What is the motivation to build measure theory?

Question

I started reading about measure theory on wikipedia, and downloaded some PDFs, but they all start defining things that I can understand, but can't imagine the motivation to define these things.

'Integration theory' started with the desire to find the area of graphs by a process of infintie summation.

With this analogy, what were mathematicians expecting to create when defining metric spaces and other things related to it? What's the central problem that created all this? I'm trying to understand Lebesgue integration but the article starts with these metric spaces definitions...

Answer 1

The central idea (as is true of much of Mathematics) is to generalize some very familiar concepts to more abstract notions. In particular, there are some things that the notions of length, area, volume, etc. have in common. For one example, if you have nothing, then what you have has no length, area, or volume. For another example, if you have two (or more, up to countably-infinitely-many) distinct objects with clearly-defined length/area/volume, then what you have should also have a clearly-defined length/area/volume, found by simply adding the lengths/areas/volumes of the individual objects.

The purpose of Measure Theory is to capture these common characteristics, and allow them to be applied to versions of "measure" that are unspecified or less intuitive/visualizable. In particular, it lets us (attempt to) answer certain questions, such as: "Can we measure every subset of the real line in a way that corresponds with length for intervals and acts like measurements 'should'?"

It turns out that the answer is: "Well...maybe...if we take certain things for granted." So, what seems like a simple question turns out to be very deep, indeed! In fact, in order for certain principles of measure theory to make sense, one must take certain things for granted. For example, if one doesn't assume that a countable union of disjoint finite sets is countable, then it is completely consistent with Measure Theory that the real number line is of length $0$! If you're curious about the details of the crazy-sounding claims, let me know, and I will explain and/or provide links to clear it up.

Answer 2

One of the primary motivations (there are many, I'm sure) for developing measure theory has to do with the failure of the theory of Riemann integration to behave nicely under taking pointwise limits of functions. In particular, it is well-known that pointwise convergence of a sequence of functions does not translate to the interchange of limits and integrals; for example, if $$ f_n(x) = \begin{cases} 1 & x\in[n,n+1]\\ 0 & \text{otherwise} \end{cases} $$ then as $n\to\infty$, $f_n$ converges pointwise to the zero function. If we take its integral, $$ \int_{-\infty}^\infty f_n(x)~dx = 1, $$ which leads to $$ \lim_{n\to\infty} \int_{-\infty}^\infty f_n(x)~dx = 1 \neq 0 = \int_{-\infty}^\infty \lim_{n\to\infty} f_n(x)~dx. $$ This inability to interchange a limit and an integral can be extremely inconvenient. One can interchange a limit and a Riemann integration under certain circumstances like uniform convergence, but this is usually far too restrictive of a condition. What measure-theoretic integration gives you is a way to do this interchange provided you have pointwise convergence and much weaker conditions than uniform convergence; in exchange, it forces you to throw out sets of measure zero. In addition, once the theory is developed you suddenly have extremely powerful tools to take limits of integrals you did not have before (e.g. Lebesgue dominated convergence theorem).

This simplification of the issue of convergence of integrals then allows one to make the great shift in analytical thinking of considering metric spaces of functions that are nicely behaved with respect to integration, in the sense that they are complete under an appropriately defined integral norm. Function spaces had been around (the continuous functions under supremum norm), but before measure theory it was not possible to precisely define a norm on a function space using an integral (at best one could define a seminorm). With measure theory, analysis drifted away from the study of functions and their continuity/differentiability properties and toward the study of function spaces.

This shift in ideas presided over decades of major mathematical and scientific change, leading to the notion of a Hilbert and Banach space and becoming deeply intertwined with the emerging field of quantum mechanics. It allowed mathematicians to introduce the tools of linear algebra to analysis (which today we call functional analysis) and look at inherently infinite-dimensional problems (such as PDEs) in a concrete framework. Today function spaces remain a central element of the toolbox of virtually all mathematicians, and measure theory is at the heart of its foundation.

If this doesn't motivate measure theory for you, then virtually nothing will :)

Answer 3

One possible motivation for measure theory (at least this motivated me, I'm not sure about the historical order of things) is the Banach Tarski Paradox: It is "possible" to decompose a sphere into finitely many parts that can than be reassembled to two spheres having the same volume!

When I first heard this I was completely confused. How should this be possible, isn't this a contradiction? It then turns out that you have to look at things more carefully and check what hidden assumptions you have. For example one usually thinks that it should be possible to associate a quantity (a positive real number) called "volume" to every subset of $\mathbb{R}^3$ which also should not change under "moving" the set (i.e. translation, rotation and taking the mirror image). You further want it to behave "additively" under (disjoint) union. It turns out that this is not possible in this generality. But of course you want to save as much as you can of this intuitively true thing and that's where measure theory comes in.

Answer 4

I took a course in the basics of Lebesgue measure this semester, and it seems to me that the focus of measure theory is to assign a number to subsets of $\mathbb R^n$ which describe the size of the set. In one dimension, this is length. In two, area and so on. Intuitively, we want to say that this is easy, that you can just add up the size of the parts and so on. However, there are some sets which are not so nicely behaved, so that finding their size is difficult. To illustrate one such counter-intuitive result that is not quite as sophisticated as Banach-Tarski, I'd like to show you an example of a set with uncountably many points, where the set has Lebesgue measure $0$ - the Cantor Middle-Thirds set.

To construct this set, begin with the interval $[0,1]$. From this set, remove the middle third, namely, the interval $[1/3, 2/3]$. The remaining piece are two sub-intervals of length 1/3rd each. Now, from each of these, remove the middle third, leaving 4 subintervals. We continue this process indefinitely.

For an illustration of this set, check out this image on wikipedia showing the construction process.

http://en.wikipedia.org/wiki/Cantor_set#mediaviewer/File:Cantor_set_in_seven_iterations.svg

Now, first we show this set is not empty. To see this, consider a point like $1/3$, which is not deleted in the first step, and is too far away to be deleted at any other step. Clearly this point remains in the set.

Next, we show the set has an uncountably infinite number of points. To do this, take a look at the picture linked. Notice that at each step, each piece is cut into two pieces by removing the middle third. To each point that remains in the set, we assign a $0$ to a point if it is on the left of the removed interval, and a $1$ if it is on the right. Clearly if a point is eventually deleted, it is denoted by a finite string of 1's and 0's, which is not unique, as every point in the interval shares that same "address" if you will. However, points in the Cantor set will have an infinite string of 1's and 0's. So there are an infinite list of 1's and 0's for each non-removed points, with two options for each. This means the number of possibilities for strings are $2^{\mathbb N}$, which is strictly greater than $|\mathbb N|$. Since $\mathbb N$ is countably infinite by definition, the Cantor set, being larger, must be uncountably infinite.

Now we show that this set has measure 0, if we take the naive definition that measure is thought of as "length." To do this, we devise an infinite series to represent the length of the removed parts, and subtract it from the length of the starting interval, which is obviously 1, since it is a unit interval. Now, at the first step, we removed 1 interval, at the second, two intervals, the third, four intervals and so on. So at the $k$-th step, we removed $2^{k-1}$ intervals. The length of the first interval was $1/3$, each of the second two were $1/9$, the third $1/27$ and so on. So each of the intervals was of length $\frac {1}{3^k}$. So this means that the total area deleted was $1/3 \sum_{k=0}^{\infty} (2/3)^k$. This is a basic geometric series, so we know that it evaluates to $\frac {1}{1-{\frac{2}{3}}}=3$. Multiplying by the $1/3$ out in front, we see that the deleted intervals have length 1! This means that the Cantor set itself has length $1-1=0$

Now, this notion rests on some naive notions, but I assure you that under the proper rigorous treatments, the results still hold. For a full treatment of the foundations of measure theory, you could check out Capinski's book Measure, Integral and Probability.