Proposition 8.23, Kechris (Classical Descriptive Set Theory) meager $F_\sigma$ set contains $A\Delta U$.

Question

Let $X$ be a topological space and $A\subseteq X$. Then the following statements are equivalent:

I) $A$ has the Baire property (BP);

II) $A=G\cup M$, where $G$ is $G_\delta$ and $M$ is meager;

III) $A=F\setminus M$ where $F$ is $F_\sigma$ and $M$ is meager.

In the proof of $I)\Rightarrow II)$, it is noted that for $U$ open we have a meager $F_\sigma$ set $F$ such that $A\Delta U\subseteq F$.

I do not see why such a set exists in general, or how to proof the existence.

Can you help me out? Thanks in advance.

Answer 1

If $A$ has the Baire property, there is an open set $U$ such that $M=A\mathrel{\triangle}U$ is meagre. Thus, there is a countable family $\mathscr{N}$ of nowhere dense sets such that $M=\bigcup\mathscr{N}$. For each $N\in\mathscr{N}$ the set $\operatorname{cl}N$ is closed and nowhere dense, so

$$F=\bigcup\{\operatorname{cl}N:N\in\mathscr{N}\}$$

is a meagre $F_\sigma$ set, and clearly $M\subseteq F$.