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I have to show that for a family of maps $X = \{X_i\}_{i \in I}$ from a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ into a measure space $(E, \mathcal{E})$, where $\mathcal{E}^I$ is the product sigma algebra of $E^I$ the following holds:

$X: \Omega \rightarrow E^I$ is $\mathcal{F}-\mathcal{E}^I$ measurable $\iff$ $\forall i \in I: X_i: \Omega \rightarrow E$ is $\mathcal{F}-\mathcal{E}$ measurable.

I have trouble proving the direction $\Leftarrow$ and doubt that it holds. If it does hold, can anybody point me into the right direction?

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Consider $\mathcal B := \{B\in \mathcal E^I: X^{-1}(B)\in\mathcal F\}$. We can show that $\mathcal B$ is a $\sigma$-algebra ($X^{-1}$ commutes with the set operations) which contains the product $\prod_{i\in I}A_i$ where only finitely many $A_i$ are different from $E$ (because the "coordinates" maps are meausrable). But these sets generate the product $\sigma$-algebra.