# 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? – Celine Harumi Aug 23 '19 at 20:30
• @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ? – Gabriel Romon Aug 23 '19 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. – Celine Harumi Aug 23 '19 at 20:38
• @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof. – Gabriel Romon Aug 23 '19 at 20:40
• But, $E$ is constituted of elements in $A$ such that $f_n \rightarrow f$ does not converge uniformly, right? – Celine Harumi Aug 23 '19 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 $\{\mid f_n-f\mid \}\subseteq E$? – Celine Harumi Aug 23 '19 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 $$A^c \cap B = \emptyset \iff B \subseteq A$$ – Brian Moehring Aug 23 '19 at 21:05
• Very nice! Now, it's all clear to me! – Celine Harumi Aug 23 '19 at 21:11