Modelli Clamfim Integrali Multipli 7 ottobre 2015

CLAMFIM Bologna Modell 1 @ Clamfm Integral Multpl 7 ottobre 2015 professor Danele Rtell danele.rtell@unbo.t 1/30?

roduct of σ-algebras Let (Ω 1, A 1, µ 1 ), (Ω 2, A 2, µ 2 ) two measure spaces. ut Ω := Ω 1 Ω 2. We want to buld a measure µ on Ω whch agrees wth the gven measures on Ω 1 and Ω 2. Thus we have to provde the doman,.e. a sutable σ-algebra on Ω. The product σ-algebra A turns out to be the mnmum σ-algebra on Ω whch contans rectangles A 1 A 2 wth A 1 A 1 and A 2 A 2 So ths σ-algebra s generated by the collecton of rectangular sets R = {A 1 A 2 A 1 A 1, A 2 A 2 } 2/30?

Theorem roduct σ-algebra A 1 A 2 s the mnmum σ-algebra such that projectons r 1 : Ω Ω 1, r 1 (ω 1, ω 2 ) = ω 1 r 2 : Ω Ω 2, r 2 (ω 1, ω 2 ) = ω 2 are measurable 3/30?

The most relevant stuaton for applcaton s when Ω 1 = Ω 2 = R, A 1 = A 2 = B Borel σ-algebra n R. Borel set n the plane can be generated n two equvalent ways. 4/30?

The most relevant stuaton for applcaton s when Ω 1 = Ω 2 = R, A 1 = A 2 = B Borel σ-algebra n R. Borel set n the plane can be generated n two equvalent ways. Theorem σ-algebras generated by R = {B 1 B 2 B 1, B 2 B}, I = {I 1 I 2 I 1, I 2 ntervals} are the same. 4/30?

roduct measure: Gudo Fubn Vence 1879, New York 1943 To buld produtct measure we need to work wth σ-fnte measure space (Ω 1, A 1, µ 1 ), (Ω 2, A 2, µ 2 ). 5/30?

roduct measure: Gudo Fubn Vence 1879, New York 1943 To buld produtct measure we need to work wth σ-fnte measure space (Ω 1, A 1, µ 1 ), (Ω 2, A 2, µ 2 ). Ths s the case for Lebesgue measure and for probablty measures. We wll use µ to ndcate product measure on product σ-algebra A 1 A 2. 5/30?

roduct measure: Gudo Fubn Vence 1879, New York 1943 To buld produtct measure we need to work wth σ-fnte measure space (Ω 1, A 1, µ 1 ), (Ω 2, A 2, µ 2 ). Ths s the case for Lebesgue measure and for probablty measures. We wll use µ to ndcate product measure on product σ-algebra A 1 A 2. Frst step: we mpose natural condton: µ (A 1 A 2 ) = µ(a 1 )µ(a 2 ) 5/30?

roduct measure: Gudo Fubn Vence 1879, New York 1943 To buld produtct measure we need to work wth σ-fnte measure space (Ω 1, A 1, µ 1 ), (Ω 2, A 2, µ 2 ). Ths s the case for Lebesgue measure and for probablty measures. We wll use µ to ndcate product measure on product σ-algebra A 1 A 2. Frst step: we mpose natural condton: µ (A 1 A 2 ) = µ(a 1 )µ(a 2 ) To assgn a measure to non rectangular sets we need to ntroduce the noton of secton of a subset A of Ω 1 Ω 2 5/30?

If A Ω 1 Ω 2 and f ω 2 Ω 2 secton of foot ω 2 s the subset of Ω 1 A ω2 = {ω 1 Ω 1 (ω 1, ω 2 ) A} 6/30?

If A Ω 1 Ω 2 and f ω 2 Ω 2 secton of foot ω 2 s the subset of Ω 1 A ω2 = {ω 1 Ω 1 (ω 1, ω 2 ) A} If A Ω 1 Ω 2 and f ω 1 Ω 1 secton of foot ω 1 s the subset of Ω 2 A ω1 = {ω 2 Ω 2 (ω 1, ω 2 ) A} 6/30?

If A Ω 1 Ω 2 and f ω 2 Ω 2 secton of foot ω 2 s the subset of Ω 1 A ω2 = {ω 1 Ω 1 (ω 1, ω 2 ) A} If A Ω 1 Ω 2 and f ω 1 Ω 1 secton of foot ω 1 s the subset of Ω 2 Theorem A ω1 = {ω 2 Ω 2 (ω 1, ω 2 ) A} If A A 1 A 2 then for any ω 2 Ω 2 we have that A ω2 A 1 and for any ω 1 Ω 1 we have that A ω1 A 2 6/30?

Defnton If A A 1 A 2 defne µ (A) = Ω 2 µ 1 (A ω2 ) dµ 2 (ω 2 ) ( ) 7/30?

Defnton If A A 1 A 2 defne µ (A) = µ 1 (A ω2 ) dµ 2 (ω 2 ) Ω 2 Theorem If µ 1, µ 2 are σ-fnte measures, then functons ( ) ω 2 µ 1 (A ω2 ), ω 1 µ 2 (A ω1 ) are, respectvely measurable wth respect to A 2, A 1 and 7/30?

Defnton If A A 1 A 2 defne µ (A) = µ 1 (A ω2 ) dµ 2 (ω 2 ) Ω 2 Theorem If µ 1, µ 2 are σ-fnte measures, then functons ( ) ω 2 µ 1 (A ω2 ), ω 1 µ 2 (A ω1 ) are, respectvely measurable wth respect to A 2, A 1 and µ 2 (A ω1 ) dµ 1 (ω 1 ) = µ 1 (A ω2 ) dµ 2 (ω 2 ) Ω 1 Ω 2 7/30?

Theorem Set functon µ ntroduced n ( ) s a measure. It s unque snce any other measure whch concdes wth µ on rectangles s egual to µ on product σ-algebra A 1 A 2 8/30?

Fubn Theorem on nested ntegrals If f L 1 (Ω 1 Ω 2 ) then functons ω 1 f(ω 1, ω 2 ) dµ 2 (ω 2 ), ω 2 f(ω 1, ω 2 ) dµ 1 (ω 1 ) Ω 2 Ω 1 belong respectvely to L 1 (Ω 1 ), L 1 (Ω 2 ) and the followng formula holds 9/30?

Fubn Theorem on nested ntegrals If f L 1 (Ω 1 Ω 2 ) then functons ω 1 f(ω 1, ω 2 ) dµ 2 (ω 2 ), Ω 2 ω 2 Ω 1 f(ω 1, ω 2 ) dµ 1 (ω 1 ) belong respectvely to L 1 (Ω 1 ), L 1 (Ω 2 ) and the followng formula holds Ω 1 Ω 2 f(ω 1, ω 2 ) d (µ 1 µ 2 ) (ω 1, ω 1 ) = 9/30?

Fubn Theorem on nested ntegrals If f L 1 (Ω 1 Ω 2 ) then functons ω 1 f(ω 1, ω 2 ) dµ 2 (ω 2 ), Ω 2 ω 2 Ω 1 f(ω 1, ω 2 ) dµ 1 (ω 1 ) belong respectvely to L 1 (Ω 1 ), L 1 (Ω 2 ) and the followng formula holds f(ω 1, ω 2 ) d (µ 1 µ 2 ) (ω 1, ω 1 ) = Ω 1 Ω ( 2 ) f(ω 1, ω 2 ) dµ 2 (ω 2 ) dµ 1 (ω 1 ) = Ω 2 Ω 1 9/30?

Fubn Theorem on nested ntegrals If f L 1 (Ω 1 Ω 2 ) then functons ω 1 f(ω 1, ω 2 ) dµ 2 (ω 2 ), Ω 2 ω 2 Ω 1 f(ω 1, ω 2 ) dµ 1 (ω 1 ) belong respectvely to L 1 (Ω 1 ), L 1 (Ω 2 ) and the followng formula holds f(ω 1, ω 2 ) d (µ 1 µ 2 ) (ω 1, ω 1 ) = Ω 1 Ω ( 2 ) f(ω 1, ω 2 ) dµ 2 (ω 2 ) dµ 1 (ω 1 ) = Ω 1 Ω ( 2 ) f(ω 1, ω 2 ) dµ 1 (ω 1 ) dµ 2 (ω 2 ) Ω 1 Ω 2 9/30?

Integral dopp Se A R 2 e x R, la sezone d pede x d A è l sottonseme A x d R: A x := {y R (x, y) A} 10/30?

Integral dopp Se A R 2 e x R, la sezone d pede x d A è l sottonseme A x d R: A x := {y R (x, y) A} Se y R la sezone d pede y d A è: A y := {x R (x, y) A} 10/30?

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Secton s Theorem Let A R 2 measurable. Then f x R, secton A x s a.e. measurable. Moreover x m(a x ) s a measurable functon and m(a) = m(a x )dx R 12/30?

Area del cercho A = {(x, y) x 2 + y 2 1} A x x 13/30?

Area del cercho A = {(x, y) x 2 + y 2 1} A x x A x = [ 1 x 2, 1 x 2] 13/30?

Area del cercho A = {(x, y) x 2 + y 2 1} A x x A x = [ 1 x 2, 1 x 2] = m(a) = 13/30? 1 1 m(a x )dx

qund m(a) = 2 1 1 1 1 x2 dx = 4 1 x2 dx = 4 π 4 = π 0 14/30?

Teorema d Fubn G. Fubn: Il teorema d rduzone per gl ntegral dopp Rend. Sem. Mat. Torno, vo1.9, 1949. Sa A R 2 msurable e sa f L(A). Sa S l sottonseme d R dove le y-sezon d A hanno msura postva S = {x R m(a x ) > 0} Allora A f(x, y)dxdy = S ( ) f(x, y)dy dx A x 15/30?

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Example. If A = { (x, y) R 2 0 x 1, x 2 y x + 1 } evaluate: xy dx dy A 17/30?

1.5 1.0 0.5 18/30?

A xy dx dy = 1 0 ( 1+x x 2 ) xydy dx 19/30?

A xy dx dy = 1 0 ( 1+x x 2 ) xydy dx = 1 0 x [ y 2 2 ] 1+x x 2 dx 19/30?

A xy dx dy = 1 0 A xy dx dy = 1 2 ( 1+x x 2 1 0 ) xydy dx = 1 0 [ y 2 x 2 ] 1+x ( x 5 + x 3 + 2x 2 + x ) dx x 2 dx 19/30?

A xy dx dy = 1 0 ( 1+x x 2 1 ) xydy dx = 1 0 x [ y 2 2 ] 1+x x 2 dx A xy dx dy = 1 2 0 ( x 5 + x 3 + 2x 2 + x ) dx = 5 8 19/30?

Example Gven A = { (x, y) R 2 y 0, y x + 3, y 2x + 3 }. Evaluate: y dx dy A Integraton doman: trangle wth vertces n ( 3 2, 0), (3, 0), (0, 3). 3.0 2.5 2.0 1.5 1.0 0.5 2 20/30?

We frst ntegrate n x and then n y : 3 0 ( 3 y y 3 2 y dx ) dy 21/30?

We frst ntegrate n x and then n y : 3 0 ( 3 y y 3 2 y dx ) dy = 3 0 ( 9 2 y 3 2 y2 ) dy 21/30?

We frst ntegrate n x and then n y : 3 0 ( 3 y y 3 2 y dx ) dy = 3 0 ( 9 2 y 3 ) 2 y2 dy = 27 4 21/30?

Change of varable Let A an open subset n R m. ϕ : A R m s a regular mappng f: (a) ϕ s njectve (b) ϕ has contnuous partal dervatves 22/30?

Change of varable Let A an open subset n R m. ϕ : A R m s a regular mappng f: (a) ϕ s njectve (b) ϕ has contnuous partal dervatves (c) det I ϕ (x) 0 for any x A where the Jacob matrx s... ϕ 1 (x)... I ϕ (x) = (ϕ 1 ϕ n )... ϕ = 2 (x)... x 1 x n............ ϕ m (x)... 22/30?

Example olar Coordnate n the plane ut A = (0, + ) [0, 2π). ϕ : A R 2 ϕ(ρ, ϑ) := (ρ cos ϑ, ρ sn ϑ) s a regular mappng. 23/30?

Example olar Coordnate n the plane ut A = (0, + ) [0, 2π). ϕ : A R 2 ϕ(ρ, ϑ) := (ρ cos ϑ, ρ sn ϑ) s a regular mappng. The Jacob determnant s det I ϕ (ρ, ϑ) = det cos ϑ ρ sn ϑ sn ϑ ρ cos ϑ = ρ > 0 23/30?

Rotaton n the plane Fx α (0, 2π) ϕ : A R 2 ϕ(x, y) := (x cos α y sn α, x sn α + y cos α) s a regular mappng. The Jacob determnant s 1 24/30?

Theorem Let ϕ : A R m a regular mappng wth A measurable. Let f : A R a measurable functon. f L(A) f and only f x f(ϕ(x)) det I ϕ (x) s summable on ϕ 1 (A). In such stuaton we have: f(u)du = f(ϕ(x)) det I ϕ (x) dx A ϕ 1 (A) 25/30?

Gauss Integral We use Fubn Theorem and polar coordnate to show that G := + e x2 dx = π 26/30?

Gauss Integral We use Fubn Theorem and polar coordnate to show that G := + e x2 dx = π Frst step R 2 e (x2 +y2) dxdy = R e x2 dx R e y2 dy = G 2 26/30?

Second step We evaluate G 2 usng polar coordnates e (x2 +y2) dxdy = ρ e ρ2 dρdϑ = 2π R 2 (0,+ ) [0,2π) 0 ρ e ρ2 dρ 27/30?

Second step We evaluate G 2 usng polar coordnates e (x2 +y2) dxdy = ρ e ρ2 dρdϑ = 2π R 2 (0,+ ) [0,2π) thus R 2 e (x2 +y2) dxdy = 2π [ e ρ2 2 ] ρ= ρ=0 = π 0 ρ e ρ2 dρ 27/30?

Fnd Lebesgue measure of A := {(x, y) R 2 x 2 xy + y 2 1} 28/30?

Fnd Lebesgue measure of A := {(x, y) R 2 x 2 xy + y 2 1} Use a π x = 1 u 1 v 2 2 4 rotaton y = 1 u + 1 v 2 2 28/30?

Fnd Lebesgue measure of A := {(x, y) R 2 x 2 xy + y 2 1} Use a π x = 1 u 1 v 2 2 4 rotaton y = 1 u + 1 v 2 2 x 2 xy + y 2 1 s transformed nto 1 2 (u v)2 1 2 (u + v)(u v) + 1 2 (u + v)2 1 28/30?

Fnd Lebesgue measure of A := {(x, y) R 2 x 2 xy + y 2 1} Use a π x = 1 u 1 v 2 2 4 rotaton y = 1 u + 1 v 2 2 x 2 xy + y 2 1 s transformed nto 1 2 (u v)2 1 2 (u + v)(u v) + 1 2 (u + v)2 1 u 2 2 + 3v2 2 1 28/30?

Why does t work? It s not a lucky guess. We have to start wth an undefned rotaton of α { x = cos α u sn α v y = sn α u + cos α v put t nto the equaton whch defnes our doman gettng (u sn α + v cos α) 2 (u cos α v sn α)(u sn α + v cos α) + (u cos α v sn α) 2 1 29/30?

Why does t work? It s not a lucky guess. We have to start wth an undefned rotaton of α { x = cos α u sn α v y = sn α u + cos α v put t nto the equaton whch defnes our doman gettng (u sn α + v cos α) 2 (u cos α v sn α)(u sn α + v cos α) + (u cos α v sn α) 2 1 Then choose the α whch force to vansh the rectangular term uv. Ths term s (sn 2 α cos 2 α)uv so α = π 4 makes the job done 29/30?

Last change of varable, smlar to polar coordnates u = 2 ρ cos ϑ v = 2 3 ρ sn ϑ whch transforms u2 2 + 3v2 2 1 nto ρ2 1 and whch Jacob determnant s ρ so that 2 3 30/30?

Last change of varable, smlar to polar coordnates u = 2 ρ cos ϑ v = 2 3 ρ sn ϑ whch transforms u2 2 + 3v2 2 1 nto ρ2 1 and whch Jacob determnant s ρ so that 2 3 l 2 (A) = 2π 0 1 0 2 3 ρ dρdϑ = 2π 3 30/30?