# Borel sets and metric exterior measure (abstract measure spaces)

An exterior measure $\mu_*$ on $X$ is called a metric exterior measure if it satisfies $\mu_* (A\cup B)=\mu_*(A) + \mu_*(B)$ whenever the distance between $A$ and $B$ is positive.

Now this theorem says that if $\mu_*$ is a metric exterior measure on a metric space $X$, then the Borel sets in $X$ are measurable.

The proof proceeds by mentioning that it suffices to prove that closed sets in $X$ are Caratheodory measurable in view of the definition of a Borel set.

I am not quite sure why the last statement is true. I know that if we can show that a closed set $F$ is Caratheodory measurable then so it it's complement (i.e. an open set). But a Borel set is an element of the Borel sigma algebra that is the intersection of all sigma algebras that contain the open sets. Certainly, Borel sets are not necessarily open so how will I use the fact that a closed set is Caratheodory measurable?

Thanks - help appreciated!

Let $S$=set of all Caratheodory measurable sets. Then observe that $S$ a $\sigma$-algebra on $X$. Hence, if you can show that every open set is measurable, then $S$ is $\sigma$-algebra containing open sets. Thus, Borel $\sigma$-algebra will be be contained in $S$ (because Borel $\sigma$-algebra is the intersection of all $\sigma$-algebras containing open sets). So every Borel set is measurable.
Because If we show that closed set is Cartheodary measurable as Cartheodary measurable sets form $$\sigma$$-algebra, that means closed under complement.
SO open set are Cartheodary measurable. and Borel set is the intersection of $$\sigma$$-algebra containing open sets.