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My proof of very basic measure theory theorem

  1. Aug 26, 2011 #1
    Hi. I have a proof of a very basic measure theory theorem related to the definition of a measure, and would like to ask posters if the proof is wrong.

    Theorem: If [itex]E[/itex] is measurable, then [itex]\overline{E}[/itex] is measurable and conversely.

    My Proof:
    Let's try the converse version first.

    [itex]m(E)=m(E \cap \overline{E})+m(E \cap E)[/itex]
    [itex]=m(E \cap \overline{E})+m(E)[/itex]
    So [itex]m(E \cap \overline{E})=0[/itex]. By this we've shown that [itex]\overline{E}[/itex] is measurable. Converse is true by similar method.

    [itex]m(\overline{E})=m(\overline{E} \cap \overline{E})+m(\overline{E} \cap E)[/itex]
    [itex]=m(\overline{E})+m(E \cap \overline{E})[/itex]
    [itex]=m(\overline{E})+0=m(\overline{E}).[/itex]
     
    Last edited: Aug 26, 2011
  2. jcsd
  3. Aug 26, 2011 #2
    That is in fact, completely wrong. In order to show that [itex]\overline{E}[/itex] is measurable, you have to show that for any set A, [itex]m(A) = m(A \cap \overline{E}) + m(A \cap \overline{\overline{E}})[/itex]. You seem to be trying to do this only in the case where A=E, which is not sufficient.
     
  4. Aug 26, 2011 #3

    micromass

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    And the converse is not true. The thing is that all closed sets are measurable. So [itex]\overline{E}[/itex] is always measurable. But E doesn't need to be.
     
  5. Aug 26, 2011 #4
    I took the meaning of the bar to be compliment, rather than closure. gunitinug, can you confirm that that's what the notation means?
     
  6. Aug 26, 2011 #5

    micromass

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    Aah, yes. That would make sense...
     
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