# $E$ is finite measurable set, if $f_n$ converges in measure to $f$ then $\dfrac{1}{f_n}$ converges in measure to $\dfrac{1}{f}$

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.

• Is there a missing condition, like $f$ must be bounded away from zero? (Consider a sequence of constant functions, $f_n=\frac1n$, and $f=0$.) – grand_chat May 21 at 17:50
• This is not true, as suggested in the example of grand_chat. – Jingeon An May 21 at 17:52
• This has been addressed here: math.stackexchange.com/questions/3685367/… – Oliver Diaz May 21 at 21:45
• @grand_chat oh sorry, I miss this condition f_n(x) and f(x) are both not equal to zero, for all x in E. – FactorY May 22 at 0:19
• But with this condition, I can't manage to get the correct answer. – FactorY May 22 at 0:39

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.