# Why is a Borel algebra a $\sigma$-algebra?

My book defines a Borel algebra $\mathcal{B}^n$ as the $\sigma$-algebra generated by the open sets of $\mathbb{R}^n$.

In order to be a $\sigma$-algebra: If $A$ is a set in a $\sigma$-algebra, so is its complement $A^c$. Suppose $X$ is an open set in $\mathcal{B}^n$, then $X^c$ is closed. But wouldn't $X^c \notin \mathcal{B}^n$ because it is closed? Is it possible to get closed sets from open ones with the properties of a $\sigma$-algebra?

• Note the use of the word "generated". It means that $\mathcal{B}^n$ is the smallest $\sigma$-algebra containing all the open sets - which is, by definition, a $\sigma$-algebra (and it contains the closed sets as well) Commented Dec 24, 2013 at 3:55

When your book says that $\mathcal{B}^{n}$ is generated by the open sets of $\mathbb{R}^{n}$, what that means is that each open set is contained, but then we also put in any non-open sets required to maintain the $\sigma$-algebra properties, but no more.
So, in this way, $\mathcal{B}^{n}$ is trivially a $\sigma$-algebra, as we've forced it to be one in our definition.
By definition, the $\sigma$-algebra generated by a collection $\mathcal{A}$ of subsets of $X$ is the smallest $\sigma$-algebra which contains $\mathcal{A}$; in general, it's not going to be equal to $\mathcal{A}$, but will be much larger. To see that such a $\sigma$-algebra exists, simply consider the intersection of all $\sigma$-algebras containing $\mathcal{A}$ (which is not empty, since the power set is such an object). It's routine to show that this is a $\sigma$-algebra, and is certainly the smallest possible one.
Now in the case of the Borel $\sigma$-algebra, it's going to contain all open sets, all closed sets, all $F_{\sigma}$ sets, all $G_{\delta}$, $F_{\sigma\delta}$, $G_{\delta\sigma}$, and many more kinds of sets.