Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)

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1 Lecture 3 ( ) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of function f. Let < < b < + be two rel numbers nd f : [, b] R ny function tht my not be continuous or bounded. The finite k +1-tuple of points D = ( 0, 1,..., k ) from the intervl [, b] is clled prtition of [, b] if = 0 < 1 < 2 <... < k = b. These points divide the intervl [, b] into intervls I i = [ i 1, i ]. We denote by I i the length of intervl I i : I i = i i 1 [, b] = b. Clerly I i = ( 1 0 ) + ( 2 1 ) ( k k 1 ) = b = [, b]. Norm of prtition D is the mximum length of n intervl of the prtition nd is denoted by λ: λ = λ(d) = mx 1 i k I i. Prtition of n intervl [, b] with points is pir (C, D) wher D = ( 0, 1,..., k ) is prtition of [, b] nd k-tuple C = (c 1, c 2,..., c k ) consists of c i I i (i.e. i 1 c i i ). Riemnn sum corresponding to the function f nd prtition with points (D, C) is defined s R(f, D, C) = I i f(c i ) = ( i i 1 )f(c i ). If f 0 on [, b], it is the sum of k rectngles I i [0, f(c i )] whose union pproximtes figure U(, b, f). However, Riemnn sum is defined for every function f, regrdless of its sign on [, b]. The following definition ws introduced by Bernhrd Riemnn ( ). Definition (first definition of Riemnn integrl, Riemnn). We sy tht f : [, b] R hs Riemnn integrl I R on [, b] if for every ε > 0 there exists δ > 0 such tht for ech prtition of [, b] with points (D, C) such tht λ(d) < δ the following holds: I R(f, D, C) < ε 1

2 Therefore, we require I R, vlues ± re not llowed (lthough, it is possible to define them). If there is such number I, we write I = (R) nd sy tht f is Riemnn integrble on the intervl [, b]. We will work with the clss of ll Riemnn integrble functions R(, b) := {f f is defined nd Riemnn integrble on [, b]}. Thus, the first definition of the Riemnn integrl cn be summrized by the formul lim R(f, D, C) R. λ(d) 0 We understnd the limit here s defined in the definition bove; s symbol, we defined only limit of sequence nd of function in point. For the second, equivlent, definition of the integrl we will need few more concepts. For f : [, b] R nd prtition D = ( 0, 1,..., k ) of intervl [, b] we define lower nd upper Riemnn sum, respectively, (even though they were introduced by Drboux) s where s(f, D) = I i m i, nd S(f, D) = f I i M i, m i = x I i f(x) nd M i = sup x I i f(x) I i = [ i 1, i ] These sums re lwys defined s(f, D) R { } nd S(f, D) R {+ } Lower nd upper Riemnn integrl, respectively, of function f on the intervl [, b] is defined s nd sup({s(f, D) : D is prtition of [, b]}), ({S(f, D) : D is prtition of [, b]}). These terms re lwys defined nd we hve f, f R = R {, + }. Definition (second definition of Riemnn integrl, Drboux). We sy tht f : [, b] R hs t [, b] Riemnn integrl, if 2 f(x) dx R.

3 This common vlue, if it exists, we denote by nd we cll it the Riemnn integrl of f on the intervl [, b]. The two definitions re equivlent: they give the sme clsses of Riemnn integrble functions nd the sme vlue of the Riemnn integrl, if defined. Clim (unbounded functions hve no integrl). If the f : [, b] R function is not bounded then it does not hve Riemnn integrl on [, b], ccording to both definitions. When D = ( 0, 1,..., k ) D = (b 0, b 1,..., b l ) re prtitions of [, b] nd D D, tht is for every i = 0, 1,..., k there exists j, such tht i = b j (therefore k l), we sy tht D is refinement of D or tht D refines D. Lemm. If f : [, b] R nd D, D re two prtitions of [, b], nd D refines D, s(f, D ) s(f, D) nd S(f, D ) S(f, D). f Proof. Considering the definition of s(f, D) S(f, D) nd the fct tht D cn be creted from D by dding points, it is enough to prove both inequlities in sitution where D = ( 0 = < 1 = b) D = ( 0 = < 1 < 2 = b). According to the definition of im f, we hve Then m 0 = 0 x 1 f(x) 0 x 1 f(x) = m 0 1 x 2 f(x) = m 1 s(f, D ) = ( 1 0)m 0 + ( 2 1)m 1 ( 1 0)m 0 + ( 2 1)m 0 = ( 2 0)m 0 = (b )m 0 = s(f, D). Proof of the inequlity S(f, D ) S(f, D) is similr. Corollry. When f : [, b] R nd D, D re two prtitions [, b], then s(f, D) S(f, D ). 3

4 Proof. Let E = D D be common refinement of both prtitions. According to the previous lemm we hve s(f, D) s(f, E) S(f, E) S(f, D ) More precisely, the first nd lst inequlity follow from the previous lemm, nd the middle one from the definition of upper nd lower sum. Theorem (lower integrl does not exceed upper). Let f : [, b] R, m = x b f(x), M = sup x b f(x) nd D, D be two prtitions of intervl [, b]. Then the following inequlities hold: m(b ) s(f, D) f f S(f, D ) M(b ). Proof. The first nd lst inequlity re the specil cses of the previous lemm. The second nd penultimte inequlity comes stright from the definition of the lower nd upper integrl s supremum or imum respectively. According to the corollry, ech element is set of lower sums whose supremum is f smller or equl to ech element of the upper sum set whose im is f. Using the definition of imum (the lrgest lower bound) nd supremum (the smllest upper bound) we get the middle inequlity: For ech prtition D, s(f, D) the lower bound of the set of upper sums, tht is, s(f, D) f, nd so f is the upper bound of the set of lower sums, thus f f. Theorem (Integrbility criterion). Let f : [, b] R. Then f R(, b) ε > 0 D : 0 S(f, D) s(f, D) < ε. In other words, f hs Riemnn s integrl if nd only if for every ε > 0 there exists prtition of D of intervl [, b] such tht its upper Riemnn sum is greter thn the corresponding lower Riemnn sum by less thn ε. Proof. We ssume tht f hs R. integrl on [, b], i.e., f R. Let ε > 0 be given. By definition of the lower nd upper integrls, there re prtitions E 1 nd E 2 so tht s(f, E 1 ) > f ε 2 = f ε 2 S(f, E 2) < f + ε 2 = f + ε 2. 4

5 According to the lemm, these inequlities lso pply fter replcing E 1 nd E 2 with their joint refinement D = E 1 E 2. Summing up both inequlities we will get S(f, D) s(f, D) < f + ε ( 2 + f + ε ) = ε. 2 Given ε > 0 we tke prtition of D stisfying the condition. According to the definition of the lower nd upper integrl we get f S(f, D) < s(f, D) + ε f + ε, thus f f < ε. This inequlity is vlid for every ε > 0, so ccording to the previous sttement we hve f R. Then f hs R. integrl on [, b]. Exmple (bounded function without integrl). A function f : [0, 1] {0, 1} defined s f(α) = 1 when α is rtionl number, nd f(α) = 0, when α is irrtionl, is clled Dirichlet function, nd does not hve Riemnn integrl on [0, 1], lthough bounded. Ech positive-length intervl contins points where f hs vlue of 0, s well s points tht hve vlue of 1. Then s(f, D) = 0 nd S(f, D) = 1 for every prtition of D nd therefore <