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If, for $1\le i \le n, 1\le j \le m(i)$, we have that $X_{ij}$ are independent and $f_i:\mathbb R^{m(i)}\to\mathbb R$ are measurable, then $f_i(X_{i,1},...,X_{i,m(i)})$ are independent.

Proof. Let $\mathcal F_{i,j}=\sigma(X_{ij})$ and $\mathcal G_i=\sigma(\cup_j\mathcal F_{ij})$. Since $f_i(X_{i,1},...,X_{i,m(i)})\in\mathcal G_i$, the desired result follows from Theorem 2.1.9 and Exercise 2.1.1.

My first question is that in the Theorem, does "measurable" mean "Borel measurable".

My second question is how to show that $f_i(X_{i,1},...,X_{i,m(i)})\in\mathcal G_i$. Let's suppose $B\in\mathcal B(\mathbb R)$, since $f_i$ is Borel measurable, then $f^{-1}_i(B)\in\mathcal B(\mathbb R^n)$. Why is it that $\{\omega\in\Omega:(X_{i1}(\omega),...,X_{i,m(i)}(\omega))\in f^{-1}_i(B)\} \in \mathcal G_i$?

For more details, see Probability: Theory and Examples, Durrett, pg. 55.

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    $\begingroup$ Yes measurable here probably means Borel-Borel measurable since it isn't obvious whether the random variables claimed to be independent are measurable if the $f_i$ are only known to be Lebesgue-Borel measurable. $\endgroup$
    – Mason
    Jan 14 at 7:32
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    $\begingroup$ in this context it mean Borel-measurable, but the theorem holds for any measurable functions between arbitrary measurable spaces, by example if $X_i:\Omega \to V$ and $f:V^n \to W$, where $V^n$ is given the product $\sigma $-algebra $\endgroup$
    – Masacroso
    Jan 14 at 8:25

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