1. Not finding help here? Sign up for a free 30min 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!

Convergence of lebesgue integral

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]fn(x)=n\chi_{(0,n^{-1})}(x)-n\chi_{(-n^{-1},0)}(x)[/tex] -1<x<1
    Prove that [tex]\int fn(x)dx\rightarrow\int f(x)[/tex]
    a. Using direct computation
    b. Use dominated convergence



    3. The attempt at a solution
    a. How to do direct computation?????

    b. Using dominated convergence
    If x is in (0,1/n), then fn(x)=<1-0=1
    If x is in (-1/n,0), then fn(x)=<0-1=-1
    If x is in neither , then fn(x)=0
    Thus |fn(x)| is dominated by |g| where g(x)=1 , a constant function on (-1,1)
    g(x) is in L1, since g(x) is Riemann integreable on (-1,1) it must also be Lebesgue integreable.
    Thus, by dominated convergence theorem:
    "limit of integration of fn" is equal to "integration of limit of fn"= integration of f(x)=0 as n goes to +infinity.
    Thus,

    [tex]\int fn(x)dx\rightarrow\int f(x)\rightarrow0 [/tex] QED
     
  2. jcsd
  3. Dec 3, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    [itex]\int f_n[/itex] isn't that hard to comptue... What have you tried?
     
  4. Dec 3, 2008 #3
    Can we say that
    [tex]\int fn(x)=\int n\chi_{(0,n^{-1})}(x) +\int -n\chi_{(-n^{-1},0)}(x)[/tex] -1<x<1 (**)
    I think the above is the crucial step. What exactly do we need for the above to hold? I guess as long as each term on the right is integreable?

    Assume (**) holds, then let u be the lebesgue measure
    [tex]\int n\chi_{(0,n^{-1})}(x) +\int -n\chi_{(-n^{-1},0)}(x)=n*u(0,n^{-1})-n*u(-n^{-1},0)=n*1/n-n*(1/n)=0[/tex]

    Is my answer to part (a) posted here , together with my answer to part (b) posted in the first post,both completely correct? =)
     
    Last edited: Dec 3, 2008
  5. Dec 5, 2008 #4
    Can someone please tell me if what i did is right?
     
  6. Dec 5, 2008 #5

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Yup, that's good.

    Your answer for part (b) is incorrect however. f_n isn't dominated by 1. Try to sketch f_1, f_2, and f_3 to get a feel of what's going on.
     
  7. Dec 5, 2008 #6
    If x is in (0,1/n), then fn(x)<n*1/n-0=1
    If x is in (-1/n,0), then fn(x)>0-n*1/n=-1
    If x is in neither , then fn(x)=0
    Thus |fn(x)| is dominated by |g| where g(x)=1, I really can't see what's wrong?
     
  8. Dec 5, 2008 #7

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    No, if x is in (0,1/n), then f_n(x)=n; and if x is in (-1/n,0), then f_n(x)=-n.
     
  9. Dec 5, 2008 #8
    ah!
    so it's dominated by |n|, which is not in L1, so we can;t use DCT at all
     
  10. Dec 5, 2008 #9

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    What do you mean?

    What you want to do is find a nice integrable function g on [-1,1] that dominates all the f_n's. It doesn't look like you've tried to do this.
     
  11. Dec 5, 2008 #10
    if x is in (0,1/n), then f_n(x)=n; and if x is in (-1/n,0), then f_n(x)=-n., and if x is in neither, f_n(x)=0
    thus, |f_n(x)|<|n| for all x on (-1,1) , but|n| is unbounded as n goes to infinity
    we cannot find a g(x) in L1 such that |fn|<|g| for all n and all x in (-1,1).
     
  12. Dec 5, 2008 #11

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    That doesn't mean you can't fit a g in between f_n and n. I don't immediately see a good candidate though; and I'm willing to guess that you're right, dct probably won't work. You can use a generalized version of the dct (say, the one mentioned in your other thread), but that's kind of stupid though, because it amounts to basically doing what you did for part (a).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Convergence of lebesgue integral
  1. Lebesgue integration (Replies: 5)

  2. Lebesgue integral (Replies: 0)

  3. Lebesgue integral (Replies: 2)

Loading...