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!
Best Answer
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$.