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Want to prove that limit of integrals diverges

  1. Feb 26, 2010 #1
    Suppose we know that [itex]f_n\to f[/itex] when [itex]n\to\infty[/itex] (in some sense), and that

    [tex]
    \int f(x)dx
    [/tex]

    diverges. Can it be deduced that

    [tex]
    \lim_{n\to\infty} \int f_n(x)dx
    [/tex]

    diverges too?
     
  2. jcsd
  3. Feb 26, 2010 #2
    If the convergence of the sequence is uniform on the interval of integration then yes
     
  4. Feb 26, 2010 #3
    I see now that this seems to follow from the Fatou's lemma

    I thought first it wouldn't be relevant, because I wanted to consider complex functions, but if [itex]f[/itex] is complex, and the integral diverges, then it means that at least one positive or negative part, of real or imaginary part, diverges.

    Then, for example if the [itex](\textrm{Re}(f))^+[/itex] diverges, then

    [tex]
    \lim_{n\to\infty} \int \textrm{Re}(f_n(x))^+ dx = \underset{n\to\infty}{\textrm{lim inf}} \int \textrm{Re}(f_n(x))^+ dx \geq \int \textrm{Re}(f(x))^+ dx = \infty
    [/tex]

    I guess that's it.
     
  5. Feb 26, 2010 #4
    But I don't think that my original problem would be solved yet, because my original problem looks more like this:

    Suppose [itex]X[/itex] is some measure space such that [itex]\mu(X)=\infty[/itex] and [itex]X_1\subset X_2\subset\cdots[/itex] is some subset sequence such that [itex]\mu(X_k)<\infty[/itex] for all k and such that

    [tex]
    \bigcup_{k=1}^{\infty} X_k = X
    [/tex]

    No I know that point wisely [itex]f_n(x)\to f(x)[/itex], and that

    [tex]
    \int\limits_X |f(x)| d\mu(x) =\infty
    [/tex]

    and I want to prove that

    [tex]
    \lim_{n\to\infty} \lim_{k\to\infty} \int\limits_{X_k} f_n(x) d\mu(x)
    [/tex]

    diverges.

    So is it certain, that the integrals of [itex]f_n[/itex] could not start converging for cancellation reasons?
     
  6. Feb 26, 2010 #5
    I've been "writing while thinking" so it could be that my posts are little confused.
     
  7. Feb 26, 2010 #6

    quasar987

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    Science Advisor
    Homework Helper
    Gold Member

    First, since for each n, the map

    [tex]\nu(A)=\int_Af_nd\mu[/tex]

    is a measure, we have by continuity, that

    [tex]
    \lim_{n\to\infty} \lim_{k\to\infty} \int\limits_{X_k} f_n(x) d\mu(x)=\lim_{n\to\infty}\int_Xf_nd\mu
    [/tex]

    So your question amounts to: is there a sequence of integrable functions f_n that converge pointwise to a non integrable function f but such that [itex]\lim_n\int_Xf_n[/itex] converges.

    The answer is provided by the old example of X=R, f(x)=sin(x)/x. It is fairly easy to show that [itex]\int_{\mathbb{R}}|f|=+\infty[/itex] by finding an appropriate sequence of step function bounded above by |f| and whose area makes up a diverging series.

    But it is known that the sequence of (integrable) functions

    [tex]f_n(x):=\chi_{[-n,n]}\frac{\sin(x)}{x}[/tex]

    (which converge pointwise to f) are such that

    [tex]\lim_{n\to\infty}\int_{\mathbb{R}}f_n=\pi[/tex]

    (more difficult).

    So, here is your counter-example.
     
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