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Thread: Borel algebra over the real line

  1. #1
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    Borel algebra over the real line

    Show that the Borel algebra over $\displaystyle \mathbb {R} $, denoted $\displaystyle \mathbb {B} _ \mathbb {R} $ is generated by the set of $\displaystyle \xi= \{ [a, \infty ) : a \in \mathbb {R} \} $

    Set up:

    I know that $\displaystyle \mathbb {B} _ \mathbb {R} = \mathbb {M} (U)$, where $\displaystyle \mathbb {M} (U) $ is the smallest $\displaystyle \sigma $-Algebra generated by U, and U is the family of all open sets in $\displaystyle \mathbb {R} $. So by a theorem that I know, if I can prove that $\displaystyle E \in \mathbb {M} (F) $, then I have $\displaystyle \mathbb {M} (E) \subset \mathbb {M} (F) $. My plan is to prove that $\displaystyle \xi \in \mathbb {M}(U) $ and vice versa.

    Proof.

    Claim: $\displaystyle \mathbb {M} ( \xi ) \subseteq \mathbb {M} (U) $

    Now, pick an element $\displaystyle [a, \infty ) \in \xi $,
    and write $\displaystyle [a, \infty ) = \bigcup _{r>0} [a,a+r) = \bigcup _{r>0} \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r ) $

    Note that $\displaystyle (a- \frac {1}{n} ,a+r) \in U \subset \mathbb {M}(U) \ \ \ \ \ \forall n \in \mathbb {N}$, so $\displaystyle \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r ) \in \mathbb {M}(U)$.

    Therefore $\displaystyle \bigcup _{r>0} \bigcap _{n=1}^{ \infty } (a- \frac {1}{n} , a+r ) \in \mathbb {M}(U) $

    So I have $\displaystyle \xi \in \mathbb {M}(U)$, which implies that $\displaystyle \mathbb {M} ( \xi ) \subset \mathbb {M} (U) $

    On the other hand, pick $\displaystyle (a,b) \in U $, then $\displaystyle (a,b) = \bigcap _{a<r<b} \bigcup _{n=1}^{ \infty } [r+ \frac {1}{n} , \infty ) \in \mathbb {M} ( \xi ) $ for the same reason as above.

    Therefore I have $\displaystyle \mathbb {M}(U) \subset \mathbb {M}( \xi ) $.

    Is this correct? Thank you.
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  2. #2
    Moo
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    Hello,

    For the first part, it would have been shorter to say that $\displaystyle [a,\infty)$ is the complement of $\displaystyle (-\infty,a)$, which can be written as $\displaystyle \bigcup_{n> 0} (a+n,a)$, which belongs to $\displaystyle \mathbb{M}(U)$
    But what you did is very correct too !

    On the other hand, pick $\displaystyle (a,b) \in U$, then $\displaystyle (a,b) = \bigcap _{a<r<b} \bigcup _{n=1}^{ \infty } [r+ \frac {1}{n} , \infty ) \in \mathbb {M} ( \xi )$ for the same reason as above.
    Hmm no.
    Because the intersection is equal to $\displaystyle (b,\infty)$ or something like that.

    Let :
    $\displaystyle A_n=\left[a-\frac 1n,\infty\right)$ and $\displaystyle B=[b,\infty)$

    Consider $\displaystyle C=\left(\bigcup_{n>0} A_n\right)\cap B^c$


    Can you finish it ?


    Anyway, good work you've done !
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