Sorry, that I didn't get you right.
> But my dilema ;) is that how come f() is Riemann integrable
I just thought, that it is clear how from your explanation about non-Riemann integrability of g():
> the higher/upper aproximation sum ( S(f,T) )
> and the lower aprox.sum ( s(f,T) )
> for g to be integrable inf{ S(g,T) } = sup{ s(g,T }
> should be true,
> but s(g,t)=0 (const) and S(g,t)=1 (const)
> so g() not Riemann integrable.)
to get the Riemann integrability for f(). But now, it doesn't seem to me clear at all! Well, if you are SURE that f() is Riemann integrable, then the explanation could be the following:
as soon as you make discretization (a=x1<x2<...<xN=b) of the interval [a,b] finer and finer, the amount of intervals [x_i,x_{i+1}], where rational numbers are with big denominater q, will dominate the other intervals, and thus
inf{ S(g,T) } = sup{ s(g,T } = 0
This is perhaps a way to show that f() is integrable. But by the way, I dimly remember that the function which is discontinous only in the countable number of points is Riemann integrable, and f() is definetely such function! For getting the value of the Riemann integral you can go this way:
function is Riemann integrable -> Lebesque integrable, and easy to show that the Lebesque integral over arbitrary finite interval is zero -> since if the function is Riemann integrable then the Riemann integral=Lebesque integral, and thus in our case 0!
So, that's it ;) How to show that discontinious function in countable number of points is Riemann integrable is good question, but I think it is done in some Calculus book. If you can find it out, then I think you could make an attachment to the entry Riemann's integral. |
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