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Math Help - Help with theorem

  1. #1
    Junior Member TheProphet's Avatar
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    Help with theorem

    I need help with understand a theorem.

    Theorem Let  (F,\mathcal{F}) = (\mathbb{R},\mathcal{B}(\mathbb{R})), and  (E,\mathcal{E}) be any measurable space. An indicator  1_{A} on  E is measurable if and only if  A \in \mathcal{E} .

    Proof: If  1_{A} is measurable then A = \{\omega \, : \, 1_{A}(\omega) = 1 \} is in  \mathcal{E} . I guess we can see this by noticing that  \{1\} = \bigcap_{n=1}^{\infty} \left(1-\frac{1}{n},1+\frac{1}{n}\right) , which is in  \mathcal{F} and thus also  1_{A}^{-1}(\{1\}) \in \mathcal{E} by definition.

    Now, conversely assume  A \in \mathcal{E} . Then if  B \in \mathcal{F}, \, 1_{A}^{-1}(B) = A \, \mbox{ if } 1 \in B \mbox{ and } 1_{A}^{-1}(B) = A^{c} \mbox{ if } 1 \notin B .
    Clearly  A^{c} \in \mathcal{E} because  \mathcal{E} is a  \sigma -algebra. But shouldn't it be  1_{A}^{-1}(B) = A if  B = \{1\} ? By taking for example  B = (-2,2) you get a lot of points where  1_{A} \neq 1 .
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  2. #2
    Super Member girdav's Avatar
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    Re: Help with theorem

    You missed some cases, for example when both 0 and 1 are in B (that's what the example B=(-2,2) shows), and when neither 0 nor 1 is in B. You have to consider the four cases.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Help with theorem

    Case (i):  \{1\} \in B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = A

    Case (ii):  \{1\} \notin B, \; \{0\} \in B \; \Rightarrow 1_{A}^{-1}(B) = A^{c}

    Case (iii):  \{1\} \in B, \; \{0\} \in B \; \Rightarrow 1_{A}^{-1}(B) = E

    Case (iv):  \{1\} \notin B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = \emptyset

    Since  \mathcal{E} is a \sigma-algebra, all these sets are in \mathcal{E} . So 1_{A} is measurable.
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