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Definition of Borel on $\mathbb R$ ($\mathcal B(\mathbb R)$) is that it's the $\sigma$-algebra generated by all open sets in $\mathbb R$.

OK, if I take some open set like $C = (0,1)$, by definition of $\sigma$-algebra the complement of $C$ must also be in $\mathcal B(\mathbb R)$, so something like $(-\infty, 0] \cup [1, \infty)$ must be in $\mathcal B(\mathbb R)$. Is that right?

  • 3
    Yes, that's right.2012-02-17
  • 1
    Clearly, any closed set is in $\mathcal B(\mathbb R)$.2012-02-17

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