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..for any measurable function $T:\mathbb{R}^2 \rightarrow \mathbb{R}$. I am confused with how to define this independence.

$X_i$ are independent if $\sigma(X_i) = X_i^{-1}(\mathcal{B}(\mathbb{R}))$ are independent families of sets within $\sigma(\bigcup\sigma(X_i))$. Per definition, $\sigma: P(\Omega) \rightarrow P(\Omega)$ for some universe $\Omega$. So, all $\sigma(X_i)$ should be sigma-algebras on the same universe for this to work, but doesn't $\sigma(T(X_1,X_2))$ has as its universe $P(\Omega)\times P(\Omega)$?

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    $\begingroup$ Let $Y$ be the random variable $Y=T(X_1,X_2)$. Then for an $\omega\in\Omega$ we define $Y(\omega)=T(X_1,X_2)(\omega):=T(\ X_1(\omega)\ ,\ X_2(\omega)\ )$. (This must be extracted from the context and it is usually the case, not something like $(\omega_1,\omega_2)\to T(\ X_1(\omega_1)\ ,\ X_2(\omega_2)\ )$. $\endgroup$
    – dan_fulea
    Nov 15 '19 at 19:34
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I have no idea what you mean by universe, but here is the proof of why a triple $(X_1, X_2, X_3)$ of random variables imply the independence of $(X_1, X_2)$ and $X_3.$

By definition, $P(X_1\in A_1, X_2 \in A_2, X_3 \in A_3) = P(X_1 \in A_1, X_2 \in A_2) P(X_3 \in A_3).$ Thus, if $Y = (X_1, X_2),$ what we have is that $P(Y \in R, X_3 \in A) = P(Y \in R) P(X_3 \in A),$ for all measurable rectangles $R.$ We can apply Dynkin's theorem or Monotone class theorem and we are done.

To finish your question, remark that if $Z_1$ and $Z_2$ are independent random objects, so are $\varphi(Z_1)$ and $\psi(Z_2)$ for any measurable functions $\varphi$ and $\psi.$

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  • $\begingroup$ I suppose the "universe" is the space of definition, the probability space (as a set, or rather the set extracted from it by applying the forgetful functor which neglects the measurables and the probability needed to define the independence). $\endgroup$
    – dan_fulea
    Nov 15 '19 at 19:36
  • $\begingroup$ "Universe" as a set on which measure space (e.g. sigma-algebra) is defined. I guess I've seen it been called that on Wiki or in some book. Anyway, I guess your post together with @dan_fulea 's comment answered my question. $\endgroup$ Nov 15 '19 at 19:45

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