# Almost uniform convergence implies convergence in measure

Let $$(A,\mathcal{F},\mu)$$ be finite measure space and $$\{f_n\}$$ a sequence of finite real measurable functions so that $$f_n\rightarrow f$$ a.e. We say $$f_n\rightarrow f$$ almost uniformly if $$\epsilon>0$$, there is $$E\subseteq A$$ such that $$f_n \rightarrow f$$ uniformly on $$E^c$$ and $$\mu(E)<\epsilon$$.

I want to show that $$f_n\rightarrow f$$ almost uniformly implies convergence in $$\mu$$. For this, suppose not. Then $$\exists \eta,\epsilon>0:\forall N\in \mathbb{N}:\exists n>N:\mu(\mid f_n-f\mid\geq\epsilon)\geq \eta,$$ i.e., for infinitely many points $$n\in \mathbb{N}$$. From the definition of almost uniform convergence, $$\exists E:\mu(E)<\eta$$ and $$f_n\rightarrow f$$ uniformly on $$E^c$$. Contradiction.

Question

It seems intuitive to me. But how to deduce this contradiction precisely? I know that if $$x\in E$$, then it must satify the negation of uniform convergence which is $$\exists \epsilon>0:\forall N\in \mathbb{N}:\exists n>N:\mid f_n-f\mid\geq\epsilon.$$ Now, $$x$$ may not be in $$\{f_n \text{ does not converge in measure to } f \}$$ if $$\mu(\mid f_n(x)-f(x)\mid\geq\epsilon)<\eta$$. So I conclude that $$\{f_n \text{ does not converge in measure to } f \}\subseteq E$$ implying that $$\eta>\mu(E)\geq \eta$$; a contradiction.

My argument seems right but also very inefficient. How could you express this idea as clean as possible?

Thanks!

Here's a way without going for contradiction: Let $$\epsilon >0$$ be fixed and consider some $$\delta >0$$.

By almost uniform convergence, there exists some $$E$$ with $$\mu(E)\leq \delta$$ and some $$N$$ such that $$n\geq N\implies \forall x\in E^c, |f_n(x)-f(x)|< \epsilon$$.

For $$n\geq N$$, note the inclusion $$(|f_n-f|\geq \epsilon) \subset E$$, hence $$\mu((|f_n-f|\geq \epsilon))\leq \mu(E)\leq \delta$$.

Hence $$\forall \delta>0, \exists N, n\geq N \implies \mu((|f_n-f|\geq \epsilon))\leq \delta$$. Thus $$\mu((|f_n-f|\geq \epsilon)) \to 0$$.

• Thanks! The inclusion $\{\mid f_n-f\mid\geq \epsilon\}\subset E$ that I don't understand. What is this infinity-norm that you wrote? Aug 23, 2019 at 20:30
• @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ? Aug 23, 2019 at 20:32
• It has something to do with the fact that uniform convergence implies pointwise convergence? And then the complement of the latter is contained in the complement of the former. Aug 23, 2019 at 20:38
• @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof. Aug 23, 2019 at 20:40
• But, $E$ is constituted of elements in $A$ such that $f_n \rightarrow f$ does not converge uniformly, right? Aug 23, 2019 at 20:42

This is proved more succinctly by a direct proof.

Given $$\varepsilon > 0,$$ let $$E$$ be as in the definition of almost-uniform convergence. Then there is some $$N$$ such that $$n \geq N$$ implies $$\mu(E^c \cap \{|f_n-f|\geq\varepsilon\}) = 0$$ What does this imply for $$\mu(|f_n-f|\geq \varepsilon)?$$

• How to conclude that $\{\mid f_n-f\mid \}\subseteq E$? Aug 23, 2019 at 20:55
• @Danmat Forget about $E$ for a second. On $E^c,$ the functions converge uniformly. This means that there will be some $N$ such that $n \geq N$ implies $|f_n - f| < \varepsilon.$ In particular, on $E^c,$ $$n \geq N \implies \{|f_n - f| \geq \varepsilon\} = \emptyset.$$ This directly means $$E^c \cap \{|f_n - f| \geq \varepsilon\} = \emptyset.$$ If you want, then you can use the property of sets that $$A^c \cap B = \emptyset \iff B \subseteq A$$ Aug 23, 2019 at 21:05
• Very nice! Now, it's all clear to me! Aug 23, 2019 at 21:11