FUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (


 Gertrude Phoebe Little
 2 years ago
 Views:
Transcription
1 FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions tht re comptible with Lebesgue mesure nd then define Lebesgue integrl on these functions. Recll: f is continuous if nd only if f (O) is open for ll O open. Also, recll tht O = I k f (O) = f (I k ). Continuous functions re well behved for even Riemnnintegrl; hence we will nturlly be comptible with mesurble sets. We will consider lrger clss of functions: Definition. Let f : R #, where F. f is clled (Lebesgue) mesurble if for ll R, the set {x : f(x) > } F. Fct. Let f : R #, F. The following re equivlent: i) f is mesurble. ii) {x : f(x) } F for ll R. iii) {x : f(x) < } F for ll R. iv) {x : f(x) } F for ll R. Furthermore i)iv) imply v) {x : f(x) = } F for ll R #. Proof. (i) (ii): First, note tht {x : f(x) } = f ([, ]). Also, since [, ] = n ( n, ], it follows tht (ii) (iii): ( ) {x : f(x) } = f ( n, ] = n= n= f ( ( n, ] ) F. {x : f(x) < } = f ((, )) = f (R \ [, ]) = \ f ([, ]) F. (iii) (iv) nd (iv) (i) re similr nd left s n exercise. For (v), if R, then {f(x) = } = f ({}) = f ((, ]) f ([, ]) F. If =, {f(x) = } = {f(x) > n} F. n=
2 2 xmples.. Let f : [0, ] R be function such tht { 0 if x Q f(x) = if x Q c. Then, f ( 2, ) = {x [0, ] : f(x) > 2 } = Qc [0, ] F, if =, f (, ) = F, if =, then f (, ) = [0, ] F. It follows tht, for ll R, {x [0, ] : f(x) > } F; hence, f is mesurble. 2. Let f(x) be the piecewise function defined s 0 if x x f(x) = 2 if < x < 2 if x = 2 x if x > If < 0, then f (, ) = {x : f(x) > } = (, 2 ) F. If = 0, then f (0, ) = {x : f(x) > 0} = (, 0) (0, 2) F. If 0 < <, then f (, ) = (, ) (, 2 ) F. If 2, then f (, ) = F. Hence, f is mesurble. Fct.2 f : R, F, is mesurble if nd only if for ll O B, f (O) F. Proof. xercise. Corollry. Continuous functions re mesurble. Proof. xercise. Corollry.2 Let f : R be mesurble function on. i) If D, D F, then f D nd f \D re mesurble on. ii) If f = g lmost everywhere on, then g is mesurble on. Proof. (i) Since f : R be mesurble on, {x D : f(x) > } = (f D ) (, ) = D f (, ) F (ii) Let A = {x : f(x) g(x)}. Note tht m(a) = 0; hence A F. Then, {x : g(x) > } = {x A : g(x) > } {x \ A : g(x) > } = {x A : g(x) > } {x \ A : f(x) > } F. }{{}}{{} F (f \A ) (, ) F Fct.3 Let F nd f, g : R be mesurble functions nd finite lmost everywhere. Then: i) αf + βg is mesurble on for ll α, β R. ii) f 2 is mesurble on (hence, fg is mesurble on ). iii) is mesurble on. f
3 3 iv) f is mesurble on. Proof. (i) First, we show tht αf is mesurble on for ll α R. If α = 0, then αf = 0. Hence αf is mesurble. If α > 0, then { {x : (αf)(x) > } = x : f(x) > } F. α Similrly for α < 0. Hence αf is mesurble, nd so is βg. Therefore, it is enough to show tht f + g is mesurble on. Notice tht, {x : f(x) + g(x) > } = {x : f(x) > g(x)} Now, there exists r x Q such tht f(x) > r x > g(x). Hence {x : f(x) > g(x)} = r Q({x : f(x) > r} {x : r > g(x)} ) F. }{{}}{{} F F (ii) First, observe tht fg = 2 [(f + g)2 f 2 g 2 ]. Therefore it is enough to show tht f mesurble implies f 2 is mesurble. Now, we hve {x : (f 2 )(x) > } = {x : [f(x)] 2 > } = {x : f(x) > } {x : f(x) < } F. }{{}}{{} F F (iii) If = 0, then {x : (x) > } = {x : f(x) > 0} F. So ssume tht > 0. Then f {x : f } { (x) > = x : f(x) < } F. (iv) Follows from {x : f (x) > } = {x : f(x) > } = {x : f(x) > } {x : f(x) < } F. Definition. For functions f, g : R, define (f g)(x) = mx{f(x), g(x)} (f g)(x) = min{f(x), g(x)} f + (x) = f(x) 0 f (x) = f(x) 0 Notice tht, with the definitions bove, we hve tht f = f + + f nd f = f + f. Corollry. If f, g : R, F is mesurble on, then: i) f +, f re mesurble on. ii) (f g)(x) nd (f g)(x) re mesurble on. Proof. xercise. Convention. For the rest of this chpter f : R. will lwys be mesurble. Definition. Let A R be set, then then the function { if x A χ A (x) = 0 otherwise
4 4 is clled the chrcteristic function of the set A. sy Fct. F if nd only if χ is mesurble function. For, observe tht, χ { if (, ) = otherwise Hence χ is mesurble (on ). If F, R, then χ is not mesurble. Question: If f nd g re mesurble functions. Wht bout f g? The nswer is No in generl. xmple. Recll tht f : [0, ] [0, 2] defined by f(x) = x + ϕ(x), where φ is the Cntor Lebesgue function, is strictly incresing. Then, by Fct 9 in Chpter.II, there exists A F such tht f(a) F. xtend f continuously in strictly incresing mnner onto R. Sy g : R R such tht g [0,] = f. Then g is continuous. Next, consider χ A nd define h(x) := (χ A g )(x). Now h ( 2, ) ( ) ( ) = (χ A g ) 2, = (g χ A ) 2, ( ( )) = g χ A 2, = g(a) = f(a) F. Theorem. Let f, g be mesurble functions on their respective domins. If g is continuous, then g f is mesurble. Proof. Define h(x) := (g f)(x). If O R is Borel set, then we hve h (O) = (g f) (O) = f (g (O)). Since g is continuous, g (O) B. Hence f (g (O)) F. Recll tht, for given collection of functions {f n }, f : R, we sy tht f n f pointwise if nd only if for ll x, for ll ɛ > 0, there exists N(ɛ, x) such tht whenever n N we hve f n (x) f(x) < ɛ. Also, f n f uniformly if nd only if for ll ɛ > 0, there exists N(ɛ) such tht whenever n N we hve f n (x) f(x) < ɛ for ll x. Definition. Let {f n } be sequence of relvlued functions on mesurble set. We sy tht {f n } converges lmost everywhere (.e.) to f : R if there exists N with m(n) = 0 such tht f n f pointwise on \ N. Remrks. Uniform convergence pointwise convergence lmost everywhere convergence. Note tht the converse is not lwys true (xercise: Give exmples). Observtion. Let f n f pointwise, where ech f n is mesurble (on ). For x, if f(x) >, then m such tht f(x) > +. Thus, since f m n f pointwise, for lrge enough n, we will hve f n (x) > +. This mens tht, m m N such tht, n N, f n (x) > + m. ( )
5 5 Conversely, if (*) is stisfied, then f(x) = lim n f n (x) +. Hence, we hve m {x : f(x) > } = {x : ( m )( N )( n N)(f n (x) > + /m)} = ( ( { x : f n (x) > + m} )) F. m N n N Theorem.2 Let F nd f n : R be mesurble function for ll n. If f n f lmost everywhere on, then f is mesurble on. Proof. Since the set of x such tht {f n (x)} fils to converge to f(x) hs mesure zero, which is mesurble set, without loss of generlity by restricting f n s to subset of on which f n f pointwise, we cn ssume tht f n f pointwise on. Then by the observtion bove, for ll R {x : f(x) > } = ( ( { x : f n (x) > + m} )) F. m N n N Corollry. Let F, {f n } be fmily of mesurble functions on. If f n f uniformly or pointwise on, then f is mesurble on. Definition. Let f n be sequence of relvlued mesurble functions on F. Assume ech f n is bounded (i.e. there exits M > 0 such tht f n (x) < M for ll x ). Fix x nd define: f (x) = sup f n (x) n f (x) = inf n(x) n f(x) = lim sup f n (x) n f(x) = lim inf f n (x) n Then f, f, f, nd f re ll welldefined rel vlued functions on. xercise. Let f n be sequence of relvlued mesurble functions on F.. Show tht f, f, f, nd f re ll mesurble on. 2. Show tht f n f lmost everywhere on if nd only if m({x : lim sup f n (x) > lim inf f n (x)}) = 0. Question: If f is mesurble, does it imply tht f is mesurble? The nswer is NO in generl. xercise. Find n exmple of nonmesurble function f for which f is mesurble. Definition. A function f : R which is mesurble nd tkes only finitely mny vlues is clled simple function.
6 6 Fct.4 A function f : R is simple if nd only if there exists, 2,..., n, 2,... n F such tht n = k nd f(x) = k χ k (x). R nd Proof. ( ) follows from the definition. ( ) In order to show tht f is mesurble, ssume, 2,... n re its vlues. Then we hve f ({ k }) = k F. If i j, then i j nd hence i j =, nd n k =. Let n f(x) = k χ k (x). It is left s n exercise to verify tht both sides gree for ll x. Note tht the formul in the Fct.4 is simple function in cnonicl form. Of course, the sme function cn be expressed in mny different wys involving chrcteristic functions of different sets. Fct.5 Let f nd g be simple functions on F, then i) f + g is simple. ii) fg is simple. iii) f is simple (provided g 0). g Proof. xercise. Fct.6 Let F nd f : R be bounded mesurble function. Then for ll ɛ > 0, there exists simple functions φ ɛ nd ψ ɛ on such tht φ ɛ (x) f(x) ψ ɛ (x) on nd 0 (ψ ɛ φ ɛ )(x) < ɛ on. Proof. Since f is bounded there exists [, b] f(). Prtition [, b] into subintervls of length l < ɛ, sy < y 0 < y < < y n < y n = b. For i = 0,,... n let i = f ((y i, y i+ ]) F. Then Also, n n φ ɛ (x) = y i χ i (x) f(x) y i+ χ i (x) = ψ ɛ (x). i=0 i=0 n ψ ɛ (x) φ ɛ (x) = (y i+ y i )χ i (x) < ɛ. i=0 Theorem.3 (Approximtion by Simple Functions) Let F nd f : R #. Then f is mesurble on if nd only if there exists sequence {φ n } of simple functions on such tht φ n f lmost everywhere nd φ n f on. Furthermore, if f 0, then φ n s cn be chosen s φ n f lmost everywhere on. Proof. Since f = f + f nd f = f + + f without loss of generlity, we cn ssume f is nonnegtive (i.e. f 0). Therefore let f 0. Note tht ( ) prt is Theorem.2 bove.
7 ( ) Given n, let n = {x : f(x) n}. Apply the previous fct with ɛ =. So there n exists φ n nd ψ n simple functions on n such tht φ n f ψ n on n nd ψ n φ n < on n n. Now φ n f on n nd f φ n < on n n. xtend φ n to the rest of by letting φ n (x) = n on \ n. Now, we ll show tht φ n f pointwise on. If x is such tht f(x) <, then we hve f(x) < N for some N. Then, for ny n N, since x n for some n, we hve 0 f(x) φ n (x) ; hence, were done. n If x is such tht f(x) =, then tke φ n (x) = n; it follows tht pointwise. Hence φ n f pointwise. φ n (x) f(x) Corollry. Let F, function f : R is bounded nd mesurble if nd only if there exists {φ n }, simple functions on, such tht φ n f uniformly. Proof. xercise. xercises. Let F nd f : R be function. f is mesurble function if nd only if f (O) F, O B. 2. Let {f n } be sequence of Rvlued mesurble functions on F nd f : R be mesurble. f n f lmost everywhere on if nd only if m({x : lim sup f n (x) > lim inf f n (x)}) = 0. n n 3. Find function f : R, F, such tht f is mesurble wheres f is not. 7 III.2 Lebesgue Integrtion of Bounded Mesurble Functions Now tht we sorted out those functions comptible with the (Lebesgue) mesurble sets, we will proceed to define the Lebesgue integrl (of mesurble functions). Throughout we will ssume tht: F with m() <, ll functions f : R re bounded, nd ll simple functions will be in cnonicl form. Definition. Let φ : R be simple function defined s φ(x) = n i= iχ i (x). The (Lebesgue) integrl of φ, denoted by φ(x) dm, is defined s n φ(x) dm = i m( i ). Nottion: i= φ(x) dm = φ(x) dx = φ dm = φ. Remrks.. If we extend φ to R by letting φ(x) = 0 for ll x R \, then φ(x) dm = (φ χ )(x) dm. R 2. The Lebesgue integrl of step function is the sme s the Riemnn integrl of step function.
8 8 xmple. Let = [0, ] nd let φ(x) = χ Q [0,]. Then φ(x) dx = m(q [0, ]) = 0. Hence φ is Lebesgue integrble. Recll φ is not Riemnn integrble! xercise. Clculte [0,] φ, where φ(x) = χ C Q + 2χ C Q c + 3χ [0,]\C. Fct.7 Let φ, ψ : R be simple functions. Then i) (φ + bψ) = φ + b ψ for ll, b R. ii) φ ψ if φ ψ. iii) φ = φ + 2 φ where, 2 F, = 2, nd 2 =. Proof. i) Let φ(x) = n i= iχ i (x) nd ψ(x) = m j= b jχ Fj (x) be in cnonicl form. Let {A k } N be defined s A k = i F j with n pproprite ordering. Then {A k } N = { i F j } n,m i,j= is disjoint collection with N A k =. Hence N φ + bψ = ( i + bb j )χ Ak where i, j corresponds to the set A k with A k = i F j. Therefore, we hve N N N (φ + bψ) = ( i + bb j )m(a k ) = i m(a k ) + bb j m(a k ) = N i m(a k ) + b = φ + b ψ. N b j m(a k ) = n i m( i ) + b ii) Since φ ψ on, ψ φ 0 on. Also since ψ φ is simple, by i) we hve (ψ φ) 0 ψ φ. Proof of iii) is left s n exercise. i= m b j m(f j ) Recll: Given ny f : R bounded, we cn lwys find (construct) simple functions φ nd ψ such tht φ f ψ on nd 0 (ψ φ)(x) cn be mde rbitrrily smll on. Definition. Let F nd f : R be bounded function. We sy tht the function f is (Lebesgue) integrble over if { } { } sup φ : φ f on = inf ψ : ψ f on, where φ nd ψ re simple functions. The common vlue is clled the (Lebesgue) integrl of f over nd is denoted by f(x) dx = f = f dm. j=
9 9 For the simplicity of the rguments below, we will denote: inf ψ = U f nd sup φ = L f. f ψ f φ Theorem.4 Let F, m() <, nd f : R be bounded function. mesurble on if nd only f is Lebesgue integrble over Then f is Proof. ( ) Assume f is mesurble nd f M. Prtition the intervl [ M, M] into 2n subintervls of equl length. Let M = y n < y n+ < < y n < y n = M be the endpoints of these subintervls. Define k = f ([y k, y k+ ]). Then, k F for ll k = n,... n with k l = for k l, nd n m( k ) = m(). Now define φ n = n k= n y kχ k k= n where {ψ n } nd {φ n } re integrble. Then ψ Thus, inf f ψ inf f ψ nd ψ n = n k= n y k+χ k. Therefore, on the set, φ n (x) f(x) ψ n (x), ψ n nd sup f φ φ [ n ] ψ sup φ (ψ n φ n ) = (y k+ y k )χ k < M f φ n m(). k= n Since n is rbitrry, we hve L f = U f nd f is Lebesgue integrble. ( ) Given tht f is bounded on nd L f = U f, let 0 < ɛ n = for n. From these n ssumptions nd the definitions of inf / sup, we cn pick sequence of simple functions {ψ n } nd {φ n } on such tht φ n f ψ n on, nd ( ) 0 ψ n φ n < n. Let ψ n = inf n ψ n nd φ n = sup n φ n. Since ech ψ n nd φ n re mesurble, we hve tht both ψ n nd φ n re mesurble on, nd φ n f ψ n on. We need to show ψ n = φ n lmost everywhere on. To show this let A = {x : φ(x) < ψ(x)}. We will show tht m(a) = 0. To do this let A k = {x : ψ(x) φ(x) > /k}, then k A k = A. Now let B n k = {x : ψ n (x) φ n (x) > /k}. Then A k Bk n F nd, on, k(ψ n φ n ) >. So m(bk n ) = k(ψ n φ n ) = k χ B n k It follows tht m(a k ) = 0. Hence, we hve m(a) = 0. φ n. (ψ n φ n ) < k n.
10 0 Fct.8 Let f : [, b] R be bounded function. If f is Riemnn integrble, then f is Lebesgue integrble with b f dm = f(x) dx. Proof. xercise. [,b] Fct.9 Let F, m() <, f, g : R be bounded functions. Then we hve: ) (αf + βg) = α f + β g for ll α, β R. b) If f = g lmost everywhere on, then f = g. c) If f g lmost everywhere, then f g (Hence f f ). d) If there exists, b R such tht f(x) b on, then m() f bm(). e) If, 2 F, nd = 2 with 2 =, then f = f + f. 2 Proof. xercise. xercise. Let F with m() <, nd {f n } be sequence of rel vlued, bounded mesurble functions on such tht f n f uniformly. Then f n f. The uniform convergence condition in the exercise bove cnnot be relxed s the following exmple shows. xmple. Let f n : [0, ] R, n, be given by n 2 x if x [0, n ] f n (x) = n 2 x + 2n if x [ n, 2 n ] 0 otherwise, nd f 0 on [0, ]. Then, f n = 0 = f! [0,] [0,] Theorem.5 (goroff) Let F with m() <, nd {f n } be sequence of rel vlued mesurble functions on such tht f n f lmost everywhere. Then for ll ɛ > 0, there exists mesurble A ɛ, m( \ A ɛ ) < ɛ, such tht f n f uniformly on A ɛ. Proof. Given ɛ > 0 nd for n, j Z +, define n j = {x : f k (x) f(x) < /j}. k=n
11 Then n j F for ll n, j. If A such tht f n f pointwise on A (i.e. m( \ A) = 0), then for ll x A, x n j for some n, j. Hence A n. j Therefore A ( n j m n= n= j n ) n= = m(), since f n f.e. (hence m(a) = m()). Observe for fixed j, n j n+. j Hence by continuity of m, m( \ n) j = m() m(n) j nd ( ) lim m( \ j n) = m() lim m( j n) = m() m n j = 0. Consequently, for ll j, we cn find n(j) such tht n n(j) nd m( \ n) j < ɛ. Let 2 j A ɛ = j= j n(j), then A ɛ F. Then ( ) m( \ j n(j) ) = m ( \ j n(j) ) m( \ j n(j) ) < ɛ 2 = ɛ. j j= j= j= j Now, it is left s n exercise to show tht f n f uniformly on A ɛ. Remrk. By the exmple bove, if f n f pointwise on F, in generl, we cnnot expect f n f. However, under somewht mild dditionl conditions we hve positive nswer: n= Theorem.6 (Bounded Convergence Theorem) Let F, m() <, nd {f n } be sequence of uniformly bounded mesurble, rel vlued functions on. If f n f lmost everywhere, then f n f. Proof. Since {f n } is uniformly bounded, we cn find positive rel number M such tht f n (x) M. Then, by goroff s Theorem, given ɛ > 0, there exists A such tht m(\a) < ɛ nd f 4M n f uniformly on A. Then, find N lrge enough tht, for n N, f n (x) f(x) < ɛ for ll x A. Now for such n N, 2m(A) f n f = (f n f) + (f n f), nd hence, f n f A (f n f) + A \A \A (f n f) since f n f 2M. Therefore, we hve f n f < ɛ 2Mɛ m(a) + 2m(A) 4M = ɛ. A f n f + \A 2Mdm, Remrk. The BCT cn be restted s sying tht under the given conditions the lim nd cn be interchnged.
12 2 xmples.. Given { if x [/2, ] [/4, /3] [/6, /5]... f(x) = 0 otherwise Define f (x) = χ [/2,] f 2 (x) = χ [/4,/3] + χ [/2,]. f n (x) =. n χ [/(2k),/(2k+)] It is strightforwrd to show tht f n f pointwise (hence, is left s n exercise). Then {f n } is uniformly bounded sequence of mesurble functions. By the Bounded Convergence Theorem, [ n ([ ]) ] f = lim f n = lim m [0,] [0,] 2k, 2k [ n ( = lim 2k ) ] [ 2n ] = lim ( ) k+ 2k k ( ) k+ = = ln(2). k 2. Let f(x) = { 2x if x [0, ] Q x if x [0, ] Q c If we define g(x) = x on [0, ], then f = g lmost everywhere nd g is Riemnn Integrble = Lebesgue Integrble. Hence, f(x) dx = f(x) dx + f(x) dx = g(x) dx = [0,] Q c 0 2. [0,] [0,] Q xercises. Using only the definition, evlute the integrls: [0,] 2. Let C denote the Cntor Set. Find f(x) dx if [0,] x 2 if x C Q f(x) = x if x C Q c if x [0, ] \ C. 3. Prove tht f n f uniformly on A ɛ in goroff s Theorem. x dx nd [0,] x/3 dx. III.3. Lebesgue Integrl of Nonnegtive Functions In the previous section we defined the Lebesgue integrl for bounded mesurble functions. Now, we will continue with not necessrily bounded functions. We will del with this cse in
13 two steps: first we will define the Lebesgue integrl of the functions f : R +, nd then extend it to more generl functions f : R. Definition. Let F (m() = is llowed) nd f : R + be mesurble. We define fdm = sup{ hdm : h : R is bounded, mesurble, h f, m({x : h(x) 0}) < }. Remrk. () It is possible tht f = (2) Functions h : R such tht m({x : h(x) 0}) < re clled functions with finite support. The set {x : h(x) 0} is clled the support of h. 3 xercise. Find [0,] x dm. Fct.0 Let F, f, g : + be mesurble functons. Then () (αf + βg) = α f + β g. (b) If f g lmost everywhere on, then f g. (c) f = f + 2 f where = 2 nd 2 =. Proof. The result follows from the the sme fct on bounded functions; hence, it is left s n exercise. Theorem.7 (Ftou s Lemm) Let F nd {f n } be sequence of R + vlued, mesurble functions on. If f n f lmost everywhere on, then f dm lim inf f n dm. Proof. Let φ be bounded, mesurble, nonnegtive function with m({φ 0}) < such tht φ f lmost everywhere on. For ech n, define h n = f n φ = min{f n, φ}. Then: i) h n is bounded nd mesurble for ll n. ii) h n f n. iii) h n s vnish outside A = supp(φ) = {x : φ(x) 0}. iv) h n φ lmost everywhere on A. Then, by the Bounded Convergence Theorem, φ dm = φ dm = lim h n dm lim inf A A (Lst inequlity follows since lim n f n dm my not exist.) Therefore, f dm = sup φ dm lim inf f n dm. φ f f n dm. Remrks.. Strict inequlity in Ftou s Lemm is possible. 2. If nonnegtivity of f n s is removed, the conclusion of Ftou s Lemm is not vlid nymore.
14 4 xercise. Provide exmples for ech of the remrks mde bove. Theorem.8 (Monotone Convergence Theorem) Let F, nd {f n } be sequence of R + vlued mesurble functions on such tht f n f n+.e. (on ) for ll n. If f n f.e., then lim f n dm = f dm. Proof. From hypothesis, f n f n+ for ll n lmost everywhere on. Hence f n dm f dm for ll n. Therefore by Ftou s Lemm lim sup f n dm f dm lim inf Hence, it follows tht lim f n dm = f dm. Remrks.. MCT Ftou s Lemm. 2. MCT does not hold for Riemnn Integrl. 3. Monotonicity in MCT is essentil. f n dm. xmple. Let = R ([0, ]) nd f n (x) = χ [n, ) (x), n. Then f n 0 pointwise (but not monotoniclly), wheres, = R f n 0 = Corollry. Let F nd {u n } be sequence of positive rel vlued functions on. If f = u n (pointwise), then f dm = u n dm. n= n= R f. Proof. Define f n = n u k, for n. Then f n f.e. on. Apply MCT: n lim f n dm = f dm lim u k dm = u k dm = f dm. Corollry.2 Let F nd { n } F such tht = n n nd i j = if i j. Then for ny f : R + mesurble, f dm = f dm n n= Proof. Let u n = fχ n nd pply Corollry.. Theorem.9 (Chebychev s Inequlity) Let F nd f : R + be mesurble function. Then for ll λ > 0, m({x : f(x) > λ}) f dm. λ
15 Proof. Let λ = {x : f(x) > λ}. Assume first tht m( λ ) < nd let φ = λχ λ. Then 5 on λ, φ f. Hence φ dm λχ λ dm f dm = λm( λ ) m( λ ) f dm. λ Next we will ssume tht m( λ ) =. If we let λ n = λ [ n, n], then λ n λ nd m(λ n) < for ech n. Now, pplying bove cse to n λ nd φ n = λχ n λ, we obtin λm(λ) n f dm. Then, by continuity of m, λm( λ ) = λ lim m( n λ) f dm. Note tht this gives equlity on both sides since m( λ ) =. Hence, m( λ ) = f dm. λ Theorem.0 Let F nd f : R + be mesurble. Then f dm = 0 f = 0 lmost everywhere on. Proof. ( ) Assume f dm = 0 nd consider, for ny n, the set S = {x : f(x) > /n}. Then by Chebychev s Inequlity, Now, since it follows tht f = 0.e. on. m(s) n f dm = 0. {x : f(x) > 0} = n {x : f(x) > /n}, ( ) Next, ssume tht f = 0 lmost everywhere on. Let, for ll n, n = [ n, n]. On ech n let h be bounded, mesurble, nonnegtive function such tht h f, nd φ be ny bounded simple nonnegtive function with φ h (i.e. 0 φ h f.e.). Hence φ must be 0 lmost everywhere on n. Then φ = 0 h = 0. n n Since h is rbitrry, f = 0 follows from the definition. Remrk. Theorem.0 implies tht f = g.e. on F f = g. Definition. Let F. A function f : R + is clled (Lebesgue) integrble over if f dm <.
16 6 xmple. Let s show tht the function f(x) = is Lebesgueintegrble on [, ) nd clculte x 2. For, we will use MCT. First, let, for n, [, ) x 2 f n (x) = x if x n 2 0 if otherwise Hence, ech f n is bounded nd mesurble on [, ). Thus, n f n = f n = x. 2 [, ) [,n] Since is Riemnnintegrble on [, n], it is lso Lebesgue integrble on [, n] (with both x 2 integrls being equl). Hence, f [, ) n = n =. Therefore, by MCT, x 2 n x = lim f 2 n =. n [, ) [, ) Fct. Let F nd f : R + be integrble over. Then f(x) <.e. on. Proof. It is enough to show tht m({x : f(x) = }) = 0. For, first observe tht {x : f(x) = } = n {x : f(x) > n}. Now, by Chebychev s Inequlity, m({x : f(x) = }) m({x : f(x) > n}) f. n Since f <, letting n, we obtin the desired result. Theorem. (Beppo Levi s Lemm) Let {f n } be n incresing sequence of nonnegtive mesurble functions on F. If the sequence { f n} is bounded, then i) f n p f on for some nonnegtive mesurble function f : R #, ii) lim n f n = f <, nd iii) f is finite.e. Proof. For ech x, the sequence {f n (x)} is n incresing sequence of rel numbers; hence, it converges to n extended rel number. Therefore, we cn define f pointwise by f(x) = lim n f n (x), for ll x. Since f n f, by MCT, we hve f n f. Therefore, by the fct tht { f n} is bounded, we hve hve f < ; nd by Fct. bove we must lso hve tht f is finite.e. The following fct is nother importnt property of integrble functions. Theorem.2 Let f : R + be n integrble function on F. Then, for ny ɛ > 0 there exists δ > 0 such tht A, A F, with m(a) < δ, we hve A f < ɛ. Proof. If f < M for some M on, tke δ = ɛ. The the ssertion follows. M If f is rbitrry integrble function, let { f(x) if x n f n (x) = n if x > n.
17 Then {f n } is bounded sequence nd f n p f. on. So, by MCT, f n f. Hence, given ɛ > 0, there exists N such tht if n N, then [ f f n] < ɛ. In prticulr, if n = N, 2 we hve (f f N) < ɛ ɛ. Now, pick δ <, then for ny A with m(a) < δ we hve 2 2N f = (f f N ) + f N (f f N ) + N < ɛ 2 + ɛ 2 = ɛ. A A A A 7 xercise. Let f : R R + be n integrble function. Define set function ν : F [0, ] by ν(a) = fdm, A F. Show tht A i) 0 ν(a) for ll A F. ii) ν(a) ν(b) if A B, A, B F. iii) ν( ) = 0 = ν({}) for ny R. iv) ν( i=a i ) = i= ν(a i) if {A i } F is disjoint fmily of sets. v) ν(a) = 0 if m(a) = 0. Question. For f nd ν s in the exercise bove, wht cn you sy bout i) ν(i) = l(i) for ny intervl I R? ii) ν(a + α) = ν(a) for ny A F nd α R? iii) m(a) = 0 if ν(a) = 0 for A F? III.4. Lebesgue Integrl of Arbitrry Functions In this section, hving defined the (Lebesgue) integrl of nonnegtive mesurble function, we will extend it to rbitrry mesurble functions. First, recll tht (i) f = f + f, nd (ii) if f is mesurble, then so re f + nd f. Definition. A function f : R where F, is clled (Lebesgue) integrble over if both f + nd f re integrble over. In tht cse, f = f + f. Remrk. Since f + nd f re nonnegtive nd integrble, the set on which they tke s vlue is of mesure zero; hence, the set on which f my tke + or vlues is lso of mesure zero. Fct.2 If F, f : R is integrble nd A, A F with m(a) = 0, then f = f. Proof. xercise. Fct.3 Let F, f : R be mesurble nd g : R + be integrble with f g on. Then f is integrble with f g. In prticulr, f f. \A
18 8 Proof. xercise. Fct.4 If F nd f, g : R re integrble over, then () αf is integrble over nd αf = α f. (b) f + g is integrble over nd (f + g) = f + g. (c) f g on, then f g. (d) If = 2, where nd 2 re mesurble nd disjoint, then fχ nd fχ 2 re integrble with f = f + f. 2 Proof. The proof of () nd (c) re left s exercises. For (b) without loss of generlity, if necessry, by removing lrger set of mesure zero, we cn ssume tht f + g is finite on. Thus (f + g) + (f + g) = f + g = (f + f ) + (g + g ) nd (f + g) + + f + g = (f + g) + f + + g +. Since both sides re nonnegtive, we hve (f + g) + + f + g = (f + g) + f + + g +. If we reorgnize we hve [ (f + g) + (f + g) ] = Hence f + g is integrble. (f + + f ) + (g + g ). For the proof of (d), since f is integrble we know tht fχ f nd fχ 2 f. Now both fχ nd fχ 2 re integrble with χ = χ + χ 2. Theorem.3 (Dominted Convergence Theorem) Let {f n } be sequence of rel vlued mesurble functions on F. Let g : R be n integrble function over such tht f n g for ll n on. If f n f lmost everywhere on, then f n nd f re integrble over nd f n f. Proof. Since g is integrble over nd f n g, then ech f n is integrble over. But then, since f n f lmost everywhere, we lso hve f g on. Hence f is integrble over. Observe tht {g f n } is sequence of nonnegtive integrble functions on converging to g f. Similrly {g + f n } is sequence of nonnegtive functions converging to g + f. Now by pplying Ftou s Lemm to {g f n } we hve g f = (g f) lim inf (g f n ) = g lim sup f n f lim sup f n. Next, by pplying Ftou s Lemm to {g + f n } we hve f + g = (g + f) lim inf (g + f n ) = f lim inf f n. g + lim inf f n
19 9 These two inequlities imply tht f n f. xercise. For ech n, let f n (x) = n3/2 x +n 2 x 2, x [0, ]. Show tht () f n 0 lmost everywhere on [0, ]. (b) f n (x) g(x) on [0, ] for ll n where g(x) = (c) lim [0,] f n = 0. Remrk. Converse of the DCT is not vlid! xmples. Let = [0, ]. x.. Let f n (x) = x for ll n nd f(x) = x. Then, f n f, but f n f. 2. Let f n (x) = n n 3. Define x nd f(x) = x. Then f n f, but f n f. f 0 = χ [0,] f = χ [0, 2 ], f 2 = χ [ 2,] f 3 = χ [0, 3 ], f 4 = χ [ 3, 2 3 ], f 5 = χ [ 2 3,] f 6 = χ [0, 4 f 7 = χ [ 4, ], 2 f 8 = χ [ 2, 3 ], 4 f 9 = χ [ 3 4,] f 0 = χ [0, Then we hve f n 0, but f n 0 lmost everywhere. Theorem.4 (Generlized Dominted Convergence Theorem) Let {f n } be sequence of rel vlued mesurble functions on F such tht f n f lmost everywhere on. Let {g n } be sequence of rel vlued nonnegtive functions on such tht g n g lmost everywhere on. Also let f n g n for ll n on. If g n g, then f n f. Proof. xercise. Corollry. Let F nd f : R be integrble over. () If = n (disjoint), n F. Then n= f = n= n f.
20 20 (b) Given { n } F, n nd n. If A = n n, then f = lim f. A n (c) Given { n } F, n, nd n. If A = n n, then f = lim f. A n Proof. xercise. [Hint: Use Corollry. before Chebychev s Inequlity.] Corollry.2 Let R nd f : R be integrble over. Then given ɛ > 0 there exists δ > 0 such tht for ll A F, A nd m(a) < δ, then f < ɛ. Proof. xercise. A The results stted bove hve some importnt nd interesting rmifictions. Let ν(a) = f A for ll A F, where f : R R is integrble. Then by Corollry.() bove, ν A k = ν(a k ). disjoint Also ν( ) = 0, 0 ν(a), A B ν(a) ν(b). Notice tht ν is not mesure; however, it cts like mesure. Such set functions re clled signed mesures. It turns out tht the converse of Corollry.2 is lso vlid if m() <. Fct.5 Let F, m() <, nd f : R be mesurble function on. If for ll ɛ > 0 there exists δ > 0 such tht for ll A F, A nd m(a) < δ, we hve tht f < ɛ, then A f is integrble over. Proof. WLOG we cn ssume tht f 0. Tke ɛ = nd let δ 0 be the corresponding vlue such tht whenever A, A F, with m(a) < δ 0, we hve A f <. Clim: = n k (disjoint) such tht m( k ) < δ 0 for ll k n. (The proof of this clim is left s n exercise.) Hence we now hve fχ k = f < for ll k n. A k So ech fχ k is integrble. Therefore, f = n fχ k is integrble. xercises. Let f : [, b] R be bounded function. If f is Riemnnintegrble, then f is Lebesgueintegrble nd b f(x)dm = f(x)dx. [,b]
21 2. Let F with m() < nd {f n } be sequence of Rvlued, bounded mesurble functions on such tht f n f uniformly on. Then f n dm fdm. [Do not use BCT!] 3. The function f(x) = x is mesurble on [0, ]. Clculte [0,] x dm. [Notice: f is not Riemnnintegrble on [0, ].] 4. Find f(x) dm if [0,] where C is the stndrd Cntor Set. x 2 if x C Q f(x) = x if x C Q c if x [0, ] \ C 5. Let {f n } n be sequence of Rvlued functions on [0, ] defined s { 0 if x (0, n+ f n (x) = ) [, ] n x 3 2 if x [, ). n+ n Show tht ) f n 0.e. b) [0,] f n 0. c) There is no integrble function g : [0, ] R such tht f n g.e. for ll n. How would you reconcile this result with the DCT? 2 III.5. Convergence in Mesure In regrds to convergence of sequence of functions, in ddition to uniform, pointwise nd.e. convergence, we cn dd nother mode of convergence s follows. Definition. Let {f n } be sequence of rel vlued mesurble functions on F nd let ech f n be finite lmost everywhere on. For f : R mesurble nd finite lmost everywhere on, we sy tht the sequence {f n } converge in mesure on to f if for ll α > 0 Nottion: f n m f. lim m({x : f n(x) f(x) α}) = 0. Remrk. It is esy to see tht if f n u f, then f n m f. Indeed, with n dditionl condition we cn replce uniform convergence with.e. convergence by the following result. Fct.6 Let m() < nd {f n } be sequence of rel vlued mesurble functions on F, nd f : R be finite lmost everywhere. Then f n.e. f f n m f. Proof. Clerly f is mesurble on (why?). Let α > 0 nd ɛ > 0 be rbitrry. By goroff s Theorem, there exists F, F F, such tht m( \ F ) < ɛ nd f n u f on F. Hence there exists N such tht when n N we hve f n (x) f(x) < α
22 22 for ll x F. Thus if n N, m({x : f n (x) f(x) α}) m( \ F ) < ɛ. Hence m({x : f n (x) f(x) α}) 0. Remrks.. The condition m() < is essentil. (xercise: Provide counterexmple if m() =.) 2. The converse of Fct.6 is not true in generl, either. (xercise: provide n exmple). However, it s vlid long subsequence: Theorem.5 If f n m f on F, then there exists subsequence {f nk } of {f n } such tht f nk.e. f. Proof. Let ɛ k =. Applying the hypothesis, for ech k there exists N 2 k k such tht when n N k, m({x : f n (x) f(x) }) <. Let k 2 k Then k = {x : f Nk (x) f(x) > k }. m( k ) <. Hence by BorelCntelli Lemm, for lmost every x, the point x belongs to t most finitely mny k s. Tht mens for lmost every x, there exists k x such tht x k for k k x. So, for lmost every x, f nk (x) f(x) < k for ll k k x. Hence f nk p f lmost everywhere on. Remrk. In Ftou s Lemm, MCT nd DCT, the condition f n.e. f on cn be replced by f n m f on. III.6. Riemnn Integrl vs Lebesgue Integrl Recll: Let f : [, b] R be bounded, Ṗ n be tgged prt prtition. Then n S(f, Ṗn) = f(t i )(x i+ x i ) S(f, Ṗn) b i= f =: Riemnn Integrl of f on [, b]. Let s look t this process little differently. For f : [, b] R bounded function nd P n prtition of [, b], define step functions {α n } nd {β n } by n n α n (x) = u k χ [xk,x k+ ](x), β n (x) = v k χ [xk,x k+ ](x), respectively, where u k = inf x [xk,x k+ ]{f(x)} nd v k = sup x [xk,x k+ ]{f(x)}. Then, (i) α n nd β n s n pointwise (or s P n 0). (ii) For ll n, α n nd β n re Lebesgue integrble (by construction).
23 23 (iii) For ll n, α n nd β n re Riemnnintegrble since, b n L n (f) = α n = m k (x k+ x k ) nd U n (f) = b β n = n v k (x k+ x k ), nd tht L n (f) nd U n (f) re specil cses of S(f, Ṗn). We lso hve tht L n (f) nd U n (f). Therefore, f is Riemnnintegrble if nd only if lim n U n = lim n L n (= b f). In generl, since f is bounded, both sequences α n nd β n re uniformly bounded; hence, α n (x) α(x) nd β n (x) β(x) pointwise, for some α : [, b] R nd β : [, b] R. Now, the Dominted Convergence Theorem implies tht L n (f) = α n α, where α = lim α n (x) nd [,b] U n (f) = [,b] [,b] β n [,b] β, where β = lim β n (x). If α = β, then the Lebesgue Integrl of f exists nd is equl to the common vlue. [,b] [,b] Therefore, ny bounded Riemnnintegrble function is Lebesgue integrble. Fct.7 f : [, b] R is continuous t x [, b] if nd only if α(x) = β(x). Proof. xercise. Theorem.6 Let f : [, b] R be bounded. () f is Riemnn Integrble on [, b] if nd only if f is lmost everywhere continuous on [, b]. (b) If f is Riemnn Integrble on [, b], then f is Lebesgue Integrble nd b f = f. Proof. From bove observtion, if () is proven, (b) follows esily. [,b] () ( ) Since f is Riemnn Integrble, then Since α β, we hve [,b] b lim k (β α) = 0 = α k = lim b k b β k b α = b β. (β α) α = β lmost everywhere. Hence by Fct.7, f is continuous lmost everywhere on [, b]. ( ) If f is continuous lmost everywhere, then by Fct.7 for lmost every x [, b] α(x) = β(x)(= f(x)). Since α nd β re Lebesgue Integrble, α = β. [,b] [,b]
24 24 Hence we must hve tht if L n = b α n = [,b] α n nd U n = b β n = then U n L n = 0. Tht is f is Riemnn Integrble nd we must hve b ) f = lim α n (= lim β n. b b [,b] β n, Importnt Remrk. All the concepts introduced nd developed in this chpter cn be rephrsed in n rbitrry σfinite mesure spce (X, A, µ) nd for functions f : X R (with pproprite modifictions). We will leve this s n exercise.
Math 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
More informationThe Regulated and Riemann Integrals
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
More informationChapter 4. Lebesgue Integration
4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σfinite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationNOTES AND PROBLEMS: INTEGRATION THEORY
NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (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
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
More informationEntrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim
1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationarxiv:math/ v2 [math.ho] 16 Dec 2003
rxiv:mth/0312293v2 [mth.ho] 16 Dec 2003 Clssicl Lebesgue Integrtion Theorems for the Riemnn Integrl Josh Isrlowitz 244 Ridge Rd. Rutherford, NJ 07070 jbi2@njit.edu Februry 1, 2008 Abstrct In this pper,
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
More informationThis is a short summary of Lebesgue integration theory, which will be used in the course.
3 Chpter 0 ntegrtion theory This is short summry of Lebesgue integrtion theory, which will be used in the course. Fct 0.1. Some subsets (= delmängder E R = (, re mesurble (= mätbr in the Lebesgue sense,
More informationON THE CINTEGRAL BENEDETTO BONGIORNO
ON THE CINTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives
More informationA PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES
INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. emil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationWeek 7 Riemann Stieltjes Integration: Lectures 1921
Week 7 Riemnn Stieltjes Integrtion: Lectures 1921 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0
More informationA product convergence theorem for Henstock Kurzweil integrals
A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationPrinciples of Real Analysis I Fall VI. Riemann Integration
21355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will
More informationChapter 22. The Fundamental Theorem of Calculus
Version of 24.2.4 Chpter 22 The Fundmentl Theorem of Clculus In this chpter I ddress one of the most importnt properties of the Lebesgue integrl. Given n integrble function f : [,b] R, we cn form its indefinite
More informationGeneralized Riemann Integral
Generlized Riemnn Integrl Krel Hrbcek The City College of New York, New York khrbcek@sci.ccny.cuny.edu July 27, 2014 These notes present the theory of generlized Riemnn integrl, due to R. Henstock nd J.
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationarxiv: v1 [math.ca] 7 Mar 2012
rxiv:1203.1462v1 [mth.ca] 7 Mr 2012 A simple proof of the Fundmentl Theorem of Clculus for the Lebesgue integrl Mrch, 2012 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationSTUDY GUIDE FOR BASIC EXAM
STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationMath 301 Integration I
Mth 301 Integrtion I Lecture notes of Prof. Hichm Gebrn hichm.gebrn@yhoo.com Lebnese University, Fnr, Fll 20162017 http://fs2.ul.edu.lb/mth.htm http://hichmgebrn.wordpress.com 2 Introduction nd orienttion
More informationQuestion 1. Question 3. Question 4. Graduate Analysis I Chapter 5
Grdute Anlysis I Chpter 5 Question If f is simple mesurle function (not necessrily positive) tking vlues j on j, j =,,..., N, show tht f = N j= j j. Proof. We ssume j disjoint nd,, J e nonnegtive ut J+,,
More informationTheory of the Integral
Spring 2012 Theory of the Integrl Author: Todd Gugler Professor: Dr. Drgomir Sric My 10, 2012 2 Contents 1 Introduction 5 1.0.1 Office Hours nd Contct Informtion..................... 5 1.1 Set Theory:
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationChapter 6. Infinite series
Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationProf. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf
Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More information2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals
2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More information1.3 The Lemma of DuBoisReymond
28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBoisReymond We needed extr regulrity to integrte by prts nd obtin the Euler Lgrnge eqution. The following result shows tht, t lest sometimes, the extr
More informationIntroduction to Real Analysis (Math 315) Martin Bohner
ntroduction to Rel Anlysis (Mth 315) Spring 2005 Lecture Notes Mrtin Bohner Author ddress: Version from April 20, 2005 Deprtment of Mthemtics nd Sttistics, University of Missouri Roll, Roll, Missouri 654090020
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationRiemann Stieltjes Integration  Definition and Existence of Integral
 Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed
More informationFourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )
Fourier series. Preliminry mteril on inner products. Suppose V is vector spce over C nd (, ) is Hermitin inner product on V. This mens, by definition, tht (, ) : V V C nd tht the following four conditions
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationThe RiemannLebesgue Lemma
Physics 215 Winter 218 The RiemnnLebesgue Lemm The Riemnn Lebesgue Lemm is one of the most importnt results of Fourier nlysis nd symptotic nlysis. It hs mny physics pplictions, especilly in studies of
More informationPart A: Integration. Alison Etheridge
Prt A: Integrtion Alison theridge Introduction In Mods you lerned how to integrte step functions nd continuous functions on closed bounded intervls. We begin by reclling some of tht theory. Definition.
More information7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series
7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts
More information1. GaussJacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. GussJcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationAdvanced Calculus I (Math 4209) Martin Bohner
Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationChapter 5 : Continuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 216 Néhémy Lim Chpter 5 : Continuous Rndom Vribles Nottions. N {, 1, 2,...}, set of nturl numbers (i.e. ll nonnegtive integers); N {1, 2,...}, set of ll
More informationIntroduction to Some Convergence theorems
Lecture Introduction to Some Convergence theorems Fridy 4, 005 Lecturer: Nti Linil Notes: Mukund Nrsimhn nd Chris Ré. Recp Recll tht for f : T C, we hd defined ˆf(r) = π T f(t)e irt dt nd we were trying
More information