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Let $I \equiv \lbrace [- \infty, a[ : a\in \mathbb{R}\rbrace$. Is $\sigma(I)$ Borel's sigma algebra on $\mathbb{R}$?

I'm having difficulties proving these statement. I suppose it's not the Borel's sigma algebra on $\mathbb{R}$ given that the latter does not include the singleton $\lbrace -\infty\rbrace$.

I appreciate any help!

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    $\begingroup$ If $\mathcal I=\{[\infty,a)\mid a\in\mathbb R\}$ then its elements are not subsets of $\mathbb R$ so that it cannot generate the Borel $\sigma$-algebra on $\mathbb R$. For that you need $\mathcal I=\{(\infty,a)\mid a\in\mathbb R\}$ or $\mathcal I=\{(\infty,a]\mid a\in\mathbb R\}$. $\endgroup$ – drhab Apr 7 '15 at 16:42
  • $\begingroup$ Thanks! Would this implicate that $\sigma (I)$ is not a subset of Borel's sigma algebra? $\endgroup$ – Marcus Fermat Apr 7 '15 at 17:00
  • $\begingroup$ First of all: in my comment I forgot (sorry) the minussign before $\infty$. So were I wrote $[\infty,a)$, $(\infty,a)$ and $(\infty,a]$ I meant to write $[-\infty,a)$, $(-\infty,a)$ and $(-\infty,a]$. Then $\sigma(\mathcal I)$ contains sets like $[-\infty,a)$ and these sets are not subsets of $\mathbb R$ so cannot belong to the Borel-$\sigma$-algebra on $\mathbb R$. So $\sigma(\mathcal I)$ cannot be a subset of it. $\endgroup$ – drhab Apr 7 '15 at 17:12
  • $\begingroup$ Perfect, now it's crystal clear. Thank you $\endgroup$ – Marcus Fermat Apr 7 '15 at 17:14
  • $\begingroup$ You should check that the question isn't actually asking about $\bar {\Bbb R}$, which includes $\{\pm \infty\}$ $\endgroup$ – Omnomnomnom Apr 7 '15 at 18:05

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