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Sandwich theorem for lebesgue integral

  1. Dec 3, 2008 #1
    It would be helpful if there is a sandwich theorem for lebesgue integreal. Does it exist?
    Ie. if fn<=hn<=gn on measure space (X,S,u) and fn and hn are integreable and measurable , then
    g must also be in L1.

    A related claim is that if fn-> f and hn->h, does gn also necessarily converge to some g?
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 3, 2008 #2


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    Did you mean to ask if h_n (not g) is in L^1? What do you know about h_n - in particular, is it measurable? If it's just an arbitrary function then it's easy to come up with examples of nonmeasurable functions sandwiched between L^1 functions...

    On the other hand, if h_n is measurable, then it is L^1 by the monotonicity of the integral (you have to be a little careful here).

    And when you write f_n -> f, what notion of convergence are you using?

    For future reference, you should always post your ideas when you ask for help. Forum rules and such.
  4. Dec 4, 2008 #3
    Sorry , you are right.
    Ideally, if fn, gn, and hn are all nonnegative, we can use comparison theorem to show that hn is in L1. But how about the case that fn, gn, hn are simply any measurable functions?
  5. Dec 4, 2008 #4


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    Any measurable function.... is it integrable? If we have fn<=gn<=hn with fn and hn integrable, then |fn| and |hn| are integrable as well, and hence max(|fn|,|hn|) is also. What can you say about |gn| compared to this?
  6. Dec 5, 2008 #5
    The question should include fn, hn, f,h integreable
    then max(|fn|,|hn|) is also integreable,
    then |gn|<x(|fn|,|hn|), thus gn is in L1,
    But still it doesn't say "integral gn" converges to "integral g" for some g
  7. Dec 5, 2008 #6


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    That last bit follows from the dominated convergence theorem.
  8. Dec 5, 2008 #7
    To use DCT, we need to show there exists a g such that g->gn
    but this is true, since |gn|<|fn|+|hn|<|f|+|h|+2epsilon
    thus |gn| converges to |f|+|h|
    Now, suppose the space is complete, then every absolutely convergent sequence is convergent, thus there exists a g such that gn->g
    is this right?
  9. Dec 5, 2008 #8


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    Hold on - no. If you don't know that g_n converges in the first place, then you can't say anything.

    On the other hand, we do have the following result: Let {f_n} and {h_n} be sequences of integrable functions that converge a.e. to f and h (which are also integrable functions). Let {g_n} be a sequence of measurable functions such that f_n <= g_n <= h_n for all n, and such that g_n -> g pointwise. If [itex]\int f_n \to \int f[/itex] and [itex]\int h_n \to \int h[/itex], then [itex]\int g_n \to \int g[/itex].

    I don't know off the top of my head if any of the conditions can be relaxed on this. (In fact, they might even require some strengthening!)
  10. Dec 5, 2008 #9
    Wonderful, How did you prove this? if too long, what reference book did you find this theorem?
  11. Dec 5, 2008 #10


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    I did an exercise of this sort back when I took measure theory (so I might have gotten some of the details wrong). If I'm not mistaken, this might also be an exercise in Royden (in the chapter that contains the Lebesgue dominated convergence theorem for the real line; chapter 3?). Anyway, the idea is to use the dominated convergence theorem. It might also be helpful to consider the case f_n=-g_n (so |h_n| <= f_n) first.
  12. Dec 5, 2008 #11
    why do we need gn->g pointwise
    thus |gn| converges to |f|+|h|
    Last edited: Dec 5, 2008
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