Let $\Gamma$ be a countable discrete group and $\beta \Gamma$ be its Stone–Čech compactification.


My question is that 


**Does the $\sigma$-algebra generated by clopen sets in $\beta \Gamma$ equal to the Borel $\sigma$-algebra (generated by open sets) on $\beta \Gamma$?**


Note that the Stone–Čech compactification $\beta \Gamma$ is extremally disconnected i.e. the closure of every open set in it is clopen. Moreover, there is a one-to-one correspondence between subsets $A$ of $\Gamma$ and clopen subsets $\bar{A}$ of $\beta \Gamma$. 


There is also a one-to-one correspondence between finitely additive measures on $\Gamma$ and $\sigma$-additive Borel measures on $\beta \Gamma$ by Riesz representation theorem.



Sigma algebras on the Stone–Čech compactification of a countable discrete group
No.


First of all, take note that for compact zero-dimensional spaces $X$, the $\sigma$-algebra generated by all clopen sets is precisely the $\sigma$-algebra of Baire sets (recall that in a completely regular space, the $\sigma$-algebra of Baire sets is the $\sigma$-algebra generated by collection of all zero sets). A space $X$ is said to be perfectly normal if it is normal and every closed set is a $G\_{\delta}$-set.



> 
> **Theorem:** A compact Hausdorff space $X$ is perfectly normal if and only if every Borel set is a Baire set.
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> 
> 


The above Theorem is stated in the measure theory textbook by Bogachev and other sources.


On the other hand, $\beta\Gamma$ is not perfectly normal since the singletons in $\beta\Gamma\setminus\Gamma$ are never $G\_{\delta}$-sets in $\beta\Gamma$. To see this, suppose that $U$ is a non-principal ultrafilter on $\Gamma$. If $\{U\}$ were a $G\_{\delta}$-set, then there would be a decreasing sequence of sets $(R\_{n})\_{n\in\omega}$ such that $\{U\}=\bigcap\_{n\in\omega}\{V\in\beta\Gamma\mid R\_{n}\in V\}$.


Let $S\_{n}=R\_{n}\setminus R\_{n+1}$ for all $n\in\omega$. Let $S=\bigcup\_{n}S\_{2n}$. Without loss of generality, assume that $S^{c}\in U$. Let
$V$ be an ultrafilter that contains each set $R\_{n}$ and which contains $S$. Then $V\neq U$, but $V\in\bigcap\_{n\in\omega}\{V\in\beta\Gamma\mid R\_{n}\in V\}$. This is a contradiction. Therefore, the singletons in $\beta\Gamma\setminus\Gamma$ are not $G\_{\delta}$ sets in $\beta\Gamma$.


Another way of seeing that the Baire and the Borel sets do not coincide is by observing that there are only $2^{\aleph\_{0}}$ Baire sets in $\beta\Gamma$ while there are $2^{2^{\aleph\_{0}}}$ many Borel sets in $\beta\Gamma$.


