Two Baire measure, $\left\{A\subseteq X: A \text{ Baire set}\ \mu(A)=\nu(A)\right\}$ is $\sigma$-algebra 
pag 235 Barry Simon. A comprehensirve course in analysis
Let $X$ compact hausdorff space.
Let $\mu,\nu$ baire measures, and $\mu(X)=\nu(X)=1.$
Let $S:=\left\{A\subset X: A \text{ baire set and } \mu(A)=\nu(A)\right\}$
Why $S$ es $\sigma$-algebra?
I have this:
Let $(A_n)_n$ in $S$, now
$\mu(\bigcup_n A_n)=\mu(\bigcup_n (A_{n+1}\setminus (A_1\cup\cdots \cup A_{n}))=\sum_n \mu(A_{n+1}\setminus (A_1\cup\cdots \cup A_{n}))$
And similarly, $\nu(\bigcup A_n)=\sum_n \nu(A_{n+1}\setminus (A_1\cup\cdots \cup A_{n}))$
Therefore $\mu(\bigcup_n A_n)=\nu(\bigcup_n A_n).$
I need prove $\mu(A_{n+1}\setminus (A_1\cup\cdots \cup A_{n}))=\nu(A_{n+1}\setminus (A_1\cup\cdots \cup A_{n}))$
I also tried that if $A\in S$ then $A^c\in S$.
$A\in S$, then $\mu(A)=\nu(A)$
$\mu(X)=\mu(A)+\mu(A^c)$
$\nu(X)=\nu(A)+\nu(A^c)$ then $\mu(A^c)=\nu(A^c)$
I still have to prove that $S$ is closed under union finite.
How prove that if $A,B\in S$ then $\mu(A\cup B)=\nu(A\cup B)$?

It's not closed under unions.
Consider e.g. $X = \{1,2,3,4\}$ with $\mu(\{i\}) = 1/4$ for each $i$ while $\nu(\{1\}) = \nu(\{2\}) = 1/2$, $\nu(\{3\}) = \nu(\{4\})=0$.  Then $\{1,3\}$ and $\{1,4\}$ are in $S$ but $\{1,3,4\}$ is not.
