Let $(\Omega,\mathcal{F})$ be an arbitrary measurable space, and let $\mu$ and $\nu$ be two measures defined on $(\Omega,\mathcal{F})$ such that $\nu<<\mu$. Given that $\mu$ is $\sigma$-finite, can we conclude that $\nu$ has to be $\sigma$-finite?
If so, can someone please provide a rigorous proof of the same?
We can't conclude this. Let $\mu$ be the Lebesgue-measure and $\nu$ be the measure with Lebesgue-density equal to $\infty$, i.e. $\nu(A) = \begin{cases} 0 & \lambda(A) = 0 \\ \infty & \text{else}\end{cases}$.
Now the only sets that are finite with respect to $\nu$ are Lebesgue-nullsets. But a countable union of nullsets is still a nullset, so you can't fill $\mathbb{R}$ with sets of finite measure.