Let $\nu:\mathbb{X}\longrightarrow \mathbb{R}$ and additive function, where $(X,\mathbb{X},\mu)$ is a measure space. That is, $\nu(\emptyset)=0$, and for any finite disjoint family of measurable sets $\{A_k\}_{k=1}^n$ we have $\nu(\cup_{k=1}^n A_k)=\sum_{k=1}^n \nu(A_k)$.
I want to prove that $\nu$ is absolutely continuous with respect to $\mu$ if and only if $\mu(A)=0 \implies \nu(A)=0$ for all $A\in\mathbb{X}$.
Note: We say that $\nu$ is absolutely continuous with respect to $\mu$ if for every $\varepsilon>0$ there exist $\delta>0$ such that $\mu(A)<\delta \implies |\nu(A)|<\varepsilon$, for all $A\in\mathbb{X}$.
One way is easy since $\mu(A)=0<\delta_n$ for any $\delta_n>0$ such that $|\nu(A)|<1/n$, whence $\nu(A)=0$.
For sufficiency, I began by assuming that $\nu$ is not absolutely continuous and managed to prove the existence of a sequence $\{B_n\}_{n=1}^\infty$ of measurable sets, where $B=\cap_{n=1}^\infty B_n$ is such that $\mu(B)=0$ and $|\nu(B_n)|\geq\varepsilon_0$, for all $n\in\mathbb{N}$ and for some $\varepsilon_0>0$. Then I would like to have $\nu(B)\neq 0$ and contradict the hypotesis. But I can't think of how.
Note that I have not used the finite additivity of $\nu$ so far. Any help is appreciated. Thanks in advance.
The result you are trying to show seems to be false.
Consider the case of $X=\mathbb N$ with $\mathbb X=$ the set of all subsets of $X$ and $\mu$ given by $\mu(A)=\sum_{n\in A}2^{-n}$ for every $A\in\mathbb X$.
The sequence $(\mathbf 1_{\{n\}})_{n\in X}$ of indicator functions on $X$ is a linearly independant family of elements of the vector space $\mathbb R^X$ of all $\mathbb R$-valued functions on $X$. By using a basis of $\mathbb R^X$ containing each $\mathbf 1_{\{n\}}$, you can see that there exists a linear functional $\Phi$ on $\mathbb R^X$ such that $\Phi(\mathbf 1_{\{n\}})=1$ for every $n\in X$. You can then obtain an additive set function $\nu:\mathbb X\to\mathbb R$ by letting $\nu(A)=\Phi(\mathbf 1_A)$ for every $A\in\mathbb X$.
For every $A\in\mathbb X$, it is clear that the relation $\mu(A)=0$ implies $A=\varnothing$, which in turn implies $\nu(A)=0$. But as $n$ ranges over $X$, the number $\mu(\{n\})$ can get arbitrarily small while $|\nu(\{n\})|=1$, which shows that $\nu$ is not absolutely continuous with respect to $\mu$.