I want to prove that there cannot exist two measures $\mu$ and $\nu$ such that $\mu$ is finite and $\nu$ is $\sigma$-finite but not finite. And they satisfy $\varepsilon$-$\delta$-criterion (i.e. $\forall \varepsilon >0~ \exists \delta > 0: \mu(A)<\delta \Rightarrow \nu(A)<\varepsilon$) and $\nu$ is absolutely continous with respect to $\mu$.
Can someone give me a hint? I have proven that there are such measures except that they don't satisfy the $\varepsilon$-$\delta$-criterion.
Let $(\Omega ,\mathcal F)$ be the space under consideration. Since $\nu$ is sigma finite we can find an increasing sequence of measurable sets $E_n$ such that $\nu (E_n) <\infty$ for all $n$ and $\cup_n E_n=\Omega$. Now $E_n^{c}$ decreases to the empty set. Since $\mu$ is finite this implies $\mu(E_n^c) \to 0$. Hence $\nu(E_n^c) <\epsilon$ if $n$ is so large that $\mu(E_n^c) <\delta$. In particular we get $\nu(E_n^{c}) <\infty$. Hence $\nu (\Omega)=\nu (E_n)+ \nu(E_n^{c}) <\infty$ so $\nu $ is a finite measure. [$\epsilon$ can be any fixed positive number in this proof].