$X_n$ independent variables if equations hold Let $X_1,X_2,\dots ,X_n$ be random variables.
Statement 1
If for any $B_i \in \mathcal{B}^n, i=1,2,\dots,n$,
$$P\left( \overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) \right) = \prod_{i=1}^{n} P\left(X_i^{-1} \left(B_i \right) \right)$$
then $X_i$ are independent.
Statement 2
If we set $X=\left( X_1, X_2, \dots , X_n \right)$
and since
$$\overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) = X^{-1}\left(B_1 \times B_2 \times \dots X_n \right)$$
then the equation in Statement 1 is equivalent to
$$P_X \left( B_1 \times B_2 \times \dots \times B_n \right) = P_{X_1}(B_1) \times P_{X_2}(B_2) \times \dots \times P_{X_n}(B_n)$$
Does anyone have guidance on how to proceed proving these two statements?
EDIT
I want to show that if either Statement 1 OR Statement 2 is true for $B_i \in \mathcal{P}^n, i=1,2,...,n$, then $X_i$ are mutually independent.
$\mathcal{P}^n = \{ (a_1,b_1]\times(a_2,b_2]\times \dots \times (a_n,b_n]\} \cup \{ \emptyset \}$
Attempt
My idea is that if Statement 1 is true for $B_i \in \mathcal{P}^n, i=1,2,...,n$ then, then by setting some $B_i = \mathbb{R}^n \in \mathcal{P}^n$ we get that for every subset of indices $I = \{i_1, \dots, i_k \} \subset \{1,2, \dots, n\}$  we have
$$P\left( \overset{k}{\underset{m=1}{\bigcap}}X_{i_m}^{-1}\left(B_{i_m} \right) \right) = \prod_{m=1}^{k} P\left(X_{i_m}^{-1} \left(B_{i_m} \right) \right)$$
for any $B_{i_m} \in \mathcal{P}^n$.
But the above does not imply that $X_i$ are mutually independent since we need it to be true for any $B_{i_m} \in \mathcal{B}^n$. How can I proceed from here?
If Statement 2 is true I don't know how to proceed to prove independence.
 A: Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $X:(\Omega,\Sigma,\mathbb{P})\to(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ be a random vector.
Based on the assumption of independence, we can proceed as follows:
\begin{align*}
\mathbb{P}_{X}(B_{1}\times B_{2}\times\ldots\times B_{n}) & = \mathbb{P}(X^{-1}(B_{1}\times B_{2}\times\ldots\times B_{n}))\\\\
& = \mathbb{P}(\left\{\omega\in\Omega\mid X(\omega)\in B_{1}\times B_{2}\times\ldots\times B_{n}\right\})\\\\
& = \mathbb{P}(\{\omega\in\Omega \mid (X_{1}(\omega)\in B_{1})\wedge(X_{2}(\omega)\in B_{2})\wedge\ldots\wedge(X_{n}(\omega)\in B_{n})\}\\\\
& = \mathbb{P}(\{\omega\in\Omega \mid X_{1}(\omega)\in B_{1}\}\cap\ldots\cap\{\omega\in\Omega \mid X_{n}(\omega)\in B_{n}\})\\\\
& = \mathbb{P}\left(X^{-1}_{1}(B_{1})\cap X^{-1}_{2}(B_{2})\cap\ldots\cap X^{-1}_{n}(B_{n})\right)\\\\
& = \mathbb{P}(X^{-1}_{1}(B_{1}))\times\mathbb{P}(X^{-1}_{2}(B_{2}))\times\ldots\times\mathbb{P}(X^{-1}_{n}(B_{n}))\\\\
& = \mathbb{P}_{X_{1}}(B_{1})\times\mathbb{P}_{X_{2}}(B_{2})\times\ldots\times\mathbb{P}_{X_{n}}(B_{n})
\end{align*}
Hopefully this helps!
