limit_interchange-2

THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp. 15–29 INTERCHANGING TWO LIMITS Zoran Kadelburg and Milosav M. Marjanovi´ c This paper is dedicated to the memory of our illustrious professor of analysis Slobodan Aljanˇ ci´c (1922–1993). Abstract. By the use of convenient metrics, the ordered set of natural numbers plus an ideal element and the partially ordered set of all partitions of an interval plusan ideal element are converted into metric spaces. Thus, the three different types oflimit, arising in classical analysis, are reduced to the same model of the limit of a function at a point. Then, the theorem on interchange of iterated limits, valid under the condition that one of the iterated limits exists and the other one exists uniformly, isused to derive a long sequence of statements of that type that are commonly present inthe courses of classical analysis. All apparently varied conditions accompanying such statements are, then, unmasked and reduced to one and the same: one iterated limit exists and the other one exists uniformly. ZDM Subject Classification : I35; AMS Subject Classification :0 0 A 3 5 Key words and phrases : Interchange of two limits, uniform convergence, definite integral as a limit. 1. Introduction The simple concept of metric space proves as a convenient setting for a unifica- tion of different types of convergence and continuity that arise in classical analysis. For this reason, this concept finds its place in the contents of the first contemporary courses of mathematical analysis. Particularly, by replacing the absolute value with a distance function, immediate generalizations of conditions determining classicalconcepts of convergence and continuity are obtained. Due to different terms, the following three types of limit are present in the classical analysis lima n,limf(x),limσ(f,P), being respectively: the limit of a sequence, the limit of a function at a point and the limit of integral sums. To have them as particular cases of a more general conceptof limit, generalized sequences (nets) or filters are used (and they are not common topics on the list of themes for the first courses in analysis). Our idea of integrating these different types of limit is based on the procedure of introducing metric on the set Nof natural numbers together with an ideal point and on the set of all partitions of an interval together with an ideal point. Then, themodel of the limit at a point of a function mapping a metric space into another one embraces all three types of limit present in the classical analysis. Then, our main 16 Z. Kadelburg, M. M. Marjanovi´ c objective in this paper is to use (and prove) a theorem on the interchange of two limits under the condition that one of them exists and the other one exists uniformly. Relying on this theorem, we derive proofs of a (long) sequence of statements oninterchange of two limits that are commonly present in the analysis courses. The effects of such an approach are also seen in the fact that an apparent- ly varied set of conditions associated with these particular statements translates uniquely into the requirement: one of the two limits exists and the other one exists uniformly. Now we turn our attention to the history of the 19th century mathematics to recall how the concept of uniform convergence, though used explicitly, did notbecome a common property for a longer time. In his famous Course d’analyse, (l821), Cauchy (A-L. Cauchy, 1789–1857) made a misstep with respect to rigor, stating that the sum of a convergent series of continuous functions is a continuous function. In 1826, Abel (N. H. Abel, 1802–1829) restricted this statement to the case of uniform convergence, giving a correct proof of it (Jour. f¨ ur Math. 1, 311– 339), but not isolating the property of uniform convergence. Both men, Stokes (G. G. Stokes, 1819–1903) (Trans. Camb. Phil. Soc., 85, 1848) and Seidel (Ph. L. von Seidel, 1821–1896) (Abh. der Bayer. Akad. der Wiss., 1847/49) recognized thedistinction between uniform and point-wise convergence and the significance of the former in the theory of infinite series. Further on, from a letter of Weierstrass (K. Weierstrass, 1815–1897) (unpublished till 1894 (Werke)) it is seen that he musthave drawn this distinction as early as 1841. Nevertheless, Cauchy was the first to recognize ultimately the uniform convergence as a property assuring the continuity of the sum of a series of continuous functions (Comp. Rend., 36, 1853). It wasWeierstrass who used first the concept of uniform convergence as a condition for term by term integration of a series as well as for differentiation under the integral sign. Through the circle of his students the importance of this concept was madewidely known. At the end, let us note that, when preparing this paper, we used an unpublished manuscript of the second author. 2. Iterated limits LetAbe a nonempty subset of a metric space ( M 1,d1)a n d Ba nonempty subset of a metric space ( M2,d2). Let f:A×B→Mbe a mapping into a complete metric space ( M,d). Denote by X/primethe set of accumulation points of a subset Xof a metric space. Note that A/prime×B/prime⊂(A×B)/prime, which follows from (x0,y0)∈A/prime×B/prime ⇐⇒(∀ε>0){[K(x0,ε)\{x0}]∩A/negationslash=∅and [K(y0,ε)\{y0}]∩B/negationslash=∅} =⇒(∀ε>0)[K(x0,ε)×K(y0,ε)\{(x0,y0)}]∩A×B/negationslash=∅ ⇐⇒(x0,y0)∈(A×B)/prime. Interchanging two limits 17 Letx0∈A/primeandy0∈B/prime. A limit lim y→y0f(x,y)=ϕ(x)e x i s t sf o re a c h x∈Aif the following condition holds: (∀x∈A)(∀ε>0)(∃δ>0)(∀y∈B)(0<d2(y,y0)<δ=⇒d(f(x,y),ϕ(x))<ε); and one can formulate similarly the condition for the existence of lim x→x0f(x,y)= ψ(y). Let us strengthen this condition demanding that the number δdoes not depend on x. Practically, this means that the quantifier ( ∀x∈A) has to be placed after the quantifier ( ∃δ>0). Thus we obtain that lim y→y0f(x,y)=ϕ(x) exists uniformly in x∈Aif (∀ε>0)(∃δ>0)(∀x∈A)(∀y∈B)(0<d2(y,y0)<δ=⇒d(f(x,y),ϕ(x))<ε); and the condition for the existence of lim x→x0f(x,y)=ψ(y), uniformly in y∈Bis formulated similarly. Cauchy condition for the existence of the lim y→y0f(x,y)=ϕ(x), uniformly in x∈Ais (∀ε>0)(∃δ>0)(∀x∈A)(∀y/prime∈B)(∀y/prime/prime∈B) 0<d2(y/prime,y0)<δand 0 <δ2(y/prime/prime,y0)<δ=⇒d(f(x,y/prime),f(x,y/prime/prime))<ε . Since the space ( M,d) is assumed to be complete, it is clear that this condition is necessary and sufficient for the existence of lim y→y0f(x,y)=ϕ(x), uniformly in x∈A. Writing formally, lim x→x0ϕ(x) = lim x→x0( lim y→y0f(x,y)), lim y→y0ψ(x) = lim y→y0( lim x→x0f(x,y)), we call these expressions iterated limits (which, of course need not exist). In the sequel, we shall omit additional parentheses. In this context, the expression lim (x0,y0)f(x,y) is called double limit (and it may equally be non-existent). A general connection between the mentioned limits is given by the following Proposition 1. Letf:A×B→Mbe a function from a subset A×B⊂ M1×M2intoMand(x0,y0)∈A/prime×B/prime,w h e r e M1,M2andMare metric spaces. If (i) lim (x0,y0)f(x,y)=αexists, (ii)for each y∈B,lim x→x0f(x,y)=ψ(y)exists, then lim y→y0ψ(y)exists and is equal to α. 18 Z. Kadelburg, M. M. Marjanovi´ c Proof. Condition (i) implies that (∀ε>0)(∃δ>0)(∀(x,y)∈A×B) (x,y)∈K(x0,δ)×K(y0,δ)\{(x0,y0)}=⇒d(f(x,y),α)<ε 2. According to (ii), using the continuity of function d, there exists lim x→x0d(f(x,y),α)=d(ψ(y),α)/lessorequalslantε 2. Thus, (∀ε>0)(∃δ>0)(∀y∈B)y∈K(y0,δ)\{y0}=⇒d(ψ(y),α)<ε , which means that lim y→y0ψ(y)=α. Note that the conditions of the previous proposition are pretty strong, espe- cially this is the case with condition (i). Therefore, this proposition is of smallpractical value. However, it can be used to prove the non-existence of some double limits. Example 1. Consider the following three functions: f(x,y)=x−y+x 2+y2 x+y,g (x,y)=xsin1 x+y x+y,h (x,y)=xsin1 y, all three defined on the set A×B=( 0,+∞)×(0,+∞). For the function f, lim y→0f(x,y)=1 + x,f o re a c h x∈Aand lim x→0f(x,y)=−1+y, for each y∈B,a n ds o lim x→0lim y→0f(x,y)=1/negationslash=−1 = lim y→0lim x→0f(x,y). The double limit does not exist. For the function g, lim y→0g(x,y)=s i n1 x,f o re a c h x∈Aand lim x→0g(x,y)=1 ,f o r eachy∈B. Hence, lim y→0lim x→0g(x,y) = 1, and both lim x→0lim y→0g(x,y) and the double limit do not exist. For the function h, using that 0 /lessorequalslant/vextendsingle/vextendsingle/vextendsingle/vextendsinglexsin1 y/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant|x|, we obtain that lim (0,0)h(x,y)= 0. On the other hand, lim y→0h(x,y) does not exist for any x∈A, and so the re- spective iterated limit does not exist either. The other iterated limit exists and lim y→0lim x→0h(x,y)=0 . /triangle Now we shall modify the conditions of Proposition 1, demanding that one of the limits lim x→x0f(x,y) and lim y→y0f(x,y) exists, and that the other one exists uniformly. We shall conclude that the three limits, both iterated and the double one, all exist and are equal. Such a statement synthesizes then a series of classical Interchanging two limits 19 results on interchange of limit operations. In terms of generalized sequences it is called Moore Theorem (E. H. Moore, 1891–1976)1. Theorem 1. (The theorem on interchange of two limits) Letf:A×B→M be a mapping into a complete metric space, where AandBare subsets of metric spaces M1andM2, respectively, and let x0∈A/prime\A,y0∈B/prime\B.I f (i) lim x→x0f(x,y)=ψ(y)exists for each y∈B; (ii) lim y→y0f(x,y)=ϕ(x)exists uniformly in x∈A, then the three limits lim x→x0lim y→y0f(x,y),lim y→y0lim x→x0f(x,y),and lim (x0,y0)f(x,y) all exist and are equal. Proof. Letε>0 be arbitrary. According to condition (ii), we have (1) ( ∃δ>0)(∀y∈B)0<d2(y,y0)<δ=⇒(∀x∈A)d(f(x,y),ϕ(x))<ε 6. Lety∈K(y0,δ)\{y0}. Using condition (i) we obtain (2) ( ∃δ>0)(∀x∈A)0<d1(x,x0)<δ=⇒d(f(x,y),ψ(y))<ε 6. Let us take a neighborhood of the point ( x0,y0)o ft h ef o r m U=K(x0,δ)×K(y0,δ) and let points ( x/prime,y/prime), (x/prime/prime,y/prime/prime)b e l o n gt ot h es e t U\{(x0,y0)}. Using the triangle inequality, we get d(f(x/prime,y/prime),f(x/prime/prime,y/prime/prime))/lessorequalslantd(f(x/prime,y/prime),ϕ(x/prime)) +d(ϕ(x/prime),f(x/prime,y)) +d(f(x/prime,y),ψ(y)) +d(ψ(y),f(x/prime/prime,y)) +d(f(x/prime/prime,y),ϕ(x/prime/prime)) +d(ϕ(x/prime/prime),f(x/prime/prime,y/prime/prime)). By (2), the third and the fourth summand on the right-hand side are both less than ε/6, and by (1), each of the other four summands is less than ε/6. Thus, (∀(x/prime,y/prime)∈A×B)(∀(x/prime/prime,y/prime/prime)∈A×B) (x/prime,y/prime),(x/prime/prime,y/prime/prime)∈U\{(x0,y0)}=⇒d(f(x/prime,y/prime),f(x/prime/prime,y/prime/prime))<ε , and so the function fsatisfies Cauchy condition at the point ( x0,y0). Therefore, since the space Mis complete, there exists lim (x0,y0)f(x,y)=α. 1See N. Dunford and J. T. Schwartz, Linear Operators, Part I, Interscience Publishers ,N e w York, 1958. 20 Z. Kadelburg, M. M. Marjanovi´ c Using Proposition 1 and condition (i), we have lim y→y0ψ(y)=α= lim y→y0lim x→x0f(x,y), and, by (ii), lim x→x0ϕ(x)=α= lim x→x0lim y→y0f(x,y). We shall apply the obtained result to a sequence ( fn) of functions with domain Awhich is a subset of a metric space M0and codomain which is a metric space M. Then, we can consider fn(x) as a function of two variables, taking ϕ(n,x)=fn(x). Since Ncan be understood as a subset of the metric space N∗=N∪{ ∞ } with the metric d(m,n)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 m−1 n/vextendsingle/vextendsingle/vextendsingle/vextendsingle,/parenleftbigg1 ∞=0/parenrightbigg and since ∞∈N/prime, for the function ϕ:N×A→M, the fact that for each x∈A, lim n→∞ϕ(n,x)=f(x) exists, simply means that the sequence ( fn) converges to the function fon the set A. The fact that lim n→∞ϕ(n,x)=f(x)exists uniformly in x∈A, is equivalent to the fact that the sequence (fn)converges uniformly on A.The respective condition (∀ε>0)(∃δ>0)(∀n∈N)(∀x∈A)0<d(n,∞)<δ=⇒d(ϕ(n,x),f(x))<ε , can be written as (∀ε>0)(∃m∈N)(∀n∈N)(∀x∈A)n>m =⇒d(fn(x),f(x))<ε , taking that d(n,∞)=1 nand/bracketleftbigg1 δ/bracketrightbigg =m. As an example of using such treatment of functional sequences as functions of two variables, we shall derive now the well-known theorem about continuity of the limit function. Corollary 1. Let(fn)be a uniformly convergent sequence of continuous functions from a metric space M0into a complete metric space M. Then the limit function f= lim fnis also continuous. Proof. Ifx0∈M0is an isolated point, the assertion is trivial. Let x0∈M/prime 0. For the function ϕ(n,x)=fn(x),(M0⊂M0,N⊂N∗) we have also (i) lim n→∞ϕ(n,x) exists uniformly in x∈M0, and by the continuity of all the functions fnat the point x0, Interchanging two limits 21 (ii) for each n∈N, lim x→x0ϕ(n,x)=fn(x0) exists, and by Theorem 1, it follows that lim x→x0f(x) = lim x→x0lim n→∞ϕ(n,x) = lim n→∞lim x→x0ϕ(n,x) = lim n→∞fn(x0)=f(x0). In a similar way, the following assertion on termwise differentiation can be proved. Corollary 2. Let(fn)be a sequence of real-valued functions, converging on [a,b]t oaf u n c t i o n f,a n dl e t (i)fnbe a differentiable function for each n∈N, (ii)the sequence (f/prime n)converges uniformly on [a,b]. Then the function fis differentiable and the equality f/prime(x) = lim n→∞f/prime n(x) holds. The respective properties of functional series can be derived now as easy con- sequences. 3. Double sequences A function a:N×N→Ris called a double sequence . As usual, we shall denote the value of the function aat (i,j)a saij, and this sequence will be denoted by (aij). The following theorem gives sufficient conditions for the interchange of the order of summation. Theorem 2. Let(aij)be a double sequence such that: (i) (∀i)∞/summationtext j=1|aij|=bi<+∞, (ii)∞/summationtext i=1biis a convergent series. Then∞/summationtext i=1∞/summationtext j=1aij=∞/summationtext j=1∞/summationtext i=1aij. Proof. Letfi:N∗=N∪ {∞} → Rbe the function defined by fi(k)=k/summationtext j=1aij. For each i, the function fiis continuous on N∗, because lim k→∞fi(k)=∞/summationtext j=1aij=fi(∞). 22 Z. Kadelburg, M. M. Marjanovi´ c The series∞/summationtext i=1fi(k) converges uniformly according to the Weierstrass Test, because (i) (∀i)|fi(k)|/lessorequalslantbiand (ii)∞/summationtext i=1biconverges. Therefore, lim k→∞∞/summationtext i=1fi(k)=∞/summationtext i=1lim k→∞fi(k). The desired equality follows from lim k→∞∞/summationtext i=1fi(k) = lim k→∞∞/summationtext i=1k/summationtext j=1aij= lim k→∞k/summationtext j=1∞/summationtext i=1aij=∞/summationtext j=1∞/summationtext i=1aij, ∞/summationtext i=1lim k→∞fi(k)=∞/summationtext i=1∞/summationtext j=1aij. 4. The Toeplitz Limit Theorem The following theorem was proved by Toeplitz (O. Toeplitz, 1881–1940) in Prace mat.-fiz. 22 (1911), 113–119. Theorem 3. The coefficients of the matrix  a 00a01... a 0n... a10a11... a 1n... ........................ ak0ak1... a kn... ........................  are assumed to satisfy the following two conditions: (a)for each fixed n,a kn→0ask→∞, (b)there exists a constant Ksuch that for each fixed kand any n, |ak0|+|ak1|+···+|akn|<K . Then, for every null sequence (z0,z1,...,z n,...), the numbers z/prime k=ak0z0+ak1z1+···+aknzn+··· also form a null sequence. Proof. Denote ϕ(n,k)=ak0z0+ak1z1+···+aknzn. The assumption (a) implies that (a’) lim kϕ(n,k) = 0, for each n. Interchanging two limits 23 On the other hand, (b’) lim nϕ(n,k)=∞/summationtext j=0akjzj=ψ(k) exists, uniformly in k. To prove it, observe first that, by (b), the series∞/summationtext j=0akjconverges absolutely. Fix ε>0. Let j0be such that j>j 0=⇒|zj|<ε K. Then, for j>j 0,w eh a v e |ψ(k)−ϕ(n,k)|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationtext j=n+1akjzj/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant∞/summationtext j=n+1|akj|·max j|zj|<K·ε K=ε, and so lim nϕ(n,k) exists uniformly in k. Now, by Theorem 1, we obtain that lim kz/prime k= lim klim nϕ(n,k) = lim nlim kϕ(n,k)=0. Corollary 3. If, in addition to assumption (a) of Theorem 3, (∀k)ak0+ak1+···+akn→1asn→∞, andzn→z0, then z/prime k→z0ask→∞. 5. Integral as a limit Denote by Π the set of all partitions Pof a segment [ a,b]. For the partition Pgiven by a=x0<x1<···<xn=b, the number /bardblP/bardbl= max {xi−xi−1|i=1,2,...,n } is called the norm of partition P.L e t ∞be an element which does not belong to Π and let /bardbl∞/bardbl=0 . T h es e tΠ∗=Π∪{ ∞ } , together with the metric dΠ(P1,P2)=/braceleftbigg/bardblP1/bardbl+/bardblP2/bardbl,P1/negationslash=P2, 0,P 1=P2, is a metric space which will be called the partition space and will be denoted by (Π∗,dΠ). If, for two partitions P1andP2,/bardblP1/bardbl/lessorequalslant/bardblP2/bardbl, we say that P1is smaller thanP2. Note that a finer partition is also smaller, but the converse is not true. For a partition P,l e t◦ Pbe the partition which has as partitioning points all the points of partition Pand, moreover, all the midpoints of segments of P. Then /bardbl◦ P/bardbl=1 2/bardblP/bardbl. LetP0be the partition with a=x0<x1=b. Let us define a sequence ( Pn) of partitions, taking Pn+1=◦ Pn. Then /bardblPn/bardbl=b−a 2n,a n ds o lim n→∞dΠ(∞,Pn) = lim n→∞b−a 2n=0, wherefrom it follows that ∞is an accumulation point of the set Π in the space Π∗. 24 Z. Kadelburg, M. M. Marjanovi´ c Letf:[a,b]→Rbe a bounded function, s(f,P)a n d S(f,P) be its lower and its upper sum, with respect to the partition P. These sums can be considered as functions s:Π→R,S :Π→R from the space of partitions into R. The limit lim ∞s(f,P)=αexists if the condition (∀ε>0)(∃δ>0)(∀P∈Π) 0<dΠ(P,∞)<δ=⇒|α−s(f,P)|<ε holds, or, equivalently, if the condition (∀ε>0)(∃δ>0)(∀P∈Π)/bardblP/bardbl<δ=⇒|α−s(f,P)|<ε holds. The following proposition associates this limit with the lower (upper) integral of the function f. Proposition 2. Letf:[a,b]→Rbe a bounded function. Then lim ∞s(f,P) (resp. lim ∞S(f,P))e x i s t sa n di se q u a lt o supPs(f,P)(resp. infPS(f,P)). Proof. Let supPs(f,P)=I. Since the function fis bounded, there exists a constant M>0, such that for each x∈[a,b],|f(x)|/lessorequalslantM.L e t ε>0. There exists a partition P1with partitioning points a=x0<x1<···<xk=b, such that I−s(f,P1)<ε 2. Letδ=ε 8Mk.L e t P∈Π be such that /bardblP/bardbl<δ. Denote by Pthe superposition of partitions PandP1. We have now s(f,P)−s(f,P)=/summationtext ∆∈Pm∆v(∆)−/summationtext ∆∈Pm∆v(∆), where m∆=i n f{f(x)|x∈∆}andv(∆) is the length of segment ∆. Since partition Pis finer than partition P, for each ∆ ∈Pthere exists a unique ∆/prime∈Psuch that ∆ ⊂∆/prime.L e t m/prime ∆=m∆/prime,a n ds o /summationtext ∆∈Pm∆v(∆) =/summationtext ∆∈Pm/prime ∆v(∆). Now, s(f,P)−s(f,P)=/summationtext ∆∈P(m∆−m/prime ∆)v(∆). We have m∆−m/prime ∆= 0 when ∆ has no endpoints belonging to partition P1, except points aandb,a n d m∆−m/prime ∆/negationslash= 0 when an endpoint of segment ∆ belongs to the set{x1,...,x k−1}, and this endpoint does not belong to P. The number of such segments is at most 2( k−1). Since |m∆−m/prime ∆|/lessorequalslant2M,w eh a v e s(f,P)−s(f,P)/lessorequalslant2(k−1)·2M·/bardblP/bardbl<2(k−1)·2M·ε 8Mk=ε 2. Interchanging two limits 25 Hence, when /bardblP/bardbl<δ,w eh a v e I−s(f,P)/lessorequalslant(I−s(f,P)+(s(f,P)−s(f,P))<ε 2+ε 2=ε, because Pis also finer that P1and so I−s(f,P)/lessorequalslantI−s(f,P1)<ε 2. This completes the proof that lim ∞s(f,P)=I. The equality lim ∞S(f,P)=I=i n f PS(f,P) can be proved in a similar way. If the function fis integrable, then lim ∞s(f,P) = lim ∞S(f,P)=/integraldisplayb af(x)dx. For a partition P,l e tξPbe a choice function, i.e., let ξP(∆)∈∆. The sum σ(f,P)=/summationtext{f(ξP(∆))v(∆)|∆∈P} is called the integral sum of the function f, corresponding to the partition P. Since s(f,P)/lessorequalslantσ(f,P)/lessorequalslantS(f,P), for each ξP,w eh a v e lim ∞σ(f,P)=/integraldisplayb af(x)dx, for each integrable function f. We shall consider now the question of interchanging a limit and an integral. First, we need an auxiliary statement. Lemma 1. Letf:[a,b]→Randg:[a,b]→Rbe two functions, such that (∀x∈[a,b])|f(x)−g(x)|<k . Then, for each nonempty set S⊂[a,b], the inequalities |inf x∈Sf(x)−inf x∈Sg(x)|<2k, |sup x∈Sf(x)−sup x∈Sg(x)|<2k. hold true. Proof. Letx∈Sbe a point, such that f(x)<m S(f)+k, where mS(f)= inf x∈Sf(x). Then, mS(f)>f(x)−k>g(x)−2k/greaterorequalslantmS(g)−2k, where mS(g)=i n f x∈Sg(x). Similarly, mS(g)>m S(f)−2k, which proves that |mS(f)−mS(g)|<2k. The proof of the other part of the proposition is similar. 26 Z. Kadelburg, M. M. Marjanovi´ c Theorem 4. Let(fn)be a sequence of functions, integrable on the segment [a,b],a n dl e t (fn)converges uniformly to a function f. Then, the function fis integrable and the equality lim n→∞/integraldisplayb afn(x)dx=/integraldisplayb af(x)dx is valid. Proof. Letε>0 be given. Since the sequence ( fn) converges uniformly to the function f, there exists an m∈Nsuch that (∀n∈N)(∀x∈[a,b])n/greaterorequalslantm=⇒|fn(x)−f(x)|<ε 2(b−a). The lower sum of the function fnfor the partition P, s(fn,P)=/summationtext{mn(∆)v(∆)|∆∈P}, is a function from N×Π⊂N∗×Π∗intoR.F o r n/greaterorequalslantmand each P,w eh a v e |s(f.P)−s(fn,P)|/lessorequalslant/summationtext ∆∈P|mn(∆)−m(∆)|v(∆). Applying Lemma 1, we obtain |s(f,P)−s(fn,P)|<2·ε 2(b−a)(b−a)=ε, and so lim n→∞s(fn,P)=s(f,P) exists uniformly in P.F o re a c h n∈N, lim P→∞s(fn,P)=/integraldisplayb afn(x)dx exists, because fnis an integrable function. Applying Theorem 1, it follows that both iterated limits exist and are equal. Hence, we have lim Plim ns(fn,P) = lim Ps(f,P)=/integraldisplay ¯b af(x)dx, lim nlim Ps(fn,P) = lim n/integraldisplayb afn(x)dx, and/integraldisplay ¯b af(x)dx= lim n/integraldisplayb afn(x)dx, Repeating the previous arguments and using upper sums, the equality ¯/integraldisplayb af(x)dx= lim n/integraldisplayb afn(x)dx can also be proved. Interchanging two limits 27 Note that the previous theorem remains valid when the domain of the given functions is an arbitrary set A⊂R, which is Jordan measurable (C. M. E. Jordan, 1838–1922). Namely, such a set Ais bounded and there exists a segment [ a,b] such thatA⊂[a,b]. If the functions fn:A→Rare integrable, so are the functions fn:[a,b]→Rgiven by fn(x)=/braceleftbiggfn(x),x∈A, 0,x ∈[a,b]\A. If the sequence ( fn) converges uniformly to the function f:A→R, then the sequence ( fn) converges uniformly to the function f(x)=/braceleftbiggf(x),x∈A, 0,x ∈[a,b]\A, which is also integrable. According to the previous theorem, we have lim n→∞/integraldisplay Afn(x)dx= lim n→∞/integraldisplayb afn(x)dx=/integraldisplayb af(x)dx=/integraldisplay Af(x)dx. Assuming that the limit function of a convergent sequence ( fn) is integrable, the assumption of uniform convergence can be weakened. First, we formulate and prove an auxiliary statement, which is of interest on its own. Let ( fn) be a sequence of real functions, converging to a function fon a set A⊂R.L e tAbe the closure of the set AinR∗. We shall say that ( fn)converges uniformly around a point x0∈A if there exists an open neighborhood Uof the point x0, such that ( fn)c o n v e r g e s uniformly on the set U∩A. Proposition 3. Letfn,n∈N, be real functions with domain A⊂R.T h e sequence (fn)converges uniformly in x∈Aif and only if it converges uniformly around each point of the set A⊂R∗. Proof. If the sequence ( fn) converges uniformly in x∈A, then it obviously converges uniformly around each point of the set A. Conversely, suppose that ( fn) converges uniformly around each point of A. Then ( fn) is certainly point-wise convergent to a function f.I fAcontains one of the elements −∞,+∞(or both of them), choose a respective neighborhood U−∞ orU+∞(or both of them), such that ( fn) converges uniformly on A∩U−∞, resp. A∩U+∞. For any other point x∈A,l e tUxbe a neighborhood such that ( fn) converges uniformly on Ux∩A. The set A\(U−∞∪U+∞) (if−∞or +∞does not belong to A, then U−∞, resp. U+∞is an empty set) will be compact. Hence, there exist finitely many neighborhoods Ux1,U x2, ..., U xk covering the set A\(U−∞∪U+∞). Adding U−∞and/or U+∞to them, we obtain finitely many subsets of A, such that ( fn) converges uniformly on them. Since each point of the set Abelongs to one of the sets Uξ∩A,ξ∈{x1,x2,...,x k,−∞,+∞}, the sequence ( fn) converges uniformly on A. 28 Z. Kadelburg, M. M. Marjanovi´ c Roughly speaking, more points are there around which the sequence ( fn)i sn o t uniformly convergent, less regular is the convergence of ( fn). On the other hand, Proposition 3 can be formulated as follows: The sequence ( fn) does not converge uniformly in x∈Aif and only if it does not converge uniformly around a point x∈A. Hence, nonuniform convergence can always be spotted locally. The following is the Lebesgue theorem (H. L. Lebesgue, 1875–1941) on bounded convergence for the Riemann integral (G. F. B. Riemann, 1826–1866). Theorem 5. Let(fn)be a sequence of functions, integrable on [a,b], converg- ing to a function f.I f (i)the function fis integrable, (ii) (∃M∈R)(∀n∈N)(∀x∈[a,b])|fn(x)|/lessorequalslantM, (iii)the set A={x∈[a,b]|(fn)does not converge uniformly around x} i sac l o s e ds e to fL e b e s g u em e a s u r e 0, then lim n→∞/integraldisplayb afn(x)dx=/integraldisplayb af(x)dx. Proof. Letε>0 be given. According to (iii), there exists a sequence (∆ i)i∈N of open intervals covering the set A, such that /summationtext{v(∆i)|i∈N}<ε 4M. The set Ais closed and bounded, and so, by Borel-Lebesgue Theorem (F. E. J. E. Borel, 1871–1956), there exist finitely many intervals ∆ i1,...,∆ikwhich cover A, as well. Let D=∆ i1∪···∪ ∆ik. The set Dis Jordan measurable and m(D)/lessorequalslantv(∆i1)+···+v(∆ik)<ε 4M. The sequence ( fn) converges uniformly around each point of the set [ a,b]\D,a n d so, by Proposition 3, it converges uniformly on [ a,b]\D. Hence, there exists an n0∈Nsuch that (∀n∈N)(∀x∈[a,b]\D)n/greaterorequalslantn0=⇒|fn(x)−f(x)|<ε 2(b−a). Now, for n/greaterorequalslantn0,w eh a v e /vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay b af(x)dx−/integraldisplayb afn(x)dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant/integraldisplay D|f(x)−fn(x)|dx+/integraldisplay [a,b]\D|f(x)−fn(x)|dx <2M·ε 4M+ε 2(b−a)·(b−a)=ε, which proves the theorem. Interchanging two limits 29 Example 2. The sequence of functions fn(x)=nx 1+(nx)2,x ∈[0,1], converges to the function 0: [0 ,1]→R, and it does not converge uniformly only around the point 0. The set A={0}is closed and of measure 0, while (∀n∈N)(∀x∈[a,b])/vextendsingle/vextendsingle/vextendsingle/vextendsinglenx 1+(nx)2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant1 2. Hence, we have lim n→∞/integraldisplay1 0nx 1+(nx)2dx=0./triangle At the end, let us remark that we have not included several other cases of interchange of two limits: cases of differentiation and integration of integrals de- pending on parameter, equality f/prime/prime xy=f/prime/prime yx, etc., which can be treated in the same way. Zoran Kadelburg, Faculty of Mathematics, Studentski trg 16/IV, 11000 Beograd, Serbia & Montenegro E-mail :[email protected] Milosav M. Marjanovi´ c, Mathematical Institute, Kneza Mihaila 35/I, 11000 Beograd, Serbia & Montenegro E-mail :[email protected]