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

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

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

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

    Assume all $\displaystyle A_j$ are countable, then the countable union is countable, so $\displaystyle \gamma\left( \bigcup_{j \geq 1} A_j \right) = 0$.
    Also, $\displaystyle \sum_{j \geq 1} \gamma(A_j) = 0 + 0 + \ldots = 0 $, so we have equality in this case.
    If we assume that two sets $\displaystyle A_j_{0}, A_j_{1} $ are uncountable, then $\displaystyle A_{j_{0}}^{c}, A_{j_{1}}^{c}$ are countable and since $\displaystyle \bigcap_{j \geq 1} A_{j}^{c} \subset A_{j_{0}}^{c} \cap A_{j_{1}}^{c} $ we have $\displaystyle \left(\bigcup_{j \geq 1}A_{j} \right)^{c} = \bigcap_{j \geq 1}A_{j}^{c} $ is countable.
    Thus, $\displaystyle \gamma\left( \bigcup_{j \geq 1} A_j \right) = 1 $, but
    $\displaystyle \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 $\displaystyle X$ is uncountable (if it's not the case $\displaystyle \gamma$ is not well defined). In this case, you cannot have two disjoint set whose complement is countable: if $\displaystyle A$ and $\displaystyle B$ are such sets, then $\displaystyle A^c\cup B^c$ is countable but it's the complement in $\displaystyle X$ of the empty set since $\displaystyle A$ and $\displaystyle B$ 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 $\displaystyle A_j$ 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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