Convergence in $L_{\infty}$ norm implies uniform convergence

Solution 1:

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.

Solution 2:

Just a hint: The compact support doesn't matter, but the continuity does matter. It helps you to remove null sets.