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Math Help - Lebesgue integration

  1. #1
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    Lebesgue integration

    Let f be a (Lebesgue) integrable function and A be a measurable set.

    Show that:

     \int lfl\chi_A(x) \leq K*m(A) for some K

    I know that since f is integrable we can say  \int lfl \leq K , and this feels like it should be a two line proof, but I can explicitly show how to get to the next step.
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  2. #2
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    Disregard the following... (facepalm)

    Let s be a simple function, 0\le s\le |f|1_A so s=s1_A. Call such a function "good". We aim to show the integral of any good function is bounded by K\cdot m(A) so \int |f|=\sup\{\int s : \text{s good}\} is too.

    Since |f| is integrable, the set E:=\{\int t : 0\le t \le |f|, \text{t simple}\} is bounded above and we can chuse a simple function 0\le t' \le |f| whose integral is close to the supremum. Then t := t' + k is simple and greater than every function in E for large k (oopsthis part was wrong; thanks Opalg). Suppose s is a good function. Then s is in E, so t is greater than it, and it follows that t1_A is greater than it. So t \ge s for all good s and it follows that \int t \ge \int s for all good s so that \int t \ge \int |f|. Now since t is simple, t = \sum_{i=1}^N t_i 1_{B_i}, so \int t1_A = \sum t_i m(B_i\cap A) \le (\max t_i) \sum m(B_i\cap A) \le (\max t_i)\;Nm(A) and we are done.

    (Note that I've abused notation by writing E for both the set of all simple functions between 0 and |f|, and the integrals of such functions. I hope its not confusing.)
    Last edited by maddas; April 15th 2010 at 12:04 PM. Reason: oops
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  3. #3
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    Quote Originally Posted by southprkfan1 View Post
    Let f be a (Lebesgue) integrable function and A be a measurable set.

    Show that:

     \int lfl\chi_A(x) \leq K*m(A) for some K

    I know that since f is integrable we can say  \int lfl \leq K , and this feels like it should be a two line proof, but I can explicitly show how to get to the next step.
    Assuming that you want K to be a constant independent of A, this result looks wrong.

    Take f to be the function f(x) = x^{-1/2} on the interval (0,1). If A is the interval (0,1/n) then \int\!|f|\chi_A^{\mathstrut} = \int_0^{1/n}\!\!\!\! x^{-1/2}dx = \Bigl[2x^{1/2}\Bigr]_0^{1/n} = \frac2{\sqrt n}. If \int\!|f|\chi_A^{\mathstrut} \leqslant Km(A) then \frac2{\sqrt n} \leqslant\frac Kn, or \sqrt n\leqslant K/2. That clearly cannot hold for all n.
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  4. #4
    Member mabruka's Avatar
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    I think $K$ does not depend on the set $A$, but it depends on the function f.

    Since f is integrable then it is bounded almost everywhere and using integral monotonicity we get the desired conclusion.
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