A problem on almost singular measures in real analysis 

Since $\mu$ and $\nu$ are measures, for $n\in \N$, $\nu-n^{-1}\mu$ is a signed measure. Then for each $n \in \N$ there exists a Hahn Decomposition (https://en.wikipedia.org/wiki/Hahn_decomposition_theorem) for $X$ with respect to the signed measure $\nu-n^{-1}\mu$:

$$X=P_{n}\cup N_{n}.$$
Let $P=\cup_{n=1}^{\infty} P_{n}$ and $N=\cap_{n=1}^{\infty}N_{n}=P^{c}$. Then $N$ is negative for $\nu-n^{-1}\mu$ for every $n$. So we have
\[(\nu-n^{-1}\mu)(N)\leq 0.\]
Hence \[\nu(N)-n^{-1}\mu(N)\leq 0\]
and therefore, \[ 0\leq \nu(N)\leq n^{-1} \mu(N). \]
Since $\mu(N)<\infty$, letting $n\rightarrow \infty$ we have 

\[\nu(N)=0 . \] Now we consider two cases:

Case I: $\mu(P)=0$. Then from the above result we conclude that $\mu\perp \nu$ .
Case II: $\mu(P)>0$. Then there exists $n_{0}\in N$ such that $\mu(P_{n_{0}})>0$. Since $X=P_{n_0}\cup N_{n_0}$ is a Hahn decomposition with respect to $\nu-n_0^{-1}\mu$, \[ \nu(P_{n_{0}})-n_{0}^{-1} \mu(P_{n_{0}})>0, \] and hence \[ \nu(P_{n_{0}})>n_{0}^{-1} \mu(P_{n_{0}}). \]
Letting $E:=P_{n_{0}}$ and $\epsilon=n_{0}^{-1}$ we have $\nu(E)>\epsilon \mu(E)$.         $\Box$