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Homework Help: Integrals going to 0 implies functions go to 0?

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    This question is not the assignment problem but I think that if the result I mention here is true, then my assignment problem will be solved.
    Let [itex](X,\Sigma,\mu)[/itex] be a measure space.
    Suppose that [itex] (h_n)_{n=1}^\infty[/itex] is a sequence of non-negative-real-valued integrable functions and that
    \lim_{n\to \infty}\int_Xh_nd\mu=0.
    Is it true that the functions [itex]h_n[/itex] converge pointwise to [itex]0[/itex] almost-everywhere?
    3. The attempt at a solution
    My intuition is that since the [itex]h_n[/itex] are non-negative functions, we can rule out the possibility of cancellation but I don't know what else to do and maybe the result is not even true anyway.

    I'd really appreciate any help!
  2. jcsd
  3. Nov 24, 2012 #2
  4. Nov 24, 2012 #3


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    Staff: Mentor

    Consider [0,1] with Lebesgue measure and:

    h0(x)=1 everywhere -> integral is 1

    h1(x)=1 for x<1/2, h1(x)=0 for x>=1/2 -> integral is 1/2
    h2(x)=1 for x>1/2, h2(x)=0 for x=<1/2 -> integral is 1/2

    h3(x)=1 for x<1/3, h3(x)=0 for x>=1/3 -> integral is 1/3
    h4(x)=1 for 1/3<x<2/3, h4(x)=0 else -> integral is 1/3
    h5(x)=1 for x>2/3, h5(x)=0 for x<=1/3 -> integral is 1/3

    and so on. As you can see, the integral converges to 0, but hn does not converge pointwise anywhere.

    Edit: Oh, you were quicker. Well here is an explicit example that it is wrong.
  5. Nov 24, 2012 #4


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    Take a function whose graph looks like a rectangle. e.g.

    [tex]f(x) = \begin{cases} 1 & x \in [0,1] \\ 0 & x \notin [0,1] \end{cases} [/tex]

    If you make a sequence of functions that just drags the rectangle back and forth, you get a sequence of functions whose integrals are all 1, but doesn't converge pointwise to 0 in any part of the real line that the rectangle moves through infinitely often.

    Now, shrink the width of the rectangle as you drag it back and forth. You've now added in the extra feature that the integrals decrease in size: you can make it decrease to 0 if you want. But your sequence still doesn't converge pointwise to 0 in the regions that the rectangle moves through.

    If you also increase the size of the interval through which the square moves back and forth, you will have a sequence of functions that doesn't converge pointwise anywhere.

    For added fun, you can make the height of the rectangle grow without bound (but slow enough that it's area still decreases to 0), so that the sequence of functions isn't even bounded above, even though each individual function is bounded.
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