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I have tried looking at various sources and still cant understand the following:

Consider $X= \mathbb{R}$, then the $\sigma$-algebra generated by the family of closed intervals $[a,b]$ is the same as the Borel $\sigma$-algebra. So I know that to show the above we need to show the Borel $\sigma$-algebra contains all closed intervals, and to show the other inclusion we need to show that every open set (which is a countable union of open intervals) is contained in the $\sigma$-algebra generated by the closed intervals. But I don't know how to show this. Any help will be greatly appreciated.

Edit: So I seem to have made a tiny breakthrough, so to show that the $\sigma$-algebra generated by closed intervals is a subset of the Borel $\sigma$-algebra we need to show that every closed interval is some countable union of open intervals or some countable intersection of open intervals, because by definition of $\sigma$-algebra we know that open intervals are open sets and therefore any countable union/intersection of open sets will necessarily be in the Borel $\sigma$-algebra. So this would prove $\sigma([a,b]) \subset \mathbb{B}(\mathbb{R})$

To show the other way around we need to show that every open set is some union or intersection of closed intervals which is where I am currently stuck on.

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For the first part, say $A = [a, b]$. $A$ is closed in $\mathbb{R}$. So, $A^c$ is open in $\mathbb{R}$. Since Borel sigma algebra contains all open sets it contains $A^c$ and by the property of a sigma-algebra, it contains $(A^c)^c = A$. Well, you might not need this proof, depending on how you define the Borel sigma-algebra.

The inverse: You know that Every open set in R is the union of an at most countable collection of disjoint segments. Now, you take a complement of this union and you show that it takes a form of a union of closed intervals of the form $[a, b]$. The only other thing to consider is the interval of the form $[a, \infty]$ and $[-\infty, b]$.

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    $\begingroup$ Thank you! yours is a much better way of doing it than mine! $\endgroup$
    – user1314
    Apr 16, 2015 at 14:53

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