5.2 Integrable functions

We now wish to extend the definition of the integral to a wider class of functions. This class will consist of thosefunctions that are uniform limits of step functions. The requirement that these limits be uniform is crucial.Pointwise limits of step functions doesn't work, as we will see in [link] below. The initial step in carrying out this generalization is the following.

Let $[a,b]$ be a closed bounded interval, and let $\left\{h_n\right\}$ be a sequence of step functions that converges uniformly to a function $f$ on $[a,b].$ Then the sequence $\{\int h_n\}$ is a convergent sequence of real numbers.

We will show that $\{\int h_n\}$ is a Cauchy sequence in $R.$ Thus, given an $ϵ>0,$ choose an $N$ such that for any $n\ge N$ and any $x\in [a,b],$ we have

Then, for any $m$ and $n$ both $\ge N$ and any $x\in [a,b],$ we have

Therefore,

as desired.

The preceding theorem provides us with a perfectly good idea of how to define the integral of a function $f$ that is the uniform limit of a sequence of step functions.However, we first need to establish another kind of consistency result.

If $\left\{h_n\right\}$ and $\left\{k_n\right\}$ are two sequences of step functions on $[a,b],$ each converging uniformly to the same function $f,$ then

Given $ϵ>0,$ choose $N$ so that if $n\ge N,$ then $|h_n\left(x\right)-f\left(x\right)|<ϵ/\left(2(b-a)\right)$ for all $x\in [a,b],$ and such that $|f\left(x\right)-k_n\left(x\right)|<ϵ/\left(2(b-a)\right)$ for all $x\in [a,b].$ Then, $|h_n\left(x\right)-k_n\left(x\right)|<ϵ/(b-a)$ for all $x\in [a,b]$ if $n\ge N.$ So,

if $n\ge N.$ Taking limits gives

Since this is true for arbitrary $ϵ>0,$ it follows that $lim\int h_n=lim\int k_n,$ as desired.

Let $[a,b]$ be a closed bounded interval of real numbers. A function $f:[a,b]\to R$ is called integrable on $[a,b]$ if it is the uniform limit of a sequence $\left\{h_n\right\}$ of step functions.

Let $I\left(\right[a,b\left]\right)$ denote the set of all functions that are integrable on $[a,b].$ If $f\in I\left(\right[a,b\left]\right),$ define the integral of $f,$ denoted $\int f,$ by

$$\int f=lim\int h_n,$$

where $\left\{h_n\right\}$ is some (any) sequence of step functions that converges uniformly to $f$ on $[a,b].$

As in the case of step functions, we use the following notations:

$$\int f={\int }_a^{b}f={\int }_a^{b}f\left(t\right)dt.$$

REMARK Note that [link] is crucial in order that this definition be unambiguous.Indeed, we will see below that this critical consistency result is one place where uniform limits of step functions works whilepointwise limits do not. See parts (c) and (d) of [link] . Note also that it follows from this definition that ${\int }_a^{a}f=0,$ because ${\int }_a^{a}h=0$ for any step function. In fact, we will derive almost everything about the integral of a general integrable functionfrom the corresponding results about the integral of a step function. No surprise. This is the essence of mathematical analysis, approximation.