Welcome back. I hope that everybody is doing well as midterm season winds down and finals season ramps up. In engineering, mathematics, and computer science, the integral is a powerful tool. A single-variable integral can yield the area under a curve, and a double-variable integral can provide an efficient way to compute the volume under a surface. Integrals also allow us to determine accumulated changes. For example, if we have a function f(t) and want to evaluate it at some positive t = b, then we can take f(0) and add on the change in f that happens between 0 and b. To put this mathematically, we have the following:

f(b) = f(0) + ∫_{[0, b]} (df/dx) dx.

The integral that we have been using throughout this blog series (and the integral that is most popular) is the *Riemann integral*. In the following paragraphs I will try to provide a rigorous, yet understandable guide to how the Riemann integral is formally constructed.

One concept that will be key in the construction that follows is the notion of the *infimum* and *supremum* of a set. Essentially, “infimum” is a more flexible definition of “minimum” and “supremum” is a more flexible definition of “maximum.” While we call the minimum of a set to be the smallest element in that set, we will call the infimum of a set to be that set’s *greatest* *lower bound*. Similarly, we will call the supremum of a set to be that set’s *least upper bound*.

Notationally, we will write “sup A” to denote the supremum of A and we will write “inf A” to denote the infimum of A.

Making the distinction between minimum element and greatest lower bound, or maximum element and least upper bound, is important because an infinite set may *not* have a minimum or maximum element, but *may *still have a finite infimum or supremum.

Take, for example, the set, E, of all numbers of the form 1/n, with n being any integer greater than or equal to 1. Formally, we can write this set as:

E = {1/n : n = 1, 2, 3, … }.

As n becomes larger and larger, 1/n becomes closer and closer to 0. Thus, E has infinitely many elements that get arbitrarily close to zero. Yet, 0 itself is not in E, so E has no minimum element. However, E still has an infimum, namely 0. Meanwhile, 1 is the supremum of E and 1 is an element of E, so E has a maximum element that is precisely equal to its supremum.

At this point, we can begin constructing the Riemann integral. Suppose we have a single-variable function f(x) that is defined on some closed interval [a, b] on the real number line. We will also assume that f(x) is bounded on [a, b] and is continuous almost everywhere. If f(x) were not bounded, then it would blow up to positive or negative infinity at some point in [a, b], which would make its integral more difficult to define (if it even exists at all). If f(x) were wildly discontinuous (for example, if it had a jump discontinuity at every irrational point), then the Riemann integral would not exist.

Then we will subdivide [a,b] into many pieces by defining a *partition *P, where

P = {x_{0}, x_{1}, x_{2}, …, x_{n}}, where x_{0} = a and x_{n} = b.

Next, we will define the following for j ranging between 1 and n.

m_{j} = inf {f(t): x_{j-1} ≤ t ≤ x_{j}}

M_{j} = sup {f(t): x_{j-1} ≤ t ≤ x_{j}}.

To put the two definitions above into words, we are setting m_{j }to be the greatest lower bound of the set of all values of f(t) where t ranges between the endpoints of each subinterval defined by our partition P. Simultaneously, we are setting M_{j} to be the least upper bound of the set of all values of f(t) where t ranges between the endpoints of each of our subintervals.

Next, we will let ∆x_{j} = x_{j} – x_{j-1}. This will provide us with a more compact way to refer to the width of each of our subintervals.

In a blog post from this spring, we discussed that an integral can be thought of as using infinitely many infinitely thin rectangles to approximate the (signed) area under the graph of f(x). The definitions that follow will adhere to this idea.

Call U(f, P) = ∑ M_{j} ∆x_{j} the *upper Riemann sum of f* with respect to the partition P.

Call L(f, P) = ∑ m_{j} ∆x_{j} the *lower Riemann sum of f* with respect to the partition P.

(These sums are taken for 1 ≤ j ≤ n.)

We can interpret U(f, P) as approximating the area under the graph of f(x) between x = a and x = b using many thin rectangles that just barely *overestimate *that area. We can tell that U(f, P) is an overestimate because it is essentially using the maximum value of f over each of our subintervals as the height of each rectangle. Similarly, we can interpret L(f, P) as approximating the area under the graph of f(x) using many thin rectangles that just barely *underestimate *that area. Because we are using infimums of f(x) over each of our subintervals as the heights of our rectangles, it makes sense that L(f, P) is an underestimate.

If the infimum of U(f, P) over *all possible* partitions of [a, b] is equal to the supremum of L(f, P) over *all possible *partitions of [a, b], then we will denote that common limiting value as:

∫_{[a, b] }f(x) dx, otherwise known as the Riemann integral of f over [a, b].

If we keep subdividing [a, b] into finer and finer partitions P, then U(f, P) should get smaller and smaller until it creeps infinitely close to the true value of ∫_{[a, b] }f(x) dx. At the same time, L(f, P) should get ever so slightly bigger and bigger until it creeps infinitely close to the true value of the integral of f over [a, b]. Therefore, it makes sense that we define the integral to be the common value that U(f, P) and L(f, P) approach as we take partitions that become infinitely fine.

To explore more details about the Riemann integral (and alternative integrals), I highly recommend watching YoutTube videos or reading a book on Real Analysis. In Math 425a, the introductory Real Analysis course at USC, we go back through the content of Calculus I, but rigorously prove all of the most important results and theorems. The class has been quite difficult, but has also been one of the most interesting math courses I have taken. In the next blog post, I hope we can explore the Riemann integral further by proving the Fundamental Theorem of Calculus. Until then, take care.

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