Closed Set Measurable in Borel Sigma-Algebra

Theorem

Let $\left({S, \tau}\right)$ be a topological space, and let $\mathcal B \left({\tau}\right)$ be the associated Borel $\sigma$-algebra.

Let $C$ be a closed set in $S$.

Then $C$ is $\mathcal B \left({\tau}\right)$-measurable.

Proof

Since $C$ is closed, $S \setminus C$ is open.

By definition of Borel $\sigma$-algebra, $S \setminus C \in \mathcal B \left({\tau}\right)$.

By axiom $(2)$ for $\sigma$-algebras:

$S \setminus \left({S \setminus C}\right) \in \mathcal B \left({\tau}\right)$

and this set equals $C$ by Set Difference with Set Difference since $C \subseteq S$.

The result follows by definition of measurable set.

$\blacksquare$