The Regulated and Riemann Integrals

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1 Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue to the definite integrl, but they differ in the size of the collection of functions for which they re defined. For exmple, we might use the Riemnn integrl (the ordinry integrl of beginning clculus) nd consider the function {, if x is rtionl; f(x) = 1, otherwise. The Riemnn integrl 1 f(x) dx is, s we will see, undefined. But the Lebesgue integrl, which we will develop, hs no difficulty with f(x) nd indeed 1 1. There re severl properties which we wnt n integrl to stisfy no mtter how we define it. It is worth enumerting them t the beginning. We will need to check them for our different definitions. 11

2 12 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS 1.2 Bsic Properties of n Integrl We will consider the vlue of the integrl of functions in vrious collections. These collections ll hve common domin which, for our purposes, is closed intervl. These collections of functions contin the constnt functions nd re closed under the opertions of ddition nd sclr multipliction. We will cll such collection vector spce of functions. More precisely we will cll non-empty set of rel vlued functions V defined on n intervl [, b] vector spce of functions provided: 1. If f, g V then f + g V. 2. If f V nd r R then rf V. Notice tht this implies tht the constnt function is in V. All of the vector spces we consider will contin the constnt functions. Three simple exmples of vector spces of functions defined on the intervl I = [, b] re the constnt functions, the polynomil functions, nd the continuous functions. An integrl defined on vector spce of functions V is wy to ssign rel number to ech function in V nd subintervl of I. For the function f V nd the subintervl [, b] we denote this vlue by f(x) dx nd cll it the integrl of f from to b. All the integrls we consider will stisfy five bsic properties which we now enumerte. I. Linerity: For ny functions f, g V nd ny rel numbers c 1, c 2, c 1 f(x) + c 2 g(x) dx = c 1 f(x) dx + c 2 g(x) dx. II. Monotonicity: If functions f, g V stisfy f(x) g(x) for ll x then f(x) dx g(x) dx. In prticulr if f(x) for ll x then f(x) dx. III. Additivity: For ny function f V, nd ny, b, c I, c f(x) dx + c b f(x) dx.

3 1.2. BASIC PROPERTIES OF AN INTEGRAL 13 In prticulr we llow, b nd c to occur in ny order on the line nd we note tht two esy consequences of dditivity re nd b f(x) dx. IV. Constnt functions: The integrl of constnt function f(x) = C should give the re of the rectngle under its grph; C dx = C(b ). V. Finite Sets Don t Mtter: If f nd g re functions in V with f(x) = g(x) for ll x except possibly possibly finite set, then g(x) dx. Properties III, IV nd V re not vlid for ll mthemticlly interesting theories of integrtion. Nevertheless, they hold for ll the integrls we will consider so we include them in our list of bsic properties. It is importnt to note tht these re ssumptions, however, nd there re mny mthemticlly interesting theories where they do not hold. There is one dditionl property which we will need. It differs from the erlier ones in tht we cn prove tht it holds whenever the properties bove re stisfied. Proposition (Absolute Vlue). Suppose tht for ll f in some vector spce V of functions defined on the intervl [, b] we hve defined n integrl f(x) dx nd this integrl stisfies properties I nd II bove. Then for ny function f V for which f V f(x) dx f(x) dx, for ll < b in I. Proof. This follows from monotonicity nd linerity. Since f(x) f(x) for ll x we know f(x) dx f(x) dx. Likewise f(x) f(x) so b f(x) dx f(x) dx. But f(x) dx is either equl to f(x) dx or to f(x) dx. In either cse f(x) dx is greter so f(x) dx f(x) dx.

4 14 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS 1.3 Step Functions nd the Regulted Integrl The esiest functions to integrte re step functions which we now define. Definition (Step Function). A function f : [, b] R is clled step function provided there numbers = < x 1 < x 2 < < x n 1 < x n = b such tht f(x) is constnt on ech of the open intervls ]x i 1, x i [. We will sy tht the points = < x 1 < x 2 < < x n 1 < x n = b define prtition for the step function f. Note tht the definition sys tht on the open intervls ]x i 1, x i [ of the prtition f hs constnt vlue, sy c i, but it sys nothing bout the vlues t the endpoints. The vlue of f t the points x i 1 nd x i my or my or my not be equl to c i. Of course when we define the integrl this won t mtter becuse the endpoints form finite set. Since the re under the grph of positive step function is finite union of rectngles, it is pretty obvious wht the integrl should be. The i th of these rectngles hs width (x i x i 1 ) nd height c i so we should sum up the res c i (x i x i 1 ). Of course if some of the c i re negtive then the corresponding c i (x i x i 1 ) re lso negtive, but tht is pproprite since the re between the grph nd the x-xis is below the x-xis on the intervl ]x i 1, x i [. Definition (Integrl of step function). Suppose f(x) is step function with prtition = < x 1 < x 2 < < x n 1 < x n = b nd suppose f(x) = c i for x i 1 < x < x i. Then we define n c i (x i x i 1 ) i=1 Exercise Prove tht the collection of ll step functions on closed intervl [, b] is vector spce which contins the constnt functions. 2. Prove tht if = < x 1 < x 2 < < x n 1 < x n = b is prtition for step function f with vlue c i on ]x i 1, x i [ nd y = < y 1 < y 2 < < y n 1 < y m = b is nother prtition for the sme step function with vlue d j on ]y j 1, y i [ then n c i (x i x i 1 ) = i=1 m d i (y j y j 1 ). j=1

5 1.3. STEP FUNCTIONS AND THE REGULATED INTEGRAL 15 In other words the vlue of the integrl of step function depends only on the function, not on the choice of prtition. Hint: the union of the sets of points defining the two prtitions defines third prtition nd the integrl using this prtition is equl to the integrl using ech of the prtitions. 3. Prove tht the integrl of step functions s given in Definition stisfies properties I-V of 1.2. We mde the obvious definition for the integrl of step function, but in fct, we hd bsolutely no choice in the mtter if we wnt the integrl to stisfy properties I-V bove. Theorem The integrl s given in Definition is the unique rel vlued function defined on step functions which stisfies properties I-V of 1.2. Proof. Suppose tht there is nother integrl defined on step functions nd stisfying I-V. We will denote this lternte integrl s b f(x) dx. Wht we must show is tht for every step function f(x), b f(x) dx. Suppose tht f hs prtition = < x 1 < x 2 < < x n 1 < x n = b nd stisfies f(x) = c i for x i 1 < x < x i. Then from the dditivity property b n i=1 xi x i 1 f(x) dx. (1.3.1) But on the intervl [x i 1, x i ] the function f(x) is equl to the constnt function with vlue c i except t the endpoints. Since functions which re equl except t finite set of points hve the sme integrl, the integrl of f is the sme s the integrl of c i on [x i 1, x i ]. Combining this with the constnt function property we get xi x i 1 xi x i 1 c i dx = c i (x i x i 1 ).

6 16 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS If we plug this vlue into eqution (1.3.1) we obtin b n c i (x i x i 1 ) = i=1 f(x) dx. Recll the definition of uniformly converging sequence of functions. Definition (Uniform Convergence). A sequence of functions {f m } is sid to converge uniformly on [, b] to function f if for every ε > there is n M (independent of x) such tht for ll x [, b] Contrst this with the following. f(x) f m (x) < ε whenever m M. Definition (Pointwise Convergence). A sequence of functions {f m } is sid to converge pointwise on [, b] to function f if for ech ε > nd ech x [, b] there is n M x (depending on x) such tht f(x) f m (x) < ε whenever m M x. Definition (Regulted Function). A function f : [, b] R is clled regulted provided there is sequence {f m } of step functions which converges uniformly to f. Exercise Prove tht the collection of ll regulted functions on closed intervl I is vector spce which contins the constnt functions. 2. Give n exmple of sequence of step functions which converge uniformly to f(x) = x on [, 1]. Give n exmple of sequence of step functions which converge pointwise to on [, 1], but which do not converge uniformly. Every regulted function cn be uniformly pproximted s closely s we wish by step function. Since we know how to integrte step functions it is nturl to tke sequence of better nd better step function pproximtions to regulted function f(x) nd define the integrl of f to be the limit of the integrls of the pproximting step functions. For this to work we need to know tht the limit exists nd tht it does not depend on the choice of pproximting step functions.

7 1.3. STEP FUNCTIONS AND THE REGULATED INTEGRAL 17 Theorem Suppose {f m } is sequence of step functions on [, b] converging uniformly to regulted function f. Then the sequence of numbers { f m(x) dx} converges. Moreover if {g m } is nother sequence of step functions which lso converges uniformly to f then lim m f m (x) dx = lim m g m (x) dx. Proof. Let z m = f m(x) dx. We will show tht the sequence {z m } is Cuchy sequence nd hence hs limit. To show this sequence is Cuchy we must show tht for ny ε > there is n M such tht z p z q ε whenever p, q M. If we re given ε >, since {f m } is sequence of step functions on [, b] converging uniformly to f, there is n M such tht for ll x Hence whenever, p, q M f(x) f m (x) < ε 2(b ) whenever m M. ε f p (x) f q (x) < f p (x) f(x) + f(x) f q (x) < 2 2(b ) = ε b. (1.3.2) Therefore, whenever p, q M z p z q = f p (x) f q (x) dx f p (x) f q (x) dx ε b dx = ε, where the first inequlity comes from the bsolute vlue property of Proposition nd the second follows from the monotonicity property nd eqution (1.3.2). This shows tht the sequence {z m } is Cuchy nd hence converges. Now suppose tht {g m } is nother sequence of step functions which lso converges uniformly to f then for ny ε > there is n M such tht for ll x whenever m M. It follows tht f(x) f m (x) < ε nd f(x) g m (x) < ε f m (x) g m (x) f m (x) f(x) + f(x) g m (x) < 2ε. Hence, using the bsolute vlue nd monotonicity properties, we see f m (x) g m (x) dx f m (x) g m (x) dx 2ε dx = 2ε(b ),

8 18 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS for ll m M. Since ε is rbitrrily smll we my conclude tht lim m This implies f m (x) dx lim m g m (x) dx = lim f m (x) = lim m m f m (x) g m (x) dx =. g m (x) dx. This result enbles us to define the regulted integrl. Definition (The Regulted Integrl). If f is regulted function on [, b] we define the regulted integrl by lim n f n (x) dx where {f n } is ny sequence of step functions converging uniformly to f. We next need to see tht the regulted functions form lrge clss including ll continuous functions. Theorem (Continuous functions on [, b] re regulted). Every continuous function f : [, b] R is regulted function. Proof. A continuous function f(x) defined on closed intervl [, b] is uniformly continuous. Tht is, given ε > there is corresponding δ > such tht f(x) f(y) < ε whenever x y < δ. Let ε n = 1/2 n nd let δ n be the corresponding δ gurnteed by uniform continuity. Fix vlue of n nd choose prtition = < x 1 < x 2 < < x m = b with x i x i 1 < δ n. For exmple, we could choose m so lrge tht if we define x = (b )/m then x < δ n nd then we could define x i to be + i x. Next we define step function f n by f n (x) = f(x i ) for ll x [x i 1, x i [. Tht is, on ech hlf open intervl [x 1 i, x i [ we define f n to be the constnt function whose vlue is the vlue of f t the left endpoint of the intervl. The vlue of f n (b) is defined to be f(b).

9 1.4. THE FUNDAMENTAL THEOREM OF CALCULUS 19 Clerly f n (x) is step function with the given prtition. We must estimte its distnce from f. Let x be n rbitrry point of [, b]. It must lie in one of the open intervls of the prtition or be n endpoint of one of them; sy x [x i 1, x i [. Then since f n (x) = f n (x i 1 ) = f(x i 1 ) we my conclude f(x) f n (x) f(x) f(x i 1 ) < ε n becuse of the uniform continuity of f nd the fct tht x x i 1 < δ n. Thus we hve constructed step function f n with the property tht for ll x [, b] f(x) f n (x) < ε n. So the sequence {f n } converges uniformly to f nd f is regulted function. Exercise Give n exmple of continuous function on the open intervl ], 1[ which is not regulted, i.e. which cnnot be uniformly pproximted by step functions. 2. Prove tht the regulted integrl, s given in (1.3.1), stisfies properties I-V of Prove tht f is regulted function on I = [, b] if nd only if both the limits lim f(x) nd lim x c+ f(x) x c exist for every c ], b[. (See section VII.6 of Dieudonné [1]). 1.4 The Fundmentl Theorem of Clculus The most importnt theorem of elementry clculus sserts tht if f is continuous function on [, b] then its integrl f(x) dx cn be evluted by finding n ntiderivtive. More precisely, if F (x) is n nti-derivtive of f then F (b) F (). We now cn present rigorous proof of this result. We will ctully formulte the result slightly differently nd show tht the result bove follows esily from tht formultion.

10 2 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS Theorem ( ). If f is continuous function nd we define F (x) = x f(t) dt then F is differentible function nd F (x) = f(x). Proof. By definition so we need to show tht F F ( + h) F ( ) ( ) = lim ; h h F ( + h) F ( ) lim h h = f( ). or equivlently lim F ( + h) F ( ) f( ) =. h h To do this we note tht F ( + h) F ( ) h f( ) = = = x +h f(t) dt f( ) h x +h f(t) dt f( )h h +h (f(t) f( )) dt h (1.4.1) Monotonicity tells us tht when h is positive nd if h is negtive x +h x +h (f(t) f( )) dt x +h (f(t) f( )) dt In either cse we see x +h x f(t) f( ) dt = f(t) f( ) dt x +h +h (f(t) f( )) dt x +h f(t) f( ) dt f(t) f( ) dt.

11 1.4. THE FUNDAMENTAL THEOREM OF CALCULUS 21 Combining this with inequlity (1.4.1) bove we obtin F ( + h) F ( ) +h x f( ) f(t) f( ) dt. (1.4.2) h h But the continuity of f implies tht given nd ny ε > there is δ > such tht whenever t < δ we hve f(t) f( ) < ε. Thus if h < δ then f(t) f( ) < ε for ll t between nd + h. It follows tht +h f(t) f( ) dt < ε h nd hence tht +h f(t) f( ) dt < ε. h Putting this together with eqution (1.4.2) bove we hve tht F ( + h) F ( ) f( ) < ε h whenever h < δ which ws exctly wht we needed to show. Corollry Fundmentl Theorem of Clculus. If f is continuous function on [, b] nd F is ny nti-derivtive of f then F (b) F () Proof. Define the function G(x) = x f(t) dt. By Theorem (1.4.1) the derivtive of G(x) is f(x) which is lso the derivtive of F. Hence F nd G differ by constnt, sy F (x) = G(x) + C (see Corollry (.5.4)). Then F (b) F () = (G(b) + C) (G() + C) = G(b) G() = = f(x) dx f(x) dx. f(x) dx Exercise Prove tht if f : [, b] R is regulted function nd F : [, b] R is defined to by F (x) = x f(t) dt then F is continuous.

12 22 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS 1.5 The Riemnn Integrl We cn obtin lrger clss of functions for which good integrl cn be defined by using different method of compring with step functions. Suppose tht f(x) is bounded function on the intervl I = [, b] nd tht it is n element of vector spce of functions which contins the step functions nd for which there is n integrl defined stisfying properties I-V of 1.2. If u(x) is step function stisfying f(x) u(x) for ll x I then monotonicity implies tht if we cn define f(x) dx it must stisfy f(x) dx u(x) dx. This is true for every step function u stisfying f(x) u(x) for ll x I. Let U(f) denote the set of ll step functions with this property. Then if we cn define f(x) dx in wy tht stisfies monotonicity it must lso stisfy f(x) dx u U(f) u(x) dx. (1.5.1) The imum exists becuse ll the step functions in U(f) re bounded below by lower bound for the function f. Similrly we define L(f) to be the set of ll step functions v(x) such tht v(x) f(x) for ll x I. Agin if we cn define f(x) dx in such wy tht it stisfies monotonicity it must lso stisfy sup v L(f) v(x) dx f(x) dx. (1.5.2) The supremum exists becuse ll the step functions in U(f) re bounded bove by n upper bound for the function f. Putting inequlities (1.5.1) nd (1.5.2) together, we see if V is ny vector spce of bounded functions which contins the step functions nd we mnge to define the integrl of functions in V in wy tht stisfies monotonicity then this integrl must stisfy sup v L(f) v(x) dx f(x) dx u U(f) u(x) dx (1.5.3) for every f V. Even if we cn t define n integrl for f, however, we still hve the inequlities of the ends. Proposition Let f be ny bounded function on the intervl I = [.b]. Let U(f) denote the set of ll step functions u(x) on I such tht f(x) u(x) for ll x

13 1.5. THE RIEMANN INTEGRAL 23 nd let L(f) denote the set of ll step functions v(x) such tht v(x) f(x) for ll x. Then sup v L(f) v(x) dx u U(f) u(x) dx. Proof. If v L(f) nd u U(f) then v(x) f(x) u(x) for ll x I so monotonicity implies tht v(x) dx u(x) dx. Hence if { V = } v(x) dx nd U = v L(f) { } u(x) dx u U(f) then every number in the set V is less thn or equl to every number in the set U. Thus sup V U s climed It is not difficult to see tht sometimes the two ends of this inequlity re not equl (see Exercise (1.5.4) below), but if it should hppen tht sup v L(f) v(x) dx = u U(f) u(x) dx. then we hve only one choice for f(x) dx; it must be this common vlue. This motivtes the definition of the next vector spce of functions we cn integrte. Definition The Riemnn Integrl. Suppose f is bounded function on the intervl I = [.b]. Let U(f) denote the set of ll step functions u(x) on I such tht f(x) u(x) for ll x nd let L(f) denote the set of ll step functions v(x) such tht v(x) f(x) for ll x. The function f is sid to be Riemnn integrble provided sup v L(f) v(x) dx = u U(f) u(x) dx. In this cse its Riemnn integrl f(x) dx is defined to be this common vlue. Theorem A bounded function f : [, b] R is Riemnn integrble if nd only if, for every ε > there re step functions v nd u such tht v (x) f(x) u (x) for ll x [, b] nd u (x) dx v (x) dx ε.

14 24 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS Proof. Suppose the functions v L(f) nd u U(f) hve integrls within ε of ech other. Then This implies v (x) dx sup v L(f) u U(f) v(x) dx u U(f) u(x) dx sup v L(f) u(x) dx v(x) dx ε. u (x) dx. Since this is true for ll ε > we conclude tht f is Riemnn integrble. Conversely if f is Riemnn integrble then by Proposition.1.4 there exists step function u U(f) such tht u (x) dx u (x) dx u U(f) Similrly there exists step function v L(f) such tht Hence f(x) dx u (x) dx nd u nd v re the desired functions. Exercise v (x) dx < ε/2. v (x) dx < ε/2 + ε/2 = ε. u(x) dx < ε/2. 1. At the beginning of these notes we mentioned the function f : [, 1] R which hs the vlue f(x) = if x is rtionl nd 1 otherwise. Prove tht for this function sup v L(f) 1 v(x) dx = nd Hence f is not Riemnn integrble. u U(f) 1 u(x) dx = 1. There re severl fcts bout the reltion with the regulted integrl we must estblish. Every regulted function is Riemnn integrble, but there re Riemnn integrble functions which hve no regulted integrl. Whenever function hs both types of integrl the vlues gree. We strt by giving n exmple of function which is Riemnn integrble, but not regulted.

15 1.5. THE RIEMANN INTEGRAL 25 Exmple Define the function f : [, 1] R by { 1, if x = 1 n f(x) = for n Z+ ;, otherwise. Then f(x) is Riemnn integrble nd 1 but it is not regulted. Proof. We define step function u m (x) by { 1, if x 1 m u m (x) = ; f(x), otherwise. A prtition for this step function is given by = < x 1 = 1 m < x 2 = 1 m 1 < < x m 1 = 1 2 < x m = 1. Note tht u m (x) f(x). Also 1 u m(x) dx = 1/m. This is becuse it is constnt nd equl to 1 on the intervl [, 1/m] nd except for finite number of points it is constnt nd equl to on the intervl [1/m, 1]. Hence u U(f) 1 u(x) dx m Z + 1 u m (x) dx = m Z +{ 1 m } =. Also the constnt function is f(x) nd its integrl is, so sup v L(f) 1 v(x) dx. Putting together the lst two inequlities with Proposition (1.5.1) we obtin sup v L(f) 1 v(x) dx 1 u U(f) u(x) dx. So ll of these inequlities re equlities nd by definition, f is Riemnn integrble with integrl. To see tht f is not regulted suppose tht g is n pproximting step function with prtition = < x 1 < < x m = 1 nd stisfying f(x) g(x) ε for some ε >. Then g is constnt, sy with vlue c 1 on the open intervl (, x 1 ). Now there re points 1, 2 (, x 1 ) with f( 1 ) = nd f( 2 ) = 1. Then c 1 = c 1 = g( 1 ) f( 1 ) ε nd 1 c 1 = f( 2 ) g( 2 ) ε. But c c 1 c c 1 = 1 so t lest one of c 1 nd 1 c 1 must be 1/2. This implies tht ε 1/2. Tht is, f cnnot be uniformly pproximted by ny step function to within ε if ε < 1/2. So f is not regulted.

16 26 CHAPTER 1. THE REGULATED AND RIEMANN INTEGRALS Theorem (Regulted functions re Riemnn integrble). Every regulted function f is Riemnn integrble nd the regulted integrl of f is equl to its Riemnn integrl. Proof. If f is regulted function on the intervl I = [, b] then, for ny ε >, it cn be uniformly pproximted within ε by step function. In prticulr, if ε n = 1/2 n there is step function g n (x) such tht f(x) g n (x) < ε n for ll x I. The regulted integrl f(x) dx ws defined to be lim g n(x) dx. We define two other pproximting sequences of step functions for f. Let u n (x) = g n (x) + 1/2 n nd v n (x) = g n (x) 1/2 n. Then u n (x) f(x) for ll x I becuse u n (x) f(x) = 1/2 n + g n (x) f(x) since g n (x) f(x) < 1/2 n. Similrly v n (x) f(x) for ll x I becuse f(x) v n (x) = 1/2 n + f(x) g n (x) since f(x) g n (x) < 1/2 n. Since u n (x) v n (x) = g n (x) + 1/2 n (g n (x) 1/2 n ) = 1/2 n 1, u n (x) dx v n (x) dx = u n (x) v n (x) dx = 1 b dx = 2n 1 2. n 1 Hence we my pply Theorem(1.5.3) to conclude tht f is Riemnn integrble. Also lim n lim n Since for ll n we conclude tht g n (x) dx = lim g n (x) dx = lim n n v n (x) dx lim n v n (x) + 1 dx = lim 2n n u n (x) 1 dx = lim 2n n f(x) dx g n (x) dx = f(x) dx. u n (x) dx Tht is, the regulted integrl equls the Riemnn integrl. v n (x) dx, nd u n (x) dx. Theorem The set R of Riemnn integrble functions on n intervl I = [, b] is vector spce contining the vector spce of regulted functions. Proof. We hve lredy shown tht every regulted function is Riemnn integrble. Hence we need only show tht whenever f, g R nd r R we lso hve (f + g) R nd rf R. We will do only the sum nd leve the product s n exercise.

17 1.5. THE RIEMANN INTEGRAL 27 Suppose ε > is given. Since f is Riemnn integrble there re step functions u f nd v f such tht v f (x) f(x) u f (x) for x I (i.e. u f U(f) nd v f L(f)) nd with the property tht u f (x) dx v f (x) dx < ε. Similrly there re u g U(g) nd v g L(g) with the property tht This implies tht u g (x) dx v g (x) dx < ε. (u f + u g )(x) dx (v f + v g )(x) dx < 2ε. Since (u f + u g ) U(f + g) nd (v f + v g ) L(f + g) we my conclude tht u U(f+g) u(x) dx As ε > is rbitrry we conclude tht nd hence (f + g) R. Exercise u U(f+g) u(x) dx = sup v L(f+g) sup v L(f+g) v(x) dx < 2ε. v(x) dx 1. Prove tht if f nd g re Riemnn integrble functions on n intervl [, b] then so is fg. In prticulr if r R then rf is Riemnn integrble function on [, b].

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