Baire property sigma-algebra and Borel regularity

A subset of a nowhere dense set is nowhere dense since closure is monotone. A finite union of nowhere dense sets is nowhere dense: inside any nonempty open, successively refine to a nonempty open avoiding each of the finitely many closures; the final refinement avoids their union. Subsets of meagre sets retain the same covering witnesses. For a sequence of meagre sets, A1 chooses a nowhere dense covering sequence for each; the diagonal pairing of their two natural indices yields a covering sequence for the union. Thus meagre sets form an ideal closed under countable unions. The empty set has the all-empty witness sequence.

For closed F, $F\setminus \mathrm{int}F$ is closed with empty interior: a nonempty open contained in it would be contained in F and hence in its interior, a contradiction. Thus F differs from its open interior by a nowhere dense set and has the Baire property. If U is open, its boundary $\bar{U}\setminus U$ is closed nowhere dense: any nonempty open inside $\bar{U}$ must meet U, precluding containment in that difference. These statements use only the closure and interior definitions, so hold without separation axioms.

Suppose $A\mathrm{△}U$ is meagre with U open. Its complementary set differs from closed $X\setminus U$ by the same error; step 1.2 replaces that closed set by its open interior at a further nowhere dense error. Step 1.1 therefore makes the complement Baire-property. For a sequence of Baire-property $A_n$, A1 selects witnessing opens U_n. The error $({\bigcup }_n{A}_n)\mathrm{△}({\bigcup }_n{U}_n)$ is contained in ${\bigcup }_n(A_n\mathrm{△}{U}_n)$, meagre by step 1.1. Thus this class is a sigma-algebra, containing opens and all meagre sets by F1. F2's leastness puts every Borel set in it.

Enclose $A\mathrm{△}U$ in a meagre $F_{\sigma }$ set $M={\bigcup }_n\overline{N_n}$ by closing its specified nowhere dense witnesses. Put $G=U\setminus M$ and $H=\bar{U}\cup M$. Then G is $G_{\delta }$, since $G={\bigcap }_n(U\setminus \overline{N_n})$, and H is $F_{\sigma }$. We have $G\subseteq A\subseteq H$. Moreover $A\setminus G\subseteq M$ and $H\setminus A\subseteq M\cup (\bar{U}\setminus U)$, both meagre by steps 1.1–1.2. These give the two required meagre symmetric differences, also when X, U or M is empty. QED.