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I'm trying to prove the following claim:

$f_n \in C_c$, $C_c$ being the set of continuous functions with compact support, then $\mathrm{lim}_{n \rightarrow \infty} || f_n - f||_{\infty} = 0$ implies $f_n(x) \rightarrow f(x)$ uniformly.

So, according to my understanding,

$$ \mathrm{lim}_{n \rightarrow \infty} || f_n - f||_{\infty} = 0 $$ $$ \Leftrightarrow $$ $$ \forall \varepsilon > 0 \exists N: n > N \Rightarrow |f_n(x) - f(x)| \leq || f_n - f||_{\infty} < \varepsilon$$ $\mu$-almost everywhere on $X$.

Now my problem is that I don't see how uniform convergence follows from pointwise convergence $\mu$ almost everywhere. Can someone give me a hint? I guess I have to use the fact that they have compact support but I don't see how to put this together.

Thanks for your help!

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Possible Hint: What you have written on the right hand side is not only pointwise convergence a.e., but uniform convergence a.e.. (The N doesn't depend on x.) –  Jesse Madnick Jan 22 '11 at 12:43
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You were not very specific about your hypotheses - I assume you're working on $\mathbb{R}^{n}$ with Lebesgue measure.

Suppose there exists a point $x$ such that $|f_{n}(x) - f(x)| > \varepsilon$. Then there exists an open set $U$ containing $x$ such that for all $y \in U$ you have $|f_{n}(y) - f(y)| > \varepsilon$ by continuity. But this contradicts the a.e. statement you gave.

In case you don't know yet that $f$ is continuous (or rather: has a continuous representative in $L^\infty$), a similar argument shows that $(f_{n})$ is a uniform Cauchy sequence (with the sup-norm, not only the essential sup-norm), hence $f$ will be continuous (in fact, uniformly continuous).

Note that I haven't used compact support at all, just continuity.

If you're working in a more general setting (like a locally compact space), you'd have to require that the measure gives positive mass to each non-empty open set.

Finally, note that $f$ need not have compact support. It will only have the property that it will be arbitrarily small outside some compact set ("vanish at infinity" is the technical term). For instance, $\frac{1}{1+|x|}$ can easily be uniformly approximated by functions with compact support.

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The main upshot of the argument is of course: A continuous function which is a.e. zero is zero everywhere. –  t.b. Jan 22 '11 at 12:34
    
Thank you. I'm sorry for being unspecific, I meant $(X, \mu, \Sigma)$ a measure space and $L^{\infty}(X, \mu)$. Of course it's ok to assume that $\mu = \lambda$ and $X = \mathbb{R}$. I'm interested in the case where $f$ doesn't necessarily have a continuous representative in $L^{\infty}$. So because of $lim_{n \rightarrow \infty} || f_n - f||_{\infty} = 0$, $f_n$ is a Cauchy sequence in $L^{\infty}$, so $$ \forall \varepsilon > 0: \exists N: n,m > N \Rightarrow ||f_n - f_m ||_{\infty} < \varepsilon$$. So here I can apply the argument with the open set $U$, so this is uniformly convergent. –  Matt N. Jan 22 '11 at 13:20
    
I think this answered my question very well, many thanks. Also thanks for the note about locally compact spaces. The reason why I was thinking about $C_c$ is because this set is dense in $L^p$ but not in $L^{\infty}$ and I wanted to see why. The property of compact support together with continuity causes $C_c \subset L^p$, that's why I considered functions with compact support only. –  Matt N. Jan 22 '11 at 13:31
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@Matt: Yes, exactly. Just be careful that talking about continuity means that $X$ has a topology in addition to the measure space structure and the argument only works if non-empty open sets are measurable and have positive measure, so you'd better assume that $\Sigma$ is the Borel $\sigma$-algebra of some topological space. Just think about the extremely degenerate case of the Dirac measure at a point where $L^{\infty}$-convergence just means convergence at that point. –  t.b. Jan 22 '11 at 13:39
    
@Matt: Sorry, I was distracted while I was typing, so I hadn't seen your last comment before submitting mine. As I sketched: the closure of $C_{c}$ in $L^{\infty}$ is the space $C_{0}$ of continuous functions vanishing at infinity, that is to say the continuous functions such that for every $\varepsilon$ there is a compact set $K$ such that $\sup_{x \notin K} |f(x)| \lt \varepsilon$ –  t.b. Jan 22 '11 at 13:42
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Just a hint: The compact support doesn't matter, but the continuity does matter. It helps you to remove null sets.

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