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Prove the typewriter sequence does not converge pointwise.

  1. Mar 24, 2013 #1
    (Typewriter sequence) Consider the following sequence of functions on [0,1]. Let f1= X[0, 1\2], f2= X[1\2, 1], f3= X[0, 1\4], f4= X[1/4, 1\2], f5= X[1\2, 3/4], f6= X[3/4, 1], f7= X[0, 1\8], etc. Where X is the Characteristic function.

    (a) Prove that fn does not converge for any x in [0,1].

    (b) Show that this sequence of functions is a counterexample to the statement:
    If for all n, fn and f are non-negative functions on E (with λ(E)<∞) and for all A contained in E, ∫A[\sub]fn dλ, then fn converges to f Lebesgue almost everywhere.



    Attempt at solution: assume that lim n→∞fn(y)=0, which means that for all ϵ>0 there is an N, such that for all n > N, fn(y)<ϵ.
     
    Last edited: Mar 24, 2013
  2. jcsd
  3. Mar 24, 2013 #2

    Dick

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    That's not much of an attempt. Can you explain in words why fn does not converge for any x in [0,1]?
     
  4. Mar 24, 2013 #3
    I suggest drawing the graph for each of the functions and you should be able to see what is happening and why fn(x) does not converge for any x.
     
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