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Thread: Borel set

  1. #1
    MHF Contributor Siron's Avatar
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    Borel set

    How does one show the following ($\displaystyle \mathcal{R}$ denotes the one-dimensional Borel-set):
    $\displaystyle \mathcal{R} = \sigma(\mathcal{A})$
    with $\displaystyle \mathcal{A}=\{]a,b]|a,b \in \mathbb{Q}, a < b\}\cup {\emptyset}$

    I know that $\displaystyle \mathcal{R}= \sigma(\{]a,b]|-\infty<a\leq b<\infty\})$ so I think $\displaystyle \mathcal{A} \subseteq \mathcal{R}$ and if I can show that $\displaystyle \mathcal{A}$ is a $\displaystyle \pi-$ systeme it follows by the $\displaystyle \pi-\lambda$ theorem that $\displaystyle \sigma(\mathcal{A})\subset \mathcal{R}$.

    I could do the reverse implication the same way but then I need to prove that $\displaystyle \{]a,b]|-\infty<a\leq b<\infty\} \subset \sigma(\mathcal{A})$, but I'm struggling with this implication.

    Anyone?
    Thanks in advance.
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  2. #2
    MHF Contributor
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    Re: Borel set

    Hi,
    The attachment answers your problem. I just noticed you used ]a,b] for the set {x : a < x <= b}; I used (a,b] for the same set.

    Borel set-mhfborelsets.png
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