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Continuity on L^1 spaces

  1. Sep 18, 2006 #1
    Question:

    Prove that if [itex]f \in L^1(\mathbb{R},\mathcal{B},m)[/itex] and [itex]a \in \mathbb{R}[/itex] is fixed, then [itex]F(x):=\int_{[a,x]}f\mbox{d}m[/itex] is continuous. Where [itex]\mathcal{B}[/itex] is the Borel [itex]\sigma[/itex]-algebra, and [itex]m[/itex] is a measure.
     
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  3. Sep 18, 2006 #2
    I was hoping to use the following definition:

    A function [itex]f[/itex] is continuous if for any sequence [itex]x_n[/itex] such that

    [tex]x_n \rightarrow x[/tex]

    then

    [tex]F(x_n) \rightarrow F(x)[/tex]

    Does this sound like the right approach?
     
  4. Sep 18, 2006 #3

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    Maybe. Can you show that [itex]\int_a^b+\int_b^c=\int_a^c[/itex], and then that as xn->x, [itex]\int_{x_n}^x f dm[/itex]->0?
     
  5. Sep 18, 2006 #4
    Im not sure that would help. But what do I know!? :uhh:

    I was thinking that to show that F(x) was continuous I would do something like this:

    1) Fix [itex]a \in \mathbb{R}[/itex] and let [itex]x_n[/itex] be a sequence that converges to x as n approaches infinity. You know, all the regular proof setting up stuff.

    2) Use the Dominated Convergence Theorem to show that the sequence [itex]F(x_n)[/itex] converges, and finally get something like

    3) [tex]\int_{[a,x]}f_n\mbox{d}m = \lim_{n\rightarrow\infty}\int_{[a,x]}f_n\mbox{d}m[/tex]

    Hence showing that

    [tex]F(x_n) \rightarrow_{n\rightarrow\infty} F(x)[/tex]

    So basically I think using the D.C.T. is essential here. What does anyone think of this method?
     
    Last edited: Sep 18, 2006
  6. Sep 18, 2006 #5

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    I'm guessing the f_n are functions that are equal to f everywhere except in (x_n,x), where they are 0. So you need to show that the integral from [a,x_n] of f is equal to the integral over [a,x] of f_n, and that the integrals of the f_n converges to the integral of their limit, f, which you can do using the dominated convergence theorem, bounding the |f_n| above by |f|. Sounds good. My suggestion was just to show that the error (the integral over [x_n,x] of f) goes to zero as x_n goes to x.
     
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