INTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović

Transcription

1 THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp INTERCHANGING TWO LIMITS Zorn Kdelburg nd Milosv M. Mrjnović This pper is dedicted to the memory of our illustrious professor of nlysis Slobodn Aljnčić ( ). Abstrct. By the use of convenient metrics, the ordered set of nturl numbers plus n idel element nd the prtilly ordered set of ll prtitions of n intervl plus n idel element re converted into metric spces. Thus, the three different types of it, rising in clssicl nlysis, re reduced to the sme model of the it of function t point. Then, the theorem on interchnge of iterted its, vlid under the condition tht one of the iterted its exists nd the other one exists uniformly, is used to derive long sequence of sttements of tht type tht re commonly present in the courses of clssicl nlysis. All pprently vried conditions ccompnying such sttements re, then, unmsked nd reduced to one nd the sme: one iterted it exists nd the other one exists uniformly. ZDM Subject Clssifiction: I35; AMS Subject Clssifiction: 00A35 Key words nd phrses: Interchnge of two its, uniform convergence, definite integrl s it. 1. Introduction The simple concept of metric spce proves s convenient setting for unifiction of different types of convergence nd continuity tht rise in clssicl nlysis. For this reson, this concept finds its plce in the contents of the first contemporry courses of mthemticl nlysis. Prticulrly, by replcing the bsolute vlue with distnce function, immedite generliztions of conditions determining clssicl concepts of convergence nd continuity re obtined. Due to different terms, the following three types of it re present in the clssicl nlysis n, f(x), σ(f,p), being respectively: the it of sequence, the it of function t point nd the it of integrl sums. To hve them s prticulr cses of more generl concept of it, generlized sequences (nets) or filters re used (nd they re not common topics on the list of themes for the first courses in nlysis). Our ide of integrting these different types of it is bsed on the procedure of introducing metric on the set N of nturl numbers together with n idel point nd on the set of ll prtitions of n intervl together with n idel point. Then, the model of the it t point of function mpping metric spce into nother one embrces ll three types of it present in the clssicl nlysis. Then, our min

2 16 Z. Kdelburg, M. M. Mrjnović objective in this pper is to use (nd prove) theorem on the interchnge of two its under the condition tht one of them exists nd the other one exists uniformly. Relying on this theorem, we derive proofs of (long) sequence of sttements on interchnge of two its tht re commonly present in the nlysis courses. The effects of such n pproch re lso seen in the fct tht n pprently vried set of conditions ssocited with these prticulr sttements trnsltes uniquely into the requirement: one of the two its exists nd the other one exists uniformly. Now we turn our ttention to the history of the 19th century mthemtics to recll how the concept of uniform convergence, though used explicitly, did not become common property for longer time. In his fmous Course d nlyse, (l821), Cuchy (A-L. Cuchy, ) mde misstep with respect to rigor, stting tht the sum of convergent series of continuous functions is continuous function. In 1826, Abel (N. H. Abel, ) restricted this sttement to the cse of uniform convergence, giving correct proof of it (Jour. für Mth. 1, ), but not isolting the property of uniform convergence. Both men, Stokes (G. G. Stokes, ) (Trns. Cmb. Phil. Soc., 85, 1848) nd Seidel (Ph. L. von Seidel, ) (Abh. der Byer. Akd. der Wiss., 1847/49) recognized the distinction between uniform nd point-wise convergence nd the significnce of the former in the theory of infinite series. Further on, from letter of Weierstrss (K. Weierstrss, ) (unpublished till 1894 (Werke)) it is seen tht he must hve drwn this distinction s erly s Nevertheless, Cuchy ws the first to recognize ultimtely the uniform convergence s property ssuring the continuity of the sum of series of continuous functions (Comp. Rend., 36, 1853). It ws Weierstrss who used first the concept of uniform convergence s condition for term by term integrtion of series s well s for differentition under the integrl sign. Through the circle of his students the importnce of this concept ws mde widely known. At the end, let us note tht, when prepring this pper, we used n unpublished mnuscript of the second uthor. 2. Iterted its Let A be nonempty subset of metric spce (M 1,d 1 )ndb nonempty subset of metric spce (M 2,d 2 ). Let f : A B M be mpping into complete metric spce (M,d). Denote by X the set of ccumultion points of subset X of metric spce. Note tht A B (A B), which follows from (x 0,y 0 ) A B ( >0) {[K(x 0,) \{x 0 }] A nd [K(y 0,) \{y 0 }] B } = ( >0) [K(x 0,) K(y 0,) \{(x 0,y 0 )}] A B (x 0,y 0 ) (A B).

3 Interchnging two its 17 Let x 0 A nd y 0 B. A it f(x, y) =ϕ(x) existsforechx A if the following condition holds: ( x A)( >0)( δ >0)( y B)(0 <d 2 (y, y 0 ) <δ = d(f(x, y),ϕ(x)) <); nd one cn formulte similrly the condition for the existence of f(x, y) = ψ(y). Let us strengthen this condition demnding tht the number δ does not depend on x. Prcticlly, this mens tht the quntifier ( x A) hs to be plced fter the quntifier ( δ > 0). Thus we obtin tht f(x, y) =ϕ(x) y y exists uniformly in x A if 0 ( >0)( δ >0)( x A)( y B)(0 <d 2 (y, y 0 ) <δ = d(f(x, y),ϕ(x)) <); nd the condition for the existence of f(x, y) =ψ(y), uniformly in y B is x x formulted similrly. 0 Cuchy condition for the existence of the f(x, y) =ϕ(x), uniformly in y y x A is 0 ( >0)( δ >0)( x A)( y B)( y B) 0 <d 2 (y,y 0 ) <δnd 0 <δ 2 (y,y 0 ) <δ = d(f(x, y ),f(x, y )) <. Since the spce (M,d) is ssumed to be complete, it is cler tht this condition is necessry nd sufficient for the existence of f(x, y) =ϕ(x), uniformly in y y x A. 0 Writing formlly, ϕ(x) = ( f(x, y)), ψ(x) = ( f(x, y)), we cll these expressions iterted its (which, of course need not exist). In the sequel, we shll omit dditionl prentheses. In this context, the expression f(x, y) (x 0,y 0) is clled double it (nd it my eqully be non-existent). A generl connection between the mentioned its is given by the following Proposition 1. Let f : A B M be function from subset A B M 1 M 2 into M nd (x 0,y 0 ) A B,whereM 1, M 2 nd M re metric spces. If (i) f(x, y) =α exists, (x 0,y 0) (ii) for ech y B, f(x, y) =ψ(y) exists, then ψ(y) exists nd is equl to α.

4 18 Z. Kdelburg, M. M. Mrjnović Proof. Condition (i) implies tht ( >0)( δ >0)( (x, y) A B) (x, y) K(x 0,δ) K(y 0,δ) \{(x 0,y 0 )} = d(f(x, y),α) < 2. According to (ii), using the continuity of function d, there exists Thus, d(f(x, y),α)=d(ψ(y),α) 2. ( >0)( δ >0)( y B) y K(y 0,δ) \{y 0 } = d(ψ(y),α) <, which mens tht ψ(y) =α. Note tht the conditions of the previous proposition re pretty strong, especilly this is the cse with condition (i). Therefore, this proposition is of smll prcticl vlue. However, it cn be used to prove the non-existence of some double its. Exmple 1. Consider the following three functions: f(x, y) = x y + x2 + y 2 x + y, g(x, y) = x sin 1 x + y, h(x, y) =x sin 1 x + y y, ll three defined on the set A B =(0, +) (0, +). For the function f, f(x, y) =1+x, forechx A nd f(x, y) = 1+y, y 0 x 0 for ech y B, ndso f(x, y) =1 1 = f(x, y). x 0 y 0 y 0 x 0 The double it does not exist. For the function g, g(x, y) =sin 1,forechx A nd g(x, y) =1,for y 0 x x 0 ech y B. Hence, g(x, y) = 1, nd both g(x, y) nd the double y 0 x 0 x 0 y 0 it do not exist. For the function h, using tht 0 x sin 1 y x, we obtin tht h(x, y) = (0,0) 0. On the other hnd, h(x, y) does not exist for ny x A, nd so the respective iterted it does not exist either. The other iterted it exists nd y 0 h(x, y) =0. y 0 x 0 Now we shll modify the conditions of Proposition 1, demnding tht one of the its f(x, y) nd f(x, y) exists, nd tht the other one exists uniformly. We shll conclude tht the three its, both iterted nd the double one, ll exist nd re equl. Such sttement synthesizes then series of clssicl

5 Interchnging two its 19 results on interchnge of it opertions. In terms of generlized sequences it is clled Moore Theorem (E. H. Moore, ) 1. Theorem 1. (The theorem on interchnge of two its) Let f : A B M be mpping into complete metric spce, where A nd B re subsets of metric spces M 1 nd M 2, respectively, nd let x 0 A \ A, y 0 B \ B. If (i) f(x, y) =ψ(y) exists for ech y B; (ii) f(x, y) =ϕ(x) exists uniformly in x A, then the three its f(x, y), f(x, y), nd (x 0,y 0) f(x, y) ll exist nd re equl. Proof. Let >0 be rbitrry. According to condition (ii), we hve (1) ( δ >0)( y B) 0<d 2 (y, y 0 ) <δ = ( x A) d(f(x, y),ϕ(x)) < 6. Let y K(y 0,δ) \{y 0 }. Using condition (i) we obtin (2) ( δ >0)( x A) 0<d 1 (x, x 0 ) < δ = d(f(x, y),ψ(y)) < 6. Let us tke neighborhood of the point (x 0,y 0 )oftheform U = K(x 0, δ) K(y 0,δ) nd let points (x,y ), (x,y )belongtothesetu \{(x 0,y 0 )}. Using the tringle inequlity, we get d(f(x,y ),f(x,y )) d(f(x,y ),ϕ(x )) + d(ϕ(x ),f(x, y)) + d(f(x, y),ψ(y)) + d(ψ(y),f(x, y)) + d(f(x, y),ϕ(x )) + d(ϕ(x ),f(x,y )). By (2), the third nd the fourth summnd on the right-hnd side re both less thn /6, nd by (1), ech of the other four summnds is less thn /6. Thus, ( (x,y ) A B)( (x,y ) A B) (x,y ), (x,y ) U \{(x 0,y 0 )} = d(f(x,y ),f(x,y )) <, nd so the function f stisfies Cuchy condition t the point (x 0,y 0 ). Therefore, since the spce M is complete, there exists f(x, y) =α. (x 0,y 0) 1 See N. Dunford nd J. T. Schwrtz, Liner Opertors, Prt I, Interscience Publishers, New York, 1958.

6 20 Z. Kdelburg, M. M. Mrjnović Using Proposition 1 nd condition (i), we hve nd, by (ii), ψ(y) =α = ϕ(x) =α = f(x, y), f(x, y). We shll pply the obtined result to sequence (f n ) of functions with domin A which is subset of metric spce M 0 nd codomin which is metric spce M. Then, we cn consider f n (x) s function of two vribles, tking ϕ(n, x) =f n (x). Since N cn be understood s subset of the metric spce N = N {} with the metric d(m, n) = 1 m 1 ( ) 1 n, =0 nd since N, for the function ϕ: N A M, the fct tht for ech x A, ϕ(n, x) =f(x) exists, simply mens tht the sequence (f n) converges to the function f on the set A. The fct tht ϕ(n, x) =f(x) exists uniformly in x A, is equivlent to the fct tht the sequence (f n ) converges uniformly on A. The respective condition ( >0)( δ >0)( n N)( x A) 0<d(n, ) <δ = d(ϕ(n, x),f(x)) <, cn be written s ( >0)( m N)( n N)( x A) n>m = d(f n (x),f(x)) <, tking tht d(n, ) = 1 [ ] 1 n nd = m. δ As n exmple of using such tretment of functionl sequences s functions of two vribles, we shll derive now the well-known theorem bout continuity of the it function. Corollry 1. Let (f n ) be uniformly convergent sequence of continuous functions from metric spce M 0 into complete metric spce M. Then the it function f = f n is lso continuous. Proof. If x 0 M 0 is n isolted point, the ssertion is trivil. Let x 0 M 0. For the function ϕ(n, x) =f n (x), (M 0 M 0, N N ) we hve lso (i) ϕ(n, x) exists uniformly in x M 0, nd by the continuity of ll the functions f n t the point x 0,

7 Interchnging two its 21 (ii) for ech n N, ϕ(n, x) =f n (x 0 ) exists, nd by Theorem 1, it follows tht f(x) = = f n(x 0 )=f(x 0 ). ϕ(n, x) = ϕ(n, x) In similr wy, the following ssertion on termwise differentition cn be proved. Corollry 2. Let (f n ) be sequence of rel-vlued functions, converging on [, b] tofunctionf, ndlet (i) f n be differentible function for ech n N, (ii) the sequence (f n) converges uniformly on [, b]. Then the function f is differentible nd the equlity f (x) = f n(x) holds. The respective properties of functionl series cn be derived now s esy consequences. 3. Double sequences A function : N N R is clled double sequence. As usul, we shll denote the vlue of the function t (i, j) s ij, nd this sequence will be denoted by ( ij ). The following theorem gives sufficient conditions for the interchnge of the order of summtion. Theorem 2. Let ( ij ) be double sequence such tht: (i) ( i) ij = b i < +, (ii) Then i=1 j=1 b i is convergent series. ij = ij. i=1 j=1 j=1 i=1 Proof. Let f i : N = N {} R be the function defined by f i (k) = k ij. j=1 For ech i, the function f i is continuous on N, becuse f i(k) = ij = f i (). k j=1

8 22 Z. Kdelburg, M. M. Mrjnović The series f i (k) converges uniformly ccording to the Weierstrss Test, becuse Therefore, i=1 (i) ( i) f i (k) b i nd (ii) k i=1 The desired equlity follows from k i=1 i=1 f i (k) = i=1 b i converges. i=1 f i (k) = f i(k). k k ij = k i=1 j=1 f i(k) = ij. k i=1 j=1 k ij = ij, k j=1 i=1 j=1 i=1 4. The Toeplitz Limit Theorem The following theorem ws proved by Toeplitz (O. Toeplitz, ) in Prce mt.-fiz. 22 (1911), Theorem 3. The coefficients of the mtrix n n k0 k1... kn re ssumed to stisfy the following two conditions: () for ech fixed n, kn 0 s k, (b) there exists constnt K such tht for ech fixed k nd ny n, k0 + k1 + + kn <K. Then, for every null sequence (z 0,z 1,...,z n,...), the numbers lso form null sequence. Proof. Denote The ssumption () implies tht ( ) ϕ(n, k) = 0, for ech n. k z k = k0 z 0 + k1 z kn z n + ϕ(n, k) = k0 z 0 + k1 z kn z n.

9 Interchnging two its 23 On the other hnd, (b ) ϕ(n, k) = kj z j = ψ(k) exists, uniformly in k. n j=0 To prove it, observe first tht, by (b), the series kj converges bsolutely. Fix >0. Let j j=0 0 be such tht Then, for j>j 0,wehve ψ(k) ϕ(n, k) = j=n+1 j>j 0 = z j < K. kj z j j=n+1 kj mx z j <K j K =, nd so ϕ(n, k) exists uniformly in k. n Now, by Theorem 1, we obtin tht z k = ϕ(n, k) = ϕ(n, k) =0. k k n n k Corollry 3. If, in ddition to ssumption () of Theorem 3, ( k) k0 + k1 + + kn 1 s n, nd z n z 0, then z k z 0 s k. 5. Integrl s it Denote by Π the set of ll prtitions P of segment [, b]. For the prtition P given by = x 0 <x 1 < <x n = b, the number P = mx{ x i x i 1 i =1, 2,...,n} is clled the norm of prtition P. Let be n element which does not belong to Π nd let =0. ThesetΠ =Π {}, together with the metric { P1 + P 2, P 1 P 2, d Π (P 1,P 2 )= 0, P 1 = P 2, is metric spce which will be clled the prtition spce nd will be denoted by (Π,d Π ). If, for two prtitions P 1 nd P 2, P 1 P 2, we sy tht P 1 is smller thn P 2. Note tht finer prtition is lso smller, but the converse is not true. For prtition P,letP be the prtition which hs s prtitioning points ll the points of prtition P nd, moreover, ll the midpoints of segments of P. Then P = 1 2 P. Let P 0 be the prtition with = x 0 <x 1 = b. Let us define sequence (P n ) of prtitions, tking P n+1 = P n. Then P n = b 2 n,ndso d b Π(,P n ) = 2 n =0, wherefrom it follows tht is n ccumultion point of the set Π in the spce Π.

10 24 Z. Kdelburg, M. M. Mrjnović Let f :[, b] R be bounded function, s(f,p) nds(f,p) be its lower nd its upper sum, with respect to the prtition P. These sums cn be considered s functions s: Π R, S:Π R from the spce of prtitions into R. The it s(f,p) =α exists if the condition ( >0)( δ >0)( P Π) 0 <d Π (P, ) <δ = α s(f,p) < holds, or, equivlently, if the condition ( >0)( δ >0)( P Π) P <δ = α s(f,p) < holds. The following proposition ssocites this it with the lower (upper) integrl of the function f. Proposition 2. Let f :[, b] R be bounded function. Then s(f,p) (resp. S(f,P)) existsndisequltosup P s(f,p) (resp. inf P S(f,P)). Proof. Let sup P s(f,p) =I. Since the function f is bounded, there exists constnt M>0, such tht for ech x [, b], f(x) M. Let>0. There exists prtition P 1 with prtitioning points = x 0 <x 1 < <x k = b, such tht I s(f,p 1 ) < 2. Let δ =. Let P Π be such tht P <δ. Denote by P the superposition 8Mk of prtitions P nd P 1. We hve now s(f,p ) s(f,p) = m v( ) m v( ), P where m =inf{ f(x) x } nd v( ) is the length of segment. Since prtition P is finer thn prtition P, for ech P there exists unique P such tht.letm = m,ndso m v( ) = m v( ). Now, P P P s(f,p ) s(f,p) = (m m )v( ). P We hve m m = 0 when hs no endpoints belonging to prtition P 1, except points nd b, ndm m 0 when n endpoint of segment belongs to the set {x 1,...,x k 1 }, nd this endpoint does not belong to P. The number of such segments is t most 2(k 1). Since m m 2M, wehve s(f,p ) s(f,p) 2(k 1) 2M P < 2(k 1) 2M 8Mk = 2.

11 Interchnging two its 25 Hence, when P <δ,wehve I s(f,p) (I s(f,p )+(s(f,p ) s(f,p)) < =, becuse P is lso finer tht P 1 nd so I s(f,p ) I s(f,p 1 ) < 2. This completes the proof tht s(f,p) =I. The equlity S(f,P) =I =inf P S(f,P) cn be proved in similr wy. If the function f is integrble, then s(f,p) = S(f,P) = f(x) dx. For prtition P,letξ P be choice function, i.e., let ξ P ( ). The sum σ(f,p) = { f(ξ P ( ))v( ) P } is clled the integrl sum of the function f, corresponding to the prtition P. Since s(f,p) σ(f,p) S(f,P), for ech ξ P,wehve σ(f,p) = f(x) dx, for ech integrble function f. We shll consider now the question of interchnging it nd n integrl. First, we need n uxiliry sttement. Lemm 1. Let f :[, b] R nd g :[, b] R be two functions, such tht ( x [, b]) f(x) g(x) <k. Then, for ech nonempty set S [, b], the inequlities hold true. inf f(x) inf g(x) < 2k, x S x S sup x S f(x) sup g(x) < 2k. x S Proof. Let x S be point, such tht f(x) <m S (f) +k, where m S (f) = inf f(x). Then, x S m S (f) >f(x) k>g(x) 2k m S (g) 2k, where m S (g) = infg(x). Similrly, x S m S (g) >m S (f) 2k, which proves tht m S (f) m S (g) < 2k. The proof of the other prt of the proposition is similr.

12 26 Z. Kdelburg, M. M. Mrjnović Theorem 4. Let (f n ) be sequence of functions, integrble on the segment [, b], ndlet(f n ) converges uniformly to function f. Then, the function f is integrble nd the equlity is vlid. f n (x) dx = f(x) dx Proof. Let >0 be given. Since the sequence (f n ) converges uniformly to the function f, there exists n m N such tht ( n N)( x [, b]) n m = f n (x) f(x) < 2(b ). The lower sum of the function f n for the prtition P, s(f n,p)= { m n ( )v( ) P }, is function from N Π N Π into R. Forn m nd ech P,wehve s(f.p) s(f n,p) m n ( ) m( ) v( ). P Applying Lemm 1, we obtin s(f,p) s(f n,p) < 2 (b ) =, 2(b ) nd so s(f n,p)=s(f,p) exists uniformly in P.Forechn N, s(f n,p)= f n (x) dx P exists, becuse f n is n integrble function. Applying Theorem 1, it follows tht both iterted its exist nd re equl. Hence, we hve s(f n,p) = s(f,p) = f(x) dx, P n P nd s(f n,p) = f n (x) dx, n P n b f(x) dx = n f n (x) dx, Repeting the previous rguments nd using upper sums, the equlity b f(x) dx = f n (x) dx n cn lso be proved.

13 Interchnging two its 27 Note tht the previous theorem remins vlid when the domin of the given functions is n rbitrry set A R, which is Jordn mesurble (C. M. E. Jordn, ). Nmely, such set A is bounded nd there exists segment [, b] such tht A [, b]. If the functions f n : A R re integrble, so re the functions { fn (x), x A, f n :[, b] R given by f n (x) = 0, x [, b] \ A. If the sequence (f n ) converges uniformly to the function f : A R, then the sequence (f n ) converges uniformly to the function { f(x), x A, f(x) = 0, x [, b] \ A, which is lso integrble. According to the previous theorem, we hve f n (x) dx = f n (x) dx = f(x) dx = f(x) dx. A Assuming tht the it function of convergent sequence (f n ) is integrble, the ssumption of uniform convergence cn be wekened. First, we formulte nd prove n uxiliry sttement, which is of interest on its own. Let (f n ) be sequence of rel functions, converging to function f on set A R. LetA be the closure of the set A in R. We shll sy tht (f n ) converges uniformly round point x 0 A if there exists n open neighborhood U of the point x 0, such tht (f n )converges uniformly on the set U A. Proposition 3. Let f n, n N, be rel functions with domin A R. The sequence (f n ) converges uniformly in x A if nd only if it converges uniformly round ech point of the set A R. Proof. If the sequence (f n ) converges uniformly in x A, then it obviously converges uniformly round ech point of the set A. Conversely, suppose tht (f n ) converges uniformly round ech point of A. Then (f n ) is certinly point-wise convergent to function f. If A contins one of the elements, + (or both of them), choose respective neighborhood U or U + (or both of them), such tht (f n ) converges uniformly on A U, resp. A U +. For ny other point x A, letu x be neighborhood such tht (f n ) converges uniformly on U x A. The set A \ (U U + ) (if or + does not belong to A, then U, resp. U + is n empty set) will be compct. Hence, there exist finitely mny neighborhoods U x1, U x2,..., U xk covering the set A \ (U U + ). Adding U nd/or U + to them, we obtin finitely mny subsets of A, such tht (f n ) converges uniformly on them. Since ech point of the set A belongs to one of the sets U ξ A, ξ {x 1,x 2,...,x k,, +}, the sequence (f n ) converges uniformly on A. A

14 28 Z. Kdelburg, M. M. Mrjnović Roughly speking, more points re there round which the sequence (f n )isnot uniformly convergent, less regulr is the convergence of (f n ). On the other hnd, Proposition 3 cn be formulted s follows: The sequence (f n ) does not converge uniformly in x A if nd only if it does not converge uniformly round point x A. Hence, nonuniform convergence cn lwys be spotted loclly. The following is the Lebesgue theorem (H. L. Lebesgue, ) on bounded convergence for the Riemnn integrl (G. F. B. Riemnn, ). Theorem 5. Let (f n ) be sequence of functions, integrble on [, b], converging to function f. If (i) the function f is integrble, (ii) ( M R)( n N)( x [, b]) f n (x) M, (iii) the set A = { x [, b] (f n ) does not converge uniformly round x } isclosedsetoflebesguemesure0, then f n (x) dx = f(x) dx. Proof. Let >0 be given. According to (iii), there exists sequence ( i ) i N of open intervls covering the set A, such tht { v( i ) i N } < 4M. The set A is closed nd bounded, nd so, by Borel-Lebesgue Theorem (F. E. J. E. Borel, ), there exist finitely mny intervls i1,..., ik which cover A, s well. Let D = i1 ik. The set D is Jordn mesurble nd m(d) v( i1 )+ + v( ik ) < 4M. The sequence (f n ) converges uniformly round ech point of the set [, b] \ D, nd so, by Proposition 3, it converges uniformly on [, b] \ D. Hence, there exists n n 0 N such tht ( n N)( x [, b] \ D) n n 0 = f n (x) f(x) < Now, for n n 0,wehve f(x) dx f n (x) dx which proves the theorem. D < 2M 2(b ). f(x) f n (x) dx + f(x) f n (x) dx [,b]\d 4M + (b ) =, 2(b )

15 Interchnging two its 29 Exmple 2. The sequence of functions nx f n (x) =, x [0, 1], 1+(nx) 2 converges to the function 0: [0, 1] R, nd it does not converge uniformly only round the point 0. The set A = {0} is closed nd of mesure 0, while ( n N)( x [, b]) nx 1+(nx) Hence, we hve 1 nx dx = (nx) 2 At the end, let us remrk tht we hve not included severl other cses of interchnge of two its: cses of differentition nd integrtion of integrls depending on prmeter, equlity f xy = f yx, etc., which cn be treted in the sme wy. Zorn Kdelburg, Fculty of Mthemtics, Studentski trg 16/IV, Beogrd, Serbi & Montenegro E-mil: kdelbur@mtf.bg.c.yu Milosv M. Mrjnović, Mthemticl Institute, Knez Mihil 35/I, Beogrd, Serbi & Montenegro E-mil: milomr@beotel.yu