Re: f'=0 a.e => f constant ?

From: Rasmus Villemoes (burner+usenet_at_imf.au.dk)
Date: 09/01/04 

Date: Wed, 01 Sep 2004 18:19:16 +0200

"Julien Santini" <santini.julien@wanadoo.fr> writes:

> Hello
>
> How could I show that if f is absolutely continuous, almost everywhere
> differentiable on the interval (a,b) and such that f '(x)=0 a.e, then f is
> constant ?
>

How about:

Choose some x_0 in (a, b). Then f(x) - f(x_0) is the same as the
integral of f' from x_0 to x (which makes sense since f is
a.e. differentiable; ie. for almost all t the number f'(t) exists, and
this is sufficient for defining the integral of f'); but since f' is
a.e. equal to the zero function, this integral is 0. Thus f(x) =
f(x_0); so f is constant.

/Rasmus 

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