Edit: My previous answer does not work as is. I edited it so that it becomes more of a long comment. It discusses a few aspects of the case where $X$ has a topology that makes it compact totally separated, and $\mathcal A$ is generated by the algebra of clopen sets.
Let $(X,\mathcal A)$ be a measurable space. We assume $\mathcal A$ is isomorphic to the Borel algebra $\mathcal B(Y)$ of some topological space $(Y,\mathcal T)$ (as posets). We denote by $\phi:\mathcal A\to\mathcal B(Y)$ the isomorphism of posets.
Since the Borel poset of a topological space does not change upon taking the Kolmogorov quotient, we can assume $Y$ is $T_0$. We also do the equivalent construction on $X$ (take the quotient by the relation “the measurable sets containing $x$ and $x'$ are the same”), so $\mathcal A$ separates points of $X$ and $\mathcal B(Y)$ separates points of $Y$.¹
Define $\omega\mathfrak U(\mathcal A)$ (ultrafilters stable under countable intersections) the collection of all $U\subset\mathcal A$ such that
- $\varnothing\notin U$;
- $U$ is stable under countable intersections;
- if $A\subset A'$ are elements of $\mathcal A$ and $A\in U$, then $A' \in U$;
- for all $A\in\mathcal A$, either $A\in U$ or $A^\complement\in U$.
There are natural maps from $X$ and $Y$ to $\omega\mathfrak U(\mathcal A)$, namely
$$\begin{aligned}x&\mapsto\{A\in \mathcal A:x\in A\}=:U_x,
&y&\mapsto\{A\in \mathcal A:y\in\phi(A)\}=:V_y.\end{aligned}$$
What remains of my answer is the following, which we shall prove later.
Lemma.
- The maps $x\mapsto U_x$ and $y\mapsto V_y$ are injective.
- If moreover $X$ is a Stone space and $\mathcal A$ is the $\sigma$-algebra generated by its clopen sets, then
- the map $x\mapsto U_x$ is bijective;
- the induced map $i:Y\to X$ is such that $\phi(A)=i^{-1}A$.
In the latter case, since the preimage has much nicer properties than $\phi$, this gives a lot more information about how the latter behaves. For instance, it shows that if a (not necessarily countable) union of elements of $\mathcal A$ is itself in $\mathcal A$, then $\phi$ commutes with the union. It is also true for arbitrary intersections. However, this property need not hold for the inverse $\phi^{-1}$ over $\mathcal B(Y)$: to be sure to have
$$\phi^{-1}\big(\bigcup_{j\in J}B_j\big) = \bigcup_{j\in J}\phi^{-1}(B_j), $$ one needs of course all the $B_j$ and their union to be in $\mathcal B(Y)$ (so that the equation even makes sense) but also the union of the $\phi^{-1}(B_j)$ to be in $\mathcal A$. For this reason, the set of all $\phi^{-1}(B)$ for $B\in\mathcal T$ is not clearly a topology. However, if we can show it is by other means, then it must generate the $\sigma$-algebra.
Fact.
If $\mathcal S:=\{A\in\mathcal A:\phi(A)\in\mathcal T\}$ is a topology over $X$, then it generates $\mathcal A$ as a $\sigma$-algebra.
In Jochen Wengenroth's answer to Is every sigma-algebra the Borel algebra of a topology?, it is shown that this property would not hold for the Stone space $\{0,1\}^{\mathbb R}$, so to get a contradiction it would suffice to show that $\mathcal S$ is a topology.
Proof of the fact.
Define $\mathcal B'=\{ B\in\mathcal B(Y):\phi^{-1}B\in\sigma(\mathcal S)\}$. It is a $\sigma$-algebra (because $\phi^{-1}$ preserves countable unions and intersections) and it contains $\mathcal T$, so $\mathcal B(Y)\subset\mathcal B'$, and by definition it means $\mathcal A\subset\sigma(\mathcal S)\subset\mathcal A$.
It remains to prove the lemma.
Proof of the lemma: injectivity.
It should be clear that the map $X\to\omega\mathfrak U(\mathcal A)$ is well-defined, i.e. $U_x\in\omega\mathfrak U(\mathcal A)$. The fact that it is injective comes from the fact that $\mathcal A$ separates points. The same reasoning holds for $y\mapsto V_y$.
Proof of the lemma: surjectivity.
For the surjective part, write $\mathcal T_X$ for the topology on $X$ which makes it a compact totally separated space. Let $U$ be an element of $\omega\mathfrak U(\mathcal A)$. Note that by the finite intersection property for $U$, the collection of open sets $O\in(\mathcal A\setminus U)\cap\mathcal T_X$ cannot be an open cover. Choose some $x\in X$ not covered by this collection; we will show that $U=U_x$.
The fact that $U=U_x$ is equivalent to $U_x\subset U$ (by the third and fourth properties of elements of $\omega\mathfrak U(\mathcal A)$), which is equivalent to all elements $A$ of $\mathcal A$ satisfying the following property: “exactly one of $A$ or $A^\complement$ contains $x$, and the one that does belongs to $U$”. In other words, setting $\mathcal A'$ those elements, we want to have $\mathcal A'=\mathcal A$.
In fact it is not too difficult to see that $\mathcal A'$ is a $\sigma$-algebra, and it contains the clopen sets because $x$ is not in the open cover discussed above. This concludes the proof that $U=U_x$.
Proof of the lemma: the map $i$.
To be clear, this map sends $y\in Y$ to the only $x\in X$ such that $V_y=U_x$. It is clearly well-defined and injective by the first two points of the lemma. The fact that $\phi(A)$ is always the inverse image of $A$ under $i$ is a matter of unfolding the definitions:
$$ y\in\phi(A)
\Leftrightarrow A\in V_y
\Leftrightarrow A\in U_{i(y)}
\Leftrightarrow i(y)\in A
\Leftrightarrow y\in i^{-1}A. $$
¹ For every pair $x,x'\in X$ of distinct elements, there is some $A\in\mathcal A$ containing precisely one of the two.
