# A question on Corollary of Lusin's Theorem in Rudin's Real and Complex analysis

I have a question on Corollary of Lusin's Theorem in Rudin's Real and Complex analysis (3rd edition, page 56). Here Rudin explicitly requires that $|f| \leq 1$. But I can not see why this requirement is necessary. I can not even see why $|f|$ should be bounded above. Thanks.

The corollary states: Assume the hypotheses of Lusin's Theorem are satisfied and that $|f| \leq 1$. Then there is a sequence $\{g_n \}$ such that $g_n \in C_c(X), |g_n| \leq 1,$ and

$f(x) = lim_{n\rightarrow \infty} g_n(x)$ a.e.

The conditions of Lusin's theorem are that f is complex measurable, X is locally compact and Hausdorff, $\mu(A)<\infty, f(x)=0$ if $x\notin A.$

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Can you include the corollary ? – Amr Jan 6 '13 at 6:40

Part of the point of the corollary is that $|g_n|\leq 1$.
On one hand, to have any hope that $|g_n|\leq 1$, the fact that $\lim g_n(x)=f(x)$ a.e. implies that $|f(x)|\leq 1$ a.e. (and the statement could be negligibly strengthened by only requiring the bound on $f$ a.e.).
On the other hand, more importantly, it is part of the strength of the corollary that each function $g_n$ can be chosen to have the same bound as $f$. It is important enough that the sequence $(g_n)$ can be chosen to have a uniform bound. Strengthening the hypothesis is worthwhile if it gets you stronger results.
You could also consider what the corresponding statement would be if $f$ is not assumed to be bounded. In that case, you could still get a sequence of continuous functions with compact support converging a.e. to $f$ using Lusin's theorem, but of course there would be no uniform bound in general.