Suppose $X_1, \dots, X_n, Y$ are independent random variables. Prove that $X = (X_1, \dots, X_n)$ and $Y$ are independent variables.
My attempt:
Fix $A \in \mathcal{R}$ (a Borel subset of the real numbers). Define measures
$\mu_1(B) = P(Y \in A, X \in B), B \in \mathcal{R}^n$
$\mu_2(B) = P(Y \in A)P(X \in B), B \in \mathcal{R}^n$
We have to show that $\mu_1 = \mu_2$ and then we will have shown that $P(Y \in A, X \in B) = P(Y \in A)P(X \in B)$, which is what we needed to prove.
Observe that $\mathcal{R}^n = \sigma(\mathcal{P})$ where
$$\mathcal{P}:= \{A_1 \times\dots \times A_n \mid A_1 , \dots, A_n \in \mathcal{R}\}$$
By independence of $X_1, \dots, X_n, Y$, we see that $\mu_1$ and $\mu_2$ are equal on $\mathcal{P}$ and by the unicity theorem of measures they are equal on $\mathcal{R}^n$.
Is this proof correct? Is there a way to avoid the uniqueness theorem?
Best Answer
If two probability measures $P$ and $Q$ are equal on class of sets generating a sigma algebra you cannot immediately conclude that they are equal on the sigma algebra. This requires a proof unless the class you started with is an algebra. In this case sets of the form $X_1^{-1}(A_1)\cap ..\cap X_n^{-1}(A_n)$ do not form an algebra. This class of sets is closed under finite intersections and the standard argument for this is to apply Dynkin's $\pi -\lambda$ theorem: $\{A:P(A)=Q(A)\}$ is a $\lambda$ system containing the given $\pi$ system so it contains the generated sigma algebra.