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Sequence of integrable functions on [0,1]

  1. Aug 9, 2010 #1
    If you have a sequence of integrable functions [itex]\{f_n(x)\}[/itex] on [itex][0,1][/itex] which converges to a function [itex]f(x)[/itex] pointwise for every [itex]x\in [0,1][/itex] that has the following properties:

    (1) [itex]0 \leq f_n(x) \leq f(x)[/itex] for every n and every x; and

    (2) [itex]\int_0^1 f_n(x)dx = 1[/itex] for every n;

    does it necessarily follow that the limit function [itex]f[/itex] is integrable and satisfies [itex]\int_0^1 f(x) dx = 1[/itex]?

    I can't think of why this would need to be true using the standard Lebesgue convergence theorems (bounded, monotone, or dominated), since none of them seem to apply. But I can't think of a counterexample to save the life of me. Can anyone help?
    Last edited: Aug 9, 2010
  2. jcsd
  3. Aug 9, 2010 #2


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    Draw an isoceles at 0.5 with width [tex]l_n = 2^{-n}[/tex], and height [tex]h_n= 2^{n+1}[/tex] for each n. Let [tex]f_n(x)[/tex] be the curve of the triangle where it occurs except at 0.5, and 0 elsewhere. [tex]f_n(x)[/tex] is surely lebesgue integrable (but not continuous at 0.5), and:
    [tex]\int^{1}_0 f_n(x) dx = \int^{\frac{1}{2}}_0 f_n(x) dx + \int^{1}_{\frac{1}{2}} f_n(x) dx =\frac{1}{2}l_nh_n=2^{-1-n+n+1}=1[/tex]
    which is obviously true because a continuous function need not be continuous at it's end points to be integrable.

    However, this function converges pointwise to f(x) = 0. This is obvious: any point on [0,1] will surely be valued to 0 by [tex]f_n[/tex] for all n > N for some N (you can find this N explicitly for each such point if you want by a simple calculation).
    And since [tex]f_(1/2) = 0[/tex] for all n, [tex]f_n(x) \to 0[/tex] for all x as [tex]n \to \infty[/tex].

    And surely; [tex]\int^1_0 f(x) dx \not = 1[/tex].

    EDIT: Oh, I mistakingly read the first equation to be [tex]0 \leq f(x) \leq f_n(x)[/tex]. Nevermind then, but I will try to find a solution to this problem as well.
    Last edited: Aug 9, 2010
  4. Aug 10, 2010 #3
    Let [itex]g_n(x) = \inf_{k \ge n} f_k(x)[/itex] and [itex]h_n(x) = \sup_{k \le n} f_k(x)[/itex]. Then [itex]g_n(x) \le f_n(x) \le h_n(x)[/itex], and [itex](g_n), (h_n)[/itex] are both increasing sequences, and both converge pointwise to [itex]f[/itex]. By monotone convergence, [itex]1 = \lim \int f_n \le \lim \int h_n = \int f = \lim \int g_n \le \lim \int f_n = 1[/itex].

    I think it works.

    edit: The statement also follows immediately and more nicely from Fatou's lemma, so check that out.
    Last edited: Aug 10, 2010
  5. Aug 10, 2010 #4
    Thanks for the response! I know Fatou's lemma states [itex]\int \liminf f_n \leq \liminf \int f_n[/itex], and that in this case, we have [itex]\liminf f_n = f[/itex] and [itex]\liminf \int f_n = 1[/itex], which gives [itex]\int f \leq 1 < \infty[/itex], implying [itex]f[/itex] is integrable. Does the other inequality follow from monotonicity of the integral: [itex]1 = \int f_n \leq \int f[/itex]?
    Last edited: Aug 10, 2010
  6. Aug 10, 2010 #5
    It sure does.
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