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.

  • 3
    $\begingroup$ Is there a missing condition, like $f$ must be bounded away from zero? (Consider a sequence of constant functions, $f_n=\frac1n$, and $f=0$.) $\endgroup$ – grand_chat May 21 at 17:50
  • $\begingroup$ This is not true, as suggested in the example of grand_chat. $\endgroup$ – Jingeon An May 21 at 17:52
  • 1
    $\begingroup$ This has been addressed here: math.stackexchange.com/questions/3685367/… $\endgroup$ – Oliver Diaz May 21 at 21:45
  • $\begingroup$ @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. $\endgroup$ – FactorY May 22 at 0:19
  • $\begingroup$ But with this condition, I can't manage to get the correct answer. $\endgroup$ – FactorY May 22 at 0:39

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.

| cite | improve this answer | |
  • $\begingroup$ I can't guess what does a.s mean? I just think a stands for almost and s would be? $\endgroup$ – FactorY May 22 at 3:24
  • $\begingroup$ almost surely (in analysis is most common to say almost everywhere) $\endgroup$ – Oliver Diaz May 22 at 3:25

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