If is Riemann Integrable in ,and for any integer , we have . can we get the conclusion: a.e.?
(a.e. stands for almost everywhere)
Yes. Note that if for all then for any polynomial . By Wierstrass aproximation theorem, this set (polynomials) is dense in where this last is considered with the uniform norm. The functional is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So for all . It is a well known result that under this conditions .
Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
Lets suppose, without loss of generality, that is and . Since is Riemann integrable in is can be expanded in Fourier series ...
(1)
... where...
(2)
Now is...
(3)
... and is...
(4)
The conclusion is that all the integrals in (2) are null and the same is for the and ...
Kind regards
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval can be approximated uniformly on by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function
If we denote the set of all discontinuity point,then define:
then is a continuous function on ,right?
then use the conclusion above,we get the conclusion .
so a.e.
Is this proof right?
Notice that I'm using Weierstrass to expand the set of functions such that , I'm not saying anything about .
Another thing, the you defined could be not well defined if f has a jump discontinuity.
This doesn't contradict the claim sinceOk, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.