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Suppose that $f_n,f: E \rightarrow \mathbb{R}$ are measurable and finite almost everywhere.

If $f_n \rightarrow f$ in measure, and there is some $\delta > 0$ so that for every $n$ $f_n > \delta$ a.e., then $1/f_n \rightarrow 1/f$ in measure. (We are talking about Lebesgue measure).

Consider the set $$\left\{\left|\frac{1}{f_n}-\frac{1}{f}\right| > \epsilon \right\} = \{|f-f_n| > \epsilon \cdot|f_n|\cdot|f|\}.$$ Clearly by assumption we are given that $|f_n| > \delta$ so we can easily contain the set above by $\{|f-f_n| > \epsilon \cdot\delta\cdot|f|\}$. Here is where I am having trouble. I want to bound $f$ from below and use the fact that $f_n \rightarrow f$ in measure to complete the proof. Here is my attempt to bound it so far: Since $f_n \rightarrow f$ in measure, then $m(\{|f_n-f| \geq 1\}) < \eta$ when $n > N$. Then $|f| \geq ||f_n| - |f_n-f|| \geq |\delta-|f_n-f||$. I am stuck here since I cannot use $|f_n-f| \geq 1$ to get a lower bound, I would need the opposite inequality .. I think I am almost there but not quite.

Could someone please help me out?

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Hint: $$\mu (|f_n-f| >\epsilon \delta |f|)$$ $$\leq \mu (|f_n-f| >\epsilon \delta^{2}/2)+\mu (|f| \leq \delta/2).$$ Now, $$\mu (|f| \leq \delta /2) \leq \mu (|f_n-f| >\delta /2) +\mu (|f_n| <\delta)$$ $$=\mu (|f_n-f| >\delta /2) \to 0.$$

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  • $\begingroup$ Could you elaborate a bit more? I apologize, I am new to convergence in measure .. $\endgroup$ Mar 14, 2021 at 6:01
  • $\begingroup$ @Blaze When I write $\mu(A) \leq \mu(B)+\mu(C)$ you have to justify it by verifying that $A \subseteq B \cup C$. [There are two such inequalities in my answer. In both cases the verification is quite simple: in the second case use the fact that $|f_n| \leq |f_n-f| +|f|$]. $\endgroup$ Mar 14, 2021 at 6:10
  • $\begingroup$ I understand thank you. However, I am still not seeing the first verification ... $\endgroup$ Mar 14, 2021 at 6:25
  • $\begingroup$ @Blaze Suppose $|f_n-f| >\epsilon \delta |f|$. Assume that $|f| \leq \delta /2$ is false. Then conclude that $|f_n-f| >\epsilon \delta^{2}/2 $. $\endgroup$ Mar 14, 2021 at 6:34
  • $\begingroup$ Doing that we get that $|f_n-f| > \epsilon \delta^2/2$. I do not see how this contradicts anything we have so far. $\endgroup$ Mar 14, 2021 at 6:46

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