# Correct proof?

• Jul 12th 2011, 05:03 AM
durrrrrrrr
Correct proof?
Theorem:
If f is bounded in [a,b] and the number of discontinuity points of f is finite then f is integrable in [a,b].

Proof:
Using induction:
When there are 0 discontinuity points then f i continuous and therefore f is integrable in [a,b].
Suppose the claim is true when we have k or less discontinuity points, and we'll prove it for a case where we have k+1 discont. points.
Let x be a discont. point of f in [a,b]. Let e be such that x is the only discont. point in the interval [x-e,x+e].
Then there are k or less discont. points in [a,x-e], so according to the inductive assumption f is integrable in [a,x-e]. We can make the same argument for the intervals [x-e,x+e] and [x+e,b], and so since f is integrable in [a,x-e] and [x-e,x+e] and [x+e,b], f is integrable in [a,b].

Is my proof correct? Thanks
• Jul 12th 2011, 01:14 PM
Nemesis
Re: Correct proof?
Quote:

Originally Posted by durrrrrrrr
Theorem:
If f is bounded in [a,b] and the number of discontinuity points of f is finite then f is integrable in [a,b].

Proof:
Using induction:
When there are 0 discontinuity points then f i continuous and therefore f is integrable in [a,b].
Suppose the claim is true when we have k or less discontinuity points, and we'll prove it for a case where we have k+1 discont. points.
Let x be a discont. point of f in [a,b]. Let e be such that x is the only discont. point in the interval [x-e,x+e].
Then there are k or less discont. points in [a,x-e], so according to the inductive assumption f is integrable in [a,x-e]. We can make the same argument for the intervals [x-e,x+e] and [x+e,b], and so since f is integrable in [a,x-e] and [x-e,x+e] and [x+e,b], f is integrable in [a,b].

Is my proof correct? Thanks

I think it is correct but I cannot be completely sure
Nemesis