If $\nu \ll \mu$ and $\mu$ is finite, does it implies that $\nu$ is also finite?

Question

Let $\mu$ and $\nu$ be two (positive) measures, where $\mu$ is finite. Is it true that if $\nu \ll \mu$ and $\mu$ is finite, does it imply that $\nu$ is also finite? If so, how does one prove it? If not, what is a counter-example?

Answer 1

No. Take $((0,1),m)$ where $m$ is the Lebesgue measure and $\nu(dx)=\frac{1}{x}m(dx)$ for a counter-example.

Answer 2

No. For instance Lebesgue measure is absolute-continuously equivalent to $$\mu(\bullet)=\frac1{\sqrt{2\pi}}\int 1_\bullet(x)e^{-x^2/2}\,dx$$

(its density being $\sqrt{2\pi}e^{x^2/2}$). More generally, any non-zero $\sigma$-finite measure is absolute-continuously equivalent to a probability.