## The Riemann Integral

We begin with the definition of a partition.
Let $$a,b\in\real$$ and suppose $$a \lt b$$. By a partition of the interval $$[a,b]$$ we mean a collection of intervals $\mathcal{P} = \{ [x_0,x_1], [x_1,x_2], \ldots, [x_{n-1},x_n] \}$ such that $$a=x_0 \lt x_1 \lt x_2 \lt \cdots \lt x_n=b$$ and where $$n\in\N$$.
Hence, a partition $$\mathcal{P}$$ defines a finite collection of non-overlapping intervals $$I_k=[x_{k-1},x_k]$$, where $$k=1,\ldots,n$$. The norm of a partition $$\mathcal{P}$$ is defined as $\|\mathcal{P}\| = \max\{ x_1-x_0,x_2-x_1,\ldots,x_n-x_{n-1}\}.$ In other words, $$\|\mathcal{P}\|$$ is the maximum length of the intervals in $$\mathcal{P}$$. To ease our notation, we will denote a partition as $$\mathcal{P}=\{ [x_{k-1},x_k] \}_{k=1}^n$$. Let $$\mathcal{P}=\{[x_{k-1},x_k]\}_{k=1}^n$$ be a partition of $$[a,b]$$. If $$t_k\in I_k=[x_{k-1},x_k]$$ then we say that $$t_k$$ is a sample of $$I_k$$ and the set of ordered pairs $\dot{\mathcal{P}}=\{([x_{k-1},x_k],t_k)\}_{k=1}^n$ will be called a sampled partition.
Examples of sampled partitions are mid-points, right-end points, and left-end points partitions.
Now consider a function $$f:[a,b]\rightarrow\real$$ and let $$\dot{\mathcal{P}}=\{([x_{k-1},x_k],t_k)\}_{k=1}^n$$ be a sampled partition of the interval $$[a,b]$$. The Riemann sum of $$f$$ corresponding to $$\dot{\mathcal{P}}$$ is the number $S(f;\dot{\mathcal{P}}) = \sum_{k=1}^n f(t_k) (x_k - x_{k-1}).$ When $$f(x) \gt 0$$ on the interval $$[a,b]$$, the Riemann sum $$S(f;\dot{\mathcal{P}})$$ is the sum of the areas of the rectangles with height $$f(t_k)$$ and width $$(x_k-x_{k-1})$$. We now define the notion of Riemann integrability.
The function $$f:[a,b]\rightarrow\real$$ is said to be Riemann integrable if there exists a number $$L\in\real$$ such that for every $$\eps \gt 0$$ there exists $$\delta \gt 0$$ such that for any sampled partition $$\dot{\mathcal{P}}$$ that satisfies $$\|\dot{\mathcal{P}}\| \lt \delta$$ it holds that $$|S(f;\dot{\mathcal{P}})-L| \lt \eps$$.
The set of all Riemann integrable functions on the interval $$[a,b]$$ will be denoted by $$\mathcal{R}[a,b]$$.
If $$f\in\mathcal{R}[a,b]$$ then the number $$L$$ in the definition of Riemann integrability is unique.
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 $$L_1=L_2$$.
If $$f\in \mathcal{R}[a,b]$$, we call the number $$L$$ the integral of $$f$$ over $$[a,b]$$ and we denote it by $L = \int_a^b f$
Show that a constant function on $$[a,b]$$ is Riemann integrable.
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.
Prove that $$f(x)=x$$ is Riemann integrable on $$[a,b]$$.
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, $$\int_0^1 f = \frac{1}{2}$$. The next result shows that if $$f\in \mathcal{R}[a,b]$$ then changing $$f$$ at a finite number of points in $$[a,b]$$ does not affect the value of $$\int_a^b f$$. Let $$f\in \mathcal{R}[a,b]$$ and let $$g:[a,b]\rightarrow\real$$ be a function such that $$g(x)=f(x)$$ for all $$x\in [a,b]$$ except possibly at a finite number of points in $$[a,b]$$. Then $$g\in \mathcal{R}[a,b]$$ and in fact $$\int_a^bg = \int_a^b f$$. 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. We now state some properties of the Riemann integral. Suppose that $$f,g\in\mathcal{R}[a,b]$$. The following hold. 1. If $$k\in\real$$ then $$(kf)\in\mathcal{R}[a,b]$$ and $$\int_a^b kf = k\int_a^b f$$. 2. $$(f+g)\in\mathcal{R}[a,b]$$ and $$\int_a^b (f+g) = \int_a^b f + \int_a^b g$$. 3. If $$f(x)\leq g(x)$$ for all $$x\in[a,b]$$ then $$\int_a^b f\leq \int_a^b g$$. 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 $$\int_a^b f \leq \int_a^b g$$.
Properties (i), (ii), and (iii) in Theorem 7.1.8 are known as homogeneity, additivity, and monotonicity, respectively. We now give a necessary condition for Riemann integrability.
If $$f\in\mathcal{R}[a,b]$$ then $$f$$ is bounded on $$[a,b]$$.
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 $$[a,b]$$. Consider Thomae's function $$h:[0,1]\rightarrow\real$$ defined as $$h(x)=0$$ if $$x$$ is irrational and $$h(m/n)=1/n$$ for every rational $$m/n\in [0,1]$$, where $$\gcd(m,n)=1$$. In Example 5.1.7, we proved that $$h$$ is continuous at every irrational but discontinuous at every rational. Prove that $$h$$ is Riemann integrable. 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 $$\int_a^b h = 0$$. #### Exercises Suppose that $$f, g \in \mathcal{R}[a,b]$$ and let $$\alpha,\beta\in\real$$. Prove by definition that $$(\alpha f + \beta g) \in \mathcal{R}[a,b]$$. If $$f$$ is Riemann integrable on $$[a,b]$$ and $$|f(x)|\leq M$$ for all $$x\in[a,b]$$, prove that $$|\int_a^bf|\leq M(b-a)$$. Hint: The inequality $$|f(x)|\leq M$$ is equivalent to $$-M \leq f(x) \leq M$$. Then use the fact that constants functions are Riemann integrable whose integrals are easily computed. Finally, apply a theorem from this section. If $$f$$ is Riemann integrable on $$[a,b]$$ and $$(\dot{\mathcal{P}}_n)$$ is a sequence of tagged partitions of $$[a,b]$$ such that $$\|\dot{\mathcal{P}}_n\|\rightarrow 0$$ prove that \[ \int_a^b f = \lim_{n\rightarrow\infty} S(f;\dot{\mathcal{P}}_n) Hint: For each $$n\in\mathbb{N}$$ we have the real number $$s_n = S(f;\dot{\mathcal{P}}_n)$$, and we therefore have a sequence $$(s_n)$$. Let $$L=\int_a^b f$$. We therefore want to prove that $$\displaystyle\lim_{n\rightarrow\infty} s_n = L$$.
Give an example of a function $$f:[0,1]\rightarrow\real$$ that is Riemann integrable on $$[c,1]$$ for every $$c\in (0,1)$$ but which is not Riemann integrable on $$[0,1]$$. Hint: What is a necessary condition for Riemann integrability?

## Riemann Integrable Functions

To ease our notation, if $$I$$ is a bounded interval with end-points $$a \lt b$$ we denote by $$\mu(I)$$ the length of $$I$$, that is $$\mu(I)=b-a$$. Hence, if $$I=[a,b]$$, $$I=[a,b)$$, $$I=(a,b]$$, or $$I=(a,b)$$ then $$\mu(I)=b-a$$. Thus far, to establish the Riemann integrability of $$f$$, we computed a candidate integral $$L$$ and showed that in fact $$L=\int_a^b f$$. The following theorem is useful when a candidate integral $$L$$ is unknown. The proof is omitted.
A function $$f:[a,b]\rightarrow\real$$ is Riemann integrable if and only if for every $$\eps \gt 0$$ there exists $$\delta \gt 0$$ such that if $$\dprt$$ and $$\dqrt$$ are sampled partitions of $$[a,b]$$ with norm less than $$\delta$$ then $|S(f;\dprt)-S(f;\dqrt)| \lt \eps.$
Using the Cauchy Criterion, we show next that the Dirichlet function is not Riemann integrable.
Let $$f:[0,1]\rightarrow\real$$ be defined as $$f(x)=1$$ if $$x$$ is rational and $$f(x)=0$$ if $$x$$ is irrational. Show that $$f$$ is not Riemann integrable.
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 $$|S(f;\dqrt)-S(f;\dqrt)| \geq \eps_0$$.
We now state a sort of squeeze theorem for integration.
Let $$f$$ be a function on $$[a,b]$$. Then $$f\in\mathcal{R}[a,b]$$ if and only if for every $$\eps \gt 0$$ there exist functions $$\alpha$$ and $$\beta$$ in $$\mathcal{R}[a,b]$$ with $$\alpha(x)\leq f(x)\leq \beta(x)$$ for all $$x\in[a,b]$$ and $$\int_a^b (\beta-\alpha) \lt \eps$$.
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 $$\int_a^b\alpha -\frac{\eps}{3} \lt S(f;\dprt) \lt \int_a^b\beta+\frac{\eps}{3}.$$ If $$\dqrt$$ is another sampled partition with $$\|\dqrt\| \lt \delta$$ then also $$\label{eqn:s2} \int_a^b\alpha - \frac{\eps}{3} \lt S(f;\dqrt) \lt \int_a^b\beta + \frac{\eps}{3}.$$ 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 $$f\in\mathcal{R}[a,b]$$.
Step-functions, defined below, play an important role in integration theory.
A function $$s:[a,b]\rightarrow\real$$ is called a step-function on $$[a,b]$$ if there is a finite number of disjoint intervals $$I_1,I_2,\ldots,I_n$$ contained in $$[a,b]$$ such that $$[a,b] = \bigcup_{k=1}^n I_k$$ and such that $$s$$ is constant on each interval.
In the definition of a step-function, the intervals $$I_k$$ may be of any form, i.e., half-closed, open, or closed.
Let $$J$$ be a subinterval of $$[a,b]$$ and define $$\varphi_J$$ on $$[a,b]$$ as $$\varphi_J(x)=1$$ if $$x\in J$$ and $$\varphi_J(x)=0$$ otherwise. Then $$\varphi_J\in\mathcal{R}[a,b]$$ and $$\int_a^b\varphi_J = \mu(J)$$.
If $$\varphi:[a,b]\rightarrow\real$$ is a step function then $$\varphi\in\mathcal{R}[a,b]$$.
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, $$\int_a^b\varphi = \sum_{k=1}^n c_k \mu(I_k)$$.
We will now show that any continuous function on $$[a,b]$$ is Riemann integrable. To do that we will need the following.
Let $$f:[a,b]\rightarrow\real$$ be a continuous function. Then for every $$\eps \gt 0$$ there exists a step-function $$s:[a,b]\rightarrow\real$$ such that $$|f(x)-s(x)| \lt \eps$$ for all $$x\in [a,b]$$.
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 $$x\in [a,b]$$.
We now prove that continuous functions are integrable.
A continuous function on $$[a,b]$$ is Riemann integrable on $$[a,b]$$.
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.
Recall that a function is called monotone if it is decreasing or increasing.
A monotone function on $$[a,b]$$ is Riemann integrable on $$[a,b]$$.
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), $$f\in \mathcal{R}[a,b]$$.
Our las theorem is the additivity property of the integral, the proof is omitted.
Let $$f:[a,b]\rightarrow\real$$ be a function and let $$c\in (a,b)$$. Then $$f\in\mathcal{R}[a,b]$$ if and only if its restrictions to $$[a,c]$$ and $$[c,b]$$ are both Riemann integrable. In this case, $\int_a^b f = \int_a^c f + \int_c^b f$

#### Exercises

Suppose that $$f:[a,b]\rightarrow\real$$ is continuous and assume that $$f(x) \gt 0$$ for all $$x\in [a,b]$$. Prove that $$\int_a^b f \gt 0$$. Hint: A continuous function on a closed and bounded interval achieves its minimum value.
Suppose that $$f$$ is continuous on $$[a,b]$$ and that $$f(x)\geq 0$$ for all $$x\in [a,b]$$.
1. Prove that if $$\int_a^b f =0$$ then necessarily $$f(x)=0$$ for all $$x\in [a,b]$$.
2. Show by example that if we drop the assumption that $$f$$ is continuous on $$[a,b]$$ then it may not longer hold that $$f(x)=0$$ for all $$x\in [a,b]$$.
Show that if $$f:[a,b]\rightarrow\real$$ is Riemann integrable then $$|f|:[a,b]\rightarrow\real$$ is also Riemann integrable.

## The Fundamental Theorem of Calculus

Let $$f:[a,b]\rightarrow\real$$ be a function. Suppose that there exists a finite set $$E\subset [a,b]$$ and a function $$F:[a,b]\rightarrow\real$$ such that $$F$$ is continuous on $$[a,b]$$ and $$F'(x)=f(x)$$ for all $$x\in [a,b]\backslash\hspace{-0.3em} E$$. If $$f$$ is Riemann integrable then $$\int_a^b f = F(b)-F(a)$$.
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 $$\int_a^b f = F(b) - F(a)$$.
Let $$f\in\mathcal{R}[a,b]$$. The indefinite integral of $$f$$ with basepoint $$a$$ is the function $$F:[a,b]\rightarrow\real$$ defined by $F(x) := \int_a^x f$ for $$x\in [a,b]$$.
Let $$f\in\mathcal{R}[a,b]$$. The indefinite integral $$F:[a,b]\rightarrow\real$$ of $$f:[a,b]\rightarrow\real$$ is a Lipschitz function on $$[a,b]$$, and thus continuous on $$[a,b]$$.
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*}
Under the additional hypothesis that $$f\in\mathcal{R}[a,b]$$ is continuous, the indefinite integral of $$f$$ is differentiable.
Let $$f\in\mathcal{R}[a,b]$$ and let $$f$$ be continuous at a point $$c\in [a,b]$$. Then the indefinite integral $$F$$ of $$f$$ is differentiable at $$c$$ and $$F'(c) = f(c)$$.

## Riemann-Lebesgue Theorem

In this section we present a complete characterization of Riemann integrability for a bounded function. Roughly speaking, a bounded function is Riemann integrable if the set of points were it is discontinuous is not too large. We first begin with a definition of not too large.
A set $$E\subset\real$$ is said to be of measure zero if for every $$\eps \gt 0$$ there exists a countable collection of open intervals $$I_k$$ such that $E\subset\bigcup_{k=1}^\infty I_k \text{ and } \sum_{k=1}^\infty \mu(I_k) \lt \eps.$
Show that a subset of a set of measure zero also has measure zero. Show that the union of two sets of measure zero is a set of measure zero.
Let $$S\subset\real$$ be a countable set. Show that $$S$$ has measure zero.
Let $$S=\{s_1, s_2, s_3, \ldots\}$$. Consider the interval $I_k = \left(s_k - \frac{\eps}{2^{k+1}}, s_k + \frac{\eps}{2^{k+1}}\right).$ Clearly, $$s_k \in I_k$$ and thus $$S\subset\bigcup I_k$$. Moreover, $\sum_{k=1}^\infty \mu(I_k) = \sum_{k=1}^\infty \frac{\eps}{2^k} = \eps.$ As a corollary, $$\rat$$ has measure zero.
However, there exists uncountable sets of measure zero.
The Cantor set is defined as follows. Start with $$I_0 = [0,1]$$ and remove the middle third $$J_1=(\tfrac{1}{3},\tfrac{2}{3})$$ yielding the set $$I_1 = I_0 \backslash\hspace{-0.3em} J_1 = [0,\tfrac{1}{3}] \cup [\tfrac{2}{3},1]$$. Notice that $$\mu(J_1)=\frac{1}{3}$$. Now remove from each subinterval of $$I_1$$ the middle third resulting in the set $I_2 = I_1 \backslash\hspace{-0.3em} \left( (\tfrac{1}{9},\tfrac{2}{9})\cup (\tfrac{7}{9},\tfrac{8}{9}) \right) = [0,\tfrac{1}{9}]\cup [\tfrac{2}{9},\tfrac{3}{9}] \cup [\tfrac{6}{9},\tfrac{7}{9}] \cup [\tfrac{8}{9}, 1]$ The two middle thirds $$J_2=(\tfrac{1}{9},\tfrac{2}{9})\cup (\tfrac{7}{9},\tfrac{8}{9})$$ removed have total length $$\mu(J_2) = 2 \frac{1}{9}$$. By induction, having constructed $$I_{n}$$ which consists of the union of $$2^n$$ closed subintervals of $$[0,1]$$, we remove from each subinterval of $$I_n$$ the middle third resulting in the set $$I_{n+1} = I_n \backslash\hspace{-0.3em} J_{n+1}$$, where $$J_{n+1}$$ is the union of the $$2^n$$ middle third open intervals and $$I_{n+1}$$ now consists of $$2^{n+1}$$ disjoint closed-subintervals. By induction, the total length of $$J_{n+1}$$ is $$\mu(J_{n+1}) = \frac{2^n}{3^{n+1}}$$. The Cantor set is defined as $C = \bigcap_{n=1}^\infty I_n.$
We now state the Riemann-Lebesgue theorem.
Let $$f:[a,b]\rightarrow\real$$ be a bounded function. Then $$f$$ is Riemann integrable if and only if the points of discontinuity of $$f$$ forms a set of measure zero.