2

Does the Carathéodory definition of measurable set encompass the most sets possible?

By this I mean, are all the sets that are well behaved enough with pre-measure that they have all the properties that we want completely given by the Carathéodory criterion of measurability?

It's reasonable to require that measurable sets cut up measurable sets into components that sum up to the measure of the original set but it's not obvious why all sets need to be cut up in this way or why this is the most general definition of measurable set.

Why couldn't lebesgue measurable sets be some proper subset of an even less restrictive definition?

D.R.
  • 10,556
Michael
  • 21
  • 1
    Yeah where did you get Cathedory from? – Matt Samuel Mar 12 '15 at 04:05
  • 1
    Teacher always said it as Cathedory, it's kind of irrelevant nitpicking since the question is clear. – Michael Mar 12 '15 at 04:29
  • @D.R., please stop this spree of editing old questions and thus bumping them to the front page (and bumping newer question off). – Gerry Myerson Oct 21 '19 at 04:16
  • @GerryMyerson oops, sorry. I just wanted to make sure all the measureable-set related questions I found helpful got tagged that way so they are easier to find for any future people looking for answers relating to measurable sets. – D.R. Oct 21 '19 at 04:34
  • @D.R. fine. Next time, please retag three-a-day until done. – Gerry Myerson Oct 21 '19 at 06:41

2 Answers2

4

In some sense, you are asking the wrong question. We don't want the $\sigma$-algebra of Lebesgue measurable sets to encompass too many sets - in fact, we prefer as few sets as reasonably permissible to be Lebesgue measurable. The Lebesgue measurability criterion is designed to exclude bad sets. The real question should be, "Does the definition of Lebesgue measurability exclude enough bad sets to be useful?"

The most important properties that we want out of Lebesgue measurability are the following:

  1. A product of bounded intervals should be measurable. Makes sense, since it's obvious how to define the volume of a $n$-dimensional box.

  2. There should be a reasonable definition of measure so that if we translate a measurable set, it shouldn't change the measure.

The Lebesgue measurable sets encompass all products of intervals but not too many more (they form a minimal good $\sigma$-algebra containing products of intervals in some sense), and Lebesgue measure is a complete translation-invariant measure on the Lebesgue measurable sets. In fact, even better: one can prove it is the unique such measure. And as it turns out, it has several nice intuitive properties in addition: see http://en.wikipedia.org/wiki/Lebesgue_measure#Properties. This is what I would consider a positive answer to the question I raised in the first paragraph.

In principle, one could easily come up with a less restrictive definition of measurable set. Take any non-measurable set, and use it and the Lebesgue $\sigma$-algebra to generate a new $\sigma$-algebra. But then, of course, the Carathéodory criterion no longer holds by fiat, and that's something we'd really like.

The best reason I see to assume the Carathéodory criterion is that the subsets in a measurable space that satisfy the criterion automatically form a $\sigma$-algebra, so it actually makes sense to use it to define a $\sigma$-algebra. (In particular, you use it to obtain countable additivity.) There are probably measure spaces that don't use the Carathéodory criterion to define measurability, but at least for Euclidean space the Carathéodory criterion makes too much sense to replace it with a weaker criterion for measurability.

Gyu Eun Lee
  • 19,584
  • 3
    I can't help but disagree with the assertion that we don't want as many sets as measurable as possible. Isn't it the goal of measure theory to measure as many sets as possible? – Michael Mar 14 '15 at 03:20
  • 1
    No, the exact opposite. The whole issue with our naive notion of measure was that in general, subsets of Euclidean space can be (and usually are) quite pathological to the point of violating all of our intuition about volume (e.g. Vitali set). To put this another way, including more sets in our class of measurable sets makes it harder for our measure to have nice properties like translation invariance. If our goal is a measure we can work with we should only include as many sets as we need, to avoid screwing up the measure. – Gyu Eun Lee Mar 14 '15 at 08:25
  • 3
    I'm not sure I agree on your philosophy here: we know that we can't get a measure with the right properties on the entire power set of $\mathbb{R}$, but why shouldn't it be a fortunate state of affairs if we could? (As it happens, pathological examples convince us this is impossible, but why wouldn't we want to measure pathological sets if the universe were different and the axioms, therefore, said we could?) –  Jul 19 '17 at 17:53
  • Incidentally, in your fourth paragraph, it's not clear that we could extend Lebesgue measure to this new $\sigma$-algebra. –  Jul 19 '17 at 17:56
  • We might learn about the Borel measure, or its completion, in a first course on measure theory. But the reason we needed that in the first place is because the Riemann-Stieltjes integral doesn't work on a bunch of sets/functions that are entirely reasonable to measure/integrate. Saying that the "point" of measure theory is to exclude non-measurable sets is kind of like saying that the point of group theory is to exclude monoids and groupoids. It's just not. Nevermind the hundreds of domain-specific measures we read about in advanced analysis, like the Weiner measure, etc. – nomen May 28 '19 at 23:26
  • Echoing @Michael's concern; it's not clear to me why we don't want too many measurable sets. I get that it's paramount to have countable additivity, but if you could add more sets into your sigma-algebra without losing countable additivity, why wouldn't you? Don't we want the largest possible collection of "measurable" sets you can have that includes all the intervals, is a sigma-algebra, and also has countable additivity?. I just asked a question that boiled down to this: https://math.stackexchange.com/questions/3305000/is-the-sigma-algebra-generated-by-the-carath%C3%A9odory-condition-maximal – Billy Smith Jul 29 '19 at 17:19
0

The obvious reason is that "divide, measure each parts, and add up" is so essential to make a measure useful.

Intuitively, if you cut an apple into two pieces with a knife, you would expect the volumes of each parts adding up to the volume of that apple as a whole. It would be very weird if it does not.

Gyu Eun Lee is correct on this one: the point is not including as many sets as possible, the point is excluding as many sets as possible while keeping its (the idea of measures) usefulness. As Einstein once said: everything should be made as simple as possible, but not simpler.

John Z. Li
  • 129
  • 4