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Limit of measble functions

  1. Sep 11, 2011 #1
    Hi,
    1)is this true? If f_n(x) -> f(x) pointwise, then
    {x:f(x)<=a} = union{k=1^infty}intersection{n=k^infty}{x:f_n(x)<=a}.
    2)if A is measurable set, subset of reals, then is A-const set measurable?
    Thanks
     
  2. jcsd
  3. Sep 12, 2011 #2
    I'm not getting your question. Could you make it more clear.
     
  4. Sep 12, 2011 #3

    micromass

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    For the first, you want to know if

    [tex]\{f\leq a\}=\liminf\{f_n\leq a\}[/tex]

    In other words, you want to know whether [itex]f\leq a[/itex] if and only if [itex]f_n\leq a[/itex] eventually.

    This is false, take [itex]f_n(x)=\frac{1}{n}x^2[/itex], then f(x)=0 and take a=0.
    Then [itex]\{f\leq 0\}=\mathbb{R}[/itex], but [itex]\liminf\{f_n\leq 0\}=\{0\}[/itex]

    As you do notice, you do have the [itex]\supseteq[/itex] inclusion.

    (b) is true. Let [itex]\mathbb{B}[/itex] be the Borel-sigma-algebra. Then we put

    [itex]\mathcal{A}=\{B-k~\vert~B\in \mathcal{B}\}[/itex]

    Try to prove that [itex]\mathcal{A}[/itex] is a sigma-algebra which contains the open intervals...
     
  5. Sep 12, 2011 #4
    Appreciate your help, you i "proved" my first claim. Could you help to find my error?
    So, if for given x [itex]f(x) \leq a[/itex], then for [itex]\forall \varepsilon>0, \forall x[/itex] [itex]\exists k[/itex] such that [itex]\forall n\geq k[/itex]


    [tex]f(x) - \varepsilon < f_n(x) < f(x) + \varepsilon \leq a + \varepsilon [/tex]
    Since [itex]\varepsilon [/itex] was arbitrary, we're done!
     
  6. Sep 12, 2011 #5

    micromass

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    The thing is that your n depends on [itex]\varepsilon[/itex]. So if you take [itex]\varepsilon[/itex] smaller, then n will be bigger. So the argument, "[itex]\varepsilon[/itex] is arbitrary" doesn't really work here.
     
  7. Sep 12, 2011 #6
    I see it now, thanks
     
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