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So I have a proposition which states the following:

Each of the following families of sets generate the Borel $\sigma$-algebra:

1)The family of all open intervals $(a,b)$, $a,b, \in \mathbb{R}$.

2) The family of all closed intervals $[a,b]$, $a,b, \in \mathbb{R}$.

3) The family of open intervals $(a,\infty)$, $a\in \mathbb{R}$.

4) The family of closed intervals $[a,\infty)$, $a\in \mathbb{R}$.

The proof of part 1 involved us showing that every open set is a countable union of open intervals and then since every open set is contained in this $\sigma$- algebra therefore it contains the borel $\sigma$-algebra but I am not sure how to show the other way around i.e. to show that the $\sigma$-algebra generated by the family of open intervals is contained in the borel $\sigma$-algebra

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  • $\begingroup$ Simply, every open interval is a borel set, so the $\sigma$ lagebra generated by these open intervals, which is the smallest $\sigma$ algebra conatianing them, is in the borel $\sigma $ algebra . $\endgroup$ – Nizar Apr 16 '15 at 9:11
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The Borel $\sigma$-algebra is generated by all the open sets. A $\sigma$-algebra generated by a subset of the open sets must therefore be a subset of the Borel $\sigma$-algebra.

Similarly, the set of closed intervals is a subset of the Borel $\sigma$-algebra, so the $\sigma$-algebra generated by those intervals will be contained within the Borel $\sigma$-algebra. You should be able to make this work for all your other cases too.

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  • $\begingroup$ So this is what I have understood so far; Every open set in $\mathbb{R}$ is of the form $(a,b)$, $a,b \in \mathbb{R}$ and a countable union of open sets is open and a finite intersection of open sets is also open. So every open set in $\mathbb{R}$ is a countable union of open intervals as described above which means that every open set is contained in the $\sigma$- algebra generated by these open intervals and hence it contains the Borel $\sigma$-algebra and for the other way around since every open set is an open interval itself we can show the other inclusion as well. Is this correct? $\endgroup$ – user1314 Apr 16 '15 at 9:34
  • $\begingroup$ Not every open set is an open interval. All open intervals are open, all open sets are a countable union of open intervals. All unions of open sets, of any cardinality, are open, we just get lucky with the reals that we only need countable unions. The rest seems correct. $\endgroup$ – user24142 Apr 16 '15 at 9:47
  • $\begingroup$ Ahh yes that makes sense now, it is because every open interval is an open set that we can say the $\sigma$-algebra generated by open intervals is contained in the Borel $\sigma$-algebra. So for part 2 and others, again we can indeed show that every open interval is a countable union of those sets and therefore show that the sigma algebra generated is contained in the Borel algebra but how about showing the reverse direction for part 2 ? $\endgroup$ – user1314 Apr 16 '15 at 9:55
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a Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement. Borel sets are named.

For a topological space X, the collection of all Borel sets on X forms a σ-algebra, known as the Borel algebra or Borel σ-algebra. The Borel algebra on X is the smallest σ-algebra containing all open sets (or, equivalently, all closed sets) for example be have The Borel algebra on the reals is the smallest σ-algebra on R which contains all the intervals. \begin{lemma} let $ C = {(a,b) :a<b}$ then $\sigma(C)=B_{R}$ IS the Borel field generated by the family of all open intervals C. \end{lemma}

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