proof of equivalent definitions of analytic sets for measurable spaces

Let $(X,ℱ)$ be a measurable space and $A$ be a subset of $X$. For any uncountable Polish space $Y$ with Borel $\sigma$-algebra (http://planetmath.org/BorelSigmaAlgebra) $ℬ$, we show that the following are equivalent.

1. $A$ is $ℱ$-analytic (http://planetmath.org/AnalyticSet2).

2. $A$ is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a set $S\in ℱ\otimes ℬ$ onto $X$.

Here, $ℱ\otimes ℬ$ denotes the product $\sigma$-algebra (http://planetmath.org/ProductSigmaAlgebra) of $ℱ$ and $ℬ$.

(1) implies (2):Let $𝒢$ denote the paving consisting of the closed subsets of $Y$. If $A$ is $ℱ$-analytic then there exists a set $S\in {(ℱ\times 𝒢)}_{\sigma \delta }$ such that $A={\pi }_X(S)$, where ${\pi }_X:X\times Y\to X$ is the projection map (see proof of equivalent definitions of analytic sets for paved spaces).In particular, $𝒢\subseteq ℬ$ implies that $S$ is contained in the $\sigma$-algebra $ℱ\otimes ℬ$.

(2) implies (1):This is an immediate consequence of the result that projections of analytic sets are analytic.