1549901369-Elements_of_Real_Analysis__Denlinger_

362 Chapter 7 • The Riemann Integral

Theorem 7.2.4 If Q is a refinement of a partition P , then

(a) S.(f, Q) 2: S.(f, P), and (b) S(f, Q) ~ S(f, P).

That is, as we refine the partitions of [a, b], the lower Darboux sums "in-
crease" and the upper Darboux sums "decrease. "

Proof. Suppose Q is a refinement of a partition P of [a, b]. Then Q con-
tains all the points of P , and perhaps more points as well. It is sufficient to
consider the case when Q contains exactly one point not contained in P
(Exercise 2). This point will belong to exactly one of subintervals [xk-l,xk]
created by P. Denote the added point by x'k E [xk-l,xk]· Then Q
{xo,x1,x2,··· ,Xk-1,x'k,xk,··· ,xn}, and with the help of Theorem 7.1.2,

(a) S.(f, Q)

k-1

= L mi6i +inf f[xk-1, x'k](x'k - Xk-1) +inf f[x'k, xk](xk -x'k)

i=l

n

+ I: mi6i

i=k+l

k-1

2: L mi6i + mk(x'k -Xk-1) + mk(xk -x'k) +

i=l i=k+l

k-1 n

= L mi6i + mk(xk - Xk-1) + L mi6i

i=l i=k+l

n

= L mi6i = S.(f, P). Thus, S.(f, Q) 2: S.(f, P).

i=l

(b) S(f, Q)

k-1

Also,

= L Mi6i +sup f[xk-1, x'k](x'k - Xk-1) +sup f[x'k, xk](xk - x'k)

i=l

n

+ L Mi6i

i=k+l

k-1 n

~ L Mi6i + Mk(x'k - Xk-1) + Mk(Xk - x'k) + L Mi6i

i=l i=k+l

k-1 n

= L Mi6i + Mk(Xk -Xk-1) + L Mi6i

i=l i=k+l

n

= L Mi6i = S(f, P). Thus, S(f, Q) ~ S(f, P). •

i=l