Question

Problem 11.11

I have included a picture of the question (and the referenced problem 11.5), followed by definitions and theorems so you're able to use this books particular language. The information I include ranges from basic definitions to the fundamental theorems of calculus.
Problem 11.11. Show, if f : [0,1] → R is bounded and the lower integral of f is positive, then there is an open interval on which f > 0. (Compare with problems 11.5 above and

Answer 1

Given ti[a,b] - IR bounded. We prove the case in which is obtained from P by introducing one additional single point. the general Care in which i k additional obtained from p by introducing in tanto points then follows from induction. So het p=&a=%2x, < X <- =-{X6,< x; <--ancama b] be a partition of [a,b] and h e be refinement of P obtained by adding a single point, say y t(xi-1, xi) line. Q={a= x <x,<--<* l<y<xik--- LXn]. it for each 921,2-nn, iti, ma = lub {f(a) Ixtisneni? (least upper bound) or suf and Mi = sub {f(x) | x1-14XL nis m = suph Hus lacusacy} mi = sup { flas/ y sasniš. - mic mi and mic mi fas for A, BH$) CIR AT ASB.) 2) Bup Assup B inf Axint B. Also 17 UCPH) = &miari, u t & mint; +Mily-xi-1) + m{"(xi-2) u(pit) - U (Q,f) = milli-x-)-mily-21-1) - mi" (1-7) 7 Mi (Mi-% (-)-Mi (y=(-1) - Moldi-y) U(Pt)>,u(at)

similisly, it we wet misinf{f(x)) xi-saste? me = inteflas nitsas y mill-put {f(x) ) Y EX Snit. of milzmi and mi">mi L(Q,+)- Lepit) = mily-xi-1) + mi" (al-ygami (nicas) 7 mily-x-1) + mi (x,-y)-mi (xirdit). =) L(Pit) & L (Q, f) ② It is clear by Definition 11.1. 2 (af) = êmios ulat) = & mi on (where ai = M; 12-1]. 121 121 and Mizmit i=1 to n. [as infaa kupA, A $6} & midis & mini 3) LIQ,f) < u cQ,t). - from 0, ②, ③ L(Put) SLCQ,+) SU(Q,f)<u(pit) (1) wet p and Q are any partition of [a,b] Let p* be a common refinement of P. and a then by above result 11.11 (1) we get

L(P, f) < 2 (P*, f) and h(p* , f) LU (Bit). and L (p* f)<UCD*f). = L(P; f) EL(p*f) < u (p*f) £ uc Paf) i.e. IL ( Pagt ) LUB, c) Here & (pit) = ēmiani ial Lepif) = {miana Corresponding to the partition P. It mand in are boundes of t on [a,b] it me mis mism. - manis mianis mianie Mani putting i=1,2,-- n and adding all the inernalities, m(b-a) <LCP, f <u(pit) <mcb-a) this is true for each partition p of [a,b]. so Define SchL(Pit) / pin partition of [a,b]} and T = { u CP, f) | Pis partition of [a b]]. are boundred subset of IR infit and sups edisit. [by completwers ecciona 7. so by definition (113) - lower and upper Riemann integred are detind as tdm- sups and Tradna infit.

Now we state a theorem Let A and B be two nonempty subset of IR such that asb tata , bt B. then A has sup and B has intimun and supa sinfo. Now let p and a be two partition of [a, b] (Arbitrary, then by 11.1l (6) L(pit) EU (Qit) & partition P and a. supss inft [by above stated theorem]. (fdna If da. Herve proved.