1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    [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.

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted