Let $(X,M)$ be a measurable space, and $M=\langle A \rangle$ in which $A$ is an algebra on $X$. Suppose that $v$ is a signed measure and $m$ is a positive measure on $(X,M)$. Now, can we say: $v$ is absolutely continuous with respect to $m$ if $v(E)=0$ for every $E$ in $A$ for which $m(E)=0$?
In other words, in the definition of absolutely continuous, can we replace $M$ by the algebra $A$ that generates $M$? If not, please give a counterexample.
As GEdgar pointed out, this is not true. Let $m$ be the Lebesgue measure on $\mathbb{R}$ and $A$ be the algebra generated by the half-open intervals $[a,b)$. Notice a property of these intervals: for every such interval $I$ and every $x\in\mathbb{R}$ there is $\epsilon>0$ such that the entire set $[x,x+\epsilon)$ is either contained in $I$ or is contained in its complement. This property passes to the algebra $A$ (just take the minimum of the epsilons involved). Consequently, every nonempty set in $A$ has positive Lebesgue measure. The implication $$ \forall E\in A: \quad m(E)=0\implies v(E)=0 $$ holds vacuously, no matter what $v$ is. This gives a negative answer to the question.