Lemma 7

Every Lusin subspace $A$ of a metrizable space $X$ is a Borel set in $X$.

By Proposition 13 of no. 5 there exist a sieve C and a continuous bijection f of L(C) onto A. With the notation of Lemma 5 of no. 6, for each integer n and each c ∈ C_n, let g_n(c) denote the subspace f(q_n(c)) of X; it is a Lusin subspace and therefore a Souslin subspace of X. As c runs through C_n, the sets g_n(c) are pairwise disjoint, because f is bijective; hence, by Theorem 2 of no. 6, there is a family c → g'_n(c) (c ∈ C_n) of Borel sets in X, pairwise disjoint and such that g'_n(c) ⊃ g_n(c) for all c ∈ C_n. Replacing g'_n(c) by its intersection with the closure g_n(c) of g_n(c) in X if necessary, we may suppose that g'_n(c) ⊂ g_n(c). Let c_{n-1}, c_{n-2}, ..., c_0 denote the images of c in C_{n-1}, C_{n-2}, ..., C_0, under the surjections

respectively; and let h_n(c) denote the intersection of the sets

g'_n(c),\ g'_{n-1}(c_{n-1}),\ ...\ ,\ g'_0(c_0).

Since q_i(c_i) ⊃ q_n(c) for 0 ≤ i ≤ n − 1, h_n(c) contains g_n(c); it is clear also that h_n(c) is a Borel set and is contained in g_n(c), and that as c runs through C_n, the h_n(c) are mutually disjoint sets; finally, by construction, for each c' ∈ C_{n+1} we have h_{n+1}(c') ⊂ h_n(p_n(c')). Let then B_n be the union of the sets h_n(c) as c runs through C_n; B_n is a Borel set, and B_{n+1} ⊂ B_n; also B_n contains the union of the sets g_n(c) (c ∈ C_n), which is A. Let B be the intersection of the decreasing sequence of sets B_n; B is a Borel set and contains A. We shall show that B = A, and this will complete the proof.

Let x be a point of B. Then, for each integer n, there exists a unique c ∈ C_n such that x ∈ h_n(c); let us denote this c by c_n(x). The sequence (c_n(x))_{n≥0} belongs to L(C). The decreasing sequence (g_n(c_n(x))) converges by definition to a point a ∈ A; the sequence of closures of these sets also converges to a in X, hence a fortiori so does the sequence (h_n(c_n(x))). Now x belongs to all the sets h_n(c_n(x)), therefore x = a ∈ A. Lemma 7 is thus proved, and with it Theorem 3.