Re: A Question on Luzin's Theorem

From: Martin Vaeth (vaeth_at_mathematik.uni-wuerzburg.de)
Date: 08/23/04 

Date: Mon, 23 Aug 2004 16:30:02 +0000 (UTC)

Li Yi <l-o-w-a-i@163.com> wrote:
>
> Since [tex]g(x) = \frac{f(x)}{1+|f(x)|}[/tex] ([tex]f(x) =
> \frac{g(x)}{1-|g(x)|}[/tex])
> we can suppose $f(x)$ is bounded without loss of generality. There
> exists measurable simple functions $\{\phi_k(x)\}$ converges to $f(x)$
> uniformly on $E$

> My question is: Where is Egorov's Theorem used in the proof above?

Egorov's theorem is not needed here, because one can arrange
*uniform* convergence of the sequence \phi_k in this particular situation.
However, in general (i.e. if you do not want to use the "trick" to work
with a bounded function or if you consider functions with values in a
Banach space) the measurability of f implies by definition only that a
*pointwise* (a.e.) convergent sequence \phi_k exists - then it is natural
to apply Egorov's theorem to get uniform convergence on a
"sufficiently large portion" of E.

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