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I am reading these notes on the measure which say (pg4) that the Lebesgue measure on $[0,1]$ can be got by considering the algebra: $$A=\{\text{all finite unions of subintervals}\}$$ firstly does this include both open and closed intervals? And secondly this family generates the Borel sigma-algebra on $[0,1]$ is there any easy way to show this?

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  • $\begingroup$ Yes, that's probably the intention of the author: intervals are just subsets that contain everything between any two members. Some people take $A$ = finite unions of "half open" intervals.The issue is $A$ being closed under complementation. $\endgroup$ – Tim kinsella Jul 28 '17 at 18:02
  • $\begingroup$ The smallest $\sigma$-algebra containing all subintervals of $[0,1]$ is the Borel $\sigma$-algebra, where subinterval can be taken to mean open, closed, or half-open interval. To show this, use the fact that every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, and that every open interval is a countable union of closed intervals. $\endgroup$ – Reveillark Jul 28 '17 at 18:02
  • $\begingroup$ To your second question, the Borel algebra is generated by the open intervals since they form a basis for the topology on the interval. $\endgroup$ – Tim kinsella Jul 28 '17 at 18:03
  • $\begingroup$ The author claims that $A$ is an algebra, so probably he means to include both open and closed subintervals. If he only used open intervals, for example, then $A$ would not be an algebra: the complement of an open interval is not open. Alternatively, he could use only intervals of the form $[a,b)$, $(-\infty, b)$, or $[a, \infty)$. $\endgroup$ – Bungo Jul 28 '17 at 18:03
  • $\begingroup$ @Reveillark just to clarify do you mean that every open subset of $\Bbb{R}$ is a countable union of disjoint open intervals, rather then a countable disjoint union? $\endgroup$ – Quantum spaghettification Jul 28 '17 at 18:53
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Yes, it contains all finite and infinite open, closed, half-open, half-closed intervals of $\mathbb{R}$, is the way I read it.

Note that all intervals are Borel (open intervals are open, closed intervals are closed, and $[a,b) = [a,b] \cap [0, b)$ is in the $\sigma$-algebra generated by the open sets. So $\mathcal{A} \subset Bor(\mathbb{R})$, and so $\sigma(\mathcal{A})$ (its generated $\sigma$-algebra) is a subset of $Bor(\mathbb{R})$.
So $\sigma(\mathcal{A}) \subseteq Bor(\mathbb{R})$.

But also all open sets $O$ are in $\sigma(\mathcal{A})$, as open sets of $[0,1]$ are unions of at most countably many open intervals ,all of which lie in $\mathcal{A}$. So $Bor(\mathbb{R}) \subseteq \sigma(\mathcal{A})$ so we have equality.

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  • $\begingroup$ Please note that in the question, we are working inside $[0,1]$, so there are no infinite intervals. $\endgroup$ – Nate Eldredge Jul 28 '17 at 23:30
  • $\begingroup$ @NateEldredge thx, I changed it. $\endgroup$ – Henno Brandsma Jul 29 '17 at 6:30

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