Bernoulli product measure Let $\Omega=\{0,1\}^\mathbb{N}$ and $\mathcal{A}$ the sigma-algebra generated by the  cylinders sets $\{w\in\Omega\vert \forall s \in S, w_s=\epsilon_s\}$ with $S\subset\mathbb{N}$ finite and $\epsilon_s\in\{0,1\}$. Let $p\in(0,1)$. We take product measure with density $p$ on $(\Omega,\mathcal{A})$:$\mu=\prod_{n\in\mathbb{N}} \mu_n$ where $\mu_n$ is Bernoulli measure on $\{0,1\}$ given by $\mu_n(w_n=0)=1-p$, $\mu_n(w_n=1)=p$.
I would like to prove that for all $\epsilon>0$ and $B\in\mathcal{A}$, there is a family of cylinders $(C_i)_{1\leq i \leq N}$ such as $\mu\left(B\Delta \bigcup_{1\leq i\leq N} C_i\right)<\epsilon$, where $\Delta$ is the symmetric difference. Actually, I'm also looking for a simple proof of the existence of the product measure. I would very much appreciate any help / references!
 A: Of course the existence of the product measure is a standard thing in probability, following from the Kolmogorov Extension Theorem. Here you could cheat: $\Omega$ is compact; you could define the product measure as a weak* limit of finite product measures, more or less.
If I'm not missing anything the bit about the symmetric differences is immediate from a simple fact about measure theory. Say $A$ is the class of finite unions of cylinders. Then $A$ is an algebra (if this is not obvious, first convince yourself that $A$ is the same as the class of all sets that depend on finitely many coordinates). So what you ask about follows from this:
Lemma. Suppose $(X,\mathcal A, \mu)$ is a probability space. Suppose $A$ is an algebra that generates $\mathcal A$. Then for every $E\in\mathcal A$ and $\epsilon>0$ there exists $F\in A$ with $$\mu(E\Delta F)<\epsilon.$$
Proof: Let $B$ be the set of all $E\in\mathcal  A$ that satisfy the conclusion. We only need to show that $B$ is a sigma-algebra. 
It's clear that $F\in B$ implies $F^c\in B$, since $$F^c\Delta E^c=F\Delta E.$$
And it's easy to see that $B$ is closed under finite unions (if this is not clear it should be clear after reading the rest of this proof). Hence $B$ is closed under finite intersections, and hence in showing $B$ is closed under countable unions we can restrict to disjoint sets.
Suppose $E_1,E_2,\dots\in B$ are disjoint and let $E=\bigcup_n E_n$. Let $\epsilon>0$. There exists $N$ so $$\mu\left(\bigcup_{n=N+1}^\infty E_n\right)<\epsilon/2.$$ For each $n$ there exists $F_n\in A$ with $$\mu(E_n\Delta F_n)<2^{-(n+1)}.$$Let $F=\bigcup_{n=1}^NF_n$, and note that $$E\Delta F\subset\bigcup_{n=1}^N(E_n\Delta F_n)\cup\bigcup_{n=N+1}^\infty E_n.$$
