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Proving f = 0 almost everywhere

  1. Apr 10, 2015 #1
    I am working on a problem##^{(1)}## in Measure & Integration (chapter on Product Measures) like this:

    Suppose that ##f## is real-valued and integrable with respect to 2-dimensional Lebesgue measure on ##[0, 1]^2## and also

    ##\int_{0}^{a} \int_{0}^{b} f(x, y) dy dx = 0##
    for all ##a, b \in [0, 1].## Show that ##f = 0## almost everywhere.
    Here are what I only know of this problem:

    (1) The mapping is ##f : [0, 1] \times [0, 1] \longrightarrow \Bbb R##

    (2) Since we need to prove ##f = 0## a.e., and borrowing from set builder of convergence a.e., I think I need to prove this:

    ##\mu (\{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\}) = 0.##

    Since this problem is under chapter on Product Measures involving iterated integral, I suspect I need to use the Fubini Theorem, but I do not know how to do it. Any hints or helps would be very much appreciated. Thank you very much for your time and effort.


    ______________________________________________________________________________

    (1) Richard F. Bass' http://homepages.uconn.edu/~rib02005/real-analysis-04nov2011.pdf [Broken] 2nd. edition, chapter 11: Product Measure, Exercise 11.6, page 88.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 10, 2015 #2

    RUber

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    Assume that f is non-zero on a set that is not measure zero. Can you have the integral be zero? (yes) However, you can partition that set into the set where f is positive and the set where f is negative. It is impossible for the integral to be zero over both sets.
    Apply that same rationale to your argument with the limits of integration a and b.
     
  4. Apr 10, 2015 #3
    Thank you to RUber for your speedy response. Allow me though to ask you one question:

    Am I in the right direction on (2)? In other words, are you giving me the hints above within the framework of ##\mu (\{(x, y) \in [0, 1]^2 \mid f(x, y) \neq 0 \}) = 0##?

    Thank you again for your time and effort.
     
  5. Apr 10, 2015 #4

    RUber

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    That's right. I propose proof by contradiction.
     
  6. Apr 10, 2015 #5

    WWGD

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    To add a bit to RUber's idea, illustrating, assume you have a small square s * (with sides parallel to the x-axis an y-axis respectfully) with measure>0 ,where ## f \neq 0 ## , and so that ##f == 1## in the left half and ## f == -1## on the right half. Remember that for simple functions f with values ##c_i##: ,## \int f = \int (\Sigma c_i \ chi_{c_i} ) = \Sigma \ c_i m( x: f(x)=c_i) ## . Now remember that the integral must be zero over _all_ partitions.

    * Just to illustrate; obviously this set does not have to be a square; it doesn't even have to contain a square of measure >0..
     
    Last edited: Apr 10, 2015
  7. Apr 11, 2015 #6
    Here are what I managed to come up after feedbacks from you two, but I am still lost after the first two steps:

    (1) By way of contradiction, assume that ##f = 0## a.e. is not true, meaning that ##\mu (\{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\}) \neq 0.##
    (2) Let ##A = \{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\} ##, that is ##f(A) \neq 0##
    (3) ...
    (4) ...

    Any further hints would be very much appreciated. Thank you for your time.
     
  8. Apr 11, 2015 #7

    WWGD

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    Basically, look for a rectangle ## a \times b ## , for some specific values of ##a,b ## that will contain a subset of the set of measure non-zero where the positive part does not cancel out the negative part in the integral. You can use a rectangle for S and the characteristic function as a motivating example (ultimately using simple functions to approximate your measurable function may be necessary.) so that the rectangle ## [0,a] \times [0,b] ## intersects S at S' in such a way that ## \int_{S'} f \neq 0 ## , following up on RUber's idea. Remember that the integral of f is assumed to be 0 over _all_ rectangles ##[0,a] \times [0,b] ; 0<a,b<1 ##.
     
  9. Apr 11, 2015 #8

    Dick

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    Did you people, WWGD and RUber actually solve this problem? I think it's a bit harder than you are thinking. You (A.Magnus) should easily be able to show given the premises that the integral over ANY rectangle in the domain is zero. This isn't a Riemann integral, it's a Lebesgue integral. You could imagine that you would start it a function that's 1 over half the domain and -1 over half the domain and then somehow scramble it up so EVERY rectangle contains a equal amount of 1 and -1. That's what you need to show is impossible. I think you need something like http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem or http://en.wikipedia.org/wiki/Lebesgue's_density_theorem.
     
    Last edited: Apr 12, 2015
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