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

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

    Now, conversely assume $\displaystyle A \in \mathcal{E} $. Then if $\displaystyle 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 $\displaystyle A^{c} \in \mathcal{E} $ because $\displaystyle \mathcal{E} $ is a $\displaystyle \sigma $-algebra. But shouldn't it be $\displaystyle 1_{A}^{-1}(B) = A $ if $\displaystyle B = \{1\} $ ? By taking for example $\displaystyle B = (-2,2) $ you get a lot of points where $\displaystyle 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 $\displaystyle 0$ and $\displaystyle 1$ are in $\displaystyle B$ (that's what the example $\displaystyle B=(-2,2)$ shows), and when neither $\displaystyle 0$ nor $\displaystyle 1$ is in $\displaystyle B$. 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): $\displaystyle \{1\} \in B, \; \{0\} \notin B \; \Rightarrow 1_{A}^{-1}(B) = A $

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

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

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

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