Bourbaki

Exercise 20, § 6

General Topology, IX, p. 124

Bourbaki marks this one as harder than the rest.

Let X,X′X, X' be two metrizable spaces. For every countable even (resp. odd) ordinal α\alpha, one says that a mapping ff from XX into X′X' is Borel of class α\alpha if, for every closed subset F′F' of X′X', f−1(F′)f^{-1}(F') belong to Fα(X)\mathfrak{F}_\alpha(X) (resp. Gα(X)\mathfrak{G}_\alpha(X)) (IX, p. 119, exerc. 6); then, for every open subset G′G' of X′X', f−1(G′)f^{-1}(G') belong to Gα(X)\mathfrak{G}_\alpha(X) (resp. Fα(X)\mathfrak{F}_\alpha(X)). The Borel functions of class 0 are the continuous functions. α)\alpha) For the characteristic function φA\varphi_A of a subset AA of XX to be of class α\alpha, it is necessary and sufficient that AA belong to Fα∩Gα\mathfrak{F}_\alpha \cap \mathfrak{G}_\alpha. b) Let ff be a mapping of class α\alpha from XX into X′X'; prove that for every subset B′∈Fβ(X′)B' \in \mathfrak{F}_\beta(X') (resp. B′∈Gβ(X′)B' \in \mathfrak{G}_\beta(X')), f−1(B′)f^{-1}(B') belong to Fα+β(X)\mathfrak{F}_{\alpha+\beta}(X) (resp. Gα+β(X)\mathfrak{G}_{\alpha+\beta}(X)). (Reason by transfinite induction on β\beta.) c) Let ff be a mapping of class α\alpha from XX into X′X', gg a mapping of class β\beta from X′X' into a metrizable space X′′X''; prove that g∘fg \circ f is a mapping of class α+β\alpha + \beta. d) For every countable even (resp. odd) ordinal α\alpha, let (An)(A_n) be a sequence of subsets of XX belonging to Gα\mathfrak{G}_\alpha (resp. Fα\mathfrak{F}_\alpha) and forming a covering of XX; if a mapping ff from XX into X′X' is such that its restriction to each of the AnA_n is of class α\alpha, then ff is of class α\alpha. Analogous result for a finite sequence of subsets of XX belonging to Fα\mathfrak{F}_\alpha (resp. Gα\mathfrak{G}_\alpha) and forming a covering of XX.

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.