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.


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?



2 Answers 2


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$.

  • $\begingroup$ 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? $\endgroup$ Commented Aug 23, 2019 at 20:30
  • $\begingroup$ @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ? $\endgroup$ Commented Aug 23, 2019 at 20:32
  • $\begingroup$ 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. $\endgroup$ Commented Aug 23, 2019 at 20:38
  • $\begingroup$ @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof. $\endgroup$ Commented Aug 23, 2019 at 20:40
  • $\begingroup$ But, $E$ is constituted of elements in $A$ such that $f_n \rightarrow f$ does not converge uniformly, right? $\endgroup$ Commented 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 $\mu(|f_n-f|\geq \varepsilon)?$

  • $\begingroup$ How to conclude that $\{\mid f_n-f\mid \}\subseteq E$? $\endgroup$ Commented Aug 23, 2019 at 20:55
  • $\begingroup$ @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 $$A^c \cap B = \emptyset \iff B \subseteq A$$ $\endgroup$ Commented Aug 23, 2019 at 21:05
  • $\begingroup$ Very nice! Now, it's all clear to me! $\endgroup$ Commented Aug 23, 2019 at 21:11

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