Let $(X, \mathcal{A})$ be a measure space. We can define a measure $\nu$ w.r.t. to another measure $\mu$ and a function $f:X\rightarrow \bar{\mathbb{R}}^+$ (called the density) as follows: $$\nu(A)=\int_X f \,\chi_A d\mu, \quad \forall A\in\mathcal{A}. $$
Then, it holds that $\nu$ is absolutely continuous w.r.t. $\mu$. This is essentially the "trivial" opposite direction of the Radon-Nikodym theorem, however, I still dont understand the proof.
Essentially, the proof comes down to say that if $\mu(A)=0$, then $\mu(\{x\in X:f(x)\chi_A(x)>0\})=\mu(\{x\in A:f(x)>0 \})=0$. But this implication is not clear to me. Of course the set $\{x\in A:f(x)>0 \}$ is a subset of $A$ so one could think that due to monotonicity of measures this set is a nullset, too. But since we dont know if the measurable space is complete, we do not know if every subset of a nullset is also contained in the sigma algebra, so we can actually not make a statement about that set, no?