\lecture{06}{2023-11-03}{}

% \begin{refproof}{thm:kuratowskiulam}
    \begin{enumerate}[(i)]
        \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.
    \end{enumerate}
    % TODO: fix counter
    \begin{claim} % Claim 2
        \label{thm:kuratowskiulam:c2}
        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.
    \end{claim}
    \begin{refproof}{thm:kuratowskiulam:c2}
        Suppose that $R$ is meager.
        Then by \yaref{thm:kuratowskiulam:c1b},
        we have that $C = \{x : R_x \text{ is meager }\}$ is comeager.
        \begin{itemize}
            \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.
        \end{itemize}

        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.
    \end{refproof}
    \begin{enumerate}[(i)]
        \item[(ii)]
            ``$\impliedby$''
            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 $A$ is a countable union of open rectangles.
            At least one of them, say $G \times H \subseteq A$,
            is not meager.
            By \yaref{thm:kuratowskiulam:c2},
            both $G$ and $H$ are not meager.
            Since
            $\{x\colon A_x \text{ is meager} \land M_x \text{ is meager}\}$
            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 $\lightning$.

            ``$\implies$''
            This is \yaref{thm:kuratowskiulam:c1b}.
    \end{enumerate}
\end{refproof}

\section{Borel sets} % TODO: fix chapters

\begin{definition}
    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}.
\end{definition}
\begin{remark}
    Note that all Borel sets have the Baire property.
\end{remark}

\subsection{The hierarchy of Borel sets}

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.

\iffalse % TODO Fix this!
\resizebox{\textwidth}{!}{
% https://q.uiver.app/#q=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
\[\begin{tikzcd}
	& {\Sigma_1^0} && {\Sigma^0_2} &&&& {\Sigma^0_\xi} \\
	{\Delta_1^0} && {\Delta^0_2} && {\Delta^0_3} & \ldots & {\Delta^0_\xi} && {\Delta^0_{\xi + 1}} & \ldots \\
	& {\Pi^0_1} && {\Pi_2^0} &&&& {\Pi^0_\xi}
	\arrow["\subseteq", hook, from=2-1, to=1-2]
	\arrow["\subseteq"', hook', from=2-1, to=3-2]
	\arrow[hook, from=3-2, to=2-3]
	\arrow[hook, from=1-2, to=2-3]
	\arrow[hook, from=2-3, to=1-4]
	\arrow[hook', from=2-3, to=3-4]
	\arrow[hook, from=2-7, to=1-8]
	\arrow[hook', from=2-7, to=3-8]
	\arrow[hook, from=1-4, to=2-5]
	\arrow[hook', from=3-4, to=2-5]
	\arrow[hook, from=1-8, to=2-9]
	\arrow[hook', from=3-8, to=2-9]
\end{tikzcd}\]%
}
\fi

\begin{proposition}
    Let $X$ be a metrizable space.
    Then
    \begin{enumerate}[(a)]
        \item $\Sigma^0_\eta(X) \cup \Pi^0_\eta(X) \subseteq \Delta^0_\xi(X)$
            for all $1 \le \eta < \xi < \omega_1$.
        \item $\cB(X) = \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    \begin{enumerate}[(a)]
        \item \begin{observe}
                \label{ob:sigmasuffices}
                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.
            \end{observe}
            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 \yaref{ob:sigmasuffices}
            (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
            \begin{IEEEeqnarray*}{rCl}
                \Sigma^0_\eta(X) &=&
                    \{ \bigcup_{n}  A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \eta\}\\
                    &\subseteq&
                    \{\bigcup_{n}B_n : B_n \in \Pi^0_{\beta_n}(X), \beta_n < \xi\}
                    = \Sigma^0_\xi(X).
            \end{IEEEeqnarray*}
        \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.
    \end{enumerate}

\end{proof}

\begin{example}
    % TODO move to counter examples.
    Consider the cofinite topology on $\omega_1$.
    Then the non-empty open sets of this are not $F_\sigma$.
\end{example}
