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Measurable functions

  1. Oct 17, 2013 #1
    Let [itex]E[/itex] be of finite measure and let [itex]\{ f_{n} \} _{n \geq 1} : E \rightarrow \overline{\mathbb{R}}[/itex] measurable functions, finites almost everywhere in [itex]E[/itex] such that [itex]f_{n} \rightarrow_{n \to \infty}[/itex] f almost everywhere in [itex]E[/itex]. Prove that exists a sequence [itex](E_{i})_{i \geq 1}[/itex] of measurable sets of [itex]E[/itex] such that:
    1) [itex]| E- \displaystyle \bigcup_{i=1}^{\infty} E_{i} | = 0[/itex]
    2) For every [itex]i \geq 1[/itex], [itex]f \rightrightarrows_{n \to \infty}[/itex] in [itex]E_{i}[/itex]

    Notation:
    [itex]f_{n} \rightarrow_{n \to \infty} f[/itex] means "[itex]f_{n}[/itex] converges to [itex]f[/itex] as [itex]n \to \infty[/itex]"
    [itex]f \rightrightarrows_{n \to \infty}[/itex] means "[itex]f_{n}[/itex] converges uniformly to [itex]f[/itex] as [itex]n \to \infty[/itex]"
    I don't know if there are multiple common definitions, but here is the mine:
    A function [itex]f : \mathbb{R}^{n} \rightarrow \mathbb{R}[/itex] is measurable if for all [itex]a \in \mathbb{R}[/itex]:
    [itex]\{ f > a \} = \{ x \in \mathbb{R}^{n} / f(x) > a \}[/itex] is measurable in the sense of Lebesgue.
    [itex]E \in R^{n}[/itex] is measurable in the Lebesgue sense if for every [itex]\varepsilon > 0 \exists U \in \mathcal{U} / E \in U \wedge |U-E|_{e} < \varepsilon[/itex]
    Where [itex]|[/itex] [itex]|_{e}[/itex] is the outer measure.


    Attemp/idea:
    I tried to divide [itex]E[/itex] in [itex]E_{k,n} = \{ x \in E / | f(x)-f_{n} | < \frac{1}{k} \}[/itex] but it will not help at all because something may "converge" similar in the first m terms but converge otherwise later.
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2
    Sorry, did you mistype this? I don't know what it means.
     
  4. Oct 17, 2013 #3
    This was a mistake, I meant measurable.
     
    Last edited: Oct 18, 2013
  5. Oct 18, 2013 #4
    In a previous exercise (I didn't realize it could be useful here, so I didn't post it) I had the following result:
    Let [itex]E[/itex] be a set of finite measure and [itex]\{ f_{k} \}_{k \in \mathbb{N}} : E \rightarrow \mathbb{R}[/itex] a sequence of measurable functions such that for every [itex]x \in E[/itex], exists [itex]M_{x} \in \mathbb{R}_{>0}[/itex]:
    [itex]| f_{k} (x) | \leq M_{x}, \forall k \in \mathbb{N}[/itex]
    Then for every [itex]\varepsilon > 0[/itex] exists [itex]F \subset E[/itex] closed and [itex]M \in \mathbb{R}_{>0}[/itex] such that:
    [itex]| E - F | < \varepsilon[/itex] and [itex]| f_{k} (x) | \leq M, \forall k \in \mathbb{N}, \forall x \in F[/itex]

    If there's a set where [itex]\forall M \in \mathbb{R}_{>0} \exists k \in \mathbb{N} / | f - f_{k} | > M[/itex], then this set has to have null measure, because either the functions doesn't converge here or they're not finite.

    Let's be [itex]E = H \cup Z[/itex] where [itex]\{ f_{k} \}_{k \in \mathbb{N}}[/itex] are finite and converges to [itex]f[/itex] in [itex]H[/itex] and [itex]| Z | = 0[/itex] ([itex]E-Z=H[/itex]) (They may be not finite in differents sets, but all are of zero measure and the countable union of sets of null measure has null measure).

    As [itex] H [/itex] is measurable, for every [itex] \varepsilon > 0[/itex] exists [itex] C [/itex] closed such that [itex] C \in H [/itex] and [itex] | H - C | < \varepsilon [/itex].

    I'll try with [itex]E_{k} = \{ x \in H / f_{n} (x) < k \forall n \in \mathbb{N}[/itex]

    Edit: I had a non-trivial mistake so I was trying to fix it but I failed and I have a lecture in 6 hours, so I'd better to sleep now.
     
    Last edited: Oct 18, 2013
  6. Oct 18, 2013 #5
    Getting late here -- will look at this tomorrow if someone doesn't help you sooner.
     
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