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Proof of Hahn decomposition theorem

  1. Sep 29, 2011 #1

    Fredrik

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    I'm trying to read this proof, and I'm stuck on the inequality on page 27 following the statement "It follows that every measurable subset..." Why does it hold?

    The theorem is about signed measures, i.e. functions that are like measures, but can assign both positive and negative "sizes" to sets. A measurable set is said to be positive if all its measurable subsets have a non-negative size. The term negative is defined similarly. The theorem asserts that the set X is a disjoint union of a positive set A and a negative set B. The strategy of the proof is roughly this: First find a set B and show that it's negative. Define A=X-B. Suppose that A is not positive. (This will lead to a contradiction). It's not too hard to see that A doesn't have any negative subsets, but we can still pick a subset [itex]E_0\subset A[/itex] that has a negative size. Then we cut away disjoint pieces of [itex]E_0[/itex], denoted by [itex]E_1,E_2,\dots[/itex] that have positive sizes. The goal is to show that [itex]E_0-\bigcup_{k=1}^\infty E_k[/itex] is negative.
     
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  3. Sep 29, 2011 #2

    micromass

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    Hmm, it tool me a while, but I've got it. Remember that [itex]m_k[/itex] is the smallest integer such that there is a set [itex]E_k[/itex].

    [tex]\mu (E_k)\geq \frac{1}{m_k}[/tex]

    Now take F measurable, if

    [tex]\mu(F)\geq \frac{1}{m_k-1}[/tex]

    then [itex]m_k-1[/itex] is a smaller integer and [itex]F[/itex] is another set for which

    [tex]\mu(F)\geq \frac{1}{m_k-1}[/tex]

    This is in contradiction with the choice of [itex]m_k[/itex]. Thus it must hold that

    [tex]\mu(F)<\frac{1}{m_k-1}[/tex]
     
  4. Sep 29, 2011 #3

    Fredrik

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    You sir, are awesome. That was crystal clear. Thank you.
     
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