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I am reading these notes 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 $n \geq N_{\delta}$.

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up vote 4 down vote accepted

Yes, that is right.

It is a good exercise to prove that your definition of $L^\infty$ convergence is equivalent to:

There is a set $E \subset \mathcal{M}$ with $\mu(E) = 0$ such that $f_n \to f$ uniformly on $E^c$.

The thing to notice is that your definition should be expanded as:

For all $\epsilon > 0$ there exists $N$ such that for all $n \ge N$ there exists $F \in \mathcal{M}$ such that $\mu(F) = 0$ and $|f_n(x) - f(x)| \le \epsilon$ for all $x \in F^c$.

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 $\epsilon,n$.

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Thank you @Nate Eldredge for the comment and the exercise. Shouldn't the end of your definition read $x \in F^c$ instead of $x \in F$. – Alex Aug 27 '12 at 1:11
@AlexOlssen: Indeed, thanks. Fixed. – Nate Eldredge Aug 27 '12 at 1:58

Here's an example to see the difference:

Consider the sequence of functions $f_n$, where $f_n(x)= 1$, whenever $\frac{-1}{n}<x<\frac{1}{n}$ and zero otherwise. These functions are clearly measurable. Also, the point-wise limit is the function $f$ which is zero on $R \backslash\{0\}$ and $f(0)=1$. Finally, one can show that that this converges almost uniformly (By chopping of small intervals centered around $0$), but not in $L^\infty$(Since we can do no such chopping).

Since I am in a hurry, I could only give a sketch. Perhaps I could try and clarify any doubts you have in the comments?

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Thanks a lot for the example! – Alex Aug 27 '12 at 1:15
I thought that I would spell out why the example works. To see convergence almost uniformly we can pick $N > 2 / \varepsilon$. Then $\mu(\{x: |f_n(x) - f(x)| > \varepsilon\}) < 2 * (1 / (2 / \varepsilon)) = \varepsilon$. However no matter what $N$ we pick there is still a set with positive measure where $|f_n(x) - f(x)| = 1$ so that we do not get convergence $L^{\infty}$. – Alex Aug 27 '12 at 1:27
@AlexOlssen Perfect.Couldn't have said it better myself. :) – Ravi Aug 27 '12 at 1:32

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