Let $A$ be a Boolean $\sigma$-algebra with the countable chain condition, in particular this means (See [1], Chapter 30, Corollary 1) that $A$ is complete. The following may or may not add constraint on $A$ (I am curious as to if it does), we assume that there is a normalized positive measure algebra $p$ on $A$ (i.e. such that $p(x)=0$ only if $x=0$ as well as $p(1)=1$). We can prove easily using the countable chain condition that $p\left( \bigvee_{i\in \mathcal I} x_i\right)=\sum_{i\in\mathcal I} p(x_i)$ is true for arbitrary collections of disjoint elements of $A$ (indeed at most only countably many of them are not $0$). Observe that the fact that $A$ has the countable chain condition (and therefore is complete) is a consequence of the fact that $p$ is positive and finite.
I am wondering if the following is true or if there is a counter example
There exists a countable subset $E\subseteq A$ such that the sub-algebra generated by $E$ is $A$.
Attempts :
One idea to provide a counter example is the following, let $\mathbb B([0,1[)$ be the Borel sets on $[0,1[$ and let $\lambda$ be the Lebesgue measure on $\mathbb B([0,1[)$. Then for $A,B\in \mathbb B([0,1[)$ define the equivalence relation $\lambda(A\triangle B)=0$, then if we let $A$ be the set of equivalence class of $\mathbb B([0,1[)$ modulo this equivalence relation. Denote by $[A]$ the equivalence class of $A$ and let $p([A])=\lambda(A)$. Then we can show that $p$ is positive and that $A$ is a Boolean $\sigma$-algebra with the countable chain condition. I think that the set $E=\left\{ \left[ \frac{2k}{2^n},\frac{2k+1}{2^n} \right[ : n\in\mathbb N, 0\leq k < 2^{n-1}\right\}$ might do it, this would mean that the statement is true for at least one uncountable boolean algebra satisfying the requirements.
Corollary 1 of chapter 11 of [1] might be of use, its statement is the following :
Let $A$ be a Boolean algebra generated by a set $E$, and for each finite subset $F$ of $E$, let $B_F$ be the subalgebra of $A$ generated by $F$. The family $$\{ B_F : F\subseteq E\text{ and } F \text{ is finite}\}$$ is directed, and its union is $A$.
Context :
This is related to solving this question on the space of measure algebra generated by a boolean $\sigma$-algebra with a positive normalized measure algebra on it.
References :
[1] Halmos, Paul, and Steven Givant. Introduction to Boolean algebras. Springer New York, 2009.
