Notes #5. We then define the upper and lower sums for the partition P to be, respectively, U(P,f)= M k x k. k=1. L(P,f)= m k x k.

Transcription

1 Notes #5. The Riemnn Integrl Drboux pproch Suppose we hve bounded function f on closed intervl [, b]. We will prtition this intervl into subintervls (not necessrily of the sme length) nd crete mximl nd miniml pproximtions. First, let s develop some nottion to del with these ides. Definition 5..: A prtition P of intervl [, b] is set of points = x 0 < x < x 2 < < x n < x n = b. We set I k = [x k,x k ] nd x k = l(i k ), the length of the subintervl. The norm (or mesh) of the prtition is denoted P nd is equl to mx{ x k } k. Now, we ll let M k = sup{f(t) :t I k } m k = inf{f(t) :t I k }. We then define the upper nd lower sums for the prtition P to be, respectively, U(P,f)= L(P,f)= M k x k m k x k By looking t these sums over ll possible prtitions, we produce the upper nd lower integrls, defined by f = inf P {U(P,f)} f = sup{l(p,f)} P Definition 5..2:A bounded function f on the closed nd bounded intervl [, b] is sid to be Riemnn integrble if f = f. In this cse, the integrl is written f nd we sy f R[, b].

2 Exmples: The constnt function f(x) = is Riemnn integrble on, sy [0, ], s is ny step function. However, the Dirichlet function {, if x Q g(x) = 0, otherwise is not Riemnn integrble. Definition 5..3: A prtition P is refinement of the prtition P if P P. Lemm 5..4: If P is refinement of P then L(P,f) L(P,f) U(P,f) U(P,f). Proof: The ide here is to first show tht one-point refinement stisfies the inequlity. By induction, then, n rbitrry refinement will (since every prtition is finite). Suppose P = {x 0,x,...,x n } nd crete refinement P by dding the single point x. This point must fll in the interior of one of the subintervls, sy I k. Then we hve L(P,f)= m i x i i= =(m x + + m k x k )+m k x k +(m k+ x k+ + + m n x n ) (m x + + m k x k )+m (x x k ) = L(P,f). + m (x k x )+(m k+ x k+ + + m n x n ) where m nd m re the minimum vlues for f in the two subintervls of I k creted by x. A similr rgument cn be mde to show the U(P,f) U(P,f). Since the lower sum for given prtition is never bigger thn the upper sum, the full inequlity follows. [Note: An immedite corollry of this is tht Some criteri tht provide conditions for the Riemnn integrbility of functions include the following theorems. Theorem 5..5: A bounded rel-vlued function f on [, b] is Riemnn-integrble if nd only if for ech ɛ>0 there exists prtition P of [, b] such tht f U(P,f) L(P,f) <ɛ. f.] Proof:. Suppose the inequlity holds for given ɛ>0, then 0 f f U(P,f) L(P,f) <ɛ, f R[, b].

3 . Suppose f R[, b] nd let ɛ>0. Then there re prtitions P nd P 2 such tht U(P 2,f) f<ɛ/2 nd f L(P,f) <ɛ/2. Let P = P P 2, then U(P,f) U(P 2,f) < from which the desired inequlity follows. ɛ/2 < L(P,f)+ɛ L(P,f)+ɛ, Theorem 5..6: Let f be rel-vlued function on [, b]. ) If f is continuous on [, b], then f R[, b]. b) If f is monotone on [, b], then f R[, b]. Proof: () Let f be continuous, nd ɛ>0. Choose ɛ such tht (b )ɛ <ɛ. Since f is continuous it is uniformly continuous, so there exists δ>0 such tht f(x) f(y) <ɛ whenever x y <δ. Let P = {x 0,x,...,x n } be prtition of [, b] with P <δ. Then M i m i <ɛ,so U(P,f) L(P,f)= n (M i m i ) x i ɛ x i = ɛ (b ) <ɛ. i= i= Therefore by the previous theorem, f R[, b]. (b) Left to reder. Theorem 5..7: (Composition Theorem) Let f be bounded nd f R[, b] with rn(f) [c, d]. If φ is continuous on [c, d], then φ f R[, b]. The direct proof of this theorem is rther involved, but we will see tht it follows simply from Lebesgue s Theorem, which we stte below. Also note the conditions for the composition to be Riemnn integrble it is not enough simply tht f nd φ re Riemnn integrble. See the exmple t the end of the section for two integrble functions whose composition isn t. Corollry 5..8: If f R[, b] nd p R[x], then () p f R[, b] in prticulr, f 2 R[, b], nd (b) f R[, ]. Definition A subset E R hs mesure zero if for ech ɛ > 0 there exists countble collection (possibly finite) of open intervls {I n } covering E such tht n l(i n) <ɛ. Theorem (Lebesgue s Theorem) A bounded function f :[, b] R is Riemnn integrble if nd only if the set of discontinuities of f (in [, b]) hs mesure zero.

4 The proof of this theorem will hve to wit, but let s use it nywy to prove the Composition Theorem. Proof of Theorem 5..7: We note tht in generl tht composition preserves continuity. In this cse, if f is continuous t point x [, b] nd φ is continuous t f(x), then φ f is continuous t x. Let A = {x [, b] :f is discontinuous t x} B = {y [, b] :φ f is discontinuous t y}. Since φ is continuous on [c, d] rn(f), then B A. Nowf R[, b] so by Lebesgue s Theorem the set of discontinuities of f, i.e., the set A, is set of mesure 0. Consequently, so is B. This mens (by Lebesgue s Theorem gin) tht the composition φ f is Riemnn integrble, s well. Exmple: Consider the function f :[0, ] [0, ] defined by, if x =0 f(x) = 0, if x is irrtionl if x = m n Q in lowest terms. n, This function is continuous on the irrtionls nd discontinuous t ech rtionl. So its set of discontinuities hs mesure zero nd so is Riemnn integrble. Now, let g : [0, ] [0, ] be defined by { 0, if x =0 g(x) =, otherwise Now g is continuous except t x = 0 nd so is Riemnn integrble, but g f is the Dirichlet function {, if x Q g f(x) = 0, otherwise which we know not to be Riemnn integrble. EXERCISES:. Let f(x) =x 2. Find the upper nd lower sums for f over [, 3] for ech of the prtitions below. ) P = {, 0,, 2, 3} b) P = {, 0,.5,.8,.5, 3} 2. Let f,g R[, b] nd suppose f(x) g(x) for ll x [, b]. Prove tht f g. 3. () Let f(x) 0 is continuous on [, b]. Prove: If f = 0 then f 0on[, b]. (b) Show tht the conclusion in () is flse if f is not continuous. 4. () If f R[, b] show directly tht f 2 R[, b]. (b) Give n exmple of bounded function f on [, b] for which f 2 f/ R[, b]. R[, b], but 5. Show tht the countble union of sets of mesure zero hs mesure zero.

5 2. Properties of the Riemnn Integrl The following properties of the Riemnn integrl re fmilir from undergrdute clculus. Theorem Let f,g R[, b]. Then g, kf, f nd fg re Riemnn integrble nd () (2) (3) c c ( g) = g. (kf) =k f, for ny k R. f = 0, for ny c [, b]. (4) If c (, b) then (5) f f. f = c c f. Proof: We ll leve the proofs of 2,3, nd 4 to the reder. () Write f for the set of discontinuities of f. First if f nd g re Riemnn integrble then f nd g re of mesure zero. f+g f g is lso of mesure zero, so g R[, b]. Now, let P = {x 0,x,...,x n } be prtition of [, b] nd, nlogous to the M i in our development of the integrl, we will let F k = sup{f(t) :t I k } nd G k = sup{g(t) :t I k }. Clerly, f(t)+g(t) F k + G k for ll t I k, so sup{f(t)+g(t) :t I k } F k + G k for ech k =, 2,...,n. This mens tht, for ll prtitions, P, U(P, g) U(P,f)+U(P,g). Reclling tht f,g R[, b], for ny ɛ>0 we cn find prtitions F nd G so tht U(F,f) < ɛ/2 nd U(G,g) < Let P = F G, refinement of both F nd G, then U(P, g) U(P,f)+U(P,g) U(F,f)+U(G,g) < ( g) < g + ɛ, g + ɛ. g + ɛ/2.

6 Since this is true for ny ɛ>0, we hve ( g) Similrly, we cn show tht g ( g). From these inequlities we hve g. ( g) g ( g) which implies tht ( g) = g. (5) If f R[, b] then by Corollry 5..8 f R[, b]. Let c = ± so tht f = c f. Now, cf(t) f(t), t [, b] we hve U(P,cf) U(P, f ) for ech prtition P of [, b]. Since both cf nd f re Riemnn integrble, f = c f = cf f. 3. Riemnn s pproch to the integrl In Drboux s pproch, the integrl exists if our lower nd upper sums converge to the sme vlue. In the originl development of the Riemnn integrl, the integrl exists if single sum the Riemnn sum converges. But this sum is determined not just by prtitions, but by considering every possible point within ech subintervl of the prtition s well. Definition Let f :[, b] R be bounded function nd P = {x 0,x,...,x n } prtition of [, b]. Choosing n rbitrry t k in ech I k, the sum S(P,f)= f(t k ) x k is clled Riemnn sum (with respect to P nd {t k : k =,...n}). In order to produce n integrl we need to know wht is ment by the limit of this sum. The sum is tken over ll choices of our t k nd over ll prtitions of given norm s tht norm converges to 0.

7 Definition We sy tht the limit of the Riemnn sum exists, tht is, lim P 0 S(P,f)=I if, for ech ɛ>0, there exists δ>0 such tht f(t k ) s k I <ɛ for ll prtitions P with P <δnd ll choices of the t k. Theorem Let f :[, b] R be bounded function. Then if lim S(P,f)=I P 0 exists, then f R[, b] nd I = f. Conversely, if f R[, b] then the limit exists nd equls f. 4. Some importnt theorems nd consequences In this section we look t some of the importnt theorems of integrl clculus the fundmentl theorems, the men vlue theorem, nd the chnge of vrible (substitution) theorem. Theorem (First Fundmentl Theorem of Clculus) If f R[, b] nd F is n ntiderivtive of f, then f(x) dx = F (b) F (). Proof: Let P = {x 0,x,..., x n } be prtition of [, b]. If F = f, then, by the MVT for derivtives, there exists t k Int(I k ) such tht Now, f(t k ) x k = F (x k ) F (x k )=f(t k ) x k. [F (x k ) F (x k )] = F (b) F (). But ech Riemnn sum is bounded below nd bove by L(P,f) nd U(P,f) respectively. Therefore, f F (b) F () nd since f R[, b] we must hve f = F (b) F (). f,

8 Theorem (Second Fundmentl Theorem of Clculus) Let f R[, b] nd define F by x F (x) = f(t) dt. Then F is continuous on [, b] nd if f is continuous t c [, b] then F is differentible nd F (c) =f(c). [Note: The integrbility of f does not necessrily imply the existence of n ntiderivtive F.] Theorem (Chnge of Vrible Theorem) Let g be differentible on [, b] nd g R[, b]. If f is continuous on A = g([, b]), then f(g(x))g (x) dx = g(b) g() f(t) dt. Proof: g is continuous, so A is closed nd bounded intervl. Also, f g is continuous nd g R[, b], so f g)g R[, b]. We need to consider two possibilities for A. Cse : A = {c} is single point. In this cse, g is constnt so g 0 nd both integrls bove re 0. Cse 2: A =[c, d] is n intervl. For x A define F : A R by F (x) = x g() f(t) dt. Now, since f is continuous, F (x) =f(x) for ll x A. By the chin rule for derivtives d dx F (g(x)) = F (g(x))g (x) =f(g(x))g (x). Therefore, f(g(x)) dx = F (g(b)) F (g()) = g(b) g() f(t) dt.

9 5. Improper Riemnn integrls Thus fr, we hve concerned ourselves with functions tht re bounded on closed intervl [, b]. Now let s consider functions defined on the left-open intervl (, b] which re bounded nd Riemnn integrble on [c, b] for ech c (, b], but unbounded t. (This extension is originlly due to Cuchy.) Definition Suppose f R[c, b] for ll c (, b]. The improper Riemnn integrl of f on (, b], lso denoted f, is defined to be f = lim f, c c provided this limit exists. If it does, the improper integrl is sid to be convergent. Otherwise, it is sid to be divergent. Obviously, we cn crete similr definitions for [, b) or[, c) (c, b] or(, b) or, ), etc. Exmples: () Suppose f(x) =/x on (0, ]. lim c 0 c the improper integrl dx diverges. 0 x x dx = lim c 0 ln(c) =, so (2) f(x) = ln(x) on (0, ]. lim c 0 c ln(x) dx = lim c 0( c ln(c) ) = (3) For Riemnn integrls, f R[, b] = f 2 R[, b]. This is not true for improper Riemnn integrls. Consider the function f(x) =/ x on (0, ]. f(x) dx = 2, but 0 0 (/ x) 2 dx = /x dx diverges. 0 (4) Similrly, while f R[, b] = f R[, b], the sme is not true for improper Riemnn integrls. Let, f be defined on (0, ] by Now, 0 f(x) =2x sin( x 2 ) 2 x cos( x 2 )= d dx (x2 sin( x 2 )). d f(x)dx = lim c 0 c dx (x2 sin(/x 2 )) dx = sin() lim(x 2 sin(/x 2 )) = sin(). c 0 To justify this lst limit, note tht 0 x 2 sin(/x 2 ) x 2 0sx 0. On the other hnd, 0 f(x) dx = lim c 0 c d dx x2 sin(/x 2 ) dx. To see tht this integrl diverges, note tht the zeros of f occur when tn(/x 2 )= 4. Let 0 <θ<π/2beone of the vlues for which tn(θ) =/4, then, since the period of the tngent is π, the zeros of tn(/x 2 ) stisfy /x 2 = θ + kπ, k N. Tht is, x = θ k = θ+kπ. To compute this integrl we wnt to look t the bsolute vlue of f evluted on

10 the intervl whose endpoints re two djcent zeros. Noting tht sin(α + π) = sin(α), we hve θk A k = d θk dx x2 sin(/x 2 ) dx d dx x2 sin(/x 2 ) dx Now, 0 θ k+ θ k+ = θ + kπ sin(θ + kπ) sin(θ +(k +)π) θ +(k +)π = > f(x) dx = π sin(θ + kπ) π sin(θ) k +2. A k > θ/π + k + θ/π +(k +) π sin(θ) This lst sum diverges so, too, does the integrl. k +2 = π sin(θ) [ ].

11 6. The Riemnn-Stieltjes integrl The Riemnn-Stieltjes integrl is generliztion of the Riemnn integrl tht provides, in some sense, for weighting of independent vrible. Referring to the sme definitions for prtition, etc. tht we used in developing the Riemnn integrl, we define the Riemnn-Stieltjes integrl (RSI). Definition Let α be monotone incresing nd f bounded, both rel-vlued functions on [, b]. If P = {x 0,x,..., x n } is prtition of [, b], we set α k = α(x k ) α(x k ). The upper nd lower Riemnn-Stieltjes sums (with respect to f,p, nd α) re then, respectively, U(P,f,α)= M i α k L(P,f,α)= m i α k. The upper nd lower RSIs re defined to be, respectively, fdα = inf P {U(P,f,α)} fdα = sup{l(p,f,α)}. P If fdα = fdα then f is sid to be Riemnn-Stieltjes integrble with respect to α, denoted f RS α [, b], nd the integrl is written simply fdα. We cn develop theorems for RSIs tht re nlogues to those of the Riemnn integrl. Unless otherwise specified, α is ssumed to be monotone incresing on [, b]. Theorem Suppose f is bounded on [, b], then fdα fdα. Theorem Suppose f is bounded on [, b]. Then, f RS α [, b] if nd only if for ech ɛ>0 there exists prtition P of [, b] such tht U(P,f,α) L(P,f,α) <ɛ. Theorem Suppose f is ny rel-vlued function on [, b] nd α is monotone incresing. ) If f is continuous, then f RS α [, b]. b) If f is monotone nd α is continuous, then f RS α [, b].

12 Theorem () If f,g RS α [, b], then g nd cf re in RS α [, b] for every c R nd ( g) dα = cfdα = c fdα+ fdα. gdα (b) If f RS αi [, b] for i =, 2, then f RS α +α 2 [, b] nd fd(α + α 2 )= fdα + fdα 2. (c) If f RS α [, b] nd c (, b), then f is RS integrble wrt α on [, c] nd [c, b] with fdα = c fdα+ c fdα. (d) If f,g RS α [, b] nd f(x) g(x) for ll x [, b] then fdα gdα. (e) If f(x) M for ll x [, b] nd f RS α [, b], then f RS α [, b] nd fdα f dα M[α(b) α()].