Let $E$ be a measurable set of finite measure, $f_n$ converges to $f$ in measure on $E$, $f$ is finite and $\neq0$ almost everywhere. Prove that $\dfrac{1}{f_n}$ converges in measure to $\dfrac{1}{f}$.
Here is my attempt:
For every $r > 0$, for every $\varepsilon >0$,
$\Big\{x \in E : \Big\vert \dfrac{1}{f_n} - \dfrac{1}{f} \Big\vert \ge r \Big\}$ $=\Big\{x \in E : \Big\vert \dfrac{f_n-f}{f_nf}\Big\vert \ge r \Big\} $ $=\Big\{x \in E : |f_n-f|\Big\vert \dfrac{1}{f_nf} \Big\vert \ge r \Big\} $.
Since $f_n$ converges to $f$ in measure, $\Big\{x \in E : |f_n-f| \ge r \Big\} < \varepsilon$.
But I don't see how to evaluate $\Big\vert \dfrac{1}{f_nf} \Big\vert$ as well as deal with problem.
Please give me some hints. Any help or advice is highly appreciated.
This general result is useful:
Lemma: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space. A sequence $S=\{f_n\}$ converges in measure to $f$ iff any subsequence $\{f_{n'}\}$ of $S$ has a further subsequence $\{f_{n''}\}$ that converges $\mu$-a.s. to $f$
Sketch of proof of sufficiency: First check that $f_n$ converges in measure to $f$ iff $\int|f_n-f|\wedge1\,d\mu\xrightarrow{n\rightarrow\infty}0$. This is pretty standard. Now suppose $f_n$ satisfies the hypothesis bu that $f_n$ does not converge in measure to $f$. Then there is a subsequence $\{f_{n'}\}$such that $\mu(|f_{n'}-f|>\varepsilon\}\geq\delta$ for some $\delta>0$ and $\varepsilon>0$. But then there is a subsequence $\{f_{n''}\}$ of $\{f_{n'}\}$ that converges to $f$ $\mu$-a.s. which would imply that $f_{n''}$ converges in measure to $f$. This is a contradiction.
The details can be usually find in measure theoretic books of probability (Kallenberg, Billingsley) and in many Measure theory textbooks.