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Let $X$, $Y$ be metric spaces and $f: X \to Y$. Suppose $X$ is Baire and $Y$ is separable. If $f ^{− 1} (O)$ is $F_{\sigma}$ for every open $O \subset Y$, show that $f$ is continuous in a dense $G_{\delta}$ of $X$

In the Dissertationes Matematicae (Rozprawy Matematyczne) book they prove that if $X$ is a Baire space, $Y$ is a space that has a countable base and $f: X \to Y$, then there exists a dense subset $D$ such that $f |_{D}$ is continuous (Theorem 3.6). Also, in How to show that the set of points of continuity is a $G_{\delta}$ they show that the set of continuity points of a function is $G_{\delta}$. Therefore $D$ would be $G_\delta$. But I don't know how to use the hypothesis that $f ^{− 1} (O)$ is $F_{\sigma}$ for every open $O \subset Y$. Any idea please?

user1999
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  • $f$ is continuous on a dense $G_{\delta}$ means there exists a dense $G_{\delta}$ set on which $f$ is continuous. – Kavi Rama Murthy Dec 01 '20 at 23:32
  • @KaviRamaMurthy Do you happen to have any idea how to build that $G_\delta$, to proceed with the test, would you please? – user1999 Dec 02 '20 at 00:32
  • What if $Z=X$? Then $f|_Z^{-1}(F)=f^{-1}(F).$ But if $f^{-1}(F)$ is closed in $X$ (i.e. in $Z$) for $every$ closed $F$ in $Y$ then $f$ is continuous – DanielWainfleet Dec 02 '20 at 13:39
  • @DanielWainfleet Why $f^{-1}(F)$ is closed in $X$? – user1999 Dec 02 '20 at 14:05
  • In your A : "... we have to see that $f|_Z^{-1}(F)$ is closed in $Z.$ " – DanielWainfleet Dec 03 '20 at 06:38
  • The point here is that $F_\sigma$ sets have the Baire property, i.e. they are open sets modulo a meager set (set contained in a countable union of nowhere dense sets). So if $(O_n)n$ is a countable basis of topology in $Y$, then for each $n$, $f^{-1}[O_n]$ is open up to a meager set $M_n$. If you remove the countable union $\bigcup_n M_n$ from $X$, you will be left with a comeager subset (containing dense $G\delta$ subset) of $X$ on which $f$ is continous. – user446046 Dec 03 '20 at 13:01
  • Could you please explain to me why the $F_ {\delta}$ are open sets modulo a meager set? – user1999 Dec 03 '20 at 13:16
  • You can try to show that sets with this (Baire) property form a $\sigma$-algebra that contains all open sets. In your particular $F_\sigma$ case, you only need to show that closed sets have this property and that this property is closed under taking countable unions. The latter should be straightforward. For the former, notice that for a closed set $F$, $F\setminus\mathrm{interior}F$ is nowhere dense, so meager. – user446046 Dec 03 '20 at 13:23
  • @user446046 I appreciate your suggestion. I will try to demonstrate. I would like to see a bit more detail in an answer though, thanks! – user1999 Dec 03 '20 at 13:34
  • I attempted some more detailed anwert. – user446046 Dec 03 '20 at 14:32
  • @user446046 Thank you! – user1999 Dec 03 '20 at 14:39

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Let me recall some definitions. Let $X$ be a topological space (say just a metric space). A nowhere dense set is a set $N\subseteq X$ which is not dense in any open set, i.e. for any non-empty open $U\subseteq X$ there is an non-empty open subset $V\subseteq U$ with $V\cap N=\emptyset$. A set is meager if it can be covered by a countable union of nowhere dense sets.

Topological space is Baire if it satisfies the Baire category theorem, that is, the complement of every meager set is dense. A set $B\subseteq X$ has the Baire property if it can be written as $U\triangle N$, where $U$ is open and $N$ is meager.

Suppose that $f:X\rightarrow Y$ is a map, where $X$ is Baire and $Y$ is second countable (i.e. it has a countable basis of open sets) such that for every open $U\subseteq Y$, the preimage $f^{-1}(U)$ has the Baire property. Then there is a comeager (complement of meager) subset $G\subseteq X$ such that $f\upharpoonright G$ is continuous. Such $G$ moreover contains a dense $G_\delta$ set. Indeed, let $(O_n)_n$ be a countable basis of topology of $Y$. Set $B_n:=f^{-1}(O_n)$. By the assumption, $B_n=U_n\triangle M_n$, for some open $U_n$ and meager $M_n$. Set $G:=X\setminus \bigcup_n M_n$. Clearly $G$ is comeager and for every $n$, $f^{-1}(O_n)\cap G$ is relatively open in $G$.

So to answer your question it remains to check that $F_\sigma$ sets have the Baire property. As I wrote in the comment, sets with the Baire property form a $\sigma$-algebra containing all open sets, so it contains all Borel sets. For the specific case of $F_\sigma$ sets, we first check that every closed set $F\subseteq X$ has the Baire property. Let $\mathrm{int}F$ denote the interior of $F$ (can be empty of course) and let $\partial F$ denote $F\setminus \mathrm{int}F$. It is easy to check that $\partial F$ is nowhere dense, in particular meager, so $F$ has the Baire property. So let now $F=\bigcup_n F_n$ be an $F_\sigma$ set, where each $F_n$ is closed. Then $F=(\bigcup_n \mathrm{int}F_n)\cup F\setminus \bigcup_n \mathrm{int}F_n$. We have $F\setminus\bigcup_n \mathrm{int}F_n\subseteq \bigcup_n \partial F_n$, so it is meager.

user446046
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  • Its $G$ is dense because $X$ is a Baire space and $G$ is the complement of a meager set. Why can we find that set $G_{\delta}$ contained in $G$? – user1999 Dec 03 '20 at 18:47
  • $X\setminus G$ is meager, so by definition it is contained in $N:=\bigcup_n N_n$, where each $N_n$ is a closed nowhere dense set. The set $N$ is $F_\sigma$, so the complement $X\setminus N$ is $G_\delta$ and $X\setminus N\subseteq G$. The set $X\setminus N$ is still dense by the Baire category theorem. – user446046 Dec 03 '20 at 18:54
  • If $X$ were not Baire, the only thing that is lost is the density of $G$? – user1999 Dec 04 '20 at 13:14
  • @user1999 If $X$ weren't Baire, $G$ might be (in principle) empty. – user446046 Dec 06 '20 at 08:31