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Preimage of a Lebesgue measurable set under a Lebesgue measurable function.

  1. Nov 15, 2008 #1
    I've been doing a little work with Borel measures and don't want to confuse Borel measurable functions with Lebesgue measurable functions for R^n -> R^m.

    I'm, of course, familiar with the definition that a function f:R->R is Lebesgue measurable if the preimage of intervals/open sets/closed sets/Borel sets is Lebesgue measurable.

    We also say a function between general measurable X, Y spaces is measurable if the preimage of a set in the sigma algebra corresponding to Y is in the sigma algebra corresponding to X.

    For a Lebesgue measurable function f:R->R is it necessarily true that the preimage of a Lebesgue measurable set is Lebesgue measurable? I don't see that this need necessarily be the case.

    Am I correct in thinking that a Lebesgue measurable function f:R->R is a measurable function from R with the Lebesgue sigma algebra to R with the Borel sigma algebra (and NOT the Lebesgue sigma algebra)?

    I'd be very grateful if anyone could clear up my confusion.
     
  2. jcsd
  3. Nov 15, 2008 #2

    morphism

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    That's an excellent question. The answer is yes: the common definition of a "Lebesgue measurable function" from R to R is really a Lebesgue-to-Borel measurable function. There are examples of functions from R to R which are Lebesgue-to-Borel measurable, but which fail to pull back Lebesgue measurable sets to Lebesgue measurable sets. Here's an example:

    Let g:[0,1]->R denote the Cantor function, and extend it to all of R by defining g(x)=0 if x<0 and g(x)=1 if x>1. Then set h(x)=x+g(x). This function has a lot of peculiar properties. For our purposes, we only need the following two:
    1) h is a bijection, and
    2) if C is the cantor set, then the Lebesgue measure of h(C) is 1.

    It's a standard fact that a set of positive Lebesgue measure has a nonmeasurable subset. So let W be a nonmeasurable subset of h(C). The pullback V=h-1(W) is a subset of the Cantor set, hence has measure zero.

    Finally, let f=h-1. f is a Lebesgue-to-Borel measurable function. However, f-1(V)=h(V)=W is not Lebesgue measurable.
     
  4. Nov 16, 2008 #3
    Thanks a lot for the explanation and example! It's clear now.

    The link you gave mentions the usual calculus formula for arc length doesn't work even though the cantor function is a curve differentiable almost everywhere. Is it sufficient for the curve to be Lipschitz continuous in order for the usual formula (as a Lebesgue integral) to be valid or do I need it to be Lipschitz and injective?

    Thanks again.
     
  5. Nov 16, 2008 #4

    morphism

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    To be honest I'm not very familiar with how the arc length formula carries over to the Lebesgue integral. I would guess that we need some sort of absolute continuity (which the Lipschitz condition guarantees). So you probably don't need injectivity.
     
  6. Nov 16, 2008 #5

    morphism

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    I want to make an addendum to post #2. Property (1) should have read "h is a homeomorphism (i.e. a continuous bijection with a continuous inverse)" instead of just "h is a bijection". This makes the fact that f is measurable transparent.

    And here's a fun problem for you to think about: if f:R->R is measurable and invertible, is its inverse necessarily measurable?
     
  7. Nov 17, 2008 #6
    Surely not? In your example f is measurable but but f^-1 isn't :D

    Thanks again for your help :)
     
  8. Nov 17, 2008 #7

    gel

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    I think its worth mentioning that in that example f is Lebesgue-to-Borel (and Borel-to-Borel) measurable, as is its inverse, and in fact any continuous function. It's just that f is not Lebesgue-to-Lebesgue (or Borel-to-Lebesgue) measurable.

    However I think, h=f-1 will be Lebesgue-to-Lebesgue measurable, but not Borel-to-Lebesgue measurable (exercise!).
     
    Last edited: Nov 17, 2008
  9. Nov 17, 2008 #8

    gel

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    I think thats an interesting question, but which of the different definitions of measurability are you refering to?
    Borel-Borel, Lebesgue-Borel, Lebesgue-Lebesgue. All gets a bit confusing after a while!

    Edit: if you mean Borel-Borel then the answer is yes, but the proof I know is a rather complicated one involving analytic subsets of Polish spaces, and uses the result that a set is Borel if and only if both it and its complement is analytic.
     
  10. Nov 17, 2008 #9

    morphism

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    It's kind of ironic that I didn't mention what type of measurability I was referring to, isn't it? I meant Lebesgue-Borel. And as you noted, f won't do as a counterexample.
     
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