# How Borel sets make $\sigma$-algebra in a topological space?

I am trying to wrap my head around random variables and can't prove the following questions:

How, in a topological space $(X, \mathcal{T})$, the collection of all Borel sets, say $\mathfrak{B}$, make a $\sigma$-algebra? And, Why is it the smallest $\sigma$-algebra containing $\mathcal{T}$?

In the introduction of the Wikipedia entry of Borel sets these two statements appear. So, I'm using the definition of Borel set given there:

A Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement.

I will appreciate any help. Thank you.

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A Borel set is any set in a topological space that can be formed from open sets through the iterated operations of countable union, countable intersection, and relative complement. –  Michael Greinecker Nov 23 '12 at 8:01
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## 1 Answer

It makes a $\sigma$-algebra because, directly from the definition you give, it is easy to see that it is closed under taking complements and countable intersection/union. Moreover, any $\sigma$-algebra containing $\mathcal{T}$ must contain all such sets as can be formed from the open sets in $\mathcal{T}$, so the Borel sets must be the smallest $\sigma$-algebra containing $\mathcal{T}$, as all other such $\sigma$-algebras must contain it.

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Thanks for answering. –  Sayantan Nov 23 '12 at 6:53
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