Clamfim Lesson 12 Martedì 27 Novembre 2012 Passage to the limit

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1 CLAMFIM Clamfm Lesson 12 Martedì 27 Novembre 2012 assage to the lmt professor Danele Rtell 1/22?

2 Theorem Generaton of measures Let (X, A, µ) a measure space, f : X [0 + ] nonnegatve and measurable. Then for any A A φ(a) := s a measure on A A fdµ Moreover φ s fnte,.e. φ(x) < + f and only f f L(X, µ) 2/22?

3 Yesterday we proved that A = A n, A A j = = φ(a) φ(a n ). n=1 n=1 Thus to get the thess we have to revert nequalty (A). (A) 3/22?

4 Yesterday we proved that A = A n, A A j = = φ(a) φ(a n ). (A) n=1 n=1 Thus to get the thess we have to revert nequalty (A). Frst observe that f exsts A n such that φ(a n ) = + thess s mmedate snce we have φ(a) φ(a n ). We can then lmt our proof when φ(a n ) < + for any n N. 3/22?

5 For any ε > 0 we can choose a smple functon s such that 0 s f and that A 1 s dµ A 1 f dµ ε 2, A 2 s dµ A 2 f dµ ε 2 4/22?

6 For any ε > 0 we can choose a smple functon s such that 0 s f and that s dµ f dµ ε A 1 A 1 2, s dµ f dµ ε A 2 A 2 2 Thus φ(a 1 A 2 ) = s dµ = A 1 A 2 s dµ + A 1 s dµ φ(a 1 ) + φ(a 2 ) ε A 2 4/22?

7 so we can nfer that φ(a 1 A 2 ) φ(a 1 ) + φ(a 2 ) 5/22?

8 so we can nfer that φ(a 1 A 2 ) φ(a 1 ) + φ(a 2 ) It follows that for any n N we have φ(a 1 A n ) φ(a 1 ) + + φ(a n ) ( ) 5/22?

9 so we can nfer that φ(a 1 A 2 ) φ(a 1 ) + φ(a 2 ) It follows that for any n N we have φ(a 1 A n ) φ(a 1 ) + + φ(a n ) ( ) Snce A A 1 A n from ( ) follows φ(a) n=1 φ(a n ) (B) Thess then follows from (A) and (B). 5/22?

10 assage to the lmt Consder the famly of ntervals I n = [0, n] and defne the sequence of functons f n = 1 n 1 I n 6/22?

11 assage to the lmt Consder the famly of ntervals I n = [0, n] and defne the sequence of functons f n = 1 n 1 I n Observe that f n dx = 1 but snce lm f n = 0 we cannot nterchange R n the lmt wth the ntegral lm n R f n dx R lm f ndx n 6/22?

12 Beppo Lev Theorem (monotone convergence) Let E A and (f n ) a sequence of measurable functons such that for any x E 0 f 1 (x) f 2 (x) (C) ut Then f(x) = lm n f n (x). lm n E f n dµ = E fdµ (D) 7/22?

13 Remark. If f n s decreasng Lev s theorem does not hold. In fact take f n (x) = 1 n for any x R n such way for any n N but f n (x) f(x) = 0 f n (x)dx = 8/22?

14 roof. From (C) we nfer that there exsts α [0, + ] such that f n dµ = α lm n E 9/22?

15 roof. From (C) we nfer that there exsts α [0, + ] such that f n dµ = α On the other sde snce lm n E E f n dµ α E E fdµ we have fdµ (E) 9/22?

16 Fx 0 < c < 1 and a smple functon such that 0 s f and, for any n N defne: E n = {x E f n (x) cs(x)} 10/22?

17 Fx 0 < c < 1 and a smple functon such that 0 s f and, for any n N defne: E n = {x E f n (x) cs(x)} from (C) we nfer E 1 E 2 and from (D) E = n=1 E n 10/22?

18 On the other sde for any n N we have f n dµ f n dµ c sdµ (F) E E n E n Take the lmt n n (F). Snce for the generaton measure theorem the ntegral s a countable addtve set functon we can use the theorem for the ncreasng sequences of nested sets to nfer α c sdµ. E 11/22?

19 On the other sde for any n N we have f n dµ f n dµ c sdµ (F) E E n E n Take the lmt n n (F). Snce for the generaton measure theorem the ntegral s a countable addtve set functon we can use the theorem for the ncreasng sequences of nested sets to nfer α c sdµ. Takng the lmt c 1 we see that α E E sdµ 11/22?

20 Now take the supremum for s α E fdµ Thess follows from (D), (E) and (F). (G) 12/22?

21 Analytc functons Gven a C functon f(x) we say that s real analytc n the nterval (x 0 ρ, x 0 + ρ) f t s possble to expand f(x) n Taylor seres f(x) = n=0 f (n) (x 0 ) (x x 0 ) n n! 13/22?

22 Analytc functons Gven a C functon f(x) we say that s real analytc n the nterval (x 0 ρ, x 0 + ρ) f t s possble to expand f(x) n Taylor seres f (n) (x 0 ) f(x) = (x x 0 ) n n! Examples n=0 1 1 x = n=0 x n, x < 1 13/22?

23 Analytc functons Gven a C functon f(x) we say that s real analytc n the nterval (x 0 ρ, x 0 + ρ) f t s possble to expand f(x) n Taylor seres f (n) (x 0 ) f(x) = (x x 0 ) n n! Examples n=0 1 1 x = e x = n=0 n=0 x n, x < 1 x n n!, x R, 13/22?

24 Analytc functons Gven a C functon f(x) we say that s real analytc n the nterval (x 0 ρ, x 0 + ρ) f t s possble to expand f(x) n Taylor seres f (n) (x 0 ) f(x) = (x x 0 ) n n! Examples e x = n=0 n=0 1 1 x = n=0 x n, x < 1 x n n!, x R, sn x = ( 1) n 1 x 2n 1 (2n 1)!, x R n=1 13/22?

25 Theorem If f n a seres of postve functons on X then n=1 f n dµ = X n=1 n=1 X f n dµ It s an mmedate consequence of Beppo Lev theorem 14/22?

26 Example. Evaluate 1 0 ln x x 2 1 dx 15/22?

27 Example. Evaluate We use the geometrc seres 1 ln x 0 x 2 1 dx x n 1 = 1, x < 1 to nfer 1 x n=1 1 1 x = 2 n=1 x 2(n 1) = n=0 x 2n 15/22?

28 Example. Evaluate We use the geometrc seres then use Lev s theorem 1 0 ln x x 2 1 dx = 1 ln x 0 x 2 1 dx x n 1 = 1, x < 1 to nfer 1 x n=1 1 1 x = n=1 x 2(n 1) = + ln x 1 x 2dx = n=0 n=0 1 0 x 2n ( x 2n ln x) dx. 15/22?

29 Integratng by parts yelds 1 0 ( x 2n ln x) dx = ] 1 [ x2n+1 2n + 1 ln x x 2n 2n + 1 dx (L) 16/22?

30 Integratng by parts yelds 1 0 ( x 2n ln x) dx = = ] 1 [ x2n+1 2n + 1 ln x 0 1 (2n + 1) x 2n 2n + 1 dx (L) 16/22?

31 Integratng by parts yelds 1 0 ( x 2n ln x) dx = = ] 1 [ x2n+1 2n + 1 ln x 0 1 (2n + 1) x 2n 2n + 1 dx (L) so that consderng (L), we can wrte ln x x 2 1 dx = n=0 1 (2n + 1) 2 16/22?

32 Fatou s Lemma Gven (X, A, µ), measure space, and (f n ) n sequence of measurable postve functons on X. Then lm nf f ndµ lm nf n n X X f n dµ 17/22?

33 Domnated convergence Theorem Gven a measure space (X, A, µ) and (f n ) n sequence of measurable functons such that lm f n(x) = f(x) n If there exsts a nonnegatve summable g L(X) such that for any x X and any n N f n (x) g(x) Then lm n X f n dµ = X f dµ 18/22?

34 Esercse. Evaluate lm n 1 ne nx 1 + nx dx 19/22?

35 Esercse. Evaluate lm n 1 Consder, for x [1, + ), h n (x) = ne nx 1 + nx dx n 1 + nx. 19/22?

36 Esercse. Evaluate lm n 1 Consder, for x [1, + ), h n (x) = ne nx 1 + nx dx n. It s a decreasng 1 + nx functon of x n x [1, ) snce h n(x) = (1 + nx) < 0 2 n 2 19/22?

37 Esercse. Evaluate lm n 1 Consder, for x [1, + ), h n (x) = ne nx 1 + nx dx n. It s a decreasng 1 + nx functon of x n x [1, ) snce h n(x) = < 0 Then, snce (1 + nx) 2 and sup x [1, ) lm x n 1 + nx = 0 n 2 n 1 + nx = h n(1) = n 1 + n 19/22?

38 We can nfer that so that h n (x) n 1 + n < 1 ne nx 1 + nx < e nx e x 20/22?

39 We can nfer that so that h n (x) n 1 + n < 1 ne nx 1 + nx < e nx e x In such a way we can use Domnated convergence theorem 20/22?

40 We can nfer that so that h n (x) n 1 + n < 1 ne nx 1 + nx < e nx e x In such a way we can use Domnated convergence theorem lm n 1 ne nx 1 + nx dx = 1 lm n ne nx 1 + nx dx = 20/22?

41 We can nfer that so that h n (x) n 1 + n < 1 ne nx 1 + nx < e nx e x In such a way we can use Domnated convergence theorem lm n 1 ne nx 1 + nx dx = 1 lm n ne nx 1 + nx dx = 1 0 dx = 0 20/22?

42 Dfferentaton under the ntegral sgn If [α, β], ]a, b[ R and f :]a, b[ [α, β] R verfes () for any t ]a, b[ functon x f(t, x) s summable n [α, β] 21/22?

43 Dfferentaton under the ntegral sgn If [α, β], ]a, b[ R and f :]a, b[ [α, β] R verfes () for any t ]a, b[ functon x f(t, x) s summable n [α, β] () for almost any x [α, β] functon t f(t, x) s dfferentable n ]a, b[ 21/22?

44 Dfferentaton under the ntegral sgn If [α, β], ]a, b[ R and f :]a, b[ [α, β] R verfes () for any t ]a, b[ functon x f(t, x) s summable n [α, β] () for almost any x [α, β] functon t f(t, x) s dfferentable n ]a, b[ () for any t ]a, b[ and for almost any x [α, β] there exsts g summable on [α, β] such that f (t, x) t g(x) 21/22?

45 then; s dfferentable and F (t) := β α f(t, x)dx 22/22?

46 then; s dfferentable and F (t) := F (t) = β α β α f(t, x)dx f (t, x)dx t 22/22?