Josia Pietsch 1d96095f62
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% \begin{refproof}{thm:kuratowskiulam}
\item Let $A$ be a set with the Baire property.
Write $A = U \symdif M$
for $U$ open and $M$ meager.
Then for all $x$,
we have that $A_x = U_x \symdif M_x$,
where $U_x$ is open,
and $\{x : M_x \text{ is meager}\}$ is comeager.
Therefore $\{x : U_x \text{ open } \land M_x \text{ meager }\}$
is comeager,
and for those $x$, $A_x$ has the Baire property.
% TODO: fix counter
\begin{claim} % Claim 2
For $P \subseteq X$, $Q \subseteq Y$ with the Baire
property, let $R \coloneqq P \times Q$.
Then $R$ is meager iff at least one of $P$ or $Q$ is meager.
Suppose that $R$ is meager.
Then by \yaref{thm:kuratowskiulam:c1b},
we have that $C = \{x : R_x \text{ is meager }\}$ is comeager.
\item If $P$ is meager, the statement holds trivially.
\item If $P$ is not meager,
then $P \cap C \neq \emptyset$.
For $x \in P \cap C$
we have that $R_x$ is meager
and $R_x = Q$,
hence $Q$ is meager.
On the other hand suppose that $P$ is meager.
Then $P = \bigcup_{n} F_n$ for nwd sets $F_n$.
Note that $F_n \times Y$ is nwd.
So $F_n \times Q$ is also nwd.
Hence $P \times Q$ is a countable union
of nwd sets,
so it is meager.
Let $A$ be a set with the Baire property
such that $\{x : A_x \text{ is meager}\}$ is comeager.
Let $A = U \symdif M$ for $U$ open and $M$ meager.
Towards a contradiction suppose that $A$ is not meager.
Then $U$ is not meager.
Since $X \times Y$ is second countable,
we have that $U$ is a countable union of open rectangles.
At least one of them, say $G \times H \subseteq U$,
is not meager.
By \yaref{thm:kuratowskiulam:c2},
both $G$ and $H$ are not meager.
$\{x\colon A_x \text{ is meager} \land M_x \text{ is meager}\}$
is comeager (using \yaref{thm:kuratowskiulam:c1b}),
there is $x_0 \in G$ such that $A_{x_0}$ is meager and $M_{x_0}$
is meager.
But then $H$ is meager as
H \setminus M_{x_0} \subseteq U_{x_0} \setminus M_{x_0}
\subseteq U_{x_0} \symdif M_{x_0} = A_{x_0}
and $M_{x_0}$ is meager $\lightning$.
This is \yaref{thm:kuratowskiulam:c1b}.
\item (ii) $\iff$ (iii): pass to complement.
\item $F \overset{\text{closed}}{\subseteq} X \times Y$ nwd.
$\implies \{x \in X : F_x \text{ nwd}\} $ comeager:
\item $W = F^c$ is open and dense, show that $\{x : W_x \text{ dense}\}$
is comeager.
\item $(V_n)$ enumeration of basis. Show that $U_n \coloneqq \{x : V_n \cap W_x \neq \emptyset\}$
is comeager for all $n$.
\item $U_n$ is open (projection of open) and dense ($W$ is dense, hence $W \cap ( U \times V_n) \neq \emptyset$ for $U$ open).
\item $F \subseteq X \times Y$ is nwd $\implies \{x \in X: F_x \text{ nwd}\}$ comeager.
(consider $\overline{F}$).
\item (ii) $\implies$:
$M \subseteq X \times Y$ meager $\implies \{x \in X: M_x \text{ meager}\}$ comeager
(write $M$ as ctbl. union of nwd.)
\item (i): If $A$ has the Baire Property,
then $A = U \symdif M$, $A_x = U_x \symdif M_x$,
$U_x$ open and $\{x : M_x \text{ meager}\}$ comeager
$\implies$ (i).
\item $P \subseteq X$, $Q \subseteq Y$ BP,
then $P \times Q$ meager $\iff$ $P$ or $Q$ meager.
\item $\impliedby$ easy
\item $\implies$ Suppose $P \times Q$ meager, $P$ not meager.
$\emptyset\neq P \cap \underbrace{\{x : (P \times Q)_x \text{ meager} \}}_{\text{comeager}} \ni x$.
$(P \times Q)_x = Q$ is meager.
\item (ii) $\impliedby$:
\item $A$ BP, $\{x : A_x \text{ meager}\}$ comeager.
\item $A = U \symdif M$.
\item Suppose $A$ not meager $\leadsto$ $U$ not meager
$\leadsto \exists G \times H \subseteq U$ not meager.
\item $G$ and $H$ are not meager.
\item $\exists x_0 \in G \cap \underbrace{\{x: A_x \text{ meager } \land M_x \text{ meager}\}}_\text{comeager}$.
\item $H$ meager, as
H \subseteq U_{x_0} \subseteq A_{x_0} \cup M_{x_0}.
Suppose that $A$ has the BP.
Then there is an open $U$ such that
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
Then $A = U \symdif M$.
\section{Borel sets} % TODO: fix chapters
Let $X$ be a topological space.
Let $\cB(X)$ denote the smallest $\sigma$-algebra,
that contains all open sets.
Elements of $\cB(X)$ are called \vocab{Borel sets}.
Note that all Borel sets have the Baire property.
\subsection{The hierarchy of Borel sets}
\gist{Let $\omega_1$ be the first uncountable ordinal.}{}
For every $d < \omega_1$,
we define by transfinite recursion
classes $\Sigma^0_\alpha$
and $\Pi^0_\alpha$
(or $\Sigma^0_\alpha(X)$ and $\Pi^0_\alpha(X)$ for a topological space $X$).
Let $X$ be a topological space.
Then define
\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]
\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
\{X \setminus A | A \in \Sigma^0_\alpha(X)\},
and for $\alpha > 1$
\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
Note that $\Pi_1^0$ is the set of closed sets,
$\Sigma^0_2 = F_\sigma$,
and $\Pi^0_2 = G_\delta$.
Furthermore define
\Delta^0_\alpha(X) \coloneqq \Sigma^0_\alpha(X) \cap \Pi^0_\alpha(X),
i.e.~$\Delta^0_1$ is the set of clopen sets.
\adjustbox{width=\textwidth, center}{%
% https://q.uiver.app/#q=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
& {\substack{\Sigma_1^0\\\text{open}}} && {\substack{\Sigma^0_2\\F_\sigma}} &&&& {\Sigma^0_\xi} \\
{\substack{\Delta_1^0\\\text{clopen}}} && {\Delta^0_2} && {\Delta^0_3} & \ldots & {\Delta^0_\xi} && {\Delta^0_{\xi + 1}} & \ldots \\
& {\substack{\Pi^0_1\\\text{closed}}} && {\substack{\Pi_2^0\\G_\delta}} &&&& {\Pi^0_\xi}
\arrow["\subseteq", hook, from=2-1, to=1-2]
\arrow["\subseteq"', hook, from=2-1, to=3-2]
\arrow["\subseteq"', hook, from=3-2, to=2-3]
\arrow["\subseteq", hook, from=1-2, to=2-3]
\arrow["\subseteq", hook, from=2-3, to=1-4]
\arrow["\subseteq"', hook, from=2-3, to=3-4]
\arrow["\subseteq", hook, from=2-7, to=1-8]
\arrow["\subseteq"', hook, from=2-7, to=3-8]
\arrow["\subseteq", hook, from=1-4, to=2-5]
\arrow["\subseteq"', hook, from=3-4, to=2-5]
\arrow["\subseteq", hook, from=1-8, to=2-9]
\arrow["\subseteq"', hook, from=3-8, to=2-9]
Let $X$ be a metrizable space.
\item $\Sigma^0_\eta(X) \cup \Pi^0_\eta(X) \subseteq \Delta^0_\xi(X)$
for all $1 \le \eta < \xi < \omega_1$.
\item $\cB(X) = \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
\item \begin{observe}
For all $1 \le \alpha < \beta < \omega_1$,
we have $\Pi^0_\alpha(X) \subseteq \Sigma^0_\beta(X)$
by taking ``unions'' of singleton sets.
Furthermore $\Sigma^0_\alpha(X) \subseteq \Pi^0_\beta(X)$
by passing to complements.
It suffices to show $\Sigma^0_\eta(X) \subseteq \Delta^0_\xi(X)$,
since $\Delta^0_\eta(X)$ is closed under complements.
Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$,
by the observation
(since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$
and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$).
So to prove (a) it suffices to show that for all $1 \le \eta < \xi < \omega_1$,
we have $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$.
For $\eta = 1, \xi = 2$
this holds, since every open set is $F_\sigma$.%
\footnote{Here we use that $X$ is metrizable!}
% \todo{REF}
For $\eta > 1, \xi > \eta$,
we have
\Sigma^0_\eta(X) &=&
\{ \bigcup_{n} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \eta\}\\
\{\bigcup_{n}B_n : B_n \in \Pi^0_{\beta_n}(X), \beta_n < \xi\}
= \Sigma^0_\xi(X).
\item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
We need to show that $\cB_0 = \cB(X)$.
Clearly $\cB_0 \subseteq \cB(X)$.
It suffices to notice that $\cB_0$ is a $\sigma$-algebra
containing all open sets.
Consider $\bigcup_{n < \omega} A_n$ for some $A_n \in B_0$.
Then $A_n \in \Pi^0_{\alpha_n}(X)$ for some $\alpha_n < \omega_1$.
Let $\alpha = \sup \alpha_n < \omega_1$.
Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$.
It is clear that $\cB_0$ is closed under complements.
\item It suffices to show that $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$
for all $1 \le \eta < \xi < \omega_1$.
For $\eta = 1, \xi = 2$ this holds,
since open sets of a metrizable space are $F_\sigma$.
\item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
This is a $\sigma$-algebra containing all open sets.
% TODO move to counter examples.
Consider the cofinite topology on $\omega_1$.
Then the non-empty open sets of this are not $F_\sigma$.