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: Extension of Lebesgue Convergence Theorem

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    This comes courtesy of Royden, problem 4.14.
    a.Show that under the hypothesis of theorem 17 we have [tex]\int |fn-f| \rightarrow 0[/tex]

    b.Let <fn> be a sequence of integrable functions such that [tex]fn \rightarrow f a.e.[/tex] with f integrable. Then [tex]\int |fn-f| \rightarrow 0[/tex] if and only if [tex] \int |fn| \rightarrow \int |f|[/tex]

    2. Relevant equations
    Theorem 17:
    Leg <gn> be a sequence of integrable functions which converges to an integrable function g. Let <fn> be a sequence of measurable functions such that [tex] |fn|\leq gn [/tex] and <fn> converges to f a.e. If
    [tex]\int g = lim \int gn[/tex]
    [tex]\int f = lim \int fn[/tex]

    3. The attempt at a solution
    My problem in part a is the same as the "if" in part b. The only if seems fairly trivial since |a-b|>=| |a| - |b| |. Unfortunately that is partially the problem. Every attempt I've made to go in the other direction reduces to that same inequality. My basic thought process far has been to start with
    [tex]\int |f| = lim \int |fn| [/tex] which we have by assumption, and then conclude that for some e>0 theres an N such that
    [tex]| \int |fN| - |f| |< e[/tex]

    Obviously this does not get me where I want to go. If I somehow had fn as an increasing function I coud keep going, but I see no real way to do this.
  2. jcsd
  3. Nov 8, 2008 #2
    After poking at it a bit more, if I could establish the same inequality that I'm arriving at on my last step for the portions of f and fn where f>=fn and where f <=fn then I could get it from there. (that is, split |f|-|fn| into positive and negative portions.)
  4. Nov 9, 2008 #3
    Figured it out. If we define Fn to be f if f<fn and fn if fn<f, then we have that |Fn|<|f|+|fn|, which we is integrable and the limit of it's integral is 2 times the limit of the integral of |f|.
    Since Fn converges a.e. to f, we then have that the integral of f is equal to the limit of the integral of Fn. Then we very quickly get that the integral of f-Fn is equal to the integral of |f-Fn| (since Fn is less than or equal to f, as well as fn) and for any n>N this is less than e. Then we also have that the limit of the integral of fn and Fn are the same, so the integral of fn-Fn converges, and is positive for every n, so the integral of |fn-Fn| is equal to it and less than e for n>m for some m. Then lastly we have that the integral of |f-fn|=integral of |f-Fn +Fn -fn| which is less than or equal to the integral of |f-Fn| + integral of |fn-Fn| which are less than e for each, so the whole thing is less than 2e, so it converges.

    Sorry if that wasn't clear but my brain is now fried and I'm going to bed.:smile::zzz:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook