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If h is L-integrable and f + g = h, then f and g also are

  1. Jun 20, 2006 #1
    A little question about the basics of Lebesgue integration. There is a theorem: if functions f and g are L-integrable then the function f+g is also L-integrable.

    This may be dumb, but I wish to know about the reciprocal lemma. Let h be a L-integrable function and f and g be functions such that h(x) = f(x) + g(x). Are f and g L-integrable? I suppose it is truth because I cant see how, being f a not L-integrable function, h could yet be a L-integrable one. But I cant' see the proof. Does someone got a minute to help? Thanks.
     
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  3. Jun 20, 2006 #2

    Hurkyl

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    Your problem is a common one -- you've locked yourself into one direction of thinking. You are trying to imagine functions f and g (at least one of which is not integrable), which make an integrable h.

    Why not think about it in a different direction? Try imagining h first, and then look at f and g.
     
  4. Jun 20, 2006 #3

    matt grime

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    No, of course, f and g need not be lebesgue integrable. This is a general phenomenon.

    If poperty P defined on something with addition and scalar multm and if P is additive and respects scalar mult, and if there are things without property P, then property P does not pass to summands.

    Proof:
    (deleted - as hurkyl implies, you should look for it yourself)

    as a hint, always try to add zero in a clever way
     
    Last edited: Jun 20, 2006
  5. Jun 20, 2006 #4

    mathman

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    How about f any non-L integrable function, g=-f?
     
  6. Jun 20, 2006 #5

    matt grime

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    That would be in some sense the extreme example. And it also embodies another important principal: wlog we can shift to the origin...
     
  7. Jun 20, 2006 #6
    Let f(x) be a L-integrable function, P be a fixed number and let g(x) such that

    g(x) = f(x) when f(x) belong to [-P, P]
    g(x) = P when f(x) > P.
    g(x) = -P when f(x) < - P.

    The fact is that f(x) is a L-integrable function and also we know that any constant function is L-integrable. Can we say that g(x) is L-integrable???
     
  8. Jun 21, 2006 #7

    matt grime

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    You can say that is Lebesgue integrable by inspection, and I imagine everyone would be happy with that, or at least they would have been if you hadn't indicated that you didn't see why it was, in the maths usage of the word, trivial (not the in belittling sense of the word).
     
  9. Jun 21, 2006 #8
    Can you give me some hint to prove that g is L-integrable?

    Note.- f, the L-integrable function, has bounded domain (let say [A,B]).

    Remember I am not saying that if x belongs to [-P,P] then g(x) = |f(x)|. In such case it would be obvious that g is L-integrable (because f, and therefore |f|, is L-integrable). I am saying that if f(x) belongs to [-P,P] then g(x) = |f(x)|. The set of x for which f(x) belongs to [-P,P] dont necesarily is an interval. How to prove that g is L-integrable?

    Same observation for the other two cases. I am not saying that if x > P then g(x) = P. The premise is not so easy.
     
    Last edited: Jun 21, 2006
  10. Jun 21, 2006 #9

    matt grime

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    Ah, I must have misread your post, sorry.
     
  11. Jun 21, 2006 #10
    So, reloading:

    Facts: f(x) is a L-integrable function (over [A, B]), P be a fixed (positive)number and let g(x) such that

    g(x) = f(x) when f(x) belong to [-P, P] (not x but f(x)).
    g(x) = P when f(x) > P.
    g(x) = -P when f(x) < - P.


    Can you tell me if this function g is L-integrable? At least a "yes" or "no".

    I have tried, unsuccessfully, the theorem that says that if p and q are L-integrable functions, then r = min (p, q) and s _ max (p, q) are also L-integrable.
     
  12. Jun 21, 2006 #11

    matt grime

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    Hmm, I honestly don't know 'instantly', but let's think about it a little. Right, OK, I see what to do (about 5 minutes, actually make it ten, my first, and second thoughts were off the mark).


    I think the answer is 'yes' it is integrable: we're just cutting off f if it gets too big or too small.

    using min max is a bit tricky, but I think you can do it.

    consider the functions a(x) = max(0,f(x)) and b(x)=min(0,f(x)). they are both measurable and a+b=f cos of the useful properties of zero. (I would insert a smiley if i didn't think they were ridiculous in a grown man). Now, I'll leave the rest to you: remember a and b are both measurable functions in their own right.
     
  13. Jun 21, 2006 #12
    Only a question, Matt: the functions a(x) and b(x) are directly used in your proof of the L-integrability of function g, or we must use some functions builded as a(x) and b(x) but with P instead of 0??
     
  14. Jun 21, 2006 #13
    Forget the last post, I think I got it.
    g(x) is equal to min(P, a(x)) + max (-P, b(x)), each of these two is L-integrable, then g is the sum of L-integrable functions, then is L-integrable.
    Thank you, Matt !!
     
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