I am doing a course in ergodic theory and one of the difficult problems we got to prepare for the exam was:
Does it exist a non-trivial measure which is absolutely continuous with respect to Borel measure?
Does this mean that I have to find a finite measure $m$ that is absolutely continuous to Borel measure $\mu$ which is written as $m \ll \mu$, saying that $m$ is dominated by $\mu$?
Could this be related to how Borel measure $\mu$ is abs. cont. w.r.t. Lebesgue measure $\lambda$?
Well, It is not very clear what would be "non trivial measure". By your comments it seems that you want to exclude the solution $m = \mu$.
Here is a rather general example: Let $f$ be any non-negative Borel measurable function, such that $\int f d\mu < \infty$. Define $m$ on the Borel $\sigma$-algebra by $m(E) = \int_E f d\mu$. Then $m$ is finite and absolutely continuous with respect to $\mu$.
If the measure space under consideration is such that its Borel $\sigma$-algebra is $\sigma$-finite, then, by Radon-Nikodym, the example above represents all possible finite measures $m$ that are absolutely continuous with respect to $\mu$.