Sufficient condition for absolute continuity of measures Let $X=2^{\omega}$ (the space of one-way infinite binary sequences) and $\mathcal{P}(X)$ the space of Borel probability measures on $X$. A measure $\mu \in \mathcal{P}(X)$ is absolutely continuous with respect to $\nu \in \mathcal{P}(X)$, written $\mu \ll \nu$, iff for every $\epsilon>0$ there is a $\delta>0$ such that for every Borel $B$, if $\nu(B)<\delta$ then $\mu(B) < \epsilon$.
Question 1 Does it suffice to check only cylinder sets? I.e. is it the case that $\mu \ll \nu$ iff for every $\epsilon>0$ there is a $\delta>0$ such that for every finite binary string $\sigma$ if $\nu([\sigma])<\delta$ then $\mu([\sigma]) < \epsilon$, where and $[\sigma] = \{ x \in X: \sigma \text{ is an initial segment of } x\}$?
Question 2 Is there some countable collection $\mathcal{C}$ of Borel sets such that $\mu \ll \nu$ iff for every $\epsilon>0$ there is a $\delta>0$ such that for every $B \in \mathcal{C}$, if $\nu(B)<\delta$ then $\mu(B) < \epsilon$?
Of course, an affirmative answer to Question 1 implies an affirmative answer to Question 2.
 A: Recall

Let $(\Omega,\mathcal B,m)$ a finite measure space, and $\mathcal A\subset\mathcal B$ a generating algebra. Then for each $\varepsilon>0$, we can find $A\in\mathcal A$ such that $m(A\Delta B)<\varepsilon$.

(see here for a proof).
 We shall apply this result to $\mathcal A$ the algebra generated by cylindrical sets.
 Fix $\varepsilon>0$. We can find $\delta>0$ such that for each $A\in\mathcal A$ satisfying $\nu(A)\leqslant 2\delta$ then $\mu(A)\leqslant \varepsilon$. Let $B\in\mathcal B(X)$. We can find $A\in\mathcal A$ such that $\mu(A\Delta B)+\nu(A\Delta B)<\delta$. If $\nu(B)\leqslant \delta$, then $\nu(A)\leqslant |\nu(B)-\nu(A)|+\nu(B)\leqslant 2\delta$ hence $\mu(A)\leqslant \varepsilon$. We can assume that $\delta<\varepsilon$. Then $\mu(A)\leqslant |\mu(B)-\mu(A)|+\mu(B)\leqslant 2\varepsilon$.
So if for all $\varepsilon>0$, we have can find $\delta>0$ such that if $A\in\cal A$ satisfies $\nu(A)\leqslant 2\delta$ then $\mu(A)\leqslant \varepsilon$, the same holds replacing $\cal A$ by $\mathcal{B}(X)$. 
The algebra generated by cylindrical subsets is countable. 
A: Any measure that is atomless but not absolutely continuous with respect to the Lebesgue measure witnesses that the answer to Question 1 is negative. 
If $\mu$ is atomless, it must satisfy the $\epsilon$-$\delta$ condition given for cylinders because otherwise, there is $\epsilon>0$ such that for all $n$ there is $\sigma$ of length $n$ (i.e. Lebesgue measure less than $2^{-n}$) such that $\mu([\sigma])> \epsilon$. But the collection of all such $\sigma$'s would then be an infinite finitely-branching tree and hence have a path (by König's lemma), which would be an atom for $\mu$.  
For a concrete example, take the Bernoulli $(1/3, 2/3)$ measure $\mu$ (the measure that corresponds to the process of flipping a bias coin with probability 1/3 of heads). This measure is clearly atomless, but is not absolutely continuous with respect to Lebesgue measure since it says that the Lebesgue-null set of sequences without the same number of $1$'s as $0$'s has measure $1$.
