Disjoint intervals for Riemann Integral

  • Thread starter sammycaps
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  • #1
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So the beginning of Rudin's Real and Complex Analysis states that the Riemann integral on an interval [itex][a,b][/itex] can be approximated by sums of the form [itex]\Sigma[/itex][itex]\stackrel{i=1}{n}[/itex]f(ti)m(Ei) where the Ei are disjoint intervals whose union is the whole interval.

At least when I learned it, the Riemann integral was partitioned and tags were taken from with the interval [xi, xi+1] which does not form a disjoint collection of intervals.

Can you allow tags to come only from the open interval (xi, xi+1) for some [itex]i[/itex] and from the closed intervals for others, so that we do in fact get a disjoint collection whose union is [a,b]. (this obviously can't work generally for Darboux sums are the inf and sup might not be contained in the image of the interval).
 

Answers and Replies

  • #2
mathman
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When you use closed intervals, the overlap consists of a finite number of points (zero length intervals) so it doesn't matter. The ti can be anywhere in the interval. Remember the whole point is making the intervals smaller and smaller.
 
  • #3
Bacle2
Science Advisor
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I think you actually need closed intervals in some arguments because compactness and (a.e) continuity guarantee the existence of max and min. I have seen this used to show that a.e continuity implies Riemann integrability.
 

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