Anlysis III Ben Green Mthemticl Institute, Oxford E-mil ddress: ben.green@mths.ox.c.uk
2000 Mthemtics Subject Clssifiction. Primry
Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1. Step functions 3 1.2. I of step function 4 1.3. Definition of the integrl 5 1.4. Very bsic theorems bout the integrl 5 1.5. Not ll functions re integrble 7 1.6. Improper integrls brief discussion 8 1.7. An exmple 9 Chpter 2. Bsic theorems bout the integrl 11 2.1. Continuous functions re integrble 11 2.2. Monotone functions re integrble 12 Chpter 3. Riemnn sums 15 Chpter 4. Integrtion nd differentition 19 4.1. First fundmentl theorem of clculus 19 4.2. Second fundmentl theorem of clculus 20 4.3. Integrtion by prts 21 4.4. Substitution 22 Chpter 5. Limits nd the integrl 23 5.1. Interchnging the order of limits nd integrtion 23 5.2. Interchnging the order of limits nd differentition 24 v
Prefce The objective of this course is to present rigorous theory of wht it mens to integrte function f : [, b] R. For which functions f cn we do this, nd wht properties does the integrl hve? Cn we give rigorous nd generl versions of fcts you lerned in school, such s integrtion by prts, integrtion by substitution, nd the fct tht the integrl of f is just f? We will present the theory of the Riemnn integrl, lthough the wy we will develop it is much closer to wht is known s the Drboux integrl. The end product is the sme (the Riemnn integrl nd the Drboux integrl re equivlent) but the Drboux development tends to be esier to understnd nd hndle. This is not the only wy to define the integrl. In fct, it hs certin deficiencies when it comes to the interply between integrtion nd limits, for exmple. To hndle these situtions one needs the Lebesgue integrl, which is discussed in future course. Students should be wre tht every time we write integrble we men Riemnn integrble. For exmple, lter on we will exhibit non-integrble function, but it turns out tht this function is integrble in the sense of Lebesgue. 1
CHAPTER 1 Step functions nd the Riemnn integrl 1.1. Step functions We re going to define the (Riemnn) integrl of function by pproximting it using simple functions clled step functions. Definition 1.1. Let [, b] be n intervl. A function φ : [, b] R is clled step function if there is finite sequence = x 0 x 1... x n = b such tht φ is constnt on ech open intervl (x i 1, x i ). Remrks. We do not cre bout the vlues of f t the endpoints x 0, x 1,..., x n. We cll sequence = x 0 x 1... x n = b prtition P, nd we sy tht φ is step function dpted to P. Definition 1.2. A prtition P given by = x 0... x n of P if every x i is n x j for some j. Lemm 1.1. We hve the following fcts bout prtitions: b is refinement (i) Suppose tht φ is step function dpted to P, nd if P is refinement of P, then φ is lso step function dpted to P. (ii) If P 1, P 2 re two prtitions then there is common refinement of both of them. (iii) If φ 1, φ 2 re step functions then so re mx(φ 1, φ 2 ), φ 1 + φ 2 nd λφ i for ny sclr λ. Proof. All completely strightforwrd; for (iii), suppose tht φ 1 is dpted to P 1 nd tht φ 2 is dpted to P 2, nd pss to common refinement of P 1, P 2. If X R is set, the indictor function of X is the function 1 X tking the vlue 1 for x X nd 0 elsewhere. Lemm 1.2. A function φ : [, b] R is step function if nd only if it is finite liner combintion of indictor functions of intervls (open nd closed). Proof. Suppose first tht φ is step function dpted to some prtition P, = x 0 x 1... x n = b. Then φ cn be written s weighted sum of the functions 1 (xi 1,x i) (ech n indictor function of n open intervl) nd the 3
4 1. STEP FUNCTIONS AND THE RIEMANN INTEGRAL functions 1 {xi} (ech n indictor function of closed intervl contining single point). Conversely, the indictor function of ny intervl is step function, nd hence so is ny finite liner combintion of these by Lemm 1.1. In prticulr, the step functions on [, b] form vector spce, which we occsionlly denote by L step [, b]. 1.2. I of step function It is obvious wht the integrl of step function should be. Definition 1.3. Let φ be step function dpted to some prtition P, nd suppose tht φ(x) = c i on the intervl (x i 1, x i ). Then we define n I(φ) = c i (x i x i 1 ). i=1 We cll this I(φ) rther thn φ, becuse we re going to define f for clss of functions f much more generl thn step functions. It will then be theorem tht I(φ) = φ, rther thn simply definition. Actully, there is smll subtlety to the definition. Our nottion suggests tht I(φ) depends only on φ, but its definition depended lso on the prtition P. In fct, it does not mtter which prtition one chooses. If one is pedntic nd writes n I(φ; P) = c i (x i x i 1 ) then one my esily check tht i=1 I(φ; P) = I(φ, P ) for ny refinement P of P. Now if φ is step function dpted to both P 1 nd P 2 then one my locte common refinement P nd conclude tht I(φ, P 1 ) = I(φ; P ) = I(φ, P 2 ). Lemm 1.3. The mp I : L step [, b] R is liner: I(λφ 1 + µφ 2 ) = λi(φ 1 ) + µi(φ 2 ). Proof. This is obvious on pssing to common refinement of the prtitions P 1 nd P 2 to which φ 1, φ 2 re dpted.
1.4. VERY BASIC THEOREMS ABOUT THE INTEGRAL 5 1.3. Definition of the integrl Let f : [, b] R be bounded function. We sy tht step function φ is minornt for f if f φ pointwise. We sy tht step function φ + is mjornt for f if f φ + pointwise. Definition 1.4. A function f is integrble if (1.1) sup I(φ ) = inf I(φ + ), φ φ + where the sup is over ll minornts φ f, nd the inf is over ll mjornts φ + f. These minornts nd mjornts re lwys ssumed to be step functions. We define the integrl f to be the common vlue of the two quntities in (1.1). We note tht the sup nd inf exist for ny bounded function f. Indeed if f M then the constnt function φ = M is minornt for f (so there is t lest one) nd evidently I(φ ) (b )M for ll minornts. mjornts. We note moreover tht, for ny function f, (1.2) sup I(φ ) inf I(φ + ). φ φ + A similr proof pplies to To see this, let φ f φ + be mjornts, dpted to prtitions P nd P + respectively. By pssing to common refinement we my ssume tht P = P + = P. Then it is cler from the definition of I(.) tht I(φ ) I(φ + ). Since φ, φ + were rbitrry, (1.2) follows. It follows from (1.2) tht if f is integrble then (1.3) I(φ ) f I(φ + ) whenever φ f φ + re minornts/mjornt. Remrk. If function f is only defined on n open intervl (, b), then we sy tht it is integrble if n rbitrry extension of it to [, b] is. It follows immeditely from the definition of step function (which does not cre bout the endpoints) tht it does not mtter which extension we choose. 1.4. Very bsic theorems bout the integrl In this section we ssemble some bsic fcts bout the integrl. Their proofs re ll essentilly routine, but there re some lbour-sving tricks to be exploited. Proposition 1.1. Suppose tht f is integrble on [, b]. Then, for ny c with < c < b, f is Riemnn integrble on [, c] nd on [c, b]. Moreover f = c f + c f.
6 1. STEP FUNCTIONS AND THE RIEMANN INTEGRAL Proof. Let M be bound for f, thus f(x) M everywhere. In this proof it is convenient to ssume tht (i) ll prtitions of [, b] include the point c nd tht (ii) ll minornts tke the vlue M t c, nd ll mjornts the vlue M. By refining prtitions if necessry, this mkes no difference to ny computtions involving I(φ ), I(φ + ). Now observe tht minornt φ of f on [, b] is precisely the sme thing s minornt φ (1) of f on [, c] juxtposed with minornt φ (2) of f on [c, b], nd tht I(φ ) = I(φ (1) ) + I(φ (2) ). A similr comment pplies to mjornts. Thus, since f is integrble, (1.4) sup φ Since sup φ (i) I(φ ) = sup I(φ (1) φ (1) equlity holds: sup φ (i) ) + sup φ (2) I(φ (2) ) = inf φ (1) + I(φ (1) + ) + inf φ (2) + I(φ (2) + ) = inf φ + I(φ + ). I(φ (i) ) inf (i) φ I(φ (i) + ) for i = 1, 2, we re forced to conclude tht + I(φ (i) ) = inf φ (i) + I(φ (i) + ) for i = 1, 2. Thus f is indeed integrble on [, c] nd on [c, b], nd it follows from (1.4) tht f = c f + c f. Corollry 1.1. Suppose tht f : [, b] R is integrble, nd tht [c, d] [, b]. Then f is integrble on [c, d]. Proof. On the exmple sheet. Proposition 1.2. If f, g re integrble on [, b] then so is λf + µg for ny λ, µ R. Moreover (λf + µg) = λ f + µ g. Tht is, the integrble functions on [, b] form vector spce nd the integrl is liner functionl (liner mp to R) on it. Proof. Suppose tht λ > 0. If φ f φ + re minornt/mjornt for f, then λφ λf λφ + re minornt nd mjornt for λf. Moreover I(λφ ) I(λφ + ) = λ(i(φ + ) I(φ )) cn be mde rbitrrily smll. Thus λf is integrble. Moreover inf φ+ I(λφ + ) = λ inf φ+ I(φ + ), inf φ I(λφ ) = λ sup φ I(φ ), nd so (λf) = λ f. If λ < 0 then we cn proceed in very similr mnner. We leve this to the reder. Now suppose tht φ f φ + nd ψ g ψ + re minornt/mjornts for f, g. Then φ + ψ f + g φ + + ψ + re minornt/mjornt for f + g (note these re steps functions) nd by Lemm 1.3 (linerity of I) inf I(φ + + ψ + ) = inf I(φ + ) + inf I(ψ + ) = φ +,ψ + φ + ψ + f + g,
1.5. NOT ALL FUNCTIONS ARE INTEGRABLE 7 whilst sup I(φ + ψ ) = sup I(φ ) + sup I(ψ ) = φ,ψ ψ φ It follows tht indeed f + g is integrble nd (f + g) = f + g. Tht (λf + µg) = λ f + µ g follows immeditely by combining these two fcts. f + g. Proposition 1.3. Suppose tht f nd g re integrble on [, b]. Then mx(f, g) nd min(f, g) re both Riemnn integrble, s is f. Proof. We hve mx(f, g) = g + mx(f g, 0), min(h, 0) = mx( h, 0) nd h = mx(h, 0) min(h, 0). Using these reltions nd Proposition 1.2, it is enough to prove tht if f is integrble on (, b), then so is mx(f, 0). Now the function x mx(x, 0) is order-preserving (if x y then mx(x, 0) mx(y, 0)) nd non-expnding (we hve mx(x, 0) mx(y, 0) x y, s cn be estblished by n esy cse-check, ccording to the signs of x, y). It follows tht if φ f φ + re minornt nd mjornt for f then mx(φ, 0) mx(f, 0) mx(φ +, 0) re minornt nd mjornt for mx(f, 0) (it is obvious tht they re both step functions) nd, since f is integrble, I(mx(φ +, 0)) I(mx(φ, 0)) I(φ + ) I(φ ) cn be mde rbitrrily smll. Proposition 1.4. If f, g re both integrble on [, b] nd if f g pointwise then f g. In prticulr, f f. Proof. The first prt is immeditely obvious from the fct tht ny minornt for f is lso minornt for g, nd so g = sup I(ψ ) sup I(φ ) = ψ φ where ψ rnges over minornts of g, nd φ over minornts of f. For the second prt, pply the first prt to f nd f, nd lso to f nd f, obtining ± f b f. f, 1.5. Not ll functions re integrble Exmple 1.1. There is bounded function f : [0, 1] R which is not (Riemnn) integrble. Proof. Consider the function f such tht f(x) = 1 if x Q nd 0 if x / Q. Since ny open intervl contins both rtionl points nd points which re not
8 1. STEP FUNCTIONS AND THE RIEMANN INTEGRAL rtionl, ny step function mjorising f must stisfy φ + (x) 1 except possibly t the finitely mny endpoints x i, nd hence I(φ + ) 1. Similrly ny minornt φ stisfies φ (x) 0 except t finitely mny points, nd so I(φ ) 0. This function f cnnot possibly be integrble. Remrk. Students will see in next yer s course on Lebesgue integrtion tht the Lebesgue integrl of this function does exist (nd equls 0). 1.6. Improper integrls brief discussion If one ttempts to ssign mening to the integrl of n unbounded function, or to the integrl of function over n unbounded domin, then one is trying to understnd n improper integrl. We will not ttempt to systemticlly define wht n improper integrl is, but few exmples should mke it cler wht is ment in ny given sitution. In discussing these exmples we ssume tht the integrls of well-known functions re wht you think they re this will be justified rigorously lter on, when we prove the fundmentl theorem of clculus. Exmple 1.2. Consider the function f(x) = log x. This is continuous on (0, 1), but it is not integrble there since it is not bounded (it tends to s x ). We hve 1 log xdx = [x log x x] 1 ε = 1 ε log ε ε. Therefore This will often be written s ε lim ε 0 + 1 0 1 ε log xdx = 1. log xdx = 1, but strictly speking this is not n integrl s discussed in this course. Exmple 1.3. Consider the function f(x) = 1/x 2. We hve Therefore This is invribly written K 1 1 x 2 dx = [ 1 ] K x 1 = 1 1 K. K 1 lim dx = 1. K 1 x2 1 1 dx = 1. x2
1.7. AN EXAMPLE 9 Exmple 1.4. Define f(x) to be log x if 0 < x 1, nd f(x) = 1 x 2 for x 1. Then it mkes sense to write 0 f(x)dx = 0, by which we men K lim f(x)dx = 0. K,ε 0 ε This is combintion of the preceding two exmples. Exmple 1.5. Define f(x) to be 1/x for 0 < x 1, nd f(0) = 0. Then f is unbounded s x 0. Excising the problemtic region, one cn look t I ε,ε := nd one esily computes tht 1 ε f(x)dx + ε 1 f(x)dx, I ε,ε = log ε ε. This does not necessrily tend to limit s ε, ε 0 (for exmple, if ε = ε 2 it does not tend to limit). One will often her the term Cuchy principl vlue (PV) for the limit lim ε 0 I ε,ε, which in this cse equls 0. We won t discuss principl vlues ny further in this course, nd in this cse it is not pproprite to write 1 x dx = 0; one could possibly write PV 1 1 1 xdx = 0. 1 1 Exmple 1.6. Similrly to the lst exmple, one should not write sin xdx = K 0, even though lim K sin xdx = 0 (becuse sin is n odd function). In this K K cse, lim K,K sin xdx does not exist. One could mybe write K PV but I would not be tempted to do so. sin xdx = 0, 1.7. An exmple Although we didn t mention it explicitly, it is obvious from the definition tht every step function φ on [, b] is integrble nd tht I(φ) = φ (tke the minornt nd mjornt to be φ itself). In this section we puse to nlyse slightly less trivil exmple from first principles. Proof. Exmple 1.7. The function f(x) = x is integrble on [0, 1], nd 1 0 f(x)dx = 1 2. We define explicit minornts nd mjornts. Let n be n integer to be specified lter, nd set φ (x) = i n for i i+1 n x n, i = 0, 1,..., n 1. Set φ + (x) = j j 1 n for n x j n, j = 1,..., n. Then φ f φ + pointwise, so φ, φ +
10 1. STEP FUNCTIONS AND THE RIEMANN INTEGRAL (being step functions) re minornt/mjornt for f. We hve nd I(φ ) = I(φ + ) = n 1 i=0 n j=1 i n 1 n = 1 2 (1 1 n ) j n 1 n = 1 2 (1 + 1 n ). It is cler tht, by tking n sufficiently lrge, we cn mke both of these s close to 1 2 s we like.
CHAPTER 2 Bsic theorems bout the integrl In this section we show tht the integrble functions re in rich supply. 2.1. Continuous functions re integrble Let P be prtition of [, b], = x 0 < x 1 < < x n = b. The mesh of P is defined to be mx i (x i x i 1 ). Thus if mesh(p) δ then every intervl in the prtition P hs length t most δ. To give n exmple, if [, b] = [0, 1] nd if x i = i N then the mesh is 1/N. Proof. Theorem 2.1. A continuous function f : [, b] R is integrble. Since f is continuous on closed nd bounded intervl, f is lso bounded. Suppose tht M is bound for f, so tht f(x) M for ll x [, b]. We will lso use the fct tht continuous function f is uniformly continuous. Let ε > 0, nd let δ be so smll tht f(x) f(y) ε whenever x y δ. Let P be prtition with mesh < δ. Let φ + be the step function whose vlue on (x i 1, x i ) is sup x [xi 1,x i] f(x) nd which tkes the vlue M t the points x i, nd let φ be the step function whose vlue on (x i 1, x i ) is inf x [xi 1,x i] f(x) nd which tkes the vlue M t the points x i. By construction, φ + is mjornt for f nd φ is minornt. Since continuous function on closed intervl ttins its bounds, there re ξ, ξ + [x i 1, x i ] such tht sup x [xi 1,x i] f(x) = f(ξ + ) nd inf x [xi 1,x i] f(x) = f(ξ ). For x (x i 1, x i ) we hve φ + (x) φ (x) f(ξ + ) f(ξ ) ε. Therefore φ + (x) φ (x) ε for ll except finitely mny points in [, b], nmely the points x i. It follows tht I(φ + ) I(φ ) ε(b ). Since ε ws rbitrry, this concludes the proof. We cn slightly strengthen this result, not insisting on continuity t the endpoints. This result would pply, for exmple, to the function f(x) = sin(1/x) on (0, 1). Theorem 2.2. A bounded continuous function f : (, b) R is integrble. 11
12 2. BASIC THEOREMS ABOUT THE INTEGRAL Proof. Suppose tht f M. Let ε > 0. Then f is continuous, nd hence uniformly continuous, on [ + ε, b ε]. Let δ be such tht if x, y [ + ε, b ε] nd x y δ then f(x) f(y) ε, nd consider prtition P with = x 0, + ε = x 1, b ε = x n 1, b = x n nd mesh δ. Let φ + be the step function whose vlue on (x i 1, x i ) is sup x [xi 1,x i] f(x) when i = 2,..., n 1, nd whose vlue on (x 0, x 1 ) nd (x n 1, x n ) is M. Let φ be the step function whose vlue on (x i 1, x i ) is inf x [xi 1,x i] f(x) when i = 2,..., n 1, nd whose vlue on (x 0, x 1 ) nd (x n 1, x n ) is M. Then φ f φ + pointwise. As in the proof of the previous theorem, we hve φ + (x) φ (x) ε when x (x i 1, x i ), i = 2,..., n 1. On (x 0, x 1 ) nd (x n 1, x n ) we hve the trivil bound φ + (x) φ (x) 2M. Thus I(φ + ) I(φ ) (b )ε + 2M 2ε, which cn be mde rbitrrily smll by tking ε rbitrrily smll. Proposition 2.1 (Men vlue theorem for integrls). Suppose tht f : [, b] R is continuous. Then there is some c [, b] such tht f = (b )f(c). Proof. Since f is continuous, it ttins its mximum M nd its minimum m. Moreover, the constnt function φ + = M is mjornt for f, nd the constnt function φ is minornt for f, so which implies tht m(b ) = I(φ ) f I(φ + ) = M(b ), m 1 f M. b By the intermedite vlue theorem, f ttins every vlue in [m, M], nd in prticulr there is some c such tht f(c) = 1 b f. 2.2. Monotone functions re integrble A function f : [, b] R is sid to be monotone if it is either non-decresing (mening x y implies f(x) f(y)) or non-incresing (mening x y implies f(x) f(y)).
2.2. MONOTONE FUNCTIONS ARE INTEGRABLE 13 Proof. Theorem 2.3. A monotone function f : [, b] R is integrble. By replcing f with f if necessry we my suppose tht f is monotone non-decresing, i.e. f(x) f(y) whenever x y. Since f() f(x) f(b), f is utomticlly bounded nd in fct f(x) M where M := mx( f(), f(b) ). Let ε > 0, nd tke prtition with mesh < ε 2 /4M 2. Define step functions dpted to this prtition s follows. On (x i 1, x i ), define φ + (x) = f(x i ) nd φ (x) = f(x i 1 ). Defining φ (x i ) = M nd φ + (x i ) = M. Then φ + is mjornt for f nd φ is minornt. Sy tht i is good if f(x i ) f(x i 1 ) ε, or equivlently if φ + (x) φ (x) ε for x (x i 1, x i ), nd bd otherwise. There cnnot be more thn 2M/ε bd intervls, since f increses by more thn ε over ech such intervl yet is bounded everywhere by M. The contribution to I(φ + ) I(φ ) from the bd intervls is thus t most 2M 2M ε ε 2 4M = ε, since we lwys hve φ 2 + φ 2M. The contribution from the good intervls is t most ε(b ). Thus I(φ + ) I(φ ) (b + 1)ε, which cn be mde rbitrrily smll by mking ε smll enough.
CHAPTER 3 Riemnn sums The wy in which we hve been developing the integrl is closely relted to the pproch tken by Drboux. In this chpter we discuss wht is essentilly Riemnn s originl wy of defining the integrl, nd show tht it is equivlent. This is of more thn merely historicl interest: the equivlence of the definitions hs severl useful consequences. If P is prtition nd f : [, b] R is function then by Riemnn sum dpted to P we men n expression of the form Σ(f; P, ξ) n = f(ξ j )(x j x j 1 ), j=1 where ξ = (ξ 1,..., ξ n ) nd ξ j [x j 1, x j ]. Proposition 3.1. Let f : [, b] R be bounded function. Fix sequence of prtitions P (i). For ech i, let Σ(f, P (i), ξ (i) ) be Riemnn sum dpted to P (i). Suppose tht there is some constnt c such tht, no mtter how ξ (i) is chosen, Σ(f, P (i), ξ (i) ) c. Then f is integrble nd c = f. Proof. Let ε > 0, nd let M = sup x [,b] f(x) be bound for f. Suppose tht P (i) is = x (i) 0... x (i) n i = b. For ech j, choose some point ξ (i) j [x (i) j 1, x(i) j ] such tht f(ξ (i) j ) sup (i) x [x f(x) ε. (Note tht f does not necessrily ttin its ] j 1,x(i) j supremum on this intervl.) Let φ i + be step function tking the vlue f(ξ (i) j ) + ε on (x (i) j 1, x(i) j ), nd with φ(i) + (x (i) j ) = M. Then φ(i) + is mjornt for f. It is esy to see tht I(φ (i) + ) ε(b ) + Σ(f; P (i), ξ (i) ). Tking i lrge enough tht Σ(f; P (i), ξ (i) ) c + ε, we therefore hve Since ε > 0 ws rbitrry, it follows tht I(φ (i) + ) ε(b ) + c + ε. By n identicl rgument, inf I(φ + ) c. φ + sup φ I(φ ) c. 15
16 3. RIEMANN SUMS Therefore nd so ll these quntities equl c. c sup I(φ ) inf I(φ + ) c, φ φ + This suggests tht we could use such Riemnn sums to define the integrl, perhps by tking some nturl choice for the sequences of prtitions P (i) such s = + j i (b ) (the prtition into i equl prts). However, Proposition 3.1 does not imply tht this definition is equivlent to the one we hve been using, since we x (i) j hve not shown tht the Riemnn sums converge if f is integrble. In fct, this requires n extr hypothesis. Recll tht the mesh mesh(p) of prtition is the length of the longest subintervl in P. Proposition 3.2. Let P (i), i = 1, 2,... be sequence of prtitions stisfying mesh(p (i) ) 0. Suppose tht f is integrble. Then lim i Σ(f, P (i), ξ (i) ) = f, no mtter wht choice of ξ (i) we mke. Proof. Let M be bound for f, tht is to sy f(x) M for ll x. Let ε > 0. Then there re minornt nd mjornt φ, φ + for f with (3.1) I(φ + ) ε f I(φ ) + ε. By refining the prtitions underlying φ, φ + if necessry, we my ssume tht both of these functions re dpted to some prtition P, = x 0 x 1... x n b. By replcing φ + with min(m, φ + ) nd φ with mx( M, φ ) we my ssume tht M φ φ + M pointwise. Let δ := ε/nm. We clim tht if P : = x 0 x 1... x n = b is prtition of mesh t most δ, nd if Σ(f, P, ξ ) is ny Riemnn sum ssocited to this prtition, then (3.2) 3ε + f Σ(f, P, ξ ) 3ε + A moment s reflection convinces one tht this clim implies the proposition, since ε > 0 ws rbitrry. We give the proof of the upper bound in the clim (3.2), the rgument for the lower bound being essentilly identicl. Written out in full, the Riemnn sum is Σ(f; P, ξ ) = f(ξ j)(x j x j 1). n j=1 Since φ + is mjornt for f, we evidently hve Σ(f; P, ξ ) φ + (ξ j)(x j x j 1). n j=1 f.
3. RIEMANN SUMS 17 Sy tht j is good if the intervl (x j 1, x j ) is wholly contined in one of the subintervls of P. Sy tht j is bd if this is not the cse. In this ltter scenrio, one or more of the prtitioning points x k of P lies in the interior (x j 1, x j ). This mens tht the totl length of the bd intervls is t most nδ. Since φ + M pointwise, it follows tht (3.3) Σ(f; P, ξ ) nδm + j good φ + (ξ j)(x j x j 1). However, since φ + is constnt nd equl to ξ j on (x j 1, x j ) when j is good, nd certinly t lest M on the bd intervls, which hve totl length t most nδ, we hve (3.4) I(φ + ) j good Compring (3.3), (3.4) nd using (3.1) gives φ + (ξ j)(x j x j 1) nδm. Σ(f; P, ξ ) 2nδM + I(φ + ) = 2ε + I(φ + ) 3ε + which is precisely the upper bound in the clim (3.2). f, Proposition 3.1 nd 3.2 together llow us to give n lterntive definition of the integrl. This is bsiclly Riemnn s originl definition. Proposition 3.3. Let f : [, b] R be function. Let P (i), i = 1, 2,... be sequence of prtitions with mesh(p (i) ) 0. Then f is integrble if nd only if lim i Σ(f, P (i), ξ (i) ) is equl to some constnt c, independently of the choice of ξ (i). If this is so, then f = c. Finlly, we cution tht it is importnt tht the limit must exist for ny choice of ξ (i). Suppose, for exmple, tht [, b] = [0, 1] nd tht P (i) is the prtition into i equl prts, thus x (i) j = j i for j = 1,..., i. Tke ξ(i) j = j i ; then the Riemnn sum Σ(f, P (i). ξ (i) ) is equl to S i (f) := 1 i f( j i i ). By Proposition 3.2, if f is integrble then S i (f) However, the converse is not true. Consider, for exmple, the function f introduced in the first chpter, with f(x) = 1 for x Q nd f(x) = 0 otherwise. This function j=1 f.
18 3. RIEMANN SUMS is not integrble, s we estblished in tht chpter. However, S i (f) = 1 for ll i.
CHAPTER 4 Integrtion nd differentition It is well-known fct, which goes by the nme of the fundmentl theorem of clculus tht integrtion nd differentition re inverse to one nother nd tht if f = F then f = F (b) F (). Our objective in this chpter is to prove rigorous versions of this fct. We will prove two sttements, sometimes known s the first nd second fundmentl theorems of clculus respectively, though there does not seem to be complete consensus on this mtter. 4.1. First fundmentl theorem of clculus The first thing to note is tht the sttement just given is not true without some dditionl ssumptions. Consider, for instnce, the function F : R R defined by F (0) = 0 nd F (x) = x 2 sin 1 x 2 for x 0. Then it is stndrd exercise to show tht F is differentble everywhere, with f = F given by f(0) = 0 nd f(x) = 2x sin(1/x 2 ) 2 x cos(1/x2 ). In prticulr, f is unbounded on ny intervl contining 0, nd so it hs no mjornts nd is not integrble ccording to our definition. An even worse exmple (the Volterr function) cn be constructed with f bounded, but still not integrble. will not give it here. This construction is rther elborte nd we These constructions show tht hypothesis of integrbility should be built into ny sttement of the fundmentl theorem of clculus. Theorem 4.1 (First fundmentl theorem). Suppose tht f is integrble on (, b). Define new function F : [, b] R by F (x) := x f(t)dt. Then F is continuous. Moreover, if f is continuous t c (, b) then F is differentible t c nd F (c) = f(c). Proof. The fct tht F is continuous follows immeditely from the fct tht f is bounded (which it must be, s it is integrble), sy by M. Then F (c + h) F (c) = c+h c f(t)dt 19 c+h c f(t) Mh.
20 4. INTEGRATION AND DIFFERENTIATION In fct, this rgument directly estblishes tht F is uniformly continuous (nd in fct uniformly Lipschitz). Now we turn to the second prt. Suppose tht c (, b) nd tht h > 0 is sufficiently smll tht c + h < b. We hve F (c + h) F (c) = c+h c f(t)dt. Since f is continuous t c, there is function ε(h), ε(h) 0 s h 0, such tht we hve f(t) f(c) ε(h) for ll t [c, c + h]. Therefore (4.1) F (c + h) F (c) hf(c) = c+h c (f(t) f(c)) ε(h)h. Essentilly the sme rgument works for h < 0 (in fct, exctly the sme rgument works if we interpret c+h f(t)dt in the nturl wy s c f(t)dt). Sttement c c+h (4.1) is exctly the definition of F being differentible t c with derivtive f. We note tht F is not necessrily differentible ssuming only tht f is Riemnnintegrble. For exmple if we tke the function f defined by f(t) = 0 for t 1 2 nd f(t) = 1 for t > 1 2 then f is integrble on [0, 1], nd the function F (x) = x 0 f(t)dt is given by F (x) = 0 for x 1 2 nd F (x) = x 1 2 for 1 2 x 1. Evidently, F fils to be differentible t 1 2. 4.2. Second fundmentl theorem of clculus We turn now to the second form of the fundmentl theorem. Theorem 4.2 (Second fundmentl theorem). Suppose tht F : [, b] R is continuous on [, b] nd differentible on (, b). derivtive F is integrble on (, b). Then Proof. F (t)dt = F (b) F (). Suppose furthermore tht its Let P be prtition, = x 0 < x 1 < < x n = b. We clim tht some Riemnn sum Σ(f; P, ξ) is equl to F (b) F (). By Proposition 3.2 (the hrder direction of the equivlence between integrbility nd limits of Riemnn sums), the second fundmentl theorem follows immeditely from this. The clim is n lmost immedite consequence of the men vlue theorem. By tht theorem, we my choose ξ i (x i 1, x i ) so tht F (ξ i )(x i x i 1 ) = F (x i ) F (x i 1 ). Summing from i = 1 to n gives n Σ(F ; P, ξ) = (F (x i ) F (x i 1 )) = F (b) F (). i=1
4.3. INTEGRATION BY PARTS 21 4.3. Integrtion by prts Everyone knows tht integrtion by prts sys tht fg = f(b)g(b) f()g() f g. We re now in position to prove rigorous version of this. Before doing so, we estblish lemm which we could hve proven t the strt of the course, in Section 1.4. Lemm 4.1. Suppose tht f, g : (, b) R re integrble. product fg. Then so is their Proof. Since fg = 1 4 (f + g)2 1 4 (f g)2, it is enough to show tht f 2 is integrble if f is. For ny sclr λ R we hve f 2 = (f + λ) 2 2λf λ 2, nd so it suffices to show tht (f + λ) 2 is integrble. Tking λ sufficiently lrge, we my therefore ssume f 0. Suppose tht 0 f(x) M. Let φ f φ + be minornt/mjornt for f with I(φ + ) I(φ ) ε. Replcing φ + with min(φ +, M) nd φ with mx(φ, 0), we my ssume tht φ + M nd φ 0 pointwise. Now we hve φ 2 f 2 φ 2 + pointwise, nd so φ 2, φ 2 +, being step functions, re minornt/mjornt for f. Moreover I(φ 2 +) I(φ 2 ) = I(φ 2 + φ 2 ) = I((φ + + φ )(φ + φ )) 2MI(φ + φ ) 2Mε, since φ + + φ 2M everywhere. Since ε ws rbitrry, the result follows. Remrk. In fct it is true tht if f is integrble nd ψ is continuous then ψ f is integrble, where ψ f(t) = ψ(f(t)) is the composition of f with ψ. The proof of the preceding lemm estblished this in the specil cse ψ(x) = x 2. We leve the generl cse s n exercise for the interested student. Proposition 4.1. Suppose tht f, g : [, b] R re continuous functions, differentible on (, b). Suppose tht the derivtives f, g re integrble on (, b). Then fg nd f g re integrble on (, b), nd fg = f(b)g(b) f()g() Proof. We use the second form of the fundmentl theorem of clculus, pplied to the function F = fg. We know from bsic differentil clculus tht F is differentible nd F = f g + fg. By Lemm 4.1 nd the ssumption tht f, g re f g.
22 4. INTEGRATION AND DIFFERENTIATION integrble, F is integrble on (, b). Applying the fundmentl theorem gives F (t)dt = F (b) F (), which is obviously equivlent to the stted clim. 4.4. Substitution Proposition 4.2 (Substitution rule). Suppose tht f : [, b] R is continuous nd tht φ : [c, d] [, b] is continuous on [c, d], hs φ(c) = nd φ(d) = b, nd mps (c, d) to (, b). Suppose moreover tht φ is differentible on (c, d) nd tht its derivtive φ is integrble on this intervl. Then f(x)dx = d c f(φ(t))φ (t)dt. Proof. Let us first remrk tht (f φ)(t) = f(φ(t)) is continuous nd hence integrble on [c, d]. It therefore follows from Lemm 4.1 tht f(φ(t))φ (t) is integrble s function of t, so the sttement does t lest mke sense. Since f is continuous on [, b], it is integrble. The first fundmentl theorem of clculus implies tht its ntiderivtive F (x) := x f(t)dt is continuous on [, b], differentible on (, b) nd tht F = f. Write (F φ)(t) := F (φ(t)). By the chin rule nd the fct tht φ((c, d)) (, b), F φ is differentible on (c, d), nd (F φ) = (F φ)φ = (f φ)φ. By the remrks t the strt of the proof, it follows tht (F φ) is n integrble function. By the second form of the fundmentl theorem, d f(φ(t))φ (t)dt = d c c (F φ) = (F φ)(d) (F φ)(c) = F (b) F () = F (b) = f(x)dx.
CHAPTER 5 Limits nd the integrl 5.1. Interchnging the order of limits nd integrtion Suppose we hve sequence of functions f n converging to limit function f. If this convergence is merely pointwise, integrtion need not preserve the limit. Exmple 5.1. There is sequence of integrble functions f n : [0, 1] R (in fct, step functions) such tht f n (x) 0 pointwise for ll x [0, 1] but f n = 1 for ll n. Thus lim n 1 0 f n = 1, whilst 1 0 lim n f n = 0, nd so interchnge of integrtion nd limit is not vlid in this cse. Proof. Define f n (x) to be equl to n for 0 < x < 1 n nd 0 elsewhere. However, if f n f uniformly then the sitution is much better. Theorem 5.1. Suppose tht f n : [, b] R re integrble, nd tht f n f uniformly on [, b]. Then f is lso integrble, nd Proof. lim n f n = f = lim f n. n Let ε > 0. Since f n f uniformly, there is some choice of n such tht we hve f n (x) f(x) ε for ll x [, b]. Now f n is integrble, nd so there is mjornt φ + nd minornt φ for f n with I(φ + ) I(φ ) ε. Define φ + := φ + + ε nd φ := φ ε. Then φ, φ + re minornt/mjornt for f. Moreover I( φ + ) I( φ ) 2ε(b ) + I(φ + ) I(φ ) 2ε(b ) + ε. Since ε ws rbitrry, this shows tht f is integrble. Now f n f Since f n f uniformly, it follows tht f n f (b ) sup f n (x) f(x). x [,b] lim f n n 23 f = 0,
24 5. LIMITS AND THE INTEGRAL nd hence tht This concludes the proof. lim n f n = f = lim f n. n An immedite corollry of this is tht we my integrte series term-by-term under suitble conditions. Corollry 5.1. Suppose tht φ i : [, b] R, i = 1, 2,... re integrble functions nd tht φ i (x) M i for ll x [, b], where i=1 M i <. Then the sum i φ i is integrble nd φ i = φ i. i i Proof. This is immedite from the Weierstrss M-test nd Theorem 5.1, pplied with f n = n i=1 φ i. 5.2. Interchnging the order of limits nd differentition The behviour of limits with respect to differentition is much worse thn the behviour with respect to integrtion. Exmple 5.2. There is sequence of functions f n : [0, 1] R, ech continuously differentible on (0, 1), such tht f n 0 uniformly but such tht f n does not converge t every point. Proof. Tke f n (x) = 1 n sin(n2 x). Then f n(x) = n cos(n 2 x). Tking x = π 4, we see tht if n is multiple of 4 then f n(x) = n, which certinly does not converge. If, however, we ssume tht the derivtives f n converge uniformly then we do hve useful result. Proposition 5.1. Suppose tht f n : [, b] R, n = 1, 2,... is sequence of functions with the property tht f n is continuously differentible on (, b), tht f n converges pointwise to some function f on [, b], nd tht f n converges uniformly to some bounded function g on (, b). Then f is differentible nd f = g. In prticulr, lim n f n = (lim n f n ). Proof. First note tht, since the f n re continuous nd f n g uniformly, g is continuous. Since we re lso ssuming g is bounded, it follows from Theorem 2.2 tht g is integrble. We my therefore pply the first form of the fundmentl theorem of clculus to g. Since g is continuous, the theorem sys tht if we define function F : [, b] R
5.2. INTERCHANGING THE ORDER OF LIMITS AND DIFFERENTIATION 25 by F (x) := x g(t)dt then F is differentible with F = g. By the second form of the fundmentl theorem of clculus pplied to f n, we hve x f n(t)dt = f n (x) f n (). Tking limits s n nd using the fct tht f n f pointwise, we obtin x lim n f n(t)dt = f(x) f(). However, since f n g uniformly, it follows from Theorem 5.1 tht Thus x lim n F (x) = x f n(t)dt = x g(t)dt. g(t)dt = f(x) f(). It follows immeditely tht f is differentible nd tht its derivtive is the sme s tht of F, nmely g. Remrk. Note tht the sttement of Proposition 5.1 involves only differentition. However, the proof involves considerble mount of the theory of integrtion. This is theme tht is seen throughout mthemticl nlysis. For exmple, the nice behviour of complex differentible functions (which you will see in course A2 next yer) is consequence of Cuchy s integrl formul. 5.1. As with our result bout integrls, we cn record series vrint of Proposition Corollry 5.2. Suppose we hve sequence of continuous functions φ i : [, b] R, continuously differentible on (, b), with i φ i converging pointwise. Suppose tht φ i (x) M i for ll x (, b), where i M i <. Then φ i is differentible nd ( i φ i ) = i Proof. Apply Proposition 5.1 with f n := n i=1 φ i. By the Weierstrss M-test, f n = n i=1 φ i does converge pointwise to some bounded function, which we my cll g. φ i.