Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    11
    ReputationRep:
    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:

    1. Start with some X \subset \mathbb{R} and consider its power set P(X)
    2. Find an outer measure \mu^* on P(X)
    3. Filter P(X) through the Caratheodory criterion to generate M \subset P(X)
    4. Now:

    a) M is automatically a \sigma-algebra, and
    b) \mu^* restricted to M 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 \sigma-algebra of measurable sets, then the Caratheodory criterion naturally pops out?
    Offline

    9
    I've always seen it formulated like this:

    Let \mu^* be an outer measure of a set X, and let A \subset X be a subset. If for every subset E \subset X, we have
    \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c),
    then A is a \sigma-algebra and the restriction of \mu^* to A is a measure.

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

    There's another equivalent criterion too. For every \varepsilon > 0, there exists an open set O \subset X, such that \mu^*(O \cap A^c) < \varepsilon.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Alex:)
    I've always seen it formulated like this:

    Let \mu^* be an outer measure of a set X, and let A \subset X be a subset. If for every subset E \subset X, we have
    \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c),
    then A is a \sigma-algebra and the restriction of \mu^* to A is a measure.

    Apparently, the intuition is that you have to test every subset E \subset X, rather than just X, 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.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (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 E measurable. Then note that, for all A:

    (A \cap E) \cap (A \cap E^c) = \emptyset i.e. those sets are disjoint
    A = (A \cap E) \cup (A \cap E^c) i.e. they exhaust A

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

    \mu^* (A) = \mu^* ( (A \cap E) \cup (A \cap E^c) = \mu^*(A \cap E)+\mu^*(A \cap E^c)

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

    One point though: we haven't assumed that A is itself measurable; does that matter?
    Offline

    16
    ReputationRep:
    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 \mathbb{R} 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  \mu^* 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  \mu_*? 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  E \subset X , we can define it as:

     \mu_*(E) = \mu^*(X) - \mu^*(E^c)

    Written in a form that's perhaps more familiar:

     \mu_*(E \cap X) =\mu^*(X) - \mu^*(E^c \cap X)

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

    \mu^*(E \cap X) =\mu^*(X) - \mu^*(E^c \cap X) \Rightarrow \mu^*(E \cap X)+\mu^*(E^c \cap X) =\mu^*(X)

    This is very close to Caratheodory. There's some handwaving to do regarding all sets rather than just the X 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!
    Offline

    9
    (Original post by atsruser)
    Continuing this line of thought: suppose we want to have E measurable. Then note that, for all A:

    (A \cap E) \cap (A \cap E^c) = \emptyset i.e. those sets are disjoint
    A = (A \cap E) \cup (A \cap E^c) i.e. they exhaust A

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

    \mu^* (A) = \mu^* ( (A \cap E) \cup (A \cap E^c) = \mu^*(A \cap E)+\mu^*(A \cap E^c)

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

    One point though: we haven't assumed that A is itself measurable; does that matter?
    The set A is called Carathéodory measurable if it satisfies the condition.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (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  E \subset X , we can define it as:

     \mu_*(E) = \mu^*(X) - \mu^*(E^c)

    Written in a form that's perhaps more familiar:

     \mu_*(E \cap X) =\mu^*(X) - \mu^*(E^c \cap X)

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

    \mu^*(E \cap X) =\mu^*(X) - \mu^*(E^c \cap X) \Rightarrow \mu^*(E \cap X)+\mu^*(E^c \cap X) =\mu^*(X)

    This is very close to Caratheodory. There's some handwaving to do regarding all sets rather than just the X 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:

    E \subset X

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

    A,E \subset X

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

    A,E \subset X, A \cap E = \emptyset

    then applying the same approach gives us:

    \mu^*(A) = \mu^*(E \cap A)+\mu^*(E^c \cap A) = 0+\mu^*(A) = \mu^*(A)

    which is content-free.

    In addition, it's still not clear to me what this is really saying when A 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.
    Offline

    16
    ReputationRep:
    (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:

    E \subset X

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

    A,E \subset X

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

    A,E \subset X, A \cap E = \emptyset

    then applying the same approach gives us:

    \mu^*(A) = \mu^*(E \cap A)+\mu^*(E^c \cap A) = 0+\mu^*(A) = \mu^*(A)

    which is content-free.

    In addition, it's still not clear to me what this is really saying when A 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  A, E \subset X measurable then by definition their intersection must be measurable, so you can apply the same argument to  A \cap E \subset A 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  A, E \subset X 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:

     \mu^*(A) = \inf ( \sum_n \mu(B_n)) = \inf ( \sum_n \mu(B_n \cap E)+ \sum_n \mu(B_n \cap E^c))

     \geq \inf ( \sum_n \mu(B_n \cap E))+\inf ( \sum_n \mu(B_n \cap E^c)) = \mu^*(A \cap E)+\mu^*(A \cap E^c)

    Where the infimum is taken over all sequences of measurable sets  (B_n) such that  A \subset \cup_n B_n . Now, we show the reverse inequality. Let  (B_n) be a sequence of measurable sets such that:

     \mu^*(A) +\epsilon \geq \sum_n \mu^*(B_n)

    Then:

     \mu^*(A) \leq \sum_n \mu^*(B_n)

     = \sum_n \mu^*(B_n \cap E) + \sum_n \mu^*(B_n \cap E^c) \leq  \mu^*(A \cap E)+\mu^*(A \cap E^c)+ 2\epsilon

    As  \epsilon was arbitrary, we are done.

    [Disclaimer: I cannot guarantee the above is free from mistakes]
    • Thread Starter
    Offline

    11
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    11
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.