Real Analysis Primer, Part 8: Lebesgue Integration.
August 6, 2011
Brief "Where are we going?" Motivation.
So far we’ve been talking about measure theory, which is nice to study just for the sake of measuring things, but when we talk about functions and measurements, we generally want to know the integral. The Riemann integral was wonderful because of its simplicity in defining it: you have a big sheet of wavy paper (the total area under a function) and you keep slicing the pieces until they look like rectangles — once they look sufficiently close to a rectangle, just use length-times-width, and add up all the rectangles. This is one interpretation of the Riemann integral, and, for the most part, it works wonderfully. But there are many functions where it fails miserably. This is where Lebesgue takes over.
So what’s different about Lebesgue? While Riemann cuts apart the domain, Lebesgue says, "Okay, how large is the set of x-values that have the range value ?" Specifically, we cut up the function into sets which have the same range value, then multiply that range value with the measure of the set.
As an example, take the step function defined on, say, such that for and for and elsewhere. The Lebesgue integral will say, "Where does take on the value 0? Well, which has measure of 18. Since the value is 0, we have , so this contributes nothing to the integral. Where does take on the value 1? On , which has a measure 1. So this contributes to the integral. Similarly, the function takes on a value 2 onn , this has measure 1, so this contributes to the integral. No other values are taken on, so this integral is 0 + 1 + 2 = 3."
This extremely simple example shows the difference between Riemann and Lebesgue at the most fundamental level. To do Riemann would be much easier in this case: it’s just adding up the rectangles.
There is a slight hitch at this point. Suppose we naively tried to apply what we’ve just said to a more complex function like defined, say, on . We know (because this forms a triangle) that the area should be 2. At every number between in the range, though, the function only takes on the value at a single point which has measure 0. Thus, every one of these contributes 0 to the integral; and the sum of 0’s will be 0. So the integral is 0? That’s not right!
The problem here is that we must define the Lebesgue integral for functions which take on simple functions (which take on a countable number of values) and somehow approximate more general functions. As we know, a function is measurable if and only if there is a sequence of simple functions which converge uniformly to it; and this will give us the means to integrate general (measurable) functions: we will take the integrals of each element of the sequence, and then define that limit to be the integral of the measurable function. A number of questions will arise instantly, like, "Does this work? Does this make sense? Is this well-defined?", all of which can easily be put down with the elegant slashing of our proof-swords.
So, let’s get down to it.
The Lebesgue Integral and Simple Functions.
We’ve already gone over simple functions (ones with a countable number of values, the pre-image of each of these being measurable making simple functions, themselves, measurable) so we might as well dive right in. We will begin by defining, after long last, the Lebesgue Integral for these simple functions. It will be defined exactly as we have motivated it above.
Definition. Let be a measurable simple function on a (measurable) set taking the values . Let be defined as the set of all such that ; that is, is the pre-image of . Then we define
to be exactly the sum
provided that the sum converges absolutely (that is, the sum converges with replacing ).
We call this the Lebesgue integral of over the set .
Note that we have explicitly noted that must be measurable, though this is not specifically stated in the texts I am using. I don’t feel I am losing much in assuming this (expect, perhaps, for more pathological examples such as being able to integrate some function which is non-zero only on a measurable subset of the non-measurable set). Notice that we have also assumed that our sum must converge absolutely, and not just conditionally.
[Note: At this point, it is not entirely clear to me why this should be necessary; in finite cases, we will generally not worry about it, but perhaps there is some reasoning as to why in infinite measured sets, we need this absolute qualification. Feel free to weigh in here.]
It is reasonable to want to check the well-defined-ness of this, but I will not do this here since it is available where ever Lebesgue integrals are gone over. I will also not define the integral of or for , because these should be relatively obvious.
TO BE CONTINUED.