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Product of non measurable sets

  1. Mar 28, 2007 #1
    This is non measurable right?
     
  2. jcsd
  3. Mar 29, 2007 #2

    mathman

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    What do you define as "product"? In set theory, the only defined operations are union, intersection, and complement.
     
  4. Mar 29, 2007 #3
    finite cartesian product: http://en.wikipedia.org/wiki/Cartesian_product
    And if yes, what about countable products or arbitrary infinite products?
     
  5. Mar 29, 2007 #4

    matt grime

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    So your question is:

    if A and B are sets in two measure spaces S and T, and if we place the product measure on SxT, then A,B not measurable implies AxB is not measurable.

    That seems eminently true. If AxB can be expressed as a 'word' in the base of the measurable sets on SxT, then the projection into the first variable implies A is measurable in S, surely.
     
  6. Mar 29, 2007 #5
    ok im not so familiar with the formal definition of a measure space and the product measure. I was think more of something like this: given a finite number of non measurable sets in R (S1,...,Sn), their cartesian product is measurable ( By non measurable in R I mean the outer measure, m*S = inf{sum over k (|Ik|) : {Ik} is a countable covering of S by open intervals } is not equal to the inner measure (instead you take sup and closed intervals contained inside the set). Or an equivalent way of saying a set S in R is measurable is to say that for an X in R, m*X = m8(X intersect S) + m*(X intersect S complement) (so if theres no equality its nonmeasurable). In Rn you use rectangles instead of intervals. Im not sure what you mean by 'word'.
     
  7. Mar 30, 2007 #6

    matt grime

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    then how are you going to be able to tell if any answers you get even make sense?
     
  8. Mar 30, 2007 #7

    mathman

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    From the description you are giving of your question, it doesn't look like you mean cartesian product at all, but rather intersection or union - I can't make out which. Try to get your terminology straightened out.
     
  9. Mar 30, 2007 #8
    well from what i read in a more advanced textbook, is that a measure space is a set equipped with a measure function (with certain properties) that is defined on a sigma algebra (which defines the collection of measurable sets in the space). So ya, obviously, if you define the product measure in a way that only allows products of measurable sets to be measurable sets in the product measure, then by definition its true...By I was trying to see this from more of a different view, like using the definitions of lebesgue outer and inner measure, and also I am wondering if an infinite product of nonmeasurable sets would be non measurable in this way (product or box topology kind of product).
     
  10. Mar 30, 2007 #9

    AKG

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    Except in trivial cases, you cannot define the product measure to only allow products of measurable sets to be measurable, because the union of measurable sets must be measurable, and the union of two products is generally not itself a product, e.g. [0,1]x[0,1] U [1,2]x[1,2] is not a product.

    Lebesgue measure is a measure, it's just a particular one. The domain of the Lebesgue measure is the collection of sets whose inner and outer Lebesgue measures agree, and this collection is a [itex]\sigma[/itex]-algebra. If we let M denote this [itex]\sigma[/itex]-algebra, and let m denote Lebesgue measure, then the 2-dim. Lebesgue measure is denoted mxm, and it's domain is [itex]M\otimes M[/itex]. The definition of a [itex]\otimes[/itex] product of [itex]\sigma[/itex]-algebras can be looked up, but in the case we're dealing with, it should coincide with the collection of subsets of [itex]\mathbb{R}^2[/itex] whose inner and outer 2-dim. Lebesgue measures agree.

    So you're not looking at the problem from a different perspective, you're just asking the question for one particular measure, the Lebesgue measure. In this setting, we can move from the general notion of measurability as belonging to the [itex]\sigma[/itex]-algebra to the particular notion of measurability as having equal inner and outer measures, since these two notions should be equivalent in this particular case.

    Remind me how the inner and outer 2-dim. measures are defined again.

    matt grime

    Your claim isn't true. Let A be a non-measurable set, and let B = {0}. Then AxB is measurable (with measure 0) but clearly A is not. In fact, if your claim were true, then it would prove something even stronger than the desired result. Your claim appears to say that if AxB is measurable, then so is A, which is equivalent to "if A is non-measurable, then AxB is non-measurable" which is stronger than what was desired, namely that "if both A and B are non-measurable, then AxB is non-measurable". The problem is that the projection of an intersection is not necessarily equal to the intersection of the projections. The domain of the product measure is the [itex]\sigma[/itex]-algebra generated by products of measurable sets, so a measurable set in the product space is a "word" of products of measurable sets. The projection of a product of measurable sets is clearly a measurable set. But since projection doesn't commute with intersection, the projection of a word of products of measurable sets is not necessarily a word of projections of products of measurable sets.
     
  11. Mar 31, 2007 #10
    In R2 instead of intervals for the outer and inner measure, rectangles are used (higher dimensional rectangles are used for Rn, and for an arbitrary product of subsets of R we would use the basis elements of the product topology, i guess this is most natural)

    ok about your counterexample, B is measurable (a zero set) so this doesnt disprove that if both A and B are non measurable then AxB is nonmeasurable.
     
  12. Mar 31, 2007 #11

    AKG

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    Thanks. I'll have to think about this.
    Yes, I know, that part of my post was addressed to matt grime
     
  13. Apr 13, 2009 #12
    suppose A,B is nonmeasurable. Hence A,B both have nonzero outermeasure (note: outermeasure exists for every subset of R). Assume AxB is measurable. Let C(x)={y in R: (x,y) is in AxB}, C(x)=the empty set if x is not in A, C(x)=B if x is in A. Let D={x in R: C(x) is nonmeasurable}. By Cavalieri's principle mD=0. Hence A is not a subset of D (since if it were
    it would be a zero set itself). Hence there is an element t in A such that t is not in D, this
    emplies that C(t) is measurable, and since t is in A, we get C(t)=B, hence B is measurable.
    Contradiction. Hence AxB is nonmeasurable. A similar proof shows that BxA is nonmeasurable. Note: the same proof would work if one of the sets was measurable with
    nonzero measure and the other nonmeasurable.
     
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