Re: Absolute Continuity Problem from Royden

From: mr0x (mr00xx_at_nospam.hotmail.com)
Date: 09/05/04 

Date: Sun, 05 Sep 2004 20:49:23 GMT

"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
news:u0ulj09viv348uedjhgsqf3c5dmlbgbk29@4ax.com...
> On Sat, 04 Sep 2004 22:47:58 GMT, "mr0x" <mr00xx@nospam.hotmail.com>
> wrote:
>
> >Hi, I'm having a lot of problem solving problem 17b from Chapter 5 from
> >Royden.
> >
> >It says
> >b).Let E = {x : g'(x)=0}. Then, m(g[E])=0.
> >
> >g is defined on part (a) of the question.
> >
> >The (a) of the question is
> >a). Let F be absolutely continuous on [c,d] and g be absolutely
continuous
> >with c <= g <= d on [a,b]. Then, Fog is absolutely continuous on [a,b].
> >
> >
> >I was thinking of a proof like this ->
> >Let E' be a subset of E where for each interval in E, we take only 1
point
> >from it. Then, m(E')=0 (since it's countable) and so as g[E] = g[E'] and
> >thus, m(g[E])=0.
> >
> >But, on problem 19a), it says ->
> >19a). Construct an absolutely continuous strictly monotone function g on
> >[0,1] such that g'=0 on a set of positive measure.
> >
> >Thus, in the proof I can't say m(E')=0 from this.
> >
> >
> >Any hints on what to go about it?
>
> start by noting that m(E) and m(g{E)) are not the
> same thing - that should clear up the contradictory
> parts...
>

I was thinking more about it and obviously Royden made it part (b) because
it was related to part (a). So, if we use change of variables, then \int Fog
= \int F'(g) g' = 0 and so, we can say m(g[E])=0. Since change of variables
isn't covered till the last two problems of the chapter, I didn't know if my
assumptions allow me to use it.

Anyway, I think that was it. Thanks.

> >Thanks.
> >
>
>
> ************************
>
> David C. Ullrich
>
> sorry about the inelegant formatting - typing
> one-handed for a few weeks...