From: "Daniel Giaimo" <dgiaimo@ix.netcom.com>
Subject: Re: real anal
Date: Sun, 11 Jun 2000 10:33:23 -0700
Newsgroups: sci.math
Summary: [missing]

<chuleta2099@my-deja.com> wrote in message
news:8h8qtp$vit$1@nnrp1.deja.com...
> Anyone know of a book which proves (or know the proof of) if f:[a,b]->R
> and f is lebesgue measurable, then, for every e>0, there exists a
> continuous function h such that m{x:f(x)!=h(x)}<e?

    Here is an outline of a proof for which you should be able to fill 
in the details:

    1. If f is Lebesgue measureable then there exists a set E in [a,b] s.t
       m([a,b] \ E) < e/3 and f is bounded on E.

    2. A bounded Lebesgue measureable function on a finite measure set E
       is in L^1(E)

    3. The continuous functions with compact support on E are dense in L^1(E)

    4. Convergence in L^1 implies convergence in measure.

    5. By, Egoroff's theorem, there exists a subset of E, F s.t. m(E\F) < e/3
       and continuous functions f_j s.t. f_j -> f uniformly on F.

    6. There exists a compact subset G of F s.t. m(F\G) < e/3.

    7. By 5., f is continuous on G.

    8. By Tietze's Extension Theorem, f|_G can be extended to a continuous
       function h on [a,b].

    9. m([a,b] \ G) < e, quod erat demonstrandum.

--Daniel Giaimo