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Thread: Measure Theory Problem

  1. #1
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    Measure Theory Problem

    Hi everyone
    I cannot find the solution to this problem,
    Let A a Lebesgue Measurable Set; Prove that given $\displaystyle 0 \leq b \leq m(A)$ then exists $\displaystyle B \subseteq A$ Lebesgue Measurable such that m(B) = b

    I appreciate your help
    Thanks
    everk.
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  2. #2
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    Quote Originally Posted by everk View Post
    Let A a Lebesgue Measurable Set; Prove that given $\displaystyle 0 \leq b \leq m(A)$ then exists $\displaystyle B \subseteq A$ Lebesgue Measurable such that m(B) = b
    Let $\displaystyle {\mathstrut\chi}_A$ denote the characteristic function of $\displaystyle A$ (so that $\displaystyle {\chi\mathstrut}_A(t) =1$ if $\displaystyle t\in A$, and $\displaystyle {\mathstrut\chi}_A(t) = 0$ if $\displaystyle t\notin A$). For each real number $\displaystyle x$, let $\displaystyle \displaystyle f(x) = \int_{-\infty}^x\!\!{\mathstrut\chi}_A(t)\,dt = m(A\cap(-\infty,x]).$ Show that $\displaystyle f$ is continuous and use the intermediate value theorem to deduce that there exists $\displaystyle c$ such that $\displaystyle f(c)=b.$
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Does that work if $\displaystyle A=(-\infty, 0)$?
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    Does that work if $\displaystyle A=(-\infty, 0)$?
    It will certainly need a bit of adjustment if $\displaystyle m(A) = \infty$. I think that you would then have to start by looking at $\displaystyle \displaystyle\int_x^X\!\!{\mathstrut\chi}_A(t)\,dt$, which can be made arbitrarily large if $\displaystyle x$ is small enough and $\displaystyle X$ is large enough. Then reduce $\displaystyle X$ until the integral reaches the value $\displaystyle b$.
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