Baire and Measure Please,I would like help with this problem.
We say that a subset $E$ of a topological space $X$ has the property of Baire if $E=G\Delta P$, where $G\subseteq X$ is open and $P \subseteq X$ is meager. Prove the following statements.


*

*A set $E\subset X$ has the property of Baire if and only if $E = F \Delta Q$, where $F \subset X$ is closed and $Q \subset X$ is meager.

*If $ E \subset X $ has the property of Baire, then $ E^c $ also has.

*The class ${\cal{B}}_a(X)=\left\{E \subseteq  X \ ; \ E \ \mbox {has the property of Baire} \right\}$ is a  $\sigma$-algebra.

*The $\sigma$-algebra ${\cal {B}}_a(X)$ is generated by the open subsets of $X$ with the meager subsets of $X$.

 A: Remark: notice that for any sets $A,B,C$, we have $A=B\Delta C$ iff $B=A\Delta C$ iff $C=A\Delta B$.


*

*Let's prove both implications simultaneously: Suppose $E=G\Delta P$ for $G$ open (closed) and $P$ meager. Let
$G'=\begin{cases}\overline{G}&\text{, if }G\text{ is open}\\\operatorname{int}G&\text{, if }G\text{ is closed}\end{cases}$.
Notice that, in both cases,
$$E\Delta G'\subseteq (E\Delta G)\cup (G\Delta G')=\underbrace{P}_{\text{meager}}\cup\underbrace{\partial G}_{\text{meager}}$$
(you can show that the frontier of open and closed sets is always nowhere dense), so $Q:=E\Delta G'$ is meager. By the remark above, $E=G'\Delta Q$, with $G'$ closed (open) and $Q$ meager.

*If $E$ has the Baire Property, then $E=G\Delta P$, $G$ open and $P$ meager. Since $E^c=\underbrace{G^c}_{\text{closed}}\Delta \underbrace{P}_{\text{meager}}$, then $E^c$ has the Baire Property, by item 1.

*By item 2, $\mathcal{B}_a(X)$ is closed by complements, so it remains only to show that $\mathcal{B}_a(X)$ is closed by countable unions: Suppose $\left\{E_n\right\}_{n\in\mathbb{N}}\subseteq \mathcal{B}_a(X)$. Then each $E_n$ can be written as $E_n=G_n\Delta P_n$, $G_n$ closed and $P_n$ meager. Notice that
$$\left(\bigcup_{n\in\mathbb{N}}E_n\right)\Delta\left(\bigcup_{n\in\mathbb{N}}G_n\right)\subseteq\bigcup_{n\in\mathbb{N}}(E_n\Delta G_n)=\underbrace{\bigcup_{n\in\mathbb{N}}\underbrace{P_n}_{\text{meager}}}_{\text{meager}}$$
so $Q=\left(\bigcup_{n\in\mathbb{N}}E_n\right)\Delta\left(\bigcup_{n\in\mathbb{N}}G_n\right)$ is meager. By the remark, $\bigcup_{n\in\mathbb{N}}E_n=\underbrace{\left(\bigcup_{n\in\mathbb{N}}G_n\right)}_{\text{open}}\Delta \underbrace{Q}_{\text{meager}}$ has the Baire Property.
Therefore, $\mathcal{B}_a(X)$ is a $\sigma$-algebra.

*Every element of $\mathcal{B}_a(X)$ is the symmetric difference of open and meager sets, so $\mathcal{B}_a(X)$ is contained in the $\sigma$-algebra generated by open and meager subsets of $X$. Conversely, every open/meager subset $A$ of $X$ can be written as $A=A\Delta\varnothing$, so $A$ has the Baire Property, that is, $A\in\mathcal{B}_a(X)$. This shows that $\mathcal{B}_a(X)$ is generated by the meager and open subsets of $X$.
