[proofplan] The generated sigma-algebra is generated by all inverse images $X_i^{-1}(A)$ with $A\in\mathcal E_i$. This makes each $X_i$ measurable and is contained in any sigma-algebra with that property. [/proofplan] [step:Build the candidate] Let \begin{align*} \mathcal C=\{X_i^{-1}(A): i\in I,\ A\in\mathcal E_i\}. \end{align*} Set $\mathcal F_0=\sigma(\mathcal C)$. For each $i\in I$ and each $A\in\mathcal E_i$, the inverse image $X_i^{-1}(A)$ belongs to $\mathcal F_0$. Hence every map $X_i:(\Omega,\mathcal F_0)\to(E_i,\mathcal E_i)$ is measurable. [/step] [step:Prove minimality] Let $\mathcal G$ be any sigma-algebra on $\Omega$ such that every $X_i:(\Omega,\mathcal G)\to(E_i,\mathcal E_i)$ is measurable. Then $X_i^{-1}(A)\in\mathcal G$ for every $i\in I$ and every $A\in\mathcal E_i$, so $\mathcal C\subset\mathcal G$. By minimality of generated sigma-algebras, \begin{align*} \mathcal F_0=\sigma(\mathcal C)\subset\mathcal G. \end{align*} Thus $\mathcal F_0$ is the smallest sigma-algebra making all $X_i$ measurable, which is precisely $\sigma(X_i:i\in I)$. [/step]