7. Riemann Integration

Let \(L_1\) and \(L_2\) be two real numbers satisfying the definition of Riemann integrability and let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta \gt 0\) such that \(|S(f;\dprt)-L_1| \lt \eps/2\) and \(|S(f;\dprt)-L_2| \lt \eps/2\), for all sampled partitions \(\dprt\) with \(\|\dprt\| \lt \delta\). Then, if \(\|\dprt\| \lt \delta\) it holds that \begin{align*} |L_1-L_2| &\leq |S(f;\dprt)-L_1| + |S(f;\dprt)-L_2|\\ & \lt \eps. \end{align*} By Theorem 2.2.7 this proves that \(L_1=L_2\).

Let \(f:[a,b]:\rightarrow\real\) be such that \(f(x) = C\) for all \(x\in [a,b]\) and let \(\dprt=\{([x_{k-1},x_k],t_k)\}\) be a sampled partition of \([a,b]\). Then \begin{align*} S(f;\dprt) &= \sum_{k=1}^n f(t_k) (x_k-x_{k-1}) \\ &= C \sum_{k=1}^n (x_k-x_{k-1}) \\ &= C (x_n-x_0)\\ &= C (b-a). \end{align*} Hence, with \(L=C(b-a)\), we obtain that \(|S(f;\dprt)-L|=0 \lt \eps\) for any \(\eps \gt 0\) and therefore \(\int_a^b f = C(b-a)\). This proves that \(f\) is Riemann integrable.

We consider the special case that \([a,b]=[0,1]\), the general case is similar. Let \(\dot{\mathcal{Q}}=\{([x_{k-1},x_k],q_k)\}\) be a sampled partition of \([0,1]\) chosen so that \(q_k=\tfrac{1}{2}(x_{k}+x_{k-1})\), i.e., \(q_k\) is the midpoint of the interval \([x_{k-1},x_k]\). Then \begin{align*} S(f;\dot{\mathcal{Q}}) &= \sum_{k=1}^n f(q_k) (x_k-x_{k-1}) \\[2ex] &= \sum_{k=1}^n \tfrac{1}{2} (x_k+x_{k-1})(x_k-x_{k-1})\\[2ex] &= \frac{1}{2}\sum_{k=1}^n (x_k^2-x_{k-1}^2)\\[2ex] &= \frac{1}{2} ( x_n^2 - x_0^2)\\[2ex] &= \frac{1}{2}( 1^2-0^2)\\[2ex] &= \frac{1}{2}. \end{align*} Now let \(\dprt=\{([x_{k-1},x_k]),t_k\}_{k=1}^n\) be an arbitrary sampled partition of \([0,1]\) and suppose that \(\|\dprt\| \lt \delta\), so that \((x_k-x_{k-1})\leq \delta\) for all \(k=1,2,\ldots,n\). If \(\dot{\mathcal{Q}}=\{([x_{k-1},x_k],q_k)\}_{k=1}^n\) is the corresponding midpoint sampled partition then \(|t_k-q_k| \lt \delta\). Therefore, \begin{align*} |S(f;\dprt)-S(f;\dot{\mathcal{Q}})| &= \left|\sum_{k=1}^n t_k (x_k-x_{k-1}) -q_k (x_k-x_{k-1})\right| \\[2ex] &\leq \sum_{k=1}^n |t_k-q_k| (x_k-x_{k-1})\\[2ex] & \lt \delta (1-0)\\[2ex] &= \delta. \end{align*} Hence, we have proved that for arbitrary \(\dprt\) that satisfies \(\|\dprt\| \lt \delta\) it holds that \(|S(f;\dprt)-1/2| \lt \delta\). Hence, given \(\eps \gt 0\) we let \(\delta=\eps\) and then if \(\|\dprt\| \lt \delta\) then \(|S(f;\dprt)-1/2| \lt \eps\). Therefore, \(\int_0^1 f = \frac{1}{2}\).

Let \(L=\int_a^b f\). Suppose that \(g(x)=f(x)\) except at one point \(x=c\). Let \(\dprt=\{([x_{k-1},x_k],t_k)\}\) be a sampled partition. We consider mutually exclusive cases. First, if \(c\neq t_k\) and \(c\neq x_k\) for all \(k\) then \(S(f;\dprt)=S(g;\dprt)\). If \(c=t_k\notin\{x_0,x_1,\ldots,x_n\}\) for some \(k\) then \[ S(g;\dprt)-S(f;\dprt) = (f(c)-g(c))(x_k-x_{k-1}). \] If \(c=t_k=t_{k-1}\) for some \(k\) then necessarily \(c=x_{k-1}\) and then \[ S(g;\dprt)-S(f;\dprt) = (f(c)-g(c))(x_k-x_{k-1}) + (f(c)-g(c))(x_{k-1}-x_{k-2}). \] Hence, in any case, by the triangle inequality \begin{align*} |S(g;\dprt)-S(f;\dprt)| &\leq 2 (|f(c)|+|g(c)|) \|\dprt\|\\ & = M \|\dprt\|. \end{align*} where \(M = 2 (|f(c)|+|g(c)|)\). Let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta_1 \gt 0\) such that \(|S(f;\dprt)-L| \lt \eps/2\) for all partitions \(\dprt\) such that \(\|\dprt\| \lt \delta_1\). Let \(\delta=\min\{\delta_1,\eps/(2M)\}\). Then if \(\|\dprt\| \lt \delta\) then \begin{align*} \|S(g;f) - L\| &\leq |S(g;\dprt)-S(f;\dprt)| + |S(f;\dprt)-L|\\[2ex] & \lt M\|\dprt| + \eps/2\\[2ex] & \lt M\eps/(2M) + \eps/2\\[2ex] &= \eps. \end{align*} This proves that \(g\in \mathcal{R}[a,b]\) and \(\int_a^b g = L = \int_a^b f\). Now suppose by induction that if \(g(x)=f(x)\) for all \(x\in [a,b]\) except at a \(j\geq 1\) number of points in \([a,b]\) then \(g\in \mathcal{R}[a,b]\) and \(\int_a^b g = \int_a^b f\). Now suppose that \(h:[a,b]\rightarrow\real\) is such that \(h(x)=f(x)\) for all \(x\in [a,b]\) except at the points \(c_1,c_2,\ldots,c_{j},c_{j+1}\). Define the function \(g\) by \(g(x)=h(x)\) for all \(x\in [a,b]\) except at \(x=c_{j+1}\) and define \(g(c_{j+1})=f(c_{j+1})\). Then \(g\) and \(f\) differ at the points \(c_1,\ldots,c_j\). Then by the induction hypothesis, \(g\in\mathcal{R}[a,b]\) and \(\int_a^b g = \int_a^b f\). Now \(g\) and \(h\) differ at the point \(c_{j+1}\) and therefore \(h\in \mathcal{R}[a,b]\) and \(\int_a^b h = \int_a^b g = \int_a^b f\). This ends the proof.

If \(k=0\) then \((kf)(x)=0\) for all \(x\) and then clearly \(\int_a^b kf = 0\), so assume that \(k\neq 0\). Let \(\eps \gt 0\) be given. Then there exists \(\delta \gt 0\) such that if \(\|\dprt\| \lt \delta\) then \(|S(f;\dprt)-\int_a^b f\| \lt \eps/|k|\). Now for any partition \(\dprt\), it holds that \(S(kf;\dprt) = k S(f;\dprt)\). Therefore, if \(\|\dprt\| \lt \delta\) then \begin{align*} \left|S(kf;\dprt)-k\int_a^b f \right| &= |k| \left|S(f;\dprt)-\int_a^b f\right|\\ & \lt |k| (\eps/|k|)\\ & = \eps. \end{align*} To prove (b), it is easy to see that \(S(f+g;\dprt) = S(f;\dprt) + S(g;\dprt)\). Given \(\eps \gt 0\) there exists \(\delta \gt 0\) such that \(|S(f;\dprt)-\int_a^b f| \lt \eps/2\) and \(|S(g;\dprt)-\int_a^b g| \lt \eps/2\), whenever \(\|\dprt\| \lt \delta\). Therefore, if \(\|\dprt\| \lt \delta\) we have that \begin{align*} \left|S(f+g;\dprt) - \left(\int_a^b f + \int_a^b g\right) \right| &\leq \left|S(f;\dprt)-\int_a^b f\right|+\left|S(g;\dprt)-\int_a^b g\right|\\ & \lt \eps. \end{align*} To prove (c), let \(\eps \gt 0\) be arbitrary and let \(\delta \gt 0\) be such that if \(\|\dprt\| \lt \delta\) then \begin{gather*} \int_a^b f - \eps/2 \lt S(f;\dprt) \lt \int_a^b f + \eps/2\\[2ex] \int_a^b g - \eps/2 \lt S(g;\dprt) \lt \int_a^b g + \eps/2 \end{gather*} Now, by assumption, \(S(f;\dprt)\leq S(g;\dprt)\) and therefore \[ \int_a^b f -\eps/2 \lt S(f;\dprt)\leq S(g;\dprt) \lt \int_a^b g + \eps/2. \] Therefore, \[ \int_a^b f \lt \int_a^b g + \eps. \] Since \(\eps\) is arbitrary, we can choose \(\eps_n=1/n\) and then passing to the limit we deduce that \(\int_a^b f \leq \int_a^b g\).

Let \(f\in\mathcal{R}[a,b]\) and put \(L=\int_a^b f\). There exists \(\delta \gt 0\) such that if \(\|\dprt\| \lt \delta\) then \(|S(f;\dprt)-L| \lt 1\) and therefore \(|S(f;\dprt)| \lt |L|+1\). Suppose by contradiction that \(f\) is unbounded on \([a,b]\). Let \(\prt\) be a partition of \([a,b]\), with sets \(I_1,\ldots,I_n\), and with \(\|\prt\| \lt \delta\). Then \(f\) is unbounded on some \(I_j\), i.e., for any \(M \gt 0\) there exists \(x\in I_j=[x_{j-1},x_j]\) such that \(f(x) \gt M\). Choose samples in \(\prt\) by asking that \(t_k=x_k\) for \(k\neq j\) and \(t_j\) is such that \[ |f(t_j)(x_j-x_{j-1})| \gt |L|+1 + \left|\sum_{k\neq j} f(t_k)(x_k-x_{k-1})\right|. \] Therefore, (using \(|a| = |a+b-b|\leq |a+b| + |b|\) implies that \(|a+b|\geq |a|-|b|\)) \begin{align*} |S(f;\dprt)| &= \left|f(t_j)(x_j-x_{j-1}) + \sum_{k\neq j} f(t_k) (x_k-x_{k-1}) \right|\\[2ex] &\geq \left|f(t_j)(x_j-x_{j-1}) \right| - \left|\sum_{k\neq j} f(t_k) (x_k-x_{k-1}) \right| \\[2ex] & \gt |L|+1. \end{align*} This is a contradiction and thus \(f\) is bounded on \([a,b]\).

Let \(\eps \gt 0\) be arbitrary and let \(E=\{x\in[0,1]\;:\; h(x)\geq \eps/2\}\). By definition of \(h\), the set \(E\) is finite, say consisting of \(n\) elements. Let \(\delta=\eps/(4n)\) and let \(\dprt\) be a sampled partition of \([0,1]\) with \(\|\dprt\| \lt \delta\). We can separate the partition \(\dprt\) into two sampled partitions \(\dot{\mathcal{P}}_1\) and \(\dot{\mathcal{P}}_2\) where \(\dot{\mathcal{P}}_1\) has samples in the the set \(E\) and \(\dot{\mathcal{P}}_2\) has no samples in \(E\). Then the number of intervals in \(\dot{\mathcal{P}}_1\) can be at most \(2n\), which occurs when all the elements of \(E\) are samples and they are at the endpoints of the subintervals of \(\dot{\mathcal{P}}_1\). Therefore, the total length of the intervals in \(\dot{\mathcal{P}}_1\) can be at most \(2n\delta=\eps/2\). Now \(0 \lt h(t_k)\leq 1\) for every sample \(t_k\) in \(\dot{\mathcal{P}}_1\) and therefore \(S(f;\dot{\mathcal{P}}_1)\leq 2n\delta = \eps/2\). For samples \(t_k\) in \(\dot{\mathcal{P}}_2\) we have that \(h(t_k) \lt \eps/2\). Therefore, since the sum of the lengths of the subintervals of \(\dot{\mathcal{P}}_2\) is \(\leq 1\), it follows that \(S(f;\dot{\mathcal{P}}_2)\leq \eps/2\). Hence \(0\leq S(f;\dot{\mathcal{P}})=S(f;\dot{\mathcal{P}}_1)+S(f;\dot{\mathcal{P}}_2) \lt \eps\). Thus \(\int_a^b h = 0\).

To show that \(f\) is not in \(\mathcal{R}[0,1]\), we must show that there exists \(\eps_0 \gt 0\) such that for all \(\delta \gt 0\) there exists sampled partitions \(\dprt\) and \(\dqrt\) with norm less than \(\delta\) but \(|S(f;\dprt)-S(f;\dqrt)|\geq \eps_0\). To that end, let \(\eps_0=1/2\), and let \(\delta \gt 0\) be arbitrary. Let \(n\) be sufficiently large so that \(1/n \lt \delta\). Let \(\dprt\) be a sampled partition of \([0,1]\) with intervals all of equal length \(1/n \lt \delta\) and let the samples of \(\dprt\) be rational numbers. Similarly, let \(\dqrt\) be a partition of \([0,1]\) with intervals all of equal length \(1/n\) and with samples irrational numbers. Then \(S(f;\dprt) = 1\) and \(S(f;\dqrt)=0\), and therefore \(|S(f;\dqrt)-S(f;\dqrt)| \geq \eps_0\).

If \(f\in\mathcal{R}[a,b]\) then let \(\alpha(x)=\beta(x)=f(x)\). Then clearly \(\int_a^b(\beta-\alpha)=0 \lt \eps\) for all \(\eps \gt 0\). Now suppose the converse and let \(\eps \gt 0\) be arbitrary. Let \(\alpha\) and \(\beta\) satisfy the conditions of the theorem, with \(\int_a^b(\beta-\alpha) \lt \frac{\eps}{3}\). Now, there exists \(\delta \gt 0\) such that if \(\|\dprt\| \lt \delta\) then \[ \int_a^b\alpha - \frac{\eps}{3} \lt S(\alpha;\dprt) \lt \int_a^b\alpha + \frac{\eps}{3} \] and \[ \int_a^b\beta - \frac{\eps}{3} \lt S(\beta;\dprt) \lt \int_a^b\beta + \frac{\eps}{3}. \] For any sampled partition \(\dprt\) it holds that \(S(\alpha;\dprt)\leq S(f;\dprt)\leq S(\beta;\dprt)\), and therefore \begin{equation} \int_a^b\alpha -\frac{\eps}{3} \lt S(f;\dprt) \lt \int_a^b\beta+\frac{\eps}{3}. \end{equation} If \(\dqrt\) is another sampled partition with \(\|\dqrt\| \lt \delta\) then also \begin{equation}\label{eqn:s2} \int_a^b\alpha - \frac{\eps}{3} \lt S(f;\dqrt) \lt \int_a^b\beta + \frac{\eps}{3}. \end{equation} Subtracting the two inequalities \eqref{eqn:s1}-\eqref{eqn:s2}, we deduce that \[ -\int_a^b(\beta-\alpha)-2\frac{\eps}{3} \lt S(f;\dprt) - S(f;\dqrt) \lt \int_a^b(\beta-\alpha) + 2\frac{\eps}{3}. \] Therefore, since \(\int_a^b(\beta-\alpha) \lt \frac{\eps}{3}\) it follows that \[ -\eps \lt S(f;\dprt) - S(f;\dqrt) \lt \eps. \] By the Cauchy criterion, this proves that \(f\in\mathcal{R}[a,b]\).

Let \(I_1,\ldots,I_n\) be the intervals where \(\varphi\) is constant, and let \(c_1,\ldots,c_n\) be the constant values taken by \(\varphi\) on the intervals \(I_1,\ldots,I_n\), respectively. Then it is not hard to see that \(\varphi = \sum_{k=1}^n c_k \varphi_{I_k}\). Then \(\varphi\) is the sum of Riemann integrable functions and therefore is also Riemann integrable. Moreover, \(\int_a^b\varphi = \sum_{k=1}^n c_k \mu(I_k)\).

Let \(\eps \gt 0\) be arbitrary. Since \(f\) is uniformly continuous on \([a,b]\) there exists \(\delta \gt 0\) such that if \(|x-u| \lt \delta\) then \(|f(x)-f(u)| \lt \eps\). Let \(n\in\N\) be sufficiently large so that \((b-a)/n \lt \delta\). Partition \([a,b]\) into \(n\) subintervals of equal length \((b-a)/n\), and denote them by \(I_1,I_2,\ldots,I_n\), where \(I_1=[x_0,x_1]\) and \(I_k=(x_{k-1},x_k]\) for \(1 \lt k\leq n\). Then for \(x,u \in I_k\) it holds that \(|f(x)-f(u)| \lt \eps\). For \(x\in I_k\) define \(s(x)=f(x_k)\). Therefore, for any \(x\in I_k\) it holds that \(|f(x)-s(x)| = |f(x)-f(x_k)| \lt \eps\). Since \(\bigcup_{k=1}^n I_k = [a,b]\), it holds that \(|f(x)-s(x)| \lt \eps\) for all \(x\in [a,b]\).

Suppose that \(f:[a,b]\rightarrow\real\) is continuous. Let \(\eps \gt 0\) be arbitrary and let \(\tilde\eps = (\eps/4)/(b-a)\). Then there exists a step-function \(s:[a,b]\rightarrow\real\) such that \(|f(x)-s(x)| \lt \tilde\eps\) for all \(x\in [a,b]\). In other words, for all \(x\in [a,b]\) it holds that \[ s(x)-\tilde\eps \lt f(x) \lt s(x) + \tilde\eps. \] The functions \(\alpha(x):=s(x)-\tilde\eps\) and \(\beta(x):=s(x)+\tilde\eps\) are Riemann integrable integrable on \([a,b]\), and \(\int_a^b (\beta-\alpha) = 2\tilde\eps (b-a) = \eps/2 \lt \eps\). Hence, by the Cauchy criterion, \(f\) is Riemann integrable.

Assume that \(f:[a,b]\rightarrow\real\) is increasing and that \(M=f(b)-f(a) \gt 0\) (if \(M=0\) then \(f\) is the zero function which is clearly integrable). Let \(\eps \gt 0\) be arbitrary. Let \(n\in\N\) be such that \(\frac{M(b-a)}{n} \lt \eps\). Partition \([a,b]\) into subintervals of equal length \(\Delta x=\frac{(b-a)}{n}\), and as usual let \(a=x_0 \lt x_1 \lt \cdots \lt x_{n-1} \lt x_n=b\) denote the resulting points of the partition. On each subinterval \([x_{k-1},x_k]\), it holds that \(f(x_{k-1})\leq f(x) \leq f(x_k)\) for all \(x\in [x_{k-1},x_k]\) since \(f\) is increasing. Let \(\alpha:[a,b]\rightarrow\real\) be the step-function whose constant value on the interval \([x_{k-1},x_k)\) is \(f(x_{k-1})\) and similarly let \(\beta:[a,b]\rightarrow\real\) be the step-function whose constant value on the interval \([x_{k-1},x_k)\) is \(f(x_k)\), for all \(k=1,\ldots,n\). Then \(\alpha(x)\leq f(x)\leq \beta(x)\) for all \(x\in[a,b]\). Both \(\alpha\) and \(\beta\) are Riemann integrable and \begin{align*} \int_a^b (\beta-\alpha) &= \sum_{k=1}^n [f(x_k) - f(x_{k-1})] \Delta x\\ & = (f(x_n)-f(x_{n-1}) \Delta x\\ & = M \frac{(b-a)}{n}\\ & \lt \eps. \end{align*} Hence by the Squeeze theorem for integrals (Theorem 7.2.3), \(f\in \mathcal{R}[a,b]\).

Assume for simplicity that \(E:=\{a,b\}\). Let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta \gt 0\) such that if \(\|\dprt\| \lt \eps\) then \(|S(f;\dprt)-\int_a^b f| \lt \eps\). For any \(\dprt\), with intervals \(I_k=[x_{k-1},x_k]\) for \(k=1,2,\ldots,n\), there exists, by the Mean Value Theorem applied to \(F\) on \(I_k\), a point \(u_k\in (x_{k-1},x_k)\) such that \(F(x_k) - F(x_{k-1}) = F'(u_k) (x_k-x_{k-1})\). Therefore, \begin{align*} F(b) - F(a) &= \sum_{k=1}^n F(x_k) - F(x_{k-1})\\ & = \sum_{k=1}^n f(u_k) (x_k-x_{k-1})\\ & = S(f;\dprt_u) \end{align*} where \(\dprt_u\) has the same intervals as \(\dprt\) but with samples \(u_k\). Therefore, if \(\|\dprt\| \lt \delta\) then \begin{align*} \left| F(b) - F(a) - \int_a^bf \right| &= \left| S(f;\dprt_u) - \int_a^b f\right|\\ & \lt \eps. \end{align*} Hence, for any \(\eps\) we have that \(\left| F(b) - F(a) - \int_a^bf \right| \lt \eps\) and this shows that \(\int_a^b f = F(b) - F(a)\).

For any \(w,z\in [a,b]\) such that \(w\leq z\) it holds that \begin{align*} F(z) &= \int_a^z f\\ & = \int_a^w f + \int_w^z f \\ &= F(w) + \int_w^z f \end{align*} and therefore \(F(z)-F(w) = \int_w^z f\). Since \(f\) is Riemann integrable on \([a,b]\) it is bounded and therefore \(|f(x)|\leq K\) for all \(x\in [a,b]\). In particular, \(-K \leq f(x) \leq K\) for all \(x\in [w,z]\) and thus \(-K(z-w) \leq \int_w^z f \leq K (z-w)\), and thus \begin{align*} |F(z) -F(w) | &= \left|\int_w^z f\right|\\ & \leq K |z-w| \end{align*}