Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measurable and Unif. Convergence in (a,b)

  1. Jul 26, 2012 #1

    Bacle2

    User Avatar
    Science Advisor

    Hi, All:

    If {f_n}:ℝ→ℝ are measurable and f_n-->f pointwise, then convergence is a.e. uniform. Are there any conditions we can add to have f_n-->f in some open interval (a,b)?

    Correction: convergence happens in some subset of finite measure; otherwise above not true.
     
    Last edited: Jul 26, 2012
  2. jcsd
  3. Jul 26, 2012 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Your original statement is incorrect. Example:

    fn(x) = n2x for 0≤x≤1/n
    ...... = n - n2(x-1/n) for 1/n<x<2/n
    ...... = 0 otherwise.

    fn(x) -> 0 pointwise, but is certainly not uniform, although it will be in any interval excluding [0,ε] for any ε.
     
  4. Jul 26, 2012 #3

    Bacle2

    User Avatar
    Science Advisor

    Yes, I was careless. I should have said: for any ε>0 , fn(x)→0 uniformly outside of a set of measure ε. Find Nε with 1/Nε<e. Then fn(x)→0 outside of [0,1/N).
     
    Last edited: Jul 26, 2012
  5. Jul 27, 2012 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The statement is confusing. What is the assumption and what are you trying to prove?
     
  6. Jul 27, 2012 #5

    Bacle2

    User Avatar
    Science Advisor

    The assumptions are the same: a sequence of measurable functions converge on a bounded subset of the real line. It follows, e.g., by Egorof's , that f_n-->f a.e. uniformly.

    I am wondering if we can add some conditions , either on the functions, or on the domain/range of the functions of the sequence so that the functions converge uniformly _in some interval (a,b)_ , instead of just a.e., uniformly. Basically, we can remove a set of measure zero and get a 1st category space with empty interior.
     
  7. Jul 27, 2012 #6

    Bacle2

    User Avatar
    Science Advisor

    Never mind, thanks; I found the condition I needed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Measurable and Unif. Convergence in (a,b)
  1. Uniform Convergence (Replies: 2)

  2. Sequence Convergence (Replies: 4)

Loading...