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I'm trying to prove that the Borel $\sigma$-algebra is equivalent to the smallest $\sigma$-algebra containing (a,$\infty$). The approach I'm trying to use is to show that if A$\subset$B and B$\subset$A, then A=B. So far I think I have the first part:

(a, $\infty$) = $\cup$(a, n) and since each (a, n) is a finite open interval, then they are included in the Borel $\sigma$-algebra and since a $\sigma$-algebra is closed with respect to countable union, then the smallest $\sigma$-algebra containing (a, $\infty$) $\subset$ Borel $\sigma$-algebra.

Looking at (a,b) it seems obvious that it would be contained within (a,$\infty$), but I'm not sure if it is sufficient to just say that for the second part of this proof or if there is a formal way to show that (a,b) is contained in (a,$\infty$).

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    $\begingroup$ $\sigma$-algebra are stable under finite intersection ! $\endgroup$
    – nicomezi
    Commented Nov 24, 2020 at 6:17
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    $\begingroup$ Your English seems to be quite good but some of the technical terms you use have no meaning. For example, there is no such thing as $\sigma$- algebra of the form $(a,\infty)$. $\endgroup$ Commented Nov 24, 2020 at 6:44

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Let $\mathcal A$ be the smallest $\sigma$-algebra containing sets of the form $(a, \infty)$. It remains to show that the Borel $\sigma$-algebra $\subset \mathcal A$.

Every open subset of $\mathbb R$ can be written as countable union of open intervals. Thus we are done if we show that all open intervals belongs to $\mathcal A$.

Since $(a, \infty) \in \mathcal A$, we have $(-\infty, a] \in \mathcal A$. Thus $(a, b]= (-\infty, b] \cap(a,\infty) \in \mathcal A$ and consequently $(a, b)= \cup (a, b-\frac{1}{n}]\in \mathcal A$.

Since $a,b \in \mathbb R$ are arbitary, it follows that all open intervals belongs to $\mathcal A$. As discussed above this proves that the Borel $\sigma$-algebra $\subset \mathcal A$.

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