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Math Help - sandwich measurable set

  1. #1
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    sandwich measurable set

    Suppose that A is subset of R (real line) with the property for every ε > 0 there are measurable sets B and C s.t. B⊂A⊂C and m(C\B)<ε
    Prove A is measurable
    By definition A is measurable we need to prove m(E)=m(E∩A)+m(E\A) for all E
    the ≤ is trivial enough to show ≥:
    Since C is measurable then m(E)= m(E∩C)+m(E\C)
    ≥ m(E∩A)+m(E\A)-ε (Since A is subset of C)
    then move the ε to LHS and since for every ε so, let ε->0 , obtain the result.
    is my solution right??? i thought i should use B and m(C\B)<ε somewhere. Could some one help me to check it??
    many thanks
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  2. #2
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    Re: sandwich measurable set

    How do you know that m(E\cap C) + m(E\setminus C) \ge m(E\cap A) + m(E\setminus A) - \varepsilon? You just stated what you need to prove, but did not prove it.

    You need to show that for all \varepsilon>0 and for all measurable sets E\subseteq \mathbb{R}, \left|m(E\cap A) + m(E\setminus A) - m(E) \right| < \varepsilon. So, you need to come up with a way of rewriting m(E\cap A)+m(E\setminus A) using the terms you know something about: m(E\cap B), m(E\cap C), m(E\setminus B), m(E\setminus C).
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  3. #3
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    Re: sandwich measurable set

    I tried the way u provided, but i cannot get anything relatives to m(C\B)< ε . Could you tell me the proof in details???
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  4. #4
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    Re: sandwich measurable set

    Since m(E\cap A) \le \inf\{m^*(E \cap K) \mid A \subseteq K, K\mbox{ is measurable} \} and m(E \setminus A) \le \inf\{m^*(E \setminus K) \mid K \subseteq A, K\mbox{ is measurable} \}, you can conclude, m(E \cap A) + m(E \setminus A) \le m(E \cap C) + m(E \setminus B). But, E \cap C = E \cap (B \cup (C \setminus B)), so m(E \cap C) = m(E \cap (B \cup (C\setminus B))) = m((E \cap B) \cup (E \cap (C\setminus B))). Since B,C are measurable, use the properties of a \sigma-algebra to show that E\cap B, E\cap (C\setminus B) are both measurable. Hence m((E \cap B) \cup (E \cap (C\setminus B))) = m(E \cap B) + m(E \cap (C\setminus B)).

    To recap, we have
    \begin{align*}m(E \cap A) + m(E \setminus A) & \le m(E \cap C) + m(E \setminus B) \\ & = m(E \cap B) + m(E \cap (C \setminus B)) + m(E \setminus B) \\ & = (m(E\cap B) + m(E\setminus B)) + m(E \cap (C\setminus B)) \\ & \le (m(E\cap B) + m(E\setminus B)) + m(C\setminus B) \\ & = m(E) + m(C\setminus B) \\ & < m(E) + \varepsilon\end{align*}

    This is because E \cap (C\setminus B) \subseteq C\setminus B, so m(E\cap (C\setminus B)) \le m(C \setminus B) and since B is measurable, m(E) = m(E\cap B) + m(E\setminus B).

    So, finally, we have m(E\cap A) + m(E \setminus A) - m(E) < \varepsilon.
    Last edited by SlipEternal; October 14th 2013 at 09:34 AM.
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