Disjoint intervals for Riemann Integral

[sammycaps]

So the beginning of Rudin's Real and Complex Analysis states that the Riemann integral on an interval [a,b] can be approximated by sums of the form \Sigma\stackrel{i=1}{n}f(t_i)m(E_i) where the E_i 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 [x_i, x_i+1] which does not form a disjoint collection of intervals.

Can you allow tags to come only from the open interval (x_i, x_i+1) for some i 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).

 

[mathman]

When you use closed intervals, the overlap consists of a finite number of points (zero length intervals) so it doesn't matter. The t_i can be anywhere in the interval. Remember the whole point is making the intervals smaller and smaller.

 

[Bacle2]

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.