13 The Definite Integral, $\int_a^b f$


The second part of calculus, after differentiation is integration. We want to define the definite integral of a bounded function $f$ on a closed, bounded interval $[a,b]$. In fact, however, not every function is integrable.

Mathematicians actually use several different definitions of integral. The one that you are familiar with from Calculus II is the Riemann integral, defined using Riemann sums. More common in advanced mathematics is the Lebesgue integral, which has some nice properties that Riemann integrals lack. However, every function that is Riemann integral is also Lebesgue integrable, with the same value, and Riemann integrals are easier to understand. We will study the Riemann integral, but using a definition of Riemann integral that extends naturally to the definition of Lebesgue integral.

The type of integral used in the textbook, defined using upper and lower sums, is actually the Darboux integral. It is equivalent to the usual definition of Riemann integral that you probably saw in Calculus II: $$\int_a^b f(x)\,dx = \lim_{\max(\Delta x_i)\to 0}\sum_{i=1}^nf(x_i*)\Delta x_i$$ However, the Darboux definition is easier to work with. To lead up to the definition, we start by defining partitions\dots

Definition: A partition of a closed bounded interval $[a,b]$ is a finite set $\{x_0,x_1,\dots,x_n\}$ such that $a=x_0<x_1<\cdots<x_n = b$. We will often denote the length of the $i^{\rm th}$ subinterval, $[x_{i-1},x_i]$, by $\Delta x_i$.

The Calculus II idea of Riemann integration is to take $\xi_i\in[x_{i-1},x_i]$ for $i=1,2,\dots,n$, and to use $\sum_{i=1}^nf(\xi_i)\Delta x_i$ as an approximation for $\int_a^b f(x)\,dx$. The sum is referred to as a Riemann sum. To get the exact value of the integral, you have to let the size of the subintervals approach zero. This is not really a very well-defined kind of limit. We use a different approach, using a particular choice of Riemann sum.

Note that at this point, for no good reason, our textbook starts using the notation "sup" instead of "lub" and "inf" instead of "glb". ("Sup" stands for "supremum" and "inf" stands for "infimum".) The meanings, in this context, are the same, but I would also allow sup and inf to be used on unbounded sets, with $\pm\infty$ as possible values. Note that the following definition requires that the function be bounded.

Definition: Suppose that $f$ is a bounded function on the closed, bounded interval $[a,b]$, and that $P=\{x_0,x_1,\dots,x_n\}$ is a partition of $[a,b]$. For $i=1,2,\dots,n$, let $M_i = \sup\{f(x)\,|\,x_{i-1}\le x\le x_i\}$ and let $m_i = \inf\{f(x)\,|\,x_{i-1}\le x\le x_i\}$. We define the upper sum of $f$ relative to the partition $P$ as $$U(P,f)=\sum_{i=1}^n M_i\cdot(x_i-x_{i-1})$$ and we define the lower sum of $f$ relative to the partition $P$ as $$L(P,f)=\sum_{i=1}^n m_i\cdot(x_i-x_{i-1})$$

An upper sum overestimates the integral of $f$. A lower sum underestimates the integral of $f$. The idea for the Riemann integral of $f$ is to take the inf of all possible upper sums, $\ds\inf_P\{U(P,f)\}$, and the sup of all possible lower sums $\ds\sup_P\{L(P,f)\}$. If those two values are the same, then the function is Riemann integrable and the integral of $f$ is defined to be their common value. To make this rigorous, we first need to define a refinement of a partition.

Definition: Let $P$ and $R$ be partitions of $[a,b]$. We say that $R$ is a refinement of $P$ if $R\subseteq P$.

It is not too hard to see that refining a partition can decrease the associated upper sum of a function and increase the lower sum.

Theorem: Let $P$ be a partition of $[a,b]$ and let $R$ be a refinement of $P$. Then $$L(P,f)\le L(R,f)\le U(R,f) \le U(P,f)$$.

Note that this means that the set of all possible lower sums of a function is bounded above by any individual upper sum, and the set of all possible upper sums of a function is bounded above by any individual lower sum. So we have

Theorem: Let $f$ be a bounded function on $[a,b]$, and let $Q_1$ and $Q_2$ be any partitions of $[a,b]$. Then $$L(Q_1,f)\le \sup_P\{L(P,f)\} \le \inf_P\{U(P,f)\}\le U(Q_2,f)$$

We are now ready to define the integral

Definition: Let $f$ be a bounded function on $[a,b]$. We say that $f$ is Riemann integrable on $[a,b]$ if $\ds\sup_P\{L(Q,f)\} = \inf_P\{U(P,f)\}$, and in that case we define the Riemann integral of $f$ on $[a,b]$ to be their common value. The Riemann integral of $f$ on $[a,b]$ is denoted $\int_a^b f$ or $\int_a^b f(x)\,dx$.

We should note that there are functions that are not Riemann integrable. An obvious example is Dirichlet's function $D(x)=\begin{cases}0&\mbox{ if $x$ is irrational}\\1&\mbox{ if $x$ is rational}\end{cases}$. Note that every upper sum for $D$ on the interval $[0,1]$ has value 1 while every lower sum has value 0. So, $\sup_P\{L(Q,D)\} = 0$ while $\inf_P\{U(P,f)\} = 1$. Since the values are not equal, $D$ is not Riemann integrable.

We will see that every continuous function is Riemann integrable. Also, every non-increasing or non-decreasing function is Riemann integrable.

Since the only kind of definite integral that we will cover is the Riemann (or Darboux) integral, I will generally drop "Riemann" from "Riemann integral" and "Riemann integrable."


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