Question about the Riemann-Integral definiton.

[riemann86]

Hello, I have a question about the definiton of the Riemann integral. I will explain my question with a problem.

It says that the Riemann Integral on[a,b] exists, if for any epsilon this holds:
|Riemannsum(f(x) on [a,b] )- Integralvalue|< epsilon, for any partion over [a,b] where the norm of the partion is less than delta.

Now comes my problem, let's say I want to prove that the Riemann-Integral of x on[a,b] is
(b^2-a^2)/2.

I would then choose a partion (b-a)/n, and show that for a given epsilon, I can make n as big as I want, so that every Riemann sum of this partion goes to the desired value. I would do this by first making the Riemann sum as small as possible with a given n, by choosing the left value at each subinterval, and then as big as I can for a given n, by choosing the right value at each subinterval.


Now comes my question, does this really prove that x is Riemann-integrable over [a,b], the problem is when we look at the definiton. It says that it is supposed to hold for every partion where the norm goes to zero. But I have only proved it for a partion where each subinterval as equal length (b-a)/n. Isn't this a really big problem with the definition. I mean you can choose that the norm of the partion goes to zero with very many partions, how does one solve this problem?

 

[HallsofIvy]

You say "I would do this by first making the Riemann sum as small as possible with a given n, by choosing the left value at each subinterval, and then as big as I can for a given n, by choosing the right value at each subinterval", which is the crucial point. If we call the sum you get using the left value "L_n" and choosing the right value "R_n", then using any other partition, to get "M_n", we must have $L_n\le M_n\le R_n$. If the function is "integrable" then the limits of L_n and R_n must be the same so that M_n, trapped between the two, must also converge to the same limit.

 

[riemann86]

Thanks, but how do you show that for any other partion it must lie between those two values? Maybe it is trivial, but I do not see it.
I guess I can accept if for the function x, but I would also like to see it for a another random function.
That is, can we prove that if:

We have a given partion where the norm of the partion goes to zero. For instance a partion where we have equal subinterval lengths, (b-a)/n.
Then for any epsilon, we can get a delta, so that if the norm of the partions is less than delta.
Then we have |Riemannsum - Integralvalue| < epsolon.

Now can we prove that if what I said holds for the partions with equal subinterval lengths, then this implies that the epsilon-delta conditions will also hold for every other partion we can construct in a way where we can control the norm of the partion?