On the contrary, a theorem of Rudin's real and complex analysis

A theorem of Rodin's real and complex analysissuppose $\mu$ and $\nu_1$,$\nu_2$ are measures on a $\sigma$_algebra m, and $\mu$ is positive:a)if $\nu$ is concentrated on A, so is $\left|\nu\right|$.b)if $\nu_1\perp\nu_2$, then $\left|\nu_1\right|\perp\left|\nu_2\right|$c)if $\nu_1\perp\mu$ and $\nu_2\perp\mu$,then $\nu_1+\nu_2\perp\mu$d)if $\nu_1<<\mu$ and $\nu_2<<\mu$, then $\nu_1+\nu_2<<\mu$ .e)if $\nu<<\mu$, then $\left|\nu\right|<<\mu$.f)if $\nu_1<<\mu$ and $\nu_2\perp\mu$, then $\nu_1\perp\nu_2$.g)if $\nu<<\mu$ and $\nu\perp\mu$ then $\nu$=0.I would like to prove the converse of this theorem, or for that matter, give an example.I have tried but unfortunately I could not do it. Any hints would be appreciated.