Let $\mathscr{I}$ be the family of all finite sequences in $\mathbb{N}$ and let us use Greek letters $\alpha,\beta\ldots$ to denote elements of $\mathscr{I}$. Given a paved space $(X,\mathscr{X})$, a subset of $X$ is said to be analytic over $\mathscr{X}$ iff it is the nucleus of some Suslin scheme on $\mathscr{X}$.
I'm trying to prove the following theorem:
Let $(X,\mathfrak{M},\mu)$ be a measure space and let $Y$ be a Polish space. Let $\mathscr{R}$ be the class of all rectangles $E\times F$, where $E\in\mathfrak{M}$ and $F$ is a closed subset of $Y$. If $A\subset X\times Y$ is analytic over $\mathscr{R}$, then the projection $\pi_{X}(A)$ of $A$ onto $X$ is analytic over $\mathfrak{M}$.
Let $\mathscr{C}$ be the class of closed subsets of $Y$. I have constructed a particular Suslin scheme $\gamma\mapsto C_{\gamma}\in\mathscr{C}$. The construction is done by induction on $l(\gamma)$, the length of $\gamma$. Assume $l(\gamma)=1$. Since $Y$ is separable there exist a countable dense subset $\left\{y_j\right\}$ of $Y$. For each $j\geq1$, we define $C_j$ to be the closed ball in $Y$, centered at $y_j$ with radius $\frac{1}{2}$. Note that the union of $\left\{ C_j \right\}$, for all $j\geq1$, is $Y$.
Suppose now that the sets $C_{\gamma}$ have been constructed for all $\gamma\in\mathscr{I}$ of length $\leq k$. Let $\gamma\in\mathscr{I}$ be a sequence of lenght $k+1$, and let the first $k$-terms of $\gamma$ be $\gamma_1,\ldots,\gamma_k$. We treat $C_{\gamma|_k}=C_{\gamma_1,\ldots,\gamma_k}$ as a space in its own, with the induced metric. This space is separable and thus there exist a countable dense subset $\left\{y_{\gamma_1,\ldots,\gamma_k,j}\right\}$ of $C_{\gamma|_k}$, indexed by $j$. For each $j\geq1$, we define $C_{\gamma_1,\ldots,\gamma_k,j}$ to be the closed ball in $C_{\gamma|_k}$, centered at $y_{\gamma_1,\ldots,\gamma_k,j}$ with radius $\frac{1}{2^{k+1}}$.
Let $\alpha\mapsto R_{\alpha}\in\mathscr{R}$ be a Suslin scheme and $A\subset X\times Y$ be the nucleus of it. Composing this Suslin scheme with the projection maps, we obtain two other Suslin schemes, one on $\mathfrak{M}$ and one on $\mathscr{C}$; \begin{equation*} \alpha\mapsto E_{\alpha}:=\pi_{X}(R_{\alpha})\in\mathfrak{M},\quad \alpha\mapsto F_{\alpha}:=\pi_{Y}(R_{\alpha})\in\mathscr{C}. \end{equation*} Obviously we have $R_{\alpha}=E_{\alpha}\times F_{\alpha}$ for all $\alpha\in\mathscr{I}$.
With this in mind, we can rewrite the nucleus $A$ in the form \begin{equation*} A = \bigcup_{f\in\mathbb{N}^\mathbb{N}} \bigcap_{n\geq1} E_{f|_n}\times F_{f|_n}, \end{equation*} and since $\bigcup_{g\in\mathbb{N}^\mathbb{N}} C_{g|_n}=Y$ by construction, we may write \begin{equation*} E_{f|_n}\times F_{f|_n} = \bigcup_{g\in\mathbb{N}^\mathbb{N}} \left( E_{f|_n} \times \left( F_{f|_n} \cap C_{g|_n} \right) \right) \end{equation*} for each $f$ and $n$. Hence we get \begin{align*} A &= \bigcup_{f\in\mathbb{N}^\mathbb{N}} \bigcap_{n\geq1} E_{f|_n}\times F_{f|_n} \\ &= \bigcup_{f\in\mathbb{N}^\mathbb{N}} \bigcap_{n\geq1} \bigcup_{g\in\mathbb{N}^\mathbb{N}} \left( E_{f|_n} \times \left( F_{f|_n} \cap C_{g|_n} \right) \right) \\ &= \bigcup_{f\in\mathbb{N}^\mathbb{N}} \bigcup_{g\in\mathbb{N}^\mathbb{N}} \bigcap_{n\geq1} \left( E_{f|_n} \times \left( F_{f|_n} \cap C_{g|_n} \right) \right) \end{align*} where the last equality can be easily checked.
This is the part I'm stuck at. I'm trying to show that $\pi_{X}(A)$ is analytic over $\mathfrak{M}$, but since the intersection of sets does not commute with the projection map, I can't get a nice representation of the set $\pi_{X}(A)$. I'm not even sure whether the theorem is actually true. A proof is given in the book Modern Methods in the Calculus of Variations by Fonseca and Leoni, but I seriously doubt the validity of that proof.
Any advice?