Re: Absolute continuity, another question,

On Fri, 30 Dec 2005 11:17:43 -0500, "James" <James545@xxxxxxxxx>
wrote:

>I have another question on Absolute continuity :
>
>Let f : [a,b] ---> R be a strictly increasing absolutely continuous function
>and g : [f(a),f(b)] ---> R be an absolutely continuous function. I am trying
>to prove that (g(f(x))) ' = g ' (f(x)) f ' (x) a.e. in [a,b], where g '
>(f(x)) f ' (x) is understood to be 0 if f ' (x) = 0, regardless of whether
>or not g ' (f(x)) exists.
>
>I have shown that g o f is absolutely continuous (using monotonicity of f).
>I tried to then represent g o f as the integral of its derivative, but I
>couldn't see how this would lead me to what I need to prove.

Say E = {x : f'(x) > 0}, F = {x : f'(x) = 0}. All you need to show is
that (i) g'(f(x)) _exists_ for almost every x in E (then the chain
rule shows that it equals what it should for almost every x in E)
and (ii) (g o f)'(x) = 0 for almost every x in F.

For (i), since g is differentiable almost everywhere, you only
need to show that

(iii) If A is a subset of E and m(A) > 0 then m(f(A)) > 0.

Which follows from

(iv) If A is a subset of [a,b] then m(f(A)) = int_A f'.

It's easy to see that f maps null sets to null sets, so you
can assume that A is a G_delta (a countable intersection of
open sets) in (iv). If A is open it's a countable union of
intervals, as is f(A), and (iv) is clear. Now if A is
a G_delta use dominated convergence on one side and the
fact that m is a measure on the other side. This gives (ii).

For (ii) you can assume that g is non-decreasing. Now
(iv) shows that m(f(F)) = 0. Since g maps null sets to
null sets it follows that m(g(f(F))) = 0. And hence
(iv), applied to g o f in place of f, shows that
int_F (g o f)' = 0, hence (g o f)' = 0 ae on F.

The key is (iv), there's probably a more elegant
way of organizing the rest of it.

>Any ideas/help is appreciated,
>
>James
>


************************

David C. Ullrich
.