If $X_1,X_2,X_3$ are independent, then $T(X_1,X_2)$ and $X_3$ are independent..

..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 $$P(\Omega)\times P(\Omega)$$?

• 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 $(\omega_1,\omega_2)\to T(\ X_1(\omega_1)\ ,\ X_2(\omega_2)\ )$. Nov 15 '19 at 19:34

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 $$\psi.$$