Assumptions: Let $\mu$ and $\nu$ be positive and finite measures on $(X, \Sigma)$. Suppose that $\mu$ is absolutely continuous with respect to $\nu$, so that the Radon-Nikodym derivative of $\mu$ with respect to $\nu$ exists, and is denoted $f$. Suppose that $f$ is bounded above.
Question: Could there exist a sequence of measurable sets $S_1, S_2, ... \in \Sigma$ such that $\nu(S_n) > 0$ for all $n \in \mathbb{N}^+$ and $$\frac{\mu(S_n)}{\nu(S_n)} \to \infty ~\text{ as }~ n \to \infty?$$ In other words, would the existence of such a sequence lead to a contradiction? I think so and would like to show that a contradiction would arise, but I'm not sure if my attempted proof below is right.
Attempted proof: Since $f$ is bounded above, there exists a real number $C > 0$ such that $f(x) < C, \forall x \in X$. Then $$\frac{\mu(S_n)}{\nu(S_n)} = \frac{\int_{S_n} f d\nu}{\nu(S_n)} \leq C \frac{\int_{S_n} d\nu}{\nu(S_n)} = C\frac{\nu(S_n)}{\nu(S_n)} = C,$$ so $\frac{\mu(S_n)}{\nu(S_n)}$ is bounded above by $C$, therefore $\frac{\mu(S_n)}{\nu(S_n)} \not \to \infty$. However, this argument seems to suggest that any sequence $S_1, S_2, ...$ satisfies $$\frac{\mu(S_n)}{\nu(S_n)} \leq C.$$ That is to say, all sequences of ratios of measures are bounded above by a fixed common $C$, which works for all sequences. Is this proof correct, and does this final argument (about a common $C$ working for all sequences) seem right?