n=1,2,\dots\) with \(|f^{\prime}_{n}(x)|\leq 1\) for all \(n\) and \(x\). Assume \(\lim_{n\to\infty}f_{n}(x)=h(x)\) for all \(x\). Prove that \(h:\mathbb{R}\to\mathbb{R}\) is continuous.
### 1.8 The Riemann Integral
Later we will define the integral in a formal way, but loosely speaking, the integral of a positive function \(f(x)\geq 0\) for all \(x\in[a,b]\) is just the _area_ under its graph. Our aim is to make this notion precise so that we can extend it to a wider class of functions. Even though the germ of the idea of integration goes back to Archimedes in the third century B.C. and we see ideas of integration developed in the seventeen century by I. Newton and G. Leibniz, it was B. Riemann (1826-1866) who gave his definition of an integral using _Riemann sums_. The approach in terms of upper and lower sums is due to G. Darboux (1842-1917). Subsequently, other more general approaches to integration were given by T. J. Stieltjes (1856-1894) and H. Lebesgue (1875-1941). In the following we consider bounded functions \(f\) defined on \([a,b]\), meaning that there is a constant \(M>0\) such that \(|f(x)|\leq M\) for all \(x\in[a,b]\).
**Definition 36**.: 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 \(k=1,2,\dots,n\), 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 _upper sum_, denoted by \(U(f,P)\), and _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 every partition \(P\). 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 \(L(f,P)\leq L(f,Q)\) and \(U(f,P)\geq U(f,Q)\) (Figure 1.46). Let \(\mathcal{P}\) be the collection of all possible partitions of the interval \([a,b]\), and let \[U(f)=\inf\{U(f,P):\ P\in\mathcal{P}\}\quad\text{and}\quad L(f)=\sup\{L(f,P):\ P\in \mathcal{P}\}.\] Let \(f\) be a bounded function on \([a,b]\) and \(P\) be any partition of \([a,b]\). First observe that \(U(f)\geq L(f,P)\); then one can prove that for any bounded function \(f\) on \([a,b]\) we always have \(L(f)\leq U(f)\). **Definition 37**.: 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, integrability is equivalent to the existence of partitions whose upper and lower sums are arbitrarily close together (see Proposition 15 below). **Example 49**.: Consider Dirichlet’s function \[f(x)=\left\{\begin{array}{ll}1&\text{ if }x\in\mathbb{Q}\\ 0&\text{ if }x\notin\mathbb{Q}.\end{array}\right.\] 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 Figure 1.46: Upper and lower sums density of rationals in \(\mathbb{R}\) implies that every subinterval of \(P\) contains points for which \(f(x)=1\), and thus \(U(f,P)=b-a\). Similarly \(L(f,P)=0\) because irrational numbers are also dense in \(\mathbb{R}\). Because this is the case for every possible partition, \(U(f)=b-a\) and \(L(f)=0\). The two are not equal, and therefore Dirichlet’s function is not Riemann integrable. This example also illustrates that not every bounded function is Riemann integrable. **Proposition 15** (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 \(P\) such that_ \[U(f,P)-L(f,P)<\epsilon.\] : Suppose \(f\) is Riemann integrable on \([a,b]\). Given \(\epsilon>0\), it follows from the definition of upper and lower sums that there are partitions \(P_{1}\) and \(P_{2}\) such that \[\left(\int_{a}^{b}f\,dx\right)-\frac{\epsilon}{2} Let \(P\) be a refinement of \(P_{1}\) and of \(P_{2}\). Then \[\left(\int_{a}^{b}f\,dx\right)-\frac{\epsilon}{2} which implies \(U(f,P)-L(f,P)<\epsilon\). For the proof of the converse statement, let \(\epsilon>0\). If a partition \(P*\) of \([a,b]\) exists such that \(U(f,P*)-L(f,P*)<\epsilon\), then \(U(f)-L(f)\leq U(f,P*)-L(f,P*)<\epsilon\). Since \(\epsilon\) is arbitrary, we have \(U(f)=L(f)\). **Example 50**.: Consider the following function \(f\) defined on the interval \([0,2]\) (Figure 1.47): \[f(x)=\left\{\begin{array}{ll}1&\text{ if }x=1\\ 0&\text{ if }x\neq 1.\end{array}\right.\] If \(P=\{0=x_{0} \[m_{k}=M_{k}=0\quad\text{if}\quad k\neq i\quad\text{but}\quad m_{i}=0\quad \text{and}\quad M_{i}=1.\] Figure 1.47: Riemann integrable discontinuous function Writing upper and lower sums, \[U(f,P) =\sum_{k=1}^{i-1}M_{k}(x_{k}-x_{k-1})+M_{i}(x_{i}-x_{i-1})+\sum_{k=i+ 1}^{n}M_{k}(x_{k}-x_{k-1})\] \[L(f,P) =\sum_{k=1}^{i-1}m_{k}(x_{k}-x_{k-1})+m_{i}(x_{i}-x_{i-1})+\sum_{k= i+1}^{n}m_{k}(x_{k}-x_{k-1}),\] which implies \[U(f,P)-L(f,P)=x_{i}-x_{i-1}.\] We can take \(|x_{i}-x_{i-1}|<\epsilon\) for a suitable \(\epsilon\) so that \[U(f,P)-L(f,P)<\epsilon.\] By Riemann’s criterion \(f\) is integrable. Clearly this function is discontinuous at \(x=1\). To show \(f\) is Riemann integrable, we embedded the point \(x=1\) into a very small subinterval whose length is less than \(\epsilon\). Going back to the definitions of upper and lower sums, \[U(f,P)-L(f,P)=\sum_{k=1}^{n}(M_{k}-m_{k})(x_{k}-x_{k-1}),\] where \(M_{k}\) and \(m_{k}\) are the supremum and infimum of the function over the interval \([x_{k-1},x_{k}]\). Said in another way, \(M_{k}-m_{k}\) is the variation when the domain is restricted to \([x_{k-1},x_{k}]\). Thus we expect the following relationship between continuity and integrability. **Proposition 16**.: _If \(f\) is continuous on \([a,b]\), then \(f\) is Riemann integrable on \([a,b]\)._ : To see this we apply the theorem that every continuous function on a closed and bounded (thus compact set) set is uniformly continuous there, meaning, given \(\epsilon>0\), there exists \(\delta>0\) such that \[|f(x)-f(y)|<\frac{\epsilon}{b-a}\quad\text{whenever}\quad|x-y|<\delta.\] Now choose a partition \(P\) so that \(x_{k}-x_{k-1}\) is less than \(\delta\). Also observe that on each compact subinterval \([x_{k-1},x_{k}]\), \(f\) being a continuous function means it assumes its infimum and supremum at some point there. By the Extreme Value Theorem, there exists \(t_{k},w_{k}\in[x_{k-1},x_{k}]\), such that \(M_{k}=f(t_{k})\) and \(m_{k}=f(w_{k})\). But we also have \(|t_{k}-w_{k}|<\delta\), therefore
Proof
Proof
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