Advanced Calculus fi..

Chapter 4 Integral Calculus of Functions of Several Variables 259Since both sequences on the right have limit 0, we conclude that Q,, -+ Po also.Hence f (Pn,) f (PO), f (en,) + f (PO). But0 < i d(f (pn,), f (Qn,)) i d(f (Pn,), f (PO)) + d(f (PO), f ( enl)).As k -+ GO, both terms on the right have limit 0. That is a contradiction. Hence fis uniformly continuous.THEOREM L Let y = f (x) be a continuous real function of the real variable xfor a 5 x ( b. Then the Riemann integralexists.Proof.We first observe that the interval a 5 x J b is a bounded closed set (Problem4 following Section 2.23). Hence by Theorem K, f is uniformly continuous.As in Section 4.1, we now consider subdivisions of the interval: a = xo cx, i . . . < x, = b. We let h be the mesh of the subdivision, the maximum ofA1x = XI - XO, . . . , Anx = x, - x,-, . We must show that there is a unique numberc such that for each t > 0 there is a 6 > 0 such thatfor all subdivisions of mesh h less than 6, no matter how xt, . . . , x,* are chosen inthe successive subintervals: 5 xT 5 x, for i = 1, . . . , n.To this end, we first observe that by Theorem J of Section 2.23, f has an absolutemaximum B and minimum B, so that A 5 f (x) 5 B for a 5 x 5 b. Hencefor all sums C f (x:)A,x. Thus the sums form a bounded set of real numbers.In particular, we consider Es, the set of sums arising only from those subdivisionsof mesh h less than 6, where 6 > 0. We let,. ,Thus~~s=glbE~, /f3s=l~bEs.for all subdivisions of mesh h less than 6.