Product Measures Consider the case $\Omega = \mathbb R^6 , F= B(\mathbb R^6)$ Then the projections $\ X_i(\omega) = x_i ,[ \omega=(x_1,x_2,\ldots,x_6) \in \Omega $ are random variables $i=1,\ldots,6$. Fix $\ S_n = S_0$ $\ u^{\Sigma X_i(\omega)}d^{n-\Sigma X_i(\omega)} \omega \in \Omega $,  $\ n=1,\ldots,6 $.
Choose the measure P = $\bigotimes_{i=1}^6 Q$  on ($\Omega,F$) where $Q$ denotes the measure $p\delta_1 + q\delta_0 $ on $(\mathbb R, B(\mathbb R))$ for some $p,q>0$ such that $p+q = 1$. Show that the projections $\ X_i(\omega), i=1,\ldots,6$ are mutually independent. 
Since $\ X_i(\omega)$ is a random variable then am I correct in saying that to show their independence I must show that their sigma algebras $\sigma(\ X_i(\omega))$ are independent how would I go about doing this? 
Thanks very much! 
 A: Yes, that is correct. You have to show that $\sigma(X_i)$ and $\sigma(X_j)$ are independent, when $j\neq i$ (note that I have omitted the $\omega$ in $\sigma(X_i(\omega))$, because that is not what you want). Now, recall that
$$
\sigma(X_i)=\sigma(\{X_i^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}),
$$
and hence it is enough to show that $\{X_i^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}$ and $\{X_j^{-1}(A)\mid A\in \mathcal{B}(\mathbb{R})\}$ are independent when $i\neq j$. Now, if $A\in\mathcal{B}(\mathbb{R})$ then
$$
X_i^{-1}(A)=\{(x_1,\ldots,x_6)\in\mathbb{R}\mid x_i\in A\}=\mathbb{R}\times\cdots \times A\times\cdots\times\mathbb{R},
$$
where $A$ is on the $i$'th place. If $j\neq i$, then 
$$
X_i^{-1}(A)\cap X_j^{-1}(B)=\mathbb{R}\times\cdots \times A\times B\times\cdots\times\mathbb{R},
$$
where $A$ is on the $i$'th place and $B$ is on the $j$'th place. Now
$$
P(X_i^{-1}(A)\cap X_j^{-1}(B))=Q(\mathbb{R})^{4}Q(A)Q(B)=Q(A)Q(B)=P(X_i^{-1}(A))P(X_j^{-1}(B)),
$$
and hence the events are independent for every choice of $A,B\in\mathcal{B}(\mathbb{R})$.
