On the Jayne-Rogers theorem

TL;DR

This paper extends the Jayne-Rogers theorem to a broader class of spaces, showing that $ ext{Δ}^0_2$-measurability coincides with piecewise continuity for mappings from perfectly paracompact, first-countable spaces to regular spaces.

Contribution

It generalizes the classical Jayne-Rogers theorem to include perfectly paracompact, first-countable spaces, broadening its applicability.

Findings

$ ext{Δ}^0_2$-measurability is equivalent to piecewise continuity in the new setting.

The result applies to mappings from perfectly paracompact, first-countable spaces.

Provides a unified framework extending the classical theorem.

Abstract

In 1982, J.E. Jayne and C.A. Rogers proved that a mapping $f \colon X \rightarrow Y$ of an absolute Souslin-F set X to a metric space Y is $\Delta^0_2$-measurable if and only if it is piecewise continuous. We now give a similar result for a perfectly paracompact first-countable space X and a regular space Y.

Equations

$$A^f\cap X^{\prime }\cap V_i\cap V_j\in {\mathcal{I}}_f,$$

$$X_t=\bigcup \left\{{\bigcap }_{n\in \omega }{F}_{t\hat{}\beta |n}:\beta \in {\omega }^{\omega }\right\}.$$

$$s⪯t⟺\mathrm{length}(s)<\mathrm{length}(t)\vee (\mathrm{length}(s)=\mathrm{length}(t)\wedge s{\le }_{\mathrm{lex}}t),$$

$$f^{-1}(O)\cap V_s\cap W_{\upsilon (s)}^s=A^f\cap V_s\cap W_{\upsilon (s)}^s\notin {\mathcal{I}}_f.$$

$$f^{-1}(O)\cap V_s\cap \left(W_{\upsilon (s)}^s\setminus \overline{W_{k+\upsilon (s)}^s}\right)\notin {\mathcal{I}}_f.$$

$$C\subset f^{-1}(O)\cap \left(W_{\upsilon (s)}^s\setminus \overline{W_{1+\upsilon (s)}^s}\right)$$

$$X^{\prime }=C\cap V_s\cap X_{h(s)\hat{}n}\notin {\mathcal{I}}_f.$$

Full text

In 1982, J.E. Jayne and C.A. Rogers [4] proved the following

The original proof of this theorem is long and quite complicated. Therefore Kačena, Motto Ros, and Semmes [5] presented a new short proof of Theorem 1. Moreover, they showed that Theorem 1 holds for a regular space $Y$. A further sharpening of the Jayne–Rogers theorem is connected with the study of $\mathbf{\Sigma}^{0}_{2}$-measurable mappings which are not piecewise continuous. Put this way, Solecki [7, Theorem 3.1] demonstrated that, given a $\mathbf{\Sigma}^{0}_{2}$-measurable mapping $f\colon X\rightarrow Y$ of an analytic space $X$ to a separable metric space $Y$, either $f$ is piecewise continuous or $X$ contains a closed subset $D$ homeomorphic to the Cantor set $2^{\omega}$ such that the restriction $f|D$ is not $\mathbf{\Delta}^{0}_{2}$-measurable. By [5, Corollary 6], the last statement is valid for an absolute Souslin-$\mathcal{F}$ set $X$ and a regular space $Y$.

Piecewise continuous mappings between Polish spaces have recently been investigated by R. Carroy and B.D. Miller [2].

Notation. For all undefined terms see [3].

A space $X$ is said to be a Souslin-$\mathcal{F}$ set in $Z$ if $X$ is a result of the $\mathcal{A}$-operation applied to a system of closed subsets of $Z$. In particular, a space $X$ is an absolute Souslin-$\mathcal{F}$ set if it is metrizable and a Souslin-$\mathcal{F}$ set in the completion $\tilde{X}$ of $X$ under its compactible metric.

A space $X$ is called perfectly paracompact if $X$ is paracompact and each closed subset of $X$ is of type $G_{\delta}$ in $X$. A space $X$ is called a Baire space if the intersection of countably many dense open sets in $X$ is dense. We call a space completely Baire if every closed subspace of it is Baire.

A mapping $f\colon X\rightarrow Y$ is said to be

1. $\bullet$

$\mathbf{\Delta}^{0}_{2}$-measurable if $f^{-1}(U)\in\mathbf{\Delta}^{0}_{2}(X)$ for every open set $U\subset Y$, 2. $\bullet$

$\mathbf{\Sigma}^{0}_{2}$-measurable if $f^{-1}(U)\in\mathbf{\Sigma}^{0}_{2}(X)$ for every open set $U\subset Y$, 3. $\bullet$

piecewise continuous if $X$ can be covered by a sequence $X_{0},X_{1},\ldots$ of closed sets such that the restriction $f|X_{n}$ is continuous for every $n\in\omega$.

Now we are ready to give the main result of the paper.

$\overline{D}$ is a Baire space as a closed subset of the completely Baire space $Z$. Then $D$ is not a $G_{\delta}$-subset of $\overline{D}$; otherwise $\overline{D}$ would be a set of the first category. Since $D$ is not a $\mathbf{\Delta}_{2}^{0}$-subset of $X$, $f$ is not $\mathbf{\Delta}_{2}^{0}$-measurable, a contradiction.

For the other direction, consider a piecewise continuous mapping $f\colon X\rightarrow Y$. One can readily verify that $f^{-1}(A)$ is an $F_{\sigma}$-subset of $X$ for every closed $A\subset Y$. Then $f^{-1}(U)$ is $G_{\delta}$ in $X$ for every open $U\subset Y$. Moreover, $f^{-1}(U)$ is an $F_{\sigma}$-subset of $X$ for every open $U\subset Y$ because $X$ is a perfectly paracompact space. Thus, $f$ is $\mathbf{\Delta}_{2}^{0}$-measurable. ∎

The basic Lemma 4 is given below.

To prove Theorem 2, we shall modify the technique due to Kačena, Motto Ros, and Semmes [5]. Therefore, the terminology and methods from [5] are applied.

The sets $A,B\subset Y$ are strongly disjoint if $\overline{A}\cap\overline{B}=\emptyset$.

Given a mapping $f\colon X\rightarrow Y$, let us denote by $\mathcal{I}_{f}$ the family of all subsets $A\subset X$ for which there is a set $F\in\mathbf{\Sigma}^{0}_{2}(X)$ such that $A\subset F$ and the restriction $f|F$ is piecewise continuous. In particular, $f$ is piecewise continuous if and only if $X\in\mathcal{I}_{f}$. Clearly, the family $\mathcal{I}_{f}$ is a $\sigma$-ideal.

Let $x\in X$, $X^{\prime}\subset X$, and $A\subset Y$. Define $A^{f}=f^{-1}(Y\setminus\overline{A})$. The pair $(x,X^{\prime})$ is said to be $f$-irreducible outside $A$ if for every neighborhood $V\subset X$ of $x$ we have $A^{f}\cap X^{\prime}\cap V\notin\mathcal{I}_{f}$. Otherwise we say that $(x,X^{\prime})$ is $f$-reducible outside $A$, i.e., there exist a neighborhood $V$ of $x$ and a set $F\in\mathbf{\Sigma}^{0}_{2}(X)$ such that $A^{f}\cap X^{\prime}\cap V\subset F$ and $f|F$ is piecewise continuous. Clearly, $x\in\overline{A^{f}\cap X^{\prime}}$ if $(x,X^{\prime})$ is $f$-irreducible outside $A$.

The following lemma is a slight modification of [5, Lemma 3]; in [5] it was proved for a metrizable space $X$.

1. (i)

$X^{\prime}\notin\mathcal{I}_{f}$, 2. (ii)

there exist a point $x\in\overline{X^{\prime}}\cap A^{f}$ and an open set $U\subset Y$ strongly disjoint from $A$ such that $f(x)\in U$ and the pair $(x,X^{\prime})$ is $f$-irreducible outside $U$.

(i) $\Rightarrow$ (ii): Denote by $G$ the open set $\bigcup\{O\mbox{ is open in }X\colon O\cap X^{\prime}\in\mathcal{I}_{f}\}$. By Lemma 1 we have $X^{\prime}\cap G\in\mathcal{I}_{f}$.

Assume toward a contradiction that (ii) does not hold, i.e., for every $x\in\overline{X^{\prime}}\cap A^{f}$ and every open set $U\subset Y$ strongly disjoint from $A$ such that $f(x)\in U$ we have that $(x,X^{\prime})$ is $f$-reducible outside $U$.

The intersection $Z=A^{f}\cap(\overline{X^{\prime}}\setminus G)$ is an $F_{\sigma}$-set in $X$. We claim that the restriction $f|Z$ is continuous. Suppose otherwise, so that there are $x\in Z$ and an open set $U\subset Y$ such that $f(x)\in U$ and there is no neighborhood $V$ of $x$ with $f(V\cap Z)\subset\overline{U}$. Because $f(x)\notin\overline{A}$ by the assumption, we can assume that $\overline{U}\cap\overline{A}=\emptyset$. Fix a neighborhood $V$ of $x$ satisfying $U^{f}\cap X^{\prime}\cap V\in\mathcal{I}_{f}$. By our hypothesis there is $x^{\prime}\in V\cap Z$ such that $f(x^{\prime})\notin\overline{U}\cup\overline{A}$. Using regularity of $Y$, we can find a neighborhood $U^{\prime}$ of $f(x^{\prime})$ which is strongly disjoint from $U$ and $A$. Let now $V^{\prime}$ be a neighborhood of $x^{\prime}$ given by the failure of (ii), i.e. $(U^{\prime})^{f}\cap X^{\prime}\cap V^{\prime}\in\mathcal{I}_{f}$. Since $\overline{U}\cap\overline{U^{\prime}}=\emptyset$, we have $X^{\prime}\cap V\cap V^{\prime}\subset U^{f}\cup(U^{\prime})^{f}$. Therefore $X^{\prime}\cap V\cap V^{\prime}\in\mathcal{I}_{f}$. In other words, $X^{\prime}\cap V\cap V^{\prime}\subset X^{\prime}\cap G$. But this implies $x^{\prime}\notin Z$, a contradiction. Thus $f|Z$ is continuous. Hence $X^{\prime}\cap Z\in\mathcal{I}_{f}$.

On the whole, we get $X^{\prime}=(X^{\prime}\cap Z)\cup(X^{\prime}\cap G)\in\mathcal{I}_{f}$, a contradiction with (i). ∎

For the sake of completeness, we reproduce the proof of Lemma 3 from [5].

[TABLE]

and thus $V_{i}\cap V_{j}$ contradicts the fact that $(x,X^{\prime})$ is $f$-irreducible outside $A$. ∎

Then there exist a countable subset $D=\{d_{n}\colon n\in\omega\}$ of $X$ and a sequence $\langle U_{n}\colon n\in\omega\rangle$ of disjoint open sets in $Y$ such that

1. (i)

$f(d_{n})\in U_{n}$* for every $n\in\omega$,* 2. (ii)

$D$* is homeomorphic to the space of rational numbers,* 3. (iii)

$\overline{D}\subset X$, where the bar denotes the closure in $Z$, 4. (iv)

$f(x)\notin\bigcup\{U_{n}\colon n\in\omega\}$* for every $x\in\overline{D}\setminus D$.*

[TABLE]

Clearly, $X_{t}\subset F_{t}$ and $X_{t}=\bigcup\{X_{t\hat{\>}n}\colon n\in\omega\}$ for every $t\in\omega^{<\omega}$.

Denote by $2^{<\omega}$ the set of all binary sequences of finite length. The construction will be carried out by induction with respect to the order $\preceq$ on $2^{<\omega}$ defined by

[TABLE]

where $\leq_{\mathrm{lex}}$ is the usual lexicographical order on $2^{\operatorname{length}(s)}$. We write $s\prec t$ if $s\preceq t$ and $s\neq t$.

The map $\upsilon$ assigns to each $t\in 2^{<\omega}$ the length of a string of zeros at the end of $t$; for example, $\upsilon(110100)=2$.

We will construct a sequence $\langle x_{s}\colon s\in 2^{<\omega}\rangle$ of points of $X$, a sequence $\langle W^{s}_{n}\colon n\in\omega\rangle$ of open subsets of $Z$, a sequence $\langle V_{s}\colon s\in 2^{<\omega}\rangle$ of subsets of $X$, a sequence $\langle U_{s}\colon s\in 2^{<\omega}\rangle$ of open subsets of $Y$, and a mapping $h\colon 2^{<\omega}\rightarrow\omega^{<\omega}$ such that for every $s\in 2^{<\omega}$:

1. (1)

$\langle X\cap W^{s}_{n}\colon n\in\omega\rangle$ forms a base at the point $x_{s}$ with respect to $X$ and $\overline{W^{s}_{n+1}}\subset W^{s}_{n}$ for each $n\in\omega$, 2. (2)

if $t\subset s$ then $V_{s}\subset V_{t}$, 3. (3)

$x_{s}\in X\cap\overline{V_{s}}$ and $\overline{V_{s}}\subset W^{s}_{0}$, 4. (4)

if $s=t\hat{\>}0$ then $x_{s}=x_{t}$ and $U_{s}=U_{t}$, 5. (5)

$f(x_{s})\in U_{s}$, 6. (6)

if the last digit of $s$ is 1 then $f(X\bigcap\overline{V_{s}})\cap\overline{\bigcup_{r\prec s}U_{r}}=\emptyset$, 7. (7)

$(x_{t},V_{t})$ is $f$-irreducible outside $A$ for every $t\preceq s$, where $A=\bigcup\{U_{r}\colon r\preceq s\}$, 8. (8)

the family $\{V_{t}\colon t\in 2^{n}\}$ is pairwise strongly disjoint in $X$ for every $n\in\omega$, 9. (9)

the family $\{U_{t}\colon t\preceq s\;\wedge$ (the last digit of $t$ is 1) or $t=\emptyset\}$ is pairwise strongly disjoint, 10. (10)

if $t\varsubsetneq s$ then $h(t)\subset h(s)$ and if, moreover, the last digit of $s$ is 1 then $\operatorname{length}(h(t))<\operatorname{length}(h(s))$, 11. (11)

$V_{s}\subset X_{h(s)}\subset F_{h(s)}$, 12. (12)

$\overline{V_{s\hat{\>}0}}\subset W^{s}_{1+\upsilon(s)}$ and $\overline{V_{s\hat{\>}1}}\subset W^{s}_{\upsilon(s)}\setminus\overline{W^{s}_{1+\upsilon(s)}}$.

For the base step of the induction, let $x$ and $U$ be given as in Lemma 2 applied to $X^{\prime}=X$ and $A=\emptyset$. Then put $x_{\emptyset}=x$, $U_{\emptyset}=U$, and $h(\emptyset)=\emptyset$. Choose a sequence $\langle W^{\emptyset}_{n}\colon n\in\omega\rangle$ satisfying condition (1) for $s=\emptyset$. Define $V_{\emptyset}=X\cap W^{\emptyset}_{1}$.

Assume that $x_{t}$, $W^{t}_{n}$, $V_{t}$, and $U_{t}$ have been constructed for every $t\prec s\hat{\>}0$. Let $A=\bigcup\{U_{t}\colon t\prec s\hat{\>}0\}$ and $O=Y\setminus\overline{A}$. By the inductive hypothesis, condition (7) says that $(x_{s},V_{s})$ is $f$-irreducible outside $A$. Then

[TABLE]

The decreasing sequence $\langle X\cap W^{s}_{n}\colon n\in\omega\rangle$ forms a base at $x_{s}$. Since $X$ is perfectly paracompact, every set $X\cap(W^{s}_{n}\setminus\overline{W^{s}_{n+1}})$ is $F_{\sigma}$ in $X$. Now we can find some $k>0$ with

[TABLE]

To simplify notation, we can assume that $k=1$. Since $f$ is $\mathbf{\Sigma}_{2}^{0}$-measurable, there exists a set $C$ closed in $X$ such that

[TABLE]

and $C\cap V_{s}\notin\mathcal{I}_{f}$. By construction, $V_{s}\subset X_{h(s)}=\bigcup\{X_{h(s)\hat{\>}n}\colon n\in\omega\}$. Then for some $n$ we have

[TABLE]

Put $h(s\hat{\>}0)=h(s)$ and $h(s\hat{\>}1)=h(s)\hat{\>}n$.

Next, set $x_{s\hat{\>}0}=x_{s}$, $U_{s\hat{\>}0}=U_{s}$, $W^{s\hat{\>}0}_{n}=W^{s}_{n+1}$, and $V_{s\hat{\>}0}=V_{s}\cap W^{s\hat{\>}0}_{1}$.

Claim. There are $x_{s\hat{\>}1}\in\overline{X^{\prime}}$ and $U_{s\hat{\>}1}\subset Y$ such that $f(x_{s\hat{\>}1})\in U_{s\hat{\>}1}$, $U_{s\hat{\>}1}$ is open and strongly disjoint from $A$, $(x_{t},V_{t})$ is $f$-irreducible outside $A\cup U_{s\hat{\>}1}$ for every $t\prec s\hat{\>}1$ and $(x_{s\hat{\>}1},X^{\prime})$ is $f$-irreducible outside $A\cup U_{s\hat{\>}1}$.

Proof of the Claim. Let $k=|\{t\in 2^{<\omega}\colon t\prec s\hat{\>}1\}|$. Using Lemma 2, for $\jmath=0,\ldots,k$ recursively construct $x_{\jmath}$ and $U_{\jmath}$ such that $f(x_{\jmath})\in U_{\jmath}$, $U_{\jmath}$ is open and strongly disjoint from $A\cup U_{<\jmath}$ (where $U_{<\jmath}=\emptyset$ if $\jmath=0$ and $U_{<\jmath}=\bigcup\{U_{i}\colon i<\jmath\}$ otherwise), and $(x_{\jmath},X^{\prime}\cap(U_{<\jmath})^{f})$ is $f$-irreducible outside $U_{\jmath}$. According to Lemma 3, for each $t\prec s\hat{\>}1$ there is at most one $j_{t}\in\{0,\ldots,k\}$ such that $(x_{t},V_{t})$ is $f$-reducible outside $A\cup U_{j_{t}}$. The pigeonhole principle implies that the claim is satisfied with $x_{s\hat{\>}1}=x_{\jmath^{*}}$ and $U_{s\hat{\>}1}=U_{\jmath^{*}}$ for some $\jmath^{*}\in\{0,\ldots,k\}$. $\triangle$

Choose a sequence $\langle W^{s\hat{\>}1}_{n}\colon n\in\omega\rangle$ of open subsets of $Z$ such that $\langle X\cap W^{s\hat{\>}1}_{n}\colon n\in\omega\rangle$ is a base at the point $x_{s\hat{\>}1}$ satisfying $\overline{W^{s\hat{\>}1}_{0}}\subset W^{s}_{\upsilon(s)}\setminus\overline{W^{s}_{1+\upsilon(s)}}$ and $\overline{W^{s\hat{\>}1}_{n+1}}\subset W^{s\hat{\>}1}_{n}$. Finally, set $V_{s\hat{\>}1}=X^{\prime}\cap W^{s\hat{\>}1}_{1}$. Obviously, $V_{s\hat{\>}1}\subset V_{s}$.

By construction, $\overline{V_{s\hat{\>}0}}\subset W^{s}_{1+\upsilon(s)}$ and $\overline{V_{s\hat{\>}1}}\subset W^{s}_{\upsilon(s)}\setminus\overline{W^{s}_{1+\upsilon(s)}}$. Then $\overline{V_{s\hat{\>}0}}\cap\overline{V_{s\hat{\>}1}}=\emptyset$. Together with (2), this implies condition (8). One can check that all the conditions (1)–(12) are satisfied.

Define the set $D=\{x_{s}\colon s\in 2^{<\omega}\}$. The set $f(D)$ is relatively discrete in $Y$ because $f(x_{s})\in U_{s}$ and $U_{s}\cap U_{t}=\emptyset$ whenever $x_{s}\neq x_{t}$. Clearly, the restriction $f|D$ is a bijection.

The set $D$ is countable and has a countable base at each point; hence, the space $D$ is second-countable. By the Urysohn theorem [3, Theorem 4.2.9], $D$ is metrizable. From conditions (1) and (3) it follows that $D$ has no isolated points. The Sierpiński theorem [6] implies that $D$ is homeomorphic to the space of rationals.

Let us check that $\overline{D}\subset X$. From conditions (2)–(4) it follows that $D\subset V^{*}_{n}=\bigcup\{\overline{V_{s}}\colon s\in 2^{n}\}$ for each $n\in\omega$. Since $2^{n}$ is a finite set, $V^{*}_{n}$ is closed in $Z$. Then $\overline{D}$ belongs to a closed subset $V^{*}=\bigcap\{V^{*}_{n}\colon n\in\omega\}$ of $Z$. Clearly, $D\subset X$. Given a point $x\in\overline{D}\setminus D$, fix $s(x,n)$ such that $x\in\overline{V_{s(x,n)}}$ and $s(x,n)\in 2^{n}$ for each $n$. Using conditions (2) and (8), find a unique $\alpha_{x}\in 2^{\omega}$ with $s(x,n)=\alpha_{x}|n$. Since $x\notin D$, the sequence $\alpha_{x}$ contains infinitely many digits $1$. By (10), the sequence $\langle h(\alpha_{x}|n)\colon n\in\omega\rangle$ has infinitely many distinct members. From (11) it follows that $x\in\bigcap\{\overline{V_{\alpha_{x}|n}}\colon n\in\omega\}\subset\bigcap\{F_{h(\alpha_{x}|n)}\colon n\in\omega\}\subset X$. Therefore, $\overline{D}\subset X$.

To obtain (iv), take a point $x\in\overline{D}\setminus D$. Striving for a contradiction, suppose that $f(x)\in U_{r}$ for some $r\in 2^{<\omega}$. By the above, $x\in\bigcap\{\overline{V_{\alpha_{x}|n}}\colon n\in\omega\}$. Since the sequence $\alpha_{x}$ contains infinitely many $1$, there is $n$ such that the last digit of $\alpha_{x}|n$ is $1$ and $r\prec\alpha_{x}|n$. From condition (6) it follows that $f(x)\notin\overline{U_{r}}$, a contradiction.

It remains to number the distinct elements of $D$ as $\{d_{n}\colon n\in\omega\}$. ∎