[proofplan] The coordinate maps are measurable if the product map is measurable because coordinate projections are measurable. Conversely, the inverse image of each measurable rectangle is an intersection of coordinate inverse images, and rectangles generate the product sigma-algebra. [/proofplan] [step:Product measurability implies coordinate measurability] If $X$ is measurable, then $X_i=\pi_i\circ X$, where $\pi_i:E_1\times\cdots\times E_n\to E_i$ is the coordinate projection. The projection $\pi_i$ is measurable for the product sigma-algebra, so each $X_i$ is measurable. [/step] [step:Coordinate measurability implies product measurability] Assume each $X_i$ is measurable. For a measurable rectangle $A_1\times\cdots\times A_n$ with $A_i\in\mathcal E_i$, \begin{align*} X^{-1}(A_1\times\cdots\times A_n) =\bigcap_{i=1}^n X_i^{-1}(A_i). \end{align*} Each set $X_i^{-1}(A_i)$ lies in $\mathcal F$, so the intersection lies in $\mathcal F$. Thus $X^{-1}$ sends every measurable rectangle to $\mathcal F$. [/step] [step:Extend from rectangles] The class of subsets $B$ of $E_1\times\cdots\times E_n$ such that $X^{-1}(B)\in\mathcal F$ is a sigma-algebra. Since it contains all measurable rectangles, it contains the sigma-algebra generated by them, namely $\mathcal E_1\otimes\cdots\otimes\mathcal E_n$. Hence $X$ is product-measurable. [/step]