# Proving that the two definitions of mutually independent variables are equivalent

I want to prove the following theorem

Let $\Omega$ be an at most countable set. Let $I$ be a finite or a countable set. The set of discrete random variables $\left\{ X_{i}\right\} _{i\in I}$ auf $\Omega$ is mutually independent (i.e. for every $S_i\in \mathbb{R}$ the sets $\{X_i^{-1} (A_i) \}$ are independent) if and only if for every arbitrary finite subsets of random variables $X_{i_{1}},\ldots,X_{i_{n}}$ and values $(s_{1},\ldots,s_{n})\in\prod_{k=1}^{n}X_{i_{k}}(\Omega)$ we have $$P(X_{i_{1}}=s_{1},\ X_{i_{2}}=s_{2},\ldots,\ X_{i_{n}}=s_{n})=\prod_{k=1}^{n}P(X_{i_{k}}=s_{k}).$$

One implication ($\Rightarrow$) is immediate. But on the other one I'm stuck. I managed to show $$P\left(\left\{ X_{i_{1}}\in A_{1}\right\} \cap\ldots\cap\left\{ X_{i_{n}}\in A_{n}\right\} \right)=P\left(\bigcup_{x_1\in A_1,\ldots,x_n \in A_n}\left[X_{i_{1}}^{-1}\left(x_{1}\right)\cap\ldots\cap X_{i_{n}}^{-1}\left(x_{n}\right)\right]\right).$$

But the union on the right side does not have be a union of disjoint sets, so I can't apply $\sigma$ additivity - which then would help me to use my assumption. Could you please tell me how to proceed ?

I know that the proof of this theorem is a standard one, with proofs of it books like Resnick or Gut, but both of these use measure-theoretic machinery, of which I don't know anything, so I can't follow those proofs; I would just like to prove it in the simple discrete setting.

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Call $E_{x_1,\dots,x_n}$ the event $\bigcap_{j=1}^nX_{i_j}^{-1}(\{x_j\})$. We can see that $E_{x_1,\dots,x_n}\cap E_{x'_1,\dots,x'_n}=\emptyset$ whenever $(x_1,\dots,x_n)\neq (x'_1,\dots,x'_n)$, because at least one coordinate differs.
Ah yes, of course - though I think you meant $E_{x_1,\dots,x_n}\cap E_{x'_1,\dots,x'_n}=\emptyset$ since $E_{x_1,\dots,x_n}\neq E_{x'_1,\dots,x'_n}$ doesn't yet allow one to use $\sigma$-additivity. –  user26698 Dec 30 '12 at 10:27