A question on convergence to zero of measurable sets

Question

Let $f$ be a strictly positive (almost everywhere) measurable function which is also integrable. Let $E_n$ be a sequence of measurable sets such that $\int_{E_n}f\to 0$ as $n\to\infty.$ Is it true that $\mu(E_n)\to 0$ where $\mu$ is a positive measure with respect to $f$ is integrable. The measure space is finite.

Answer 1

Yes. Suppose otherwise. Then, by passing to a subsequence, there is some $\epsilon > 0$ with $\int_{E_n} f \to 0$ but $\mu(E_n) \ge \epsilon$ for each $n$. For $k \ge 1$, let $F_k = \{x \in X : |f(x)| < \frac{1}{k}\}$. Since $\emptyset = \cap_{k \ge 1} F_k$ and $(X,\mu)$ is finite, there is some $k$ with $\mu(F_k) < \frac{\epsilon}{2}$. Then $\mu(F_k^c) \ge \mu(X)-\frac{\epsilon}{2}$ implies that for any $n \ge 1$, $\mu(E_n \cap F_k^c) \ge \frac{\epsilon}{2}$ and so $\int_{E_n} f \ge \int_{E_n \cap F_k^c} f \ge \int_{E_n \cap F_k^c} \frac{1}{k} \ge \frac{\epsilon}{2}\frac{1}{k}$. Contradiction.