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**A new proof of a theorem of Jayne and Rogers.**
*(English)*
Zbl 1217.03032

J. E. Jayne and C. A. Rogers [“First level Borel functions and isomorphisms”, J. Math. Pures Appl., IX. Sér. 61, 177–205 (1982; Zbl 0514.54026)] proved that if \(X\) is an abolute Souslin \(\mathcal{F}\)-set and \(Y\) is a metrizable space, then for every function \(f:X\to Y\) the following conditions are equivalent: (i) \(f\) is \(\mathbb G_\delta\)-measurable (i.e. preimages of open sets in \(Y\) are countable intersections of open sets in \(X\)), (ii) \(f\) is piecewise continuous (i.e. \(X\) can be covered with countably many closed subsets on each of which \(f\) is continuous). The paper under review provides a new proof of this theorem.

There is one point in the proof which deserves a comment. The theorem concerns absolute Souslin \(\mathcal{F}\)-sets. A subset of a metric space is said to be a Souslin \({\mathcal F}\)-set if it can be obtained by the Souslin operation \(\mathcal{A}\) from closed sets. A metric space is called an absolute Souslin \({\mathcal F}\)-set if it is a Souslin \({\mathcal F}\)-set in its metric completion.

This paper provides a new proof of the implication from (i) to (ii) in the case \(X\) is completely metrizable. To obtain the general result, the authors follow an argument of S. Solecki [“Decomposing Borel sets and functions and the structure of Baire class 1 functions”, J. Am. Math. Soc. 11, No. 3, 521–550 (1998; Zbl 0899.03034)] on the reduction of the case where \(X\) is an absolute Souslin \({\mathcal F}\)-set to the case where \(X\) is completely metrizable. The reduction (at page 198 of this paper) is as follows: If \(f\) is a \(\mathbb G_\delta\)-measurable not piecewise continuous function on an absolute Souslin \({\mathcal F}\)-set \(X\), then consider the \(\sigma\)-ideal \({\mathcal I}\) on the metric completion \(\widetilde X\) of \(X\) generated by those closed sets \(C\subseteq\widetilde X\) for which \(f\) is continuous on \(C\cap X\). This is a proper \(\sigma\)-ideal and the authors prove that it is locally determined. By a generalization of S. Solecki’s theorem [“Covering analytic sets by families of closed sets”, J. Symb. Log. 59, No. 3, 1022–1031 (1994; Zbl 0808.03031)] by P. Holický, L. Zajíček and M. Zelený [“A remark on a theorem of Solecki”, Commentat. Math. Univ. Carol. 46, No. 1, 43–54 (2005; Zbl 1121.03058)], there is a \(\mathbb{G}_\delta\)-subset \(G\) of \(\widetilde X\) which is contained in \(X\), does not belong to \(\mathcal{I}\) and, moreover, none of its nonempty open subsets belongs to \(\mathcal{I}\) (the latter property is implicit in the proof given by Holický, Zajíček and Zelený and follows from the fact that \(\mathcal{I}\) is locally determined). As \(G\) is an absolute \(\mathbb{G}_\delta\)-set, it is completely metrizable. The authors claim that \(f\) restricted to \(G\) is still not piecewise continuous. It is not clear, however, why this should be the case. The same problem occurs in the original argument in Solecki’s 1998 paper [loc. cit.] (in Theorem 3.1 on page 530).

In case \(X\) is separable (and this is the only case considered in Solecki’s paper), this problem can be fixed by the following argument, due to Solecki. Assume \(G=\bigcap_{n<\omega} U_n\), where \(U_n\subseteq\widetilde X\) are decreasing open sets. Enumerate the basic open subsets of \(G\) into a sequence \((V_n:n<\omega)\), with infinite repetitions. Since \(f\) is not continuous on each of the sets \(\mathrm{cl}_X(V_n)\cap U_n\) we can pick compact sets \(K_n\subseteq \mathrm{cl}_X(V_n)\cap U_n\) on which \(f\) is not continuous (e.g., a convergent sequence with its limit). Now \(G'=G\cup \bigcup_{n<\omega} K_n\) is an absolute \(\mathbb{G}_\delta\)-set, hence completely metrizable, since \(G'=\bigcap_{n<\omega} (U_n\cup \bigcup_{m<n} K_m)\). The points of discontinuity of \(f\) are dense in \(G'\), so \(f\) is not piecewise continuous on \(G'\).

The general case (when \(X\) is not necessarily separable) can be dealt with a similar argument. Let \(U_n\subseteq \widetilde X\) be as above. For each \(n<\omega\) find a maximal antichain \(\mathcal{V}_n\) of open balls in \(G\) of radius less than \(\frac{1}{n}\) (in the metric inherited from \(X\)). For each open ball \(V=B_G(x_V,r_V)\) in \(\mathcal{V}_n\) there is \(0<r_V'<r_V\) such that the closed ball \(\overline B_G(x_V,r_V')\) is not in \(\mathcal{I}\). Therefore, for each \(V\in\mathcal{V}_n\) and \(n<\omega\) there is a compact set \(K_V\subseteq \mathrm{cl}_{X}(B_G(x_V,r_B'))\cap U_n\) on which \(f\) is not continuous. Put \(K_n=\bigcup\{K_V:V\in\mathcal{V}_n\}\) and notice that it is closed in \(\widetilde X\) since the open balls \(B_{\widetilde X}(x_V,r_V)\) are pairwise disjoint for \(V\in\mathcal{V}_n\). The set \(G'=\bigcap_{n<\omega} (U_n\cup \bigcup_{m<n} K_m)\) is an absolute \(\mathbb{G}_\delta\)-set by the same argument as in the separable case and again the points of discontinuity of \(f\) are dense in \(G'\), hence \(f\) is not piecewise continuous on \(G'\).

Although the above considerations show how to obtain a \(\mathbb{G}_\delta\)-set \(G\) on which \(f\) is not piecewise continuous, the authors also need the fact that \(G\) is zero-dimensional (I was reminded of this by Motto-Ros) in order to argue that if \(f\) is \(F_\sigma\)-measurable on \(G\), then it is a pointwise limit of continuous functions. The above considerations do not fix this issue. I was informed, however, that this problem will be fixed in a forthcoming paper [“Some observations on ‘A new proof of a theorem of Jayne and Rogers’”] by M. Kačena and the authors.

There is one point in the proof which deserves a comment. The theorem concerns absolute Souslin \(\mathcal{F}\)-sets. A subset of a metric space is said to be a Souslin \({\mathcal F}\)-set if it can be obtained by the Souslin operation \(\mathcal{A}\) from closed sets. A metric space is called an absolute Souslin \({\mathcal F}\)-set if it is a Souslin \({\mathcal F}\)-set in its metric completion.

This paper provides a new proof of the implication from (i) to (ii) in the case \(X\) is completely metrizable. To obtain the general result, the authors follow an argument of S. Solecki [“Decomposing Borel sets and functions and the structure of Baire class 1 functions”, J. Am. Math. Soc. 11, No. 3, 521–550 (1998; Zbl 0899.03034)] on the reduction of the case where \(X\) is an absolute Souslin \({\mathcal F}\)-set to the case where \(X\) is completely metrizable. The reduction (at page 198 of this paper) is as follows: If \(f\) is a \(\mathbb G_\delta\)-measurable not piecewise continuous function on an absolute Souslin \({\mathcal F}\)-set \(X\), then consider the \(\sigma\)-ideal \({\mathcal I}\) on the metric completion \(\widetilde X\) of \(X\) generated by those closed sets \(C\subseteq\widetilde X\) for which \(f\) is continuous on \(C\cap X\). This is a proper \(\sigma\)-ideal and the authors prove that it is locally determined. By a generalization of S. Solecki’s theorem [“Covering analytic sets by families of closed sets”, J. Symb. Log. 59, No. 3, 1022–1031 (1994; Zbl 0808.03031)] by P. Holický, L. Zajíček and M. Zelený [“A remark on a theorem of Solecki”, Commentat. Math. Univ. Carol. 46, No. 1, 43–54 (2005; Zbl 1121.03058)], there is a \(\mathbb{G}_\delta\)-subset \(G\) of \(\widetilde X\) which is contained in \(X\), does not belong to \(\mathcal{I}\) and, moreover, none of its nonempty open subsets belongs to \(\mathcal{I}\) (the latter property is implicit in the proof given by Holický, Zajíček and Zelený and follows from the fact that \(\mathcal{I}\) is locally determined). As \(G\) is an absolute \(\mathbb{G}_\delta\)-set, it is completely metrizable. The authors claim that \(f\) restricted to \(G\) is still not piecewise continuous. It is not clear, however, why this should be the case. The same problem occurs in the original argument in Solecki’s 1998 paper [loc. cit.] (in Theorem 3.1 on page 530).

In case \(X\) is separable (and this is the only case considered in Solecki’s paper), this problem can be fixed by the following argument, due to Solecki. Assume \(G=\bigcap_{n<\omega} U_n\), where \(U_n\subseteq\widetilde X\) are decreasing open sets. Enumerate the basic open subsets of \(G\) into a sequence \((V_n:n<\omega)\), with infinite repetitions. Since \(f\) is not continuous on each of the sets \(\mathrm{cl}_X(V_n)\cap U_n\) we can pick compact sets \(K_n\subseteq \mathrm{cl}_X(V_n)\cap U_n\) on which \(f\) is not continuous (e.g., a convergent sequence with its limit). Now \(G'=G\cup \bigcup_{n<\omega} K_n\) is an absolute \(\mathbb{G}_\delta\)-set, hence completely metrizable, since \(G'=\bigcap_{n<\omega} (U_n\cup \bigcup_{m<n} K_m)\). The points of discontinuity of \(f\) are dense in \(G'\), so \(f\) is not piecewise continuous on \(G'\).

The general case (when \(X\) is not necessarily separable) can be dealt with a similar argument. Let \(U_n\subseteq \widetilde X\) be as above. For each \(n<\omega\) find a maximal antichain \(\mathcal{V}_n\) of open balls in \(G\) of radius less than \(\frac{1}{n}\) (in the metric inherited from \(X\)). For each open ball \(V=B_G(x_V,r_V)\) in \(\mathcal{V}_n\) there is \(0<r_V'<r_V\) such that the closed ball \(\overline B_G(x_V,r_V')\) is not in \(\mathcal{I}\). Therefore, for each \(V\in\mathcal{V}_n\) and \(n<\omega\) there is a compact set \(K_V\subseteq \mathrm{cl}_{X}(B_G(x_V,r_B'))\cap U_n\) on which \(f\) is not continuous. Put \(K_n=\bigcup\{K_V:V\in\mathcal{V}_n\}\) and notice that it is closed in \(\widetilde X\) since the open balls \(B_{\widetilde X}(x_V,r_V)\) are pairwise disjoint for \(V\in\mathcal{V}_n\). The set \(G'=\bigcap_{n<\omega} (U_n\cup \bigcup_{m<n} K_m)\) is an absolute \(\mathbb{G}_\delta\)-set by the same argument as in the separable case and again the points of discontinuity of \(f\) are dense in \(G'\), hence \(f\) is not piecewise continuous on \(G'\).

Although the above considerations show how to obtain a \(\mathbb{G}_\delta\)-set \(G\) on which \(f\) is not piecewise continuous, the authors also need the fact that \(G\) is zero-dimensional (I was reminded of this by Motto-Ros) in order to argue that if \(f\) is \(F_\sigma\)-measurable on \(G\), then it is a pointwise limit of continuous functions. The above considerations do not fix this issue. I was informed, however, that this problem will be fixed in a forthcoming paper [“Some observations on ‘A new proof of a theorem of Jayne and Rogers’”] by M. Kačena and the authors.

Reviewer: Marcin Sabok (Wrocław)

### MSC:

03E15 | Descriptive set theory |

54H05 | Descriptive set theory (topological aspects of Borel, analytic, projective, etc. sets) |