Show that the collection of Borel sets is the smallest $\sigma$-algebra that contains intervals of the form $[a,b)$, where $a<b$.
So I first prove that the collection $S$ of Borel sets contains those intervals. $S$ contains the closed interval $[a,b]$ and the (closed) single point $b$, so it contains the difference $[a,b)$.
Now I have to prove that any $\sigma$-algebra containing all intervals $[a,b)$ must contain all elements of $S$. If I can prove that it must contain all open intervals $(a,b)$, I'll be done because any open set in $\mathbb{R}$ is a countable union of open intervals. But how do I show that $(a,b)$ belongs to such a $\sigma$-algebra?
Best Answer
$\sigma$-algebras are closed under countable unions. You need to use this to show that any $\sigma$-algebra containing the half open intervals contains the intervals as well.
For instance $(0,1)= \bigcup_{n=1}^{\infty} [\frac{1}{n}, 1)$.