Closed Set Measurable in Borel Sigma-Algebra

Theorem

Let $\struct {S, \tau}$ be a topological space.

Let $\map \BB \tau$ be the associated Borel $\sigma$-algebra.

Let $C$ be a closed set in $\struct {S, \tau}$.

Then $C$ is $\map \BB \tau$-measurable.

Proof

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

By definition of Borel $\sigma$-algebra, $S \setminus C \in \map \BB \tau$.

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

$S \setminus \paren {S \setminus C} \in \map \BB \tau$

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

The result follows by definition of measurable set.

$\blacksquare$