# Whence the Caratheodory criterion?

#1
I'm trying to learn some measure theory (if I'd bothered to listen in the lectures of yesteryear that would be "relearn") and I want to understand the Caratheodory criterion. I think the following is true:

2. Find an outer measure on
3. Filter through the Caratheodory criterion to generate
4. Now:

a) is automatically a -algebra, and
b) restricted to is a measure

So, questions:

1. Are the claims above correct?

2. More importantly, how did Caratheodory come up with his criterion? (or was it Lebesgue, originally?)

I've been unable to find anything that does more than merely states the criterion, and shows that it works, without giving any intuition or motivation for it. Is there some nice argument that shows that if you start with an outer measure, and want to produce a -algebra of measurable sets, then the Caratheodory criterion naturally pops out?
0
5 years ago
#2
I've always seen it formulated like this:

Let be an outer measure of a set , and let be a subset. If for every subset , we have
,
then is a -algebra and the restriction of to is a measure.

Apparently, the intuition is that you have to test every subset , rather than just , which kinda makes sense.

There's another equivalent criterion too. For every , there exists an open set , such that .
0
#3
(Original post by Alex:)
I've always seen it formulated like this:

Let be an outer measure of a set , and let be a subset. If for every subset , we have
,
then is a -algebra and the restriction of to is a measure.

Apparently, the intuition is that you have to test every subset , rather than just , which kinda makes sense.
Hmm. I'm afraid you're not adding much to my intuition, I have to say.

However, having thought about it a bit, it seems to me that:

1. the only difference between an outer measure and measure is the requirement for countable additivity, rather than countable subadditivity, for disjoint sets.

2. we must therefore exclude at least those sets that break countable additivity

3. the Caratheodory criterion must target those sets (or maybe conversely, it only flags up "good" for those sets which don't break countable additivity)

Maybe this is the way that the criterion was first dreamt up - I need to try and think through the details a bit more.
0
#4
(Original post by atsruser)
1. the only difference between an outer measure and measure is the requirement for countable additivity, rather than countable subadditivity, for disjoint sets.

2. we must therefore exclude at least those sets that break countable additivity

3. the Caratheodory criterion must target those sets (or maybe conversely, it only flags up "good" for those sets which don't break countable additivity)

Maybe this is the way that the criterion was first dreamt up - I need to try and think through the details a bit more.
Continuing this line of thought: suppose we want to have measurable. Then note that, for all :

i.e. those sets are disjoint
i.e. they exhaust

then since we want countable additivity, we require that chops up in such a way that the following is true:

which is the Caratheodory criterion. Maybe that's all there is to it, apart from the formal proofs that it gives a -algebra, and so on?

One point though: we haven't assumed that is itself measurable; does that matter?
0
5 years ago
#5
The most intuitive explanation I've found of the Caratheodory criterion relates it to our usual ideas of Riemann integration. Quick recap; when integrating in we:

1) Approximate area from above
2) Approximate area from below
3) If (1) and (2) agree in the limit, we define this as the integral

Now, the outer measure is (hopefully) the natural choice for approximating the measure of a set from above. What about from below? What might we take as an inner measure ? The trick here is that we can define it in terms of our outer measure by subtracting the outer measure of the complement of the set. For a set , we can define it as:

Written in a form that's perhaps more familiar:

Now, we'd like condition (3): that the inner measure and the outer measure agree. That is:

This is very close to Caratheodory. There's some handwaving to do regarding all sets rather than just the I mentioned here, and you have to take some care with finite/infinite measures (where sigma finiteness comes in) but otherwise you're good to go.

Thus (viewed in this way) Caratheodory's criterion is a natural extension to our intuitive ideas about finding area via integration to measures - though the way it is taught is anything but natural!

Full credit should go to http://mathoverflow.net/questions/34...-measurability where I first came upon this explanation, and it's worth reading here if what I've said isn't clear. Hope it helps!
0
5 years ago
#6
(Original post by atsruser)
Continuing this line of thought: suppose we want to have measurable. Then note that, for all :

i.e. those sets are disjoint
i.e. they exhaust

then since we want countable additivity, we require that chops up in such a way that the following is true:

which is the Caratheodory criterion. Maybe that's all there is to it, apart from the formal proofs that it gives a -algebra, and so on?

One point though: we haven't assumed that is itself measurable; does that matter?
The set is called Carathéodory measurable if it satisfies the condition.
0
#7
(Original post by DJMayes)
The trick here is that we can define it in terms of our outer measure by subtracting the outer measure of the complement of the set. For a set , we can define it as:

Written in a form that's perhaps more familiar:

Now, we'd like condition (3): that the inner measure and the outer measure agree. That is:

This is very close to Caratheodory. There's some handwaving to do regarding all sets rather than just the I mentioned here
That's pretty nice, and I get the general idea - that's a fairly convincing argument for how Caratheodory (or Lebesgue or whoever) came up with the criterion. However, I'm wondering about your handwaving. You are starting from:

where there is an "obvious" concept of inner measure between , as you point out. However, more generally, don't we want to start by considering:

in which case, the same criterion applied here seems to require us to consider the "inner measure" between , which doesn't seem to be (to my mind at least) a well-defined concept. For example, if we have:

then applying the same approach gives us:

which is content-free.

In addition, it's still not clear to me what this is really saying when itself is not a measurable set (and the Caratheodory criterion doesn't seem to require that it be one).

But, yes, it's certainly a very interesting starting point for further consideration.
0
5 years ago
#8
(Original post by atsruser)
That's pretty nice, and I get the general idea - that's a fairly convincing argument for how Caratheodory (or Lebesgue or whoever) came up with the criterion. However, I'm wondering about your handwaving. You are starting from:

where there is an "obvious" concept of inner measure between , as you point out. However, more generally, don't we want to start by considering:

in which case, the same criterion applied here seems to require us to consider the "inner measure" between , which doesn't seem to be (to my mind at least) a well-defined concept. For example, if we have:

then applying the same approach gives us:

which is content-free.

In addition, it's still not clear to me what this is really saying when itself is not a measurable set (and the Caratheodory criterion doesn't seem to require that it be one).

But, yes, it's certainly a very interesting starting point for further consideration.
For a measurable set, it's a nice extension. If you have measurable then by definition their intersection must be measurable, so you can apply the same argument to to see that it must be true for all measurable sets. Sometimes, as you said, it's content free, but there we go.

Non-measurable sets are typically horrendous (the only construction of one I know of involves the axiom of choice, and whilst it's a pretty idea it's a disgusting set) but I think the result for non-measurable sets follows from the one for measurable ones. Let with A non-measurable. We consider the case where the outer measure (a well defined concept for a non-measurable set) is finite so that we can re-arrange orders of summation. Then:

Where the infimum is taken over all sequences of measurable sets such that . Now, we show the reverse inequality. Let be a sequence of measurable sets such that:

Then:

As was arbitrary, we are done.

[Disclaimer: I cannot guarantee the above is free from mistakes]
0
#9
(Original post by DJMayes)
Now, we'd like condition (3): that the inner measure and the outer measure agree. That is:
I'm going to have to put this stuff to one side for this week, but this comment has made me think: can we, in fact, show that if we create the set of subsets whose inner measure and outer measure agree, then that is precisely the same set of subsets which satisfy the Caratheodory criterion i.e. that they are equivalent formulations?
0
#10
(Original post by DJMayes)
The most intuitive explanation I've found of the Caratheodory criterion relates it to our usual ideas of Riemann integration.

...

Full credit should go to http://mathoverflow.net/questions/34...-measurability where I first came upon this explanation, and it's worth reading here if what I've said isn't clear. Hope it helps!
I'm too short of time to follow this up properly at the moment, but I've also found this:

http://math.stackexchange.com/questi...measurable-set

and I've got hold of a book by Paul Loya on measure and integration (google may be your friend), which goes into the intuition and history of these ideas in some detail (maybe too much) and which I intend to read through over the next couple of months.

As far as I can tell, the idea of restricting measurable sets to those which "split up" other sets nicely so that inner measure and outer measure are the same originally came from Lebesgue in his early work on integration, and it was extended into a test on *all* sets by Caratheodory, though I'm not sure of the history of that.
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