I am reading these notes http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ by Terry Tao. I have a question about the difference between convergence in $L^{\infty}$ and convergence almost uniformly.
Is the difference that convergence almost uniformly guarantees that you can get uniform convergence outside a set of arbitrarily small but still positive measure, while convergence $L^{\infty}$ gets uniform convergence outside a set of exactly measure zero? Formal definitions follow to make ideas precise.
Let $(X, \mathcal{M}, \mu)$ be a measure space. Let $f, f_1, f_2, \ldots$ be a measurable functions.
We say that $f_n \to f$ in $L^{\infty}$ if for all $\varepsilon > 0$ there is an $N_{\varepsilon}$ such that $|f_n(x) – f(x)| \leq \varepsilon$ $\mu$–a.e. when $n \geq N_{\varepsilon}$.
We say that $f_n \to f$ almost uniformly if for all $\varepsilon > 0$ there is a set $E \in \mathcal{M}$ with $\mu(E) \leq \varepsilon$ such that $f_n \to f$ uniformly on $E^c$. I.e., for each $\delta > 0$ there is an $N_{\delta}$ such that $|f_n(x) – f(x)| \leq \delta$ for all $x \in E^c$ when $n \geq N_{\delta}$.
Best Answer
Yes, that is right.
It is a good exercise to prove that your definition of $L^\infty$ convergence is equivalent to:
The thing to notice is that your definition should be expanded as:
That is, the set $F$ may depend on $\epsilon$ and $n$. In showing the equivalence with my statement, you have to find a single set $F$ that works for all $\epsilon,n$.