Lemma 6

In a Lusin space $X$, every Borel set is a Lusin subspace of $X$.

Let $\mathfrak{T}$ be the set of all subsets $A$ of $X$ such that both $A$ and $\complement A$ are Lusin subspaces of $X$. Since every closed set and every open set in $X$ is a Lusin subspace of $X$ (no. 4), $\mathfrak{T}$ contains all closed subsets of $X$, and the lemma will therefore be proved if we can show that $\mathfrak{T}$ is a $\sigma$-algebra on $X$. For this it is enough to show that if $(A_n)$ is a sequence of sets of $\mathfrak{T}$, then $\bigcap_n A_n$ and $\bigcup_n A_n$ are Lusin subspaces of $X$. Now we have seen in no. 4 that every countable intersection of Lusin subspaces is a Lusin subspace. On the other hand, if $B_n$ is the intersection of $A_n$ and $\bigcap_{i < n} \complement A_i$, it follows from the hypothesis and the preceding remark that $B_n$ is a Lusin subspace; and since $\bigcup_n A_n = \bigcup_n B_n$, the subspace $\bigcup_n A_n$ is a Lusin subspace by Lemma 2 of no. 4.