Nonfinite total variation $\left| v \right|$ such that we don't have the following equivalence

Question

Consider that if $\nu$ is some signed measure and $\mu$ is some positive measure, we say that $\nu$ is absolutely continuous with respect to $\mu$ if $\mu(E) = 0 \implies \nu(E) =0$. Moreover, if $\left| \nu \right| < \infty$, then we may restate the last condition as: $\forall \epsilon > 0$ there exists a $\delta > 0$ such that $$\mu(E) < \delta \implies \left| \nu \right|(E) < \epsilon.$$

I want to find a signed measure $\nu$ such that these statements are not equivalent. Perhaps the Weierstrass function, or the topologist's sin curve would be useful here?

Answer 1

Let $\mu$ be the lebesgue measure in $\mathbb{R}$ and $\nu(A)=\int_{A}|x|^{-1}dx$. Is easy to see that $\nu$ is absolutely continuous with respect to $\mu$.

For $\epsilon=1$ and every $\delta >0$ we have $\mu([-\delta /3,\delta /3])<\delta$, but $\nu([-\delta /3,\delta /3])=\infty$.