Beginning Darboux Integration, Math 317, Intro to Analysis II


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1 Beginning Droux Integrtion, Mth 317, Intro to Anlysis II Lets strt y rememering how to integrte function over n intervl. (you lerned this in Clculus I, ut mye it didn t stick.) This set of lecture notes will lst t lest two lectures. For n intervl I = [, ] sudivision of I is finite collection of closed suintervls of I = {J 1,... J n } such tht I = J 1 J 2 J n If i k then J i J k consists of t most one point. A sudivision of Is something like J 1 J 2 J 3 Let f e ounded function on I. Then f is ounded on ech J. Let M(f, J) = sup{f(x) x J} nd m(f, J) = inf{f(x) x J}. Let l(j) e the length of J. The Upper Droux sum S + (f, ) nd Lower Droux sum S (f, ) re given y S + (f, ) = J M(J, f) l(j) nd S (f, ) = J m(j, f) l(j) Exercise 1. For ny ounded function f : I R nd ny sudivision of I, S + (f, ) S (f, ) To see wht this mesures we ll fill in the following pictures: Over ech I k drw rectngle of height M k on the left nd of m k on the right. Shde these rectngle in. S + (f, ) is the re of the thing on the left, S (f, ) is the re of the thing on the left. These quntities re oth trying to mesure the re under the curve. One knows its too ig nd the other knows its too smll. These quntities re oth trying to mesure the re under the curve. One knows its too ig nd the other knows its too smll. The intuition you should hve is tht if we tke smller sudivisions then we will get etter informtion. For sudivision = {J 1, J 2,... J n } nd nother sudivision = {J 1,... J m}, is clled refinement of if for ll J there is J such tht J J. Prove the following: Exercise 2. If is refinement of nd f is ounded then show tht S (f, ) S (f, ) S + (f, ) S + (f, ) 1
2 2 Hint: You ve lredy shown the middle inequlity. Fix single J nd compre the single term m(j, f)l(j) with the the susum m(j, f), l(j ). J, J J Exercise 3. Any two sudivisions = {J k } n nd = {J j }m j=1 of single intervl I hve common sudivision. Hint: Let = {J J J ; J nd J J contins t lest two points} Mye if we tke the lest upper sum or the gretest lower sum we ll get n ctul sense of re? Definition 1. For the intervl I = [, ] nd ounded function f : [, ] R the upper Droux integrl nd the lower Droux integrl re defined y = inf{s+ (f, ) : sudivision of [, ]} (The lest of the overestimtes) = sup{s (f, ) : sudivision of [, ]} (The gretet of the underestimtes) Exercise 4. Use Exercise 2 to show tht. Definition 2. If = then f is clled Droux integrle on [, ] nd the Droux integrl is given y = = We ll prove t lest one of the following sic results in clss: Proposition 3. (1) If f nd g re ounded functions on [, ] then (2) If h is ounded on [, c] nd < < c then h(x)dx nd (3) If f is ounded on [, ] nd k > 0 then If k < 0 then k = k k = k nd nd (4) If f g re ech ounded on [, ] then nd k = k k = k h(x)dx
3 Proof of (1). The first clim is the most interesting. I will sketch the proof here. Suppose tht f nd g re ech ounded on I. We wish to prove tht. We lredy know tht the second inequlity holds y Exercise 4. Let s try to prove the first. The third will follow y similr rgument. = J 1,... J n e ny sudivision of I, l(i k ) e the length of J k, m(j k, f) = inf{f(x) x J k } m(j k, g) = inf{g(x) x J k } m(j k, f + g) = inf{(f + g)(x) x J k } Since m(j k, f + g) is the infimum, it follows tht for ll ɛ > 0 there is n x J k such tht m(j k, f + g) f(x) + g(x) < m(j k, f + g) + ɛ The other infem, m(j k, f) nd m(j k, g) re lower ounds on f(x) nd g(x) respectively do tht m(j k, f) + m(j k, g) f(x) + g(x) < m(j k, f + g) + ɛ Since this holds for ll ɛ > 0, it follows tht m(j k, f) + m(j k, g) m(j k, f + g). Adding these together we see tht S (f, ) + S (g, ) = m(j k, f)l(j k ) + Summrizing: m(j k, g)l(j k ) = m(j k, f + g)l(j k ) = S (f + g, ) (1) S (f, ) + S (f, ) S (f, ) 3 (m(j k, f) + m(j k, g))l(j k ) Since = sup{s (f, ) sudivision} for ll ɛ > 0 there re sudivisions nd such tht (2) nd (3) ɛ < S (f, ) ɛ < S (g, ) By Exercise 3 nd hve common refinement,. By Exercise 2 (4) S (f, ) S (f, ) nd S (g, ) S (g, ) Adding the left hlves of (2) nd (3) nd then using (4) 2ɛ < S (f, ) + S (g, ) < S (f, ) + S (g, )
4 4 Finlly, we use the inequlity (1). 2ɛ < S (f, ) + S (g, ) f(x) + y (1) S (f + g, ) This holds for ll ɛ > 0, so tht + f(x)+, s we climed. Exercise 5. (The proof of clim 2 of Proposition 3) If h is ounded on [, c] nd < < c then prove the following y thinking out refinements of sudivisions. () h(x)dx h(x)dx () h(x)dx h(x)dx Corollry 4 (Corollry to Proposition 3). [, ] then = (1) If f nd g re integrle functions on (f + g)(x)dx (2) If h is integrle on [, c], [, ] nd [, c] nd < < c then (3) If f is integrle on [, ] nd k R then k = k (4) If f g re ech integrle on [, ] then h(x)dx Exmple 5. Recll tht every intervl contins rtionl nd irrtionl numers. Let { 1 if x is rtionl F (x) = 1 if x is irrtionl Wht is 1 0 F (X)dx? Wht is 1 0 F (X)dx? Is this function Droux integrle on [0, 1]? The following theorem gives good test for integrility. I will provide some of the ides of the proof. Think of them s hints. You will fill in the proof s n exercise Theorem 6. A ounded function f : [, ] R is Droux integrle if nd only if for ll ɛ > 0 there is sudivision such tht S + (f, ) < ɛ + S (f, ). Proof. First we ssume integrility, so tht = =. Consider ny ɛ > 0. Since is the infimum of the upper sums, there is sudivision such tht, S + (f, ) ɛ On the other hnd since is the supremum of the lower sums there is such tht ɛ S (f, )
5 Get common refinement,. Use tht S (f, ) nd S + (f, ) re ech with in ɛ of to conclude tht they re within 2ɛ of ech other. Rewrite the proof with ɛ replced with the correct quntity to get the climed conclusion. Next we prove the converse. Assume tht for ll ɛ > 0 there is sudivision such tht S + (f, ) < ɛ + S (f, ). We need to show tht the upper nd lower integrls gree. Since is the infimum of ll the upper sums, it follows tht for ll ɛ > 0 there is such tht S + (f, ) ɛ < S + (f, ) Since is the supremum of ll the lower sums, it follows tht for ll ɛ > 0 there is such tht S (f, ) ɛ < S (f, ) Apply exercise 3 twice nd conclude tht, nd ll hve common sudivision, (3). By thinking out, convince yourself tht S + (f, (3) ) S (f, (3) ) is less thn ɛ. By thinking out convince yourself tht the distnce etween nd S + (f, (3) ) is less thn ɛ. By thinking out convince yourself tht the distnce etween nd S + (f, (3) ) is less thn ɛ. Wht does the tringle inequlity now give s n upper ound for nd? Since ɛ ws n ritrry positive numer cn you conclude tht =? 5
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