5.7 Darboux sums and Riemann integrals 345Figure 5.721P. Figure 5.721 is available for inspection. For each choice of the
points xk we have Mk < f(xk) < M1, and hence
n
(5.722) LDS(P) < 1 f(xk) Axk < UDS(P).
k-1
Therefore, for a given partition P, the different Riemann sums that can be
formed by making different choices of the points xk are all sandwiched
between the lower and upper Darboux sums. Information about
Riemann sums can therefore be gleaned from information about Darboux
sums.
The first step in our study of Darboux sums may seem to be a very
modest one. Let P be a given partition, and let P be a simple extension
of P. By this we mean that P' is exactly the same as P except that P'
contains one additional partition point, say X, which lies between two of
the partition points of P, say x2 < X < x3. The inequality
[ l.u.b. f(x)](X - X2) + [ l.u.b. f(x)j(x3 - X)
xx=<x5X XSx<x3
[ l.u.b. f(x)](X - x2) + [ l.u.b. f(x)](x3 - X) S M3 0x3
X,5z5za x2 5x;gxa
implies that UDS(P') < UDS(P). Figure 5.721 is not needed in the
proof of the inequality but may nevertheless be helpful. Consideration
of simple extensions of simple extensions of P leads to the conclusion that
if P is any extension of P (so that P contains all of the partition points of
P and perhaps also some additional ones), then UDS(P') S UDS(P).
An analogous argument, in which greatest lower bounds appear and the
inequality signs are reversed, shows that if P' is an extension of P, then
LDS(P') >--_ LDS(P). Suppose now that P1 and P2 are two given parti-
tions, and let P3 be an extension of both P1 and P2. Then
(5.723) LDS(P1) < LDS(P3) < UDS(P3) 5 UDS(P2)
and hence
(5.724) LDS(P1) < UDS(P2).
This is a key result of the theory. Let the symbols in
(5.73) L = f' f(x) dx, U = f 'f(x) dx