# Prove that $\frac{1}{f_n} \rightarrow \frac{1}{f}$ in measure.

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.

Hint: $$\mu (|f_n-f| >\epsilon \delta |f|)$$ $$\leq \mu (|f_n-f| >\epsilon \delta^{2}/2)+\mu (|f| \leq \delta/2).$$ Now, $$\mu (|f| \leq \delta /2) \leq \mu (|f_n-f| >\delta /2) +\mu (|f_n| <\delta)$$ $$=\mu (|f_n-f| >\delta /2) \to 0.$$
• @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 $|f_n| \leq |f_n-f| +|f|$]. Mar 14, 2021 at 6:10
• @Blaze Suppose $|f_n-f| >\epsilon \delta |f|$. Assume that $|f| \leq \delta /2$ is false. Then conclude that $|f_n-f| >\epsilon \delta^{2}/2$. Mar 14, 2021 at 6:34
• Doing that we get that $|f_n-f| > \epsilon \delta^2/2$. I do not see how this contradicts anything we have so far. Mar 14, 2021 at 6:46