Presentation Problems 5


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1 Presenttion Problems A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q). Use this to show tht for ny prtitions P 1 nd P 2 of [, b] tht L(f, P 1 ) U(f, P 2 ). Proof. First, we will prove tht U(f, P ) U(f, Q) nd L(f, P ) L(f, Q). Let [x k 1, x k ] be n subintervl of P nd suppose there exists z Q such tht x k 1 < z < x k. Denote M k = sup{f(x) : x [x k 1, x k ]} M k = sup{f(x) : x [x k 1, z]} M k = sup{f(x) : x [z, x k]} It follows tht M k M k nd M k M k, nd we hve tht M k k = M k (x k x k 1 ) = M k (x k z + z x k 1 ) = M k (x k z) + M k (z x k 1 ) M k (x k z) + M k(z x k 1 ) Further, we cn employ induction to prove this fct for ny finite number of points in this intervl [x k 1, x k ] tht re lso in Q. As such, we know tht the upper sum cnnot get lrger when we dd more points to prtition. Tht is U(f, P ) U(f, Q). Similrly, denote m k = inf{f(x) : x [x k 1, x k ]} m k = inf{f(x) : x [x k 1, z]} m k = inf{f(x) : x [z, x k]} 1
2 It follows tht m k m k nd m k m k, nd we hve tht m k k = m k (x k x k 1 ) = m k (x k z + z x k 1 ) = m k (x k z) + m k (z x k 1 ) m k(x k z) + m k(z x k 1 ) Then the lower sum cnnot get smller when we dd more points to prtition. Tht is L(f, P ) L(f, Q). Consider prtitions P 1 nd P 2 of [, b]. Define Q = P 1 P 2. Then P 1 Q nd P 2 Q. Then by our previously proved fct L(f, P 1 ) L(f, Q) U(f, Q) U(f, P 2 ). 2. Let f : [, b] R be bounded. Then f is integrble on [, b] if nd only if for ll ε > 0, there exists some prtition P ε of [, b] such tht U(f, P ε ) L(f, P ε ) < ε. Proof. ( ) Let > 0. Then by ssumption, there exists some prtition P such tht U(f, P ) L(f, P ) <. However, U(f) U(f, P ). L(f) L(f, P ). Then, U(f) L(f) U(f, P ) L(f, P ) <. Thus, 0 U(f) L(f) < for ll > 0, so U(f) = L(f), nd thus f is integrble. ( ) Since U(f) is the infimum of ll the upper sums, for > 0, we cn find P 1 such tht U(f, P 1 ) < U(f) + 2. Similrly, we cn find P 2 such tht L(f, P 2 ) > L(f) 2. Now, we cn define P = P 1 P 2. Then, U(f, P ) L(f, P ) U(f, P 1 ) L(f, P 2 ) <U(f) + 2 (L(f) 2 ) becuse f is integrble, U(f) = L(f) = = 2
3 3. Let f : [, b] R be bounded. Then f is integrble on [, b] if nd only if there exists some sequence of prtitions (P n ) of [, b] such tht lim U(f, P n) L(f, P n ) = 0. Proof. ( ) Let > 0, some nturl number N such tht U(f, P N ) L(f, P N ) < It follows from the result of problem number 2 tht f is integrble ( ) Assume f is integrble so P n is sequence of p gurnteed tht U(f, P n ) L(f, P n ) < 1/n then it follows from the result of problem number 2 tht there exists P n such tht lim U(f, P n ) L(f, P n ) = 0 4. Let f : [, b] R be incresing. Then f is integrble on [, b]. Proof. Since f : [, b] R is n incresing function, f() f(x) f(b) for ll x [, b]. Therefore, f is bounded on [, b]. Let d = b n. Let P n be finite set of points {x 0, x 1, x 2,..., x n } such tht x k = + kd. Then P n is prtition of [, b] becuse = x 0 < + d = x 1 < + 2d = x 2 <... < + nd = x n = b. Since f is incresing, we hve m k = M k = Therefore, U(f, P n ) = L(f, P n ) = inf x [x k 1,x k ] f(x) = f(x k 1). sup f(x) = f(x k ). x [x k 1,x k ] M k (x k x k 1 ) = b n m k (x k x k 1 ) = b n f(x k ) f(x k 1 ) 3
4 And ( b lim U(f, P n) L(f, P n ) = lim x x n ( b = lim x n ( b = lim x n ( b = lim x = lim x = 0 ( n ( n f(x k ) b n f(x k ) ) f(x k 1 ) )) f(x k 1 ) n 1 )) f(x k ) f(x k ) k=0 ) n (f(x n) f(x 0 )) ( (b )(f(b) f()) ) Therefore, by Presenttion 5 Problem 3, we hve f is integrble on [, b]. n 5. Let f : [, b] R be continuous. Then f is integrble on [, b]. Proof. Let f : [, b] R be continuous Notice tht [, b] is compct, therefore f is uniformly continuous. Arbitrrily pick > 0 δ > 0 such tht x y < δ = f(x) f(y) < /(b ) Now let P be prtition of [, b] where the distnce between ny consecutive term is smller thn δ Given prticulr subintervl [x k 1, x k ] of the P Bsed on the extreme vlue theorem, the function chieves extreme vlue somewhere in the bound. Hence M k = f(z k ) for some z k [x k 1, x k ] nd lso m k = f(y k ) for some y k [x k 1, x k ] Since x k x k 1 < δ nd z k, y k [x k 1, x k ] z k y k < δ So M k m k = f(z k ) f(y k ) < /(b ) U(f, P ) L(f, P ) = n (M k m k )(x k x k 1 ) < n /(b ) (x k x k 1 ) = /(b ) (b ) = Hence U(f, P ) L(f, P ) < Therefore f is integrble. 6. Let f, g be integrble on [, b] nd let α, β R. Then αf +βg is integrble on [, b] nd αf + βg dx = α f dx + β g dx. 4
5 Proof. Since f,g re integrble, by presenttion problem 3, we know there exists sequence of prtitions (P n ), (Q n ) s.t. lim [U(f, P n) L(f, P n )] = 0 & lim [U(g, Q n) L(g, Q n )] = 0 Then, we cn crete sequence of prtitions (K n ) = (P n Q n ). Since P n K n nd Q n K n, we know (by problem 1): L(f, P n ) L(f, K n ) U(f, K n ) U(f, P n ) L(g, Q n ) L(g, K n ) U(g, K n ) U(g, Q n ) which mens: U(f, K n ) L(f, K n ) U(f, P n ) L(f, P n ) nd U(g, K n ) L(g, K n ) U(g, Q n ) L(g, Q n ). Becuse U(f, K n ) L(f, K n ) 0 while it is less thn or equl to U(f, P n ) L(f, P n ), the lim [U(f, K n ) L(f, K n )] is squeezed to 0; it is the sme cse for lim [U(g, K n ) L(g, K n )] = 0. Using Algebrtic Limit Thm, we obtin lim U(f, K n ) = lim L(f, K n ) nd lim U(g, K n ) = lim L(g, K n ). Lemm: U(αf + βg, P ) = αx(f, P ) + βy (g, P ), where X = U if α > 0, X = L otherwise, nd Y = U if β > 0, Y = L otherwise. And vice verse for L(αf + βg, P ). Proof: We know tht if A R nd ca := {c A}, then if c 0, sup ca = c sup A nd inf ca = c inf A, but if c < 0 then sup ca = c inf A nd inf ca = c sup A. So if f(x) = g(x) for some 0 R nd g : R R, then m k = inf{f(x) x [x k 1, x k ]} = inf{ g(x) x [x k 1, x k ]} = inf{g(x) x [x k 1, x k ]} M k = sup{f(x) x [x k 1, x k ]} = sup{ g(x) x [x k 1, x k ]} = sup{g(x) x [x k 1, x k ]} L(f, P ) = U(f, P ) = m k (x k x k 1 ) = inf{g(x) x [x k 1, x k ]}(x k x k 1 ) = L(g, P ) M k (x k x k 1 ) = sup{g(x) x [x k 1, x k ]}(x k x k 1 ) = U(g, P ) And if f(x) = g(x) for some < 0 R nd g : R R, then L(f, P ) = m k (x k x k 1 ) = sup{g(x) x [x k 1, x k ]}(x k x k 1 ) = U(g, P ) U(f, P ) = M k (x k x k 1 ) = inf{g(x) x [x k 1, x k ]}(x k x k 1 ) = L(g, P ) 5
6 So we hve sclr multipliction of L nd U, with the property we re trying to show. Also on homework problem, we went over tht suprem dd, which cn be used to sy tht infim dd s well. Thus, U nd L stisfy the clim. Now we will resume the proof.. lim [U(αf + βg, K n) L(αf + βg, K n )] = lim U(αf + βg, K n) lim L(αf + βg, K n) = lim U(αf, K n) + lim U(βg, K n) lim L(αf, K n) lim L(βg, K n) Now we will cse on α nd β being greter thn or equl to zero. If α 0 nd β < 0 then lim [U(αf + βg, K n) L(αf + βg, K n )] = lim U(αf, K n) + lim U(βg, K n) lim L(αf, K n) lim L(βg, K n) = α lim U(f, K n) + β lim L(g, K n) α lim U(f, K n) β lim L(g, K n) = 0 Similrly, if α < 0 nd β 0, then lim [U(αf + βg, K n) L(αf + βg, K n )] = lim U(αf, K n) + lim U(βg, K n) lim L(αf, K n) lim L(βg, K n) = α lim L(f, K n) + β lim U(g, K n) α lim L(f, K n) β lim U(g, K n) = 0 If both re less thn 0, then lim [U(αf + βg, K n) L(αf + βg, K n )] = lim U(αf, K n) + lim U(βg, K n) lim L(αf, K n) lim L(βg, K n) = α lim L(f, K n) + β lim L(g, K n) α lim U(f, K n) β lim U(g, K n) = 0 = α( lim U(f, K n) lim L(f, K n)) β( lim U(g, K n) lim L(g, K n)) = α lim [U(f, K n) L(f, K n )] β lim [U(g, K n) L(g, K n )] = α 0 β 0 = 0 If both re greter thn 0, then lim [U(αf + βg, K n) L(αf + βg, K n )] = lim U(αf, K n) + lim U(βg, K n) lim L(αf, K n) lim L(βg, K n) = α lim U(f, K n) + β lim U(g, K n) α lim L(f, K n) β lim L(g, K n) = 0 = α( lim U(f, K n) lim L(f, K n)) + β( lim U(g, K n) lim L(g, K n)) = α lim [U(f, K n) L(f, K n )] + β lim [U(g, K n) L(g, K n )] = α 0 + β 0 = 0 6
7 Since in ll cses the limit is zero, we hve tht αf + βg is integrble. Since nd f dx = lim U(f, K n) g dx = lim U(g, K n) Given αf + βg is integrble, we know tht fdx = L(f, K n) = U(f, K n ) nd gdx = L(g, K n) = U(g, K n ). Thus, we cn sy αf + βg dx = lim U(αf + βg, K n) = α lim X(f, K n) + β lim Y (g, K n) = α f dx + β g dx since X, Y re either U or L 7. Let f, g be integrble on [, b] () If m f(x) M for ll x [, b], then m(b ) Proof. Recll tht f dx M(b ). L(f) = sup {L(f, P ) : P P([, b])} U(f) = inf {U(f, P ) : P P([, b])} Let P P([, b]) where P = n. Then by definition of supremum nd infimum, we hve By definition, we know tht L(f, P ) L(f) U(f) U(f, P ) L(f, P ) = U(f, P ) = m k (x k x k 1 ) M k (x k x k 1 ) 7
8 where m k = inf {f(x) : x [x k 1, x k ]} M k = sup {f(x) : x [x k 1, x k ]} Since M, m re upper nd lower bounds on f, we hve tht m m k M k M Then we cn see tht L(f, P ) = m k (x k x k 1 ) m (x k x k 1 ) = m(x n x 0 ) = m(b ) Similrly, we hve tht U(f, P ) = M k (x k x k 1 ) M (x k x k 1 ) = M(x n x 0 ) = M(b ) By definition of integrbility, we know tht L(f) = It follows tht m(b ) f dx = U(f) f dx M(b ) (b) If f(x) g(x) on [, b], then f dx g dx. 8
9 Proof. Since f(x) g(x), we know tht h(x) = f(x) g(x) 0 Then by prt (), we know tht h = By linerity, we know tht (f g) dx 0 (b ) = 0 It follows tht so (f g) dx = f dx f dx f dx g dx 0 g dx g dx (c) f is integrble on [, b] nd f dx f dx. Proof. First, we prove tht f is integrble Let P = {x 0, x 1,..., x n } be n rbitrry prtition of [, b]. Clim: We show tht for ny intervl = [x k 1, x k ], sup f inf f sup f inf f Proof: By the tringle inequlity, for ny x, y, Since f(x) f(y) f(x) f(y) f(x) f(y) f(x) f(y) = mx{f(x), f(y)} min{f(x), f(y)} sup f inf f then sup Ik f inf Ik f is n upper bound on f(x) f(y) nd so Since sup{ f(x) f(y) : x, y } sup sup{ f(x) f(y) : x, y } = sup f inf f f inf f 9
10 the clim holds. So we hve, for ll k = 1,..., n, sup f inf f sup f inf f (sup f inf f )(x k x k 1 ) (sup U( f, P ) L( f, P ) U(f, P ) L(f, P ) f inf f)(x k x k 1 ) By the theorem from clss, function g is integrble on [, b] if nd only if for ll > 0, there exists P P such tht U(g, P ) L(g, P ) <. We re given tht it holds for f. For rbitrry > 0, we know tht U( f, P ) L( f, P ) U(f, P ) L(f, P ) <, so it lso holds for f. Therefore, f is integrble on [, b]. Next, since f is integrble, Then by prt (b), we know tht By linerity, we hve f(x) f(x) f(x) f dx f dx f dx It follows tht f dx f dx f dx f dx f dx 8. Let (f n ) be sequence of relvlued functions on [, b] integrble on [, b]. If f n f uniformly on [, b], then f is integrble on [, b] nd lim f n dx = f dx. Proof. We will begin by showing tht f is bounded on [, b]. Since f n f uniformly, for ny > 0 there exists n N N + such tht for ll n > N, x [, b], we hve f n (x) f(x) <. Let = 1 be rbitrry nd choose n M > N tht is gurnteed by our ssertion. f n is integrble for ny function in our sequence so we lso know it is bounded. Then let us choose 10
11 B > 0 tht is gurnteed by this property, giving us f(x) < B for ll x [, b]. Then by the Tringle Inequlity, f(x) = f(x) f M (x) + f M (x) f(x) f M (x) + f M (x) f(x) f M (x) f(x) f M (x) < 1 Adding f M (x) to both sides of the outermost inequlity gives us f(x) < 1 + f M (x) This holds for ny x [, b], so we hve proven tht f is bounded s well. Let > 0 be rbitrry. Agin, since f n f uniformly, there exists n N N such tht f n (x) f(x) < 4(b ) (for ll n > N, nd ll x [, b]. Choose n M > N. Then f M is integrble by ssumption. By our integrbility criterion, we know there is prtition P P[, b] (let the number of components in the prtition be P := P ) such tht U(f M, P ) L(f M, P ) = M fm,k(x k x k 1 ) M fm,k(x k x k 1 ) = Now, since M > N, we hve tht f M (x) f(x) < 4( b) for ll x [, b]. For such x, by properties of bsolute vlues, f M (x) 4(b ) < f(x) < f M (x) + 4(b ) By the fct tht every upper integrl over function is t lest equl to the lower integrl, m fm,k 4(b ) < m f,k M f,k < M fm,k + 4(b ). By using the definitions of upper nd lower integrls, (M fm,k M fm,k)(x k x k 1 ) < U(f, P ) L(f, P ) = M f,k (x k x k 1 ) m f,k (x k x k 1 ) Combine the sums: the bove is equivlent to (M f,k m f,k )(x k x k 1 ) 11
12 By our bove chin of inequlities, (M f,k m f,k )(x k x k 1 ) (( M fm,k + (M fm,k m fm,k)(x k x k = ) ( m fm,k 4(b ) 2 4(b ) (x k x k 1 ) < As result, U(f, P ) L(f, P ) < for n rbitrry. By our integrbility criteri (Problem 2, Presenttion Set 5), f is integrble on [, b]. It remins to show tht fdx = lim f ndx. Let > 0 be rbitrry. Since f n f uniformly, there exists n N N such tht f n (x) f(x) < b for ll n N such tht n > N nd x [, b]. As result of linerity of integrtion (Problem 6, Presenttion Set 5): f n dx fdx = (f n f)dx for ll such n. Then by Problem 7, Presenttion Set 5, we obtin the bsolute vlue bounds (f n f)dx f n f dx < dx = (b ) b b = Since ws rbitrry, we conclude tht lim (f n f)dx = 0, nd by properties of bsolute vlues we cquire lim f ndx = fdx., s desired. 9. Let A R be countble. Then A hs mesure zero. (Note: the converse is not true.) Proof. Suppose A = {x n } where n = 1, 2, 3,..., Let > 0, define open intervls I n = (x n 2, x n+2 n + 2 ) where n = 1, 2, 3,..., n+2 Then the length of ech intervl l(i n ) = And A I n n=1 2 n+1 ) ) (x k x k 1 ) = 4(b ) 12
13 We get n=1 l(i n) = n=1 2 < n+1 Therefore if A R is countble, then A hs mesure zero. 10. Let {A k } n be finite collection of sets of mesure zero. Show tht n A k lso hs mesure zero. Proof. Let > 0. Since ech A k is of mesure zero, then for n, there exists collection of open intervls {O km } for ech A k such tht A k m=1 O km nd O km < n. m=1 Then n n A k m=1 O km nd O km < n( n ) = m=1 becuse ech collection of open intervls is countble, so the unions nd summtions re ll well defined. Thus n A k hs mesure zero. 11. Let f : [, b] R be continuous. Then there exists c (, b) such tht f dx = f(c)(b ). Proof. Let f : [, b] R be continuous. Define F (x) := x f(t) dt. Thus, F is continuous on [, b]. 13
14 If f is continuous t c, then F is differentible t c nd F (c) = f(c). Thus for ll c (, b), F (c) = f(c). Since F is continuous on [, b] nd differentible on (, b), we cn invoke the Men Vlue Theorem. Thus there exists c (, b) such tht F F (b) F () (c) = b ( = 1 b f(t) dt b So we hve f(t) dt ) = 1 b f(t) dt. or f(c) = F (c) = 1 b f(t) dt, f(c)(b ) = f(t) dt. 14
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