# Disjoint intervals for Riemann Integral

1. Mar 9, 2013

### 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(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 $i$ 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).

2. Mar 9, 2013

### mathman

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. Mar 9, 2013

### 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.