This is a short summary of Lebesgue integration theory, which will be used in the course.
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1 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, others re not. Generl Assumption 0.2. All the subsets E which we shll encounter in this course re mesurble. Fct 0.3. All mesurble subsets E R hve mesure (= mått m(e, which in simple cses correspond to the totl length of the set. E.g., the mesure of the intervl (, b is b (nd so is the mesure of [, b] nd [, b. Fct 0.4. Some sets E hve mesure zero, i.e., m(e = 0. True for exmple if E consists of finitely mny (or countbly mny points. ( måttet noll The expression.e. = lmost everywhere (n.ö. = nästn överllt mens tht something is true for ll x R, except for those x which belong to some set E with mesure zero. For exmple, the function { 1, x 1 f(x = 0, x > 1 is continuous lmost everywhere. The expression f n (x f(x.e. mens tht the mesure of the set x R for which f n (x f(x is zero. Think: n ll but finitely mny points (this is simplifiction.
2 CHAPTER 0. NTEGRATON THEORY 4 Nottion 0.5. R = (,, C = complex plne. The set of Riemnn integrble functions f : C ( R is n intervl such tht f(x p dx <, 1 p <, though much lrger thn the spce C( of continuous functions on, is not big enough for our purposes. This defect cn be remedied by the use of the Lebesgue integrl insted of the Riemnn integrl. The Lebesgue integrl is more complicted to define nd develop thn the Riemnn integrl, but s tool it is esier to use s it hs better properties. The min difference between the Riemnn nd the Lebesgue integrl is tht the former uses intervls nd their lengths while the ltter uses more generl point sets nd their mesures. Definition 0.6. A function f : C ( R is n intervl is mesurble if there exists sequence of continuous functions f n so tht f n (x f(x for lmost ll x (i.e., the set of points x for which f n (x f(x hs mesure zero. Generl Assumption 0.7. All the functions tht we shll encounter in this course re mesurble. Thus, the word mesurble is understood throughout (when needed. Definition 0.8. Let 1 p <, nd R n intervl. We write f L p ( if (f is mesurble nd f(x p dx <. We define the norm of f in L p ( to be ( 1/p f L p ( = f(x dx p. Physicl interprettion: p = 1 f L 1 ( = f(x dx
3 CHAPTER 0. NTEGRATON THEORY 5 = the totl mss. Probbility density if f(x 0, or size of the totl popultion. ( p = 2 f L 2 ( = f(x 2 dx 1/2 = totl energy (e.g. in n electricl signl, such s lternting current. These two cses re the two importnt ones (we ignore the rest. The third importnt cse is p =. Definition 0.9. f L ( if (f is mesurble nd there exists number M < such tht The norm of f is nd it is denoted by f(x < M.e. f L ( = inf{m : f(x M.e.}, f L ( = ess sup f(x x ( essentil supremum, väsentligt supremum. Think: f L ( = the lrgest vlue of f in if we ignore set of mesure zero. For exmple: f L ( = 1. 0, x < 0 f(x = 2, x = 0 1, x > 0 Definition CC (R = D = the set of (rel or complex-vlued functions on R which cn be differentited s mny times s you wish, nd which vnish outside of bounded intervl (such functions do exist!. CC ( = the sme thing, but the function vnish outside of. = 0 = 0 nfinitely mny derivtives
4 CHAPTER 0. NTEGRATON THEORY 6 Theorem Let R be n intervl. Then C C ( is dense in Lp ( for ll p, 1 p < (but not in L (. Tht is, for every f L p ( it is possible to find sequence f n CC ( so tht lim f n f L p ( = 0. Proof. Strightforwrd (but tkes lot of work. Theorem 0.12 (Ftou s lemm. Let f n (x 0 nd let f n (x f(x.e. s n. Then (if the ltter limit exists. Thus, [ ] lim f n(x f(xdx lim f n (xdx dx lim f n (xdx if f n 0 ( f cn hve no more totl mss thn f n, but it my hve less. Often we hve equlity, but not lwys. Ex. f n (x = { n, 0 x 1/n 0, otherwise. Homework: Compute the limits bove in this cse. Theorem 0.13 (Monotone Convergence Theorem. f 0 f 1 (x f 2 (x... nd f n (x f(x.e., then f(xdx = lim f n (xdx (. Thus, for positive incresing sequence we hve [ ] lim f n(x dx = lim f n (xdx (the mss of the limit is the limit of the msses. Theorem 0.14 (Lebesgue s dominted convergence theorem. (Extremely useful f f n (x f(x.e. nd f n (x g(x.e. nd g(xdx < (i.e., g L 1 (,
5 CHAPTER 0. NTEGRATON THEORY 7 then f(xdx = [ ] lim f n(x dx = lim f n (xdx. Theorem 0.15 (Fubini s theorem. (Very useful for multiple integrls. f f (is mesurble nd then the double integrl is well-defined, nd equl to = = J f(x, y dy dx < x y J J ( ( f(x, ydy dx y J x f(x, ydy dx f(x, ydx dy f f 0, then ll three integrls re well-defined, possibly =, nd if one of them is <, then so re the others, nd they re equl. Note: These theorems re very useful, nd often esier to use thn the corresponding theorems bsed on the Riemn integrl. Theorem 0.16 (ntegrtion by prts à l Lebesgue. Let [, b]be finite intervl, u L 1 ([, b], v L 1 ([, b], Then Proof. Since b s U(t = U( + b b u(tv (t = ( b u(tdt = t u(sds, V (t = V ( + t b u(tv (tdt = [U(tV (t] b U(tv(tdt. b t u(t v(sdsdt Fubini = ( U(b U( V ( + s b v(sds, t [, b]. ( b s u(tdt v(sds. s u(tdt = U(b U( u(tdt = U(b U(s,
6 CHAPTER 0. NTEGRATON THEORY 8 we get b u(tv (tdt = ( U(b U( V ( + b (U(b U(s v(sds = ( U(b U( V ( + U(b ( V (b V ( = U(bV (b U(V ( b U(sv(sds. b U(sv(sds Exmple Sometimes we need test functions with specil properties. Let us tke look how one cn proceed. b(t 1 t b(t = { e 1 t(1 t, 0 < t < 1 0, otherwise. Then we cn show tht b C (R, nd b is test function with compct support. Let B(t = t B(t b(sds nd norm it F(t =. B(1 1 F(t 1 t 0, t 0 F(t = 1, t 1 increse, 0 < t < 1. Further F(t + F(t 1 = 1, t R, clerly true for t 0 nd t 1. For 0 < t < 1 we check the derivtive d 1 (F(t F(1 t = dt B(1 [B (t B (1 t] = 1 [b(t b(1 t] = 0. B(1
7 CHAPTER 0. NTEGRATON THEORY 9 F(1 t F(t Let G(t = F(Nt. Then G increses from 0 to 1 on the intervl 0 t 1 N. G(t + G( 1 N t = F(Nt F(1 Nt = 1, t R. G(1/N t G(t 1/N 1
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