# Lebesgue outer measure of a set of countably many points is 0 - logic check

1. Oct 18, 2012

### dumbQuestion

So I know that the Lebesgue outer measure of a set of only countably many points is 0. An example of this is the rationals as a subset of the reals.

I want to make sure my intuition behind this is correct. The process: Now, if we are going to take the Lebesgue outer measure of the rationals, we first want to take every open set covering them, then for each such cover sum together the length of the individual intervals making up the covering, and finally if we construct a set representing the sums of these individual coverings, if we take the inf of this set that is the Lebesgue outer measure.

The intuition (as I think I understand it) So let's consider the rational numbers as a subset of the reals. First thing to notice is that this subset is just a bunch of (countably infinite) disjoint points. (The points are disjoint because between any two rationals, you can find infinitely many irrationals, and there are countably many because the cardinality of the rationals is aleph-0). So say we want to begin by constructing any arbitrary open covering of the rationals, we can do this by picking each rational q, then constructing an open set of the form (q-ε, q + ε) for some arbitrary ε>0. So our covering consists of countably many intervals like this. For our covering, the "sum" of the lengths of these countably many intervals will then be Ʃ(q+ε-(q-ε)) = Ʃ(2ε) for all the rationals q. So to find the Lebesgue outer measure, we form this set made up of all such sums, one for each individual cover, and look at the inf of that set. It seems like the inf will be the number we get as we take ε smaller and smaller, arbitrarily close to 0 (since the individual intervals in the cover could get arbitrarily small in length since we're just covering a bunch of disconnected points). So this number would just be 0. It seems intuitive that we could generalize this beyond the Rationals example, and to any subset of countably many points because you could perform a similar process on any such subset.

Does that sound even vaguely right? I'm just trying to wrap my mind around the basics of Lebesgue integral and measure theory and Lebesgue measure and want to make sure I'm on the right track.

2. Oct 18, 2012

### Fredrik

Staff Emeritus
You're on the right track, but you're doing one thing wrong: By choosing the length of each interval to be the same, you're ensuring that their sum will be infinite. So in the next step, you would have to "let infinity go to zero", which doesn't really make sense. Can you see a better way to choose the lengths of those intervals?

3. Oct 18, 2012

### dumbQuestion

Ok, I can see what you're saying here. I also felt a little bit uneasy about this. What about making the "epsilon" for each interval somehow dependent upon the rational point q we're building the interval around? That way at least the individual intervals would have distinct lengths.

Maybe I should sleep on it for the night :/ To be honest I am finding this Lebesgue integration really difficult to tackle (but it's super intruiging and I've very determinted to understand it). I am going to the library tomorrow to find some good text books on measure theory and I'm hoping that will help

4. Oct 18, 2012

### Fredrik

Staff Emeritus
Yes, that's the right idea. You will need to use that $\mathbb Q$ is countable when you do this. (The same trick will work for any countable subset of $\mathbb R$).

I agree that it's pretty difficult. The ideas behind the definitions are fairly simple, but you need to prove a lot of theorems to turn those ideas into a useful theory, and many of those theorems are pretty hard.

5. Oct 19, 2012

### HallsofIvy

Staff Emeritus
The simplest way to see that the Lebesque measure of a countable set is to use the fact that Lesbeque measure is "countably additive". If $\{U_i\}$ is a countable set of $\mu-measurable$ disjoint sets then $\mu\cup U_i= \sum \mu(U_i)$ as long as that sum converges. In particular, if A is a countable set, its measure is the sum of the measures of its singleton sets, that is, as sum of 0s.

6. Oct 19, 2012

### Fredrik

Staff Emeritus

7. Oct 19, 2012

### mathman

Since the rationals can be ordered, you can make the coverings ε/2, ε/4, ε/8, .... so the sum of the coverings (greater than the length of the union) is ε which is arbitrarily small.

8. Oct 19, 2012

### dumbQuestion

Hey I just wanted to thank you all for the helpful responses. I got up really early today and hit the library, found a lot of fantastic books on Lebesgue Integration (they are all super old from the 50's and 60's, but they were great!) I really focused and it makes so so much more sense now. I think Lebesgue integration is really cool. It's actually very inspiring to learn about, kind of helps me understand how theoretical mathematics is developed - in Lebesgue's case, it seems like he realized a problem: the old or existing methods, like Reimann integration was just not sufficient to explore functions that had odd behaviors, like having countable discontinuities. So with some clever out of the box thinking he developed a different method to handle integration. It just so happens there's a lot of theory necessary (measure theory) for Lebesgue's integral to come together.

Really cool stuff. I'm kind of shocked LEbesgue integration was never mentioned in my real analysis class. I'm only learning of it in the last few days but it seems so interesting. Just out of curiosity why isn't it the standard to teach Lebesgue integration now vs. Reimann integration in calc classes when Lebesgue integration is so much more powerful? Is it just because the Reimann integral is easier to digest? I would think at least they'd introduce Lebesgue integration though to calculus I students. You wouldn't have to make the students understand it fully just explain "this is a different approach, inistead of partitioning up the domain we partition up the range... it involves theory beyond the scope of this calc class but this approach allows us to integrate a whole class of broader functions, even ones that have infinite number of discontinuities". I bet that would peak the interest of a lot of calculus students

9. Oct 19, 2012

### Fredrik

Staff Emeritus
Yes, I think that's the only reason.

I think it would be appropriate to explain the ideas of Lebesgue integration in a calculus class. But since you can't teach it rigorously there, it can't actually be a part of the course, so I understand that many teachers will not want to cover it.

Since mathman posted the key part of the trick that I was thinking you might want to figure out on your own, I might as well provide the other details. Suppose that S is a countable subset of the real numbers. Then, by definition of countable, there exists a bijection $f:\mathbb Z^+\to S$. (I use $\mathbb Z^+$ for the set of positive integers). Let ε>0 be arbitrary. For each $n\in\mathbb Z^+$, define $I_n$ to be the open interval $\big(f(n)-\epsilon/2^n, f(n)+\varepsilon/2^n\big)$. The set $\{I_n|n\in\mathbb Z^+\}$ clearly covers S. For all n, the length of $I_n$ is $2^{-n+1}\varepsilon$, so the sum of the lengths of these intervals is
$$\sum_{n=0}^\infty 2^{-n+1}\varepsilon=2\varepsilon\sum_{n=0}^\infty 2^{-n} =2\varepsilon\frac{1}{1-\frac{1}{2}} =4\varepsilon.$$

10. Oct 19, 2012

### pwsnafu

It's worth pointing out that you can modify the Riemann integral quite simply to obtain the McShane integral, which has the same power as the Lebesgue integral.

The advantage of the measure theory approach is that once you have it under your belt, you can integrate functions whose domain isn't ℝ. With the generalised Riemann approach, it's not obvious, and it's really in the past 30 years or so that we have a full developed theory.