# The Unapologetic Mathematician

## Stone’s Representation Theorem III

We conclude our coverage of Stone’s representation theorem with a version for Boolean $\sigma$-algebra. Each such algebra $\mathcal{B}$ is isomorphic to a $\sigma$-algebra of subsets of some space, modulo a $\sigma$-ideal.

We start as we did for any Boolean algebra, by using the map $s$ sending $\mathcal{B}$ to the Boolean algebra of clopen subsets of the Stone space $S(\mathcal{B})$. This algebra is, of course, our identified base. Let $\mathcal{S}$ be the $\sigma$-ring generated by this base.

When we first dealt with measure rings, we quotiented out by the ideal of negligible sets. We don’t have a measure on $S(\mathcal{B})$, so we don’t have measurable sets, but we do have something almost as good. Just as when we discussed Baire spaces, we can use nowhere-denseness as a topological stand-in for negligibility. In fact, we’ll define a “meager” set to be any countable union of nowhere-dense sets, and we’ll let $\mathcal{N}$ be the collection of all meager sets in $\mathcal{S}$. It is straightforward to verify that $\mathcal{N}$ is a $\sigma$-ideal in $\mathcal{S}$.

A quick side note: classically, meager sets were said to be “of the first category”. Any other sets were said to be “of the second category”. This is the root of the term “category” in the Baire category theorem.

Now if $\{E_n\}$ is a sequence of clopen sets, then we can find a unique preimage $s^{-1}(E_n)\in\mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra, we can take the countable union of these elements, and then apply $s$ to the result. That is, we can define the set $\displaystyle E=s\left(\bigcup\limits_{n=1}^\infty s^{-1}(E_n)\right)$

Now, of course, we can also take the countable union of the sets themselves. If $s$ were an isomorphism of $\sigma$-algebras, then this union would be exactly $E$ itself; but we aren’t usually so lucky. Instead, I say that the difference $\displaystyle E\setminus\bigcup\limits_{n=1}^\infty E_n$

is nowhere-dense, and thus countable unions of clopen sets are clopen modulo meager sets.

Indeed, the countable union of the (open) sets $E_n$ is open, and so its complement is closed. The difference above is the intersection of the (closed) set $E$ with the (closed) complement of the union, and is thus closed. So if it were dense on some nonempty open set $U$ it would have to actually contain $U$. But since $U$ is open it must be the union of some collection of basic clopen sets, and we can take $U$ to be one of these sets. That is, $U\subseteq E$ and $U\cap E_n=\emptyset$ for each $n$. Since these relations only involve clopen sets, we find that $s^{-1}(U)\subseteq s^{-1}(E)$ and $s^{-1}(U)\cap s^{-1}(E_n)=\emptyset$. But there’s no way this can happen if $s^{-1}(E)$ is the union of the $s^{-1}(E_n)$!

So the map $s$ takes elements of $\mathcal{B}$ to clopen sets in $\mathcal{S}$, and then on to their equivalence classes in $\mathcal{S}/\mathcal{N}$, and this map commutes with countable unions. All that remains to show that this is an isomorphism $\mathcal{B}\cong\mathcal{S}/\mathcal{N}$ is to show that no two clopen sets in $\mathcal{S}$ represent the same equivalence class. That is, if $U_1$ and $U_2$ are distinct clopen sets, then $U_1\Delta U_2$ cannot be meager. Equivalently, no nonempty clopen set is meager.

But this is just a consequence of the Baire category theorem, here in its formulation for compact Hausdorff spaces. Indeed, if a clopen set in any Baire space — and a compact Hausdorff space is Baire — were the countable union of closed nowhere-dense sets, then its interior would be empty. But since it’s open its interior is itself, and thus is would have to be empty. Thus if $U_1$ and $U_2$ are clopen sets representing the same equivalence class modulo $\mathcal{N}$ then their (clopen) symmetric difference $U_1\Delta U_2$ is meager, and thus empty. That is, $U_1=U_2$.

August 20, 2010 - Posted by | Analysis, Measure Theory