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Math Help - Show set function is a measure

  1. #1
    Junior Member TheProphet's Avatar
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    Show set function is a measure

    Hi,

    Let  X be a set.
    Define  \mathcal{A} := \{ A \subset X : |A| \leq |\mathbb{N}| \mbox{ or } |A^{c}| \leq |\mathbb{N}| \}, where  | \cdot | means cardinality.

    Then  \mathcal{A} is a \sigma-algebra. Define a measure on this measurable space by  \gamma : \mathcal{A} \to \{0,1\} by
    \ \gamma(A) := \begin{cases} 0 \mbox{ if } A \mbox{ is countable}\\ 1 \mbox{ if }A^{c} \mbox{ is countable}\end{cases}

    Show that  \gamma is a measure, i.e. show that  \gamma(\emptyset) = 0 and that  \gamma is  \sigma -additive.

    That  \gamma(\emptyset) = 0 I guess comes easily since  \emptyset is countable. If  (A_j)_{j \geq 1} \subset \mathcal{A}, pairwise disjoint:

    Assume all A_j are countable, then the countable union is countable, so  \gamma\left( \bigcup_{j \geq 1} A_j \right) = 0.
    Also,  \sum_{j \geq 1} \gamma(A_j) = 0 + 0 + \ldots = 0 , so we have equality in this case.
    If we assume that two sets  A_j_{0}, A_j_{1} are uncountable, then  A_{j_{0}}^{c}, A_{j_{1}}^{c} are countable and since  \bigcap_{j \geq 1} A_{j}^{c} \subset A_{j_{0}}^{c} \cap A_{j_{1}}^{c} we have  \left(\bigcup_{j \geq 1}A_{j} \right)^{c} = \bigcap_{j \geq 1}A_{j}^{c} is countable.
    Thus,  \gamma\left( \bigcup_{j \geq 1} A_j \right) = 1 , but
     \sum_{j \geq 1} \gamma(A_j) = 2 ???

    I know I am wrong, please point out where.

    Thanks
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  2. #2
    Super Member girdav's Avatar
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    Re: Show set function is a measure

    You have to assume that X is uncountable (if it's not the case \gamma is not well defined). In this case, you cannot have two disjoint set whose complement is countable: if A and B are such sets, then A^c\cup B^c is countable but it's the complement in X of the empty set since A and B are disjoint.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Show set function is a measure

    So if there cant exist two disjoint sets whose complements are countable, then the sum of measures can be at most 1 also, and equal to one when just one  A_j is uncountable ?
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  4. #4
    Super Member girdav's Avatar
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    Re: Show set function is a measure

    Yes.
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  5. #5
    Junior Member TheProphet's Avatar
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    Re: Show set function is a measure

    Ok thank you
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