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From: r on 9 Jul 2010 08:07
Le 17/05/2010 23:46, W^3 a �crit :
> In article
> <921617200.174707.1274112540433.JavaMail.root(a)gallium.mathforum.org>,
> Maury Barbato<mauriziobarbato(a)aruba.it> wrote:
>
>> Hello,
>> is there some Lipschitz function f:[a,b]->R such that
>> for some Lebesgue measurable set A, the preimage of A
>> f^(-1)(A) is not Lebesgue measurable?
>>
>> Thank you very much for your attention.
>> My Best Regards,
>> Maury Barbato
>
> Yes, in fact f can be taken to be C^oo and strictly increasing. Let K
> be a fat Cantor set in [0,1] and let g : [0,1] -> [0, 1] be a C^oo
> function that vanishes precisely on K. Set f(x) = int_0^x g(t) dt.
> Then f is C^oo. If x< y, then f(y) - f(x) = int_x^y g(t) dt> 0; this
> follows because (x,y) must contain points not in K (K is nowhere
> dense), hence g> 0 somewhere in (x,y). It is easy to see that because
> f'(x) = g(x) = 0 on K, f(K) has measure 0. To finish let E be a
> nonmeasurable subset of K, and let A = f(E). Then A is Lebesgue
> measurable and its preimage is E.
Hello,
I am not sure of your last statement. if A=f(E), wehave not
automatically E = f^(-1)(A) but we have that E is included in f^(-1)(A).
On another hand, I thought that every continuous function was
measurable, and then is any Lipschitz function, so that the preimage of
any measurable set is itself a measurable set.
Regards
From: David C. Ullrich on 9 Jul 2010 08:53
On Fri, 09 Jul 2010 14:07:12 +0200, r <z(a)gmail.com> wrote:

>Le 17/05/2010 23:46, W^3 a �crit :
>> In article
>> <921617200.174707.1274112540433.JavaMail.root(a)gallium.mathforum.org>,
>> Maury Barbato<mauriziobarbato(a)aruba.it> wrote:
>>
>>> Hello,
>>> is there some Lipschitz function f:[a,b]->R such that
>>> for some Lebesgue measurable set A, the preimage of A
>>> f^(-1)(A) is not Lebesgue measurable?
>>>
>>> Thank you very much for your attention.
>>> My Best Regards,
>>> Maury Barbato
>>
>> Yes, in fact f can be taken to be C^oo and strictly increasing. Let K
>> be a fat Cantor set in [0,1] and let g : [0,1] -> [0, 1] be a C^oo
>> function that vanishes precisely on K. Set f(x) = int_0^x g(t) dt.
>> Then f is C^oo. If x< y, then f(y) - f(x) = int_x^y g(t) dt> 0; this
>> follows because (x,y) must contain points not in K (K is nowhere
>> dense), hence g> 0 somewhere in (x,y). It is easy to see that because
>> f'(x) = g(x) = 0 on K, f(K) has measure 0. To finish let E be a
>> nonmeasurable subset of K, and let A = f(E). Then A is Lebesgue
>> measurable and its preimage is E.
>Hello,
>I am not sure of your last statement. if A=f(E), wehave not
>automatically E = f^(-1)(A) but we have that E is included in f^(-1)(A).

A strictly increasing function is one-to-one.

>On another hand, I thought that every continuous function was
>measurable, and then is any Lipschitz function, so that the preimage of
>any measurable set is itself a measurable set.

The confusion is caused by ambiguity in the word "measurable".
A continuous function from R to R is automatically Borel
measurable; the inverse image of any Borel set is Borel.
Hence the inverse image of any Borel set is Lebesgue
measurable. But it's not true that f continuous implies
f is measurable in the sense that the inverse image
of every Lebesgue measurable set is Lebesgue
measurable.

It's too late to fix the termiology now. If X is a measure
space and f : X -> R then f is usually said to be measurable
if it's measurable with respect to the Borel algebra on R.
On the other hand, if X and Y are measure spaces then
f : X -> Y is measurable if the inverse image of every
measurable set is measurable. If we take Y = R and
consider "measurable" in Y to mean Lebesgue measurable
then these two definitions are different.



>Regards

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