Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.

show that :

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- May 3rd 2009, 11:53 PMmath43Riemann Integrable Function
Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.

show that : - May 8th 2009, 10:59 PMfardeen_gen
I googled away to glory and maybe this will help:

**Darboux theory**

EDIT: Can the experienced answerers shed light on this problem? - May 9th 2009, 09:55 AMputnam120
My advice would be to split the interval up into the following parts:

.

The integrals on the last 2 parts are the same. For the region in the middle since are assumed Riemann integrable they are bounded. So say . Thus you have that

- May 9th 2009, 05:06 PMhkito
where the $c$ coming from?

- May 9th 2009, 05:40 PMPlato
- May 9th 2009, 05:46 PMhkito
Well I don't see any in the problem. It's just saying that for not equal to x. For me it's not clear enough.

- May 9th 2009, 05:48 PMputnam120
It says that as long as x does not equal c you know that f(x)=g(x). However, you are not guaranteed that f(c)=g(c).

- May 9th 2009, 05:56 PMhkito
Okay,

so the problem is show that given that except at - May 9th 2009, 06:11 PMhkito
- May 9th 2009, 11:57 PMredsoxfan325
There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable". That's very nice for this problem, because if , then

Calculate the lower Riemann sum of . Let be a partition of and let be the segment containing . on , because it contains a point that is not , so therefore the lower Riemann sum and the whole integral is zero. - May 10th 2009, 01:19 PMputnam120
- May 10th 2009, 01:48 PMMoo
- May 10th 2009, 02:06 PMputnam120
Yes sorry for the ambiguity.

- May 10th 2009, 06:47 PMThePerfectHacker
We can prove something stronger. If is integrable and is a function that differs at countably many points distinct from such that they converge then is integrable with same value as . The nice thing is that the proof is similar as in the finite case. Because when you have convergence it means that within a a small enough neighborhood of the limit all the points lie there. So you form a partition with thin enough rectangles around these values.