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| James |
 Posted: Fri Dec 30, 2005 11:17 am |
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Guest
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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.
Any ideas/help is appreciated,
James |
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| Guest |
| Posted: Fri Dec 30, 2005 11:52 am |
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The main issue is how are we to do anything with the integral
\int_a^x g ' (f(t)) f ' (t) dt .
We would want to show this is just the Riemann-Stieltjes integral
\int_a^x g ' (f(x)) d f(x),
which in turn is the Riemann integral
\int_f(a)^f(x) g'(t) dt,
which of course is g(f(x)) - g(f(a)).
Proving these steps is much more elegant in the more general setting of
measure theory. This is mainly because one rarely needs to remember how
to spell Stieltjes. |
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| Guest |
| Posted: Fri Dec 30, 2005 3:12 pm |
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Need to know:
1. Definition of absolute-continuity of (signed) measures. \mu <<
\lambda
2. Definition of the Borel-Stieltjes measure associated with any
measurable continuous on the right function of bounded variation.
3. The Radon-Nikodym derivative.
I. A measurable, continuous on the right function of bounded variation
is absolutely continuous if and only if its associated Borel-Stieltjes
measure is absolutely continuous with respect to Lebesgue measure.
II.The Radon-Nikodym chain rule:
If a << b << c The Radon-Nikodym derivatives satisfy da/dc = da/db *
db/dc almost everywhere.
III. The following change-of-variables formula: (i think this is
right...)
Let x be Lebesgue measure. Let g be integrable, let f be monotone
increasing and absolutely continuous. Let F be the Borel-Stieltjes
measure associated with f.
\int_[a,b] g o f dF= \int_[a,b] g o f (dF/dx) dx = \int_f([a, b]) g
dx = \int_[f(a), f(b)] g dx
This can be proven by letting g = g+ - g-, and producing monotone
sequences of simple measurable functions approaching g+ and g-. Then
apply the monotone convergence theorem. |
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| David C. Ullrich |
| Posted: Sat Dec 31, 2005 10:22 am |
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Guest
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On Fri, 30 Dec 2005 11:17:43 -0500, "James" <James545@gmail.com>
wrote:
[quote:be8fd97963]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.
[/quote:be8fd97963]
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.
[quote:be8fd97963]Any ideas/help is appreciated,
James
[/quote:be8fd97963]
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David C. Ullrich |
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