Tue-12-11-2024 10:10
profe: Nicolás Sirolli
status:
tags: Teoria de la medida
Recordar:
M \mathcal{M} M σ − a ˊ l g e b r a \sigma-álgebra σ − a ˊ l g e b r a
I ∈ M ∀ I I\in\mathcal{M}\quad\forall I I ∈ M ∀ I z ∈ M ∀ z z\in\mathcal{M}\quad\forall z z ∈ M ∀ z
∃ ! μ : M → [ 0 , + ∞ ] \:\exists\:!\mu:\mathcal{M}\to[0,+\infty] ∃ ! μ : M → [ 0 , + ∞ ]
μ ( a , b ) = b − a \mu(a,b)=b-a μ ( a , b ) = b − a σ − \sigma- σ − σ − a d i t i v i d a d \sigma-aditividad σ − a d i t i v i d a d σ − r e g u l a r \sigma-regular σ − r e g u l a r
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Continuidad de la medida Sea ( A n ) n ∈ N ⊆ M mon o ˊ tona. 1. Si A 1 ⊂ A 2 ⊂ ⋯ ⊂ A n . Entonces μ ( ⋃ n ≥ 1 A n ) = lim n → ∞ μ ( A n ) 2. Si A 1 ⊃ A 2 ⊃ ⋯ ⊃ A n , y ( ∃ n 0 ) ( μ ( A n 0 ) < + ∞ ) . Entonces μ ( ⋂ n ≥ 1 A n ) = lim n → ∞ μ ( A n ) . \begin{array}{c}
&\textbf{Continuidad de la medida} \\
&\text{Sea $( A_{n} )_{n \in \mathbb{N}}\subseteq\mathcal{M}$ monótona.}\\
&\text{1. Si $A_{1}\subset A_{2}\subset\dots \subset A_{n}$. }\\
&\text{Entonces $\mu\left( \bigcup_{n\geq 1}A_{n} \right)=\underset{ n \to \infty }{ \lim }\mu(A_{n})$}\\
&\text{2. Si $A_{1}\supset A_{2}\supset\dots \supset A_{n}$, y $(\:\exists\:n_{0})(\mu(A_{n_{0}})<+\infty)$. }\\
&\text{Entonces $\mu\left( \bigcap_{n\geq1}A_{n} \right)=\underset{ n \to \infty }{ \lim }\mu(A_{n})$.}
\end{array} Continuidad de la medida Sea ( A n ) n ∈ N ⊆ M mon o ˊ tona. 1. Si A 1 ⊂ A 2 ⊂ ⋯ ⊂ A n . Entonces μ ( ⋃ n ≥ 1 A n ) = n → ∞ lim μ ( A n ) 2. Si A 1 ⊃ A 2 ⊃ ⋯ ⊃ A n , y ( ∃ n 0 ) ( μ ( A n 0 ) < + ∞ ) . Entonces μ ( ⋂ n ≥ 1 A n ) = n → ∞ lim μ ( A n ) . Dem: {\color{violet} \text{Dem:} } Dem:
Sean A ~ n \widetilde{A}_{n} A n A 1 ~ = A 1 , A n ~ = A n ∖ A n − 1 ( n ≥ 2 ) \widetilde{A_{1}}=A_{1},\widetilde{A_{n}}=A_{n}\setminus A_{n-1}\quad(n\geq 2) A 1 = A 1 , A n = A n ∖ A n − 1 ( n ≥ 2 )
A m ~ ∩ A n ~ = ∅ s i n ≠ m , y ⋃ ˙ n = 1 N A n ~ = A n \widetilde{A_{m}}\cap \widetilde{A_{n}}=\emptyset\:si\:n\neq m,y\quad \dot{\bigcup}_{n=1}^{N} \widetilde{A_{n}}=A_{n} A m ∩ A n = ∅ s i n = m , y ⋃ ˙ n = 1 N A n = A n En particular
⋃ n A n = ⋃ n ˙ A n ~ \bigcup_{n}A_{n}=\dot{\bigcup_{n}}^{}\widetilde{A_{n}} n ⋃ A n = n ⋃ ˙ A n Por la σ − \sigma- σ −
μ ( ⋃ n A n ) = ∑ n μ ( A n ~ ) = lim N → ∞ ∑ n = 1 N μ ( A n ~ ) = lim N → ∞ μ ( A N ) \mu\left( \bigcup_{n}A_{n} \right)=\sum_{n}\mu(\widetilde{A_{n}})=\lim_{ N \to \infty } \sum_{n=1}^{N} \mu(\widetilde{A_{n}})=\lim_{ N \to \infty } \mu(A_{N}) μ ( n ⋃ A n ) = n ∑ μ ( A n ) = N → ∞ lim n = 1 ∑ N μ ( A n ) = N → ∞ lim μ ( A N )
Tomo B n = A n 0 ∖ A n B_{n}=A_{n_{0}}\setminus A_{n} B n = A n 0 ∖ A n B n ⊆ B n + 1 B_{n}\subseteq B_{n+1} B n ⊆ B n + 1
⋃ B n = A n 0 ∖ ⋂ A n \bigcup B_{n}=A_{n_{0}}\setminus \bigcap A_{n} ⋃ B n = A n 0 ∖ ⋂ A n Por prop. 1:
lim n → ∞ μ ( B n ) ⏟ μ ( A n 0 ∖ A n ) = μ ( A n 0 ∖ ⋂ A n ) \lim_{ n \to \infty } \underset{\mu(A_{n_{0}}\setminus A_{n})}{\underbrace{\mu(B_{n})}} =\mu\left( A_{n_{0}}\setminus \bigcap A_{n} \right) n → ∞ lim μ ( A n 0 ∖ A n ) μ ( B n ) = μ ( A n 0 ∖ ⋂ A n ) Así,
lim n → ∞ μ ( A 0 ) − μ ( A n ) = μ ( A n 0 ) − μ ( ⋂ A n ) \lim_{ n \to \infty } \mu(A_{0})-\mu(A_{n})=\mu(A_{n_{0}})-\mu\left( \bigcap A_{n} \right) n → ∞ lim μ ( A 0 ) − μ ( A n ) = μ ( A n 0 ) − μ ( ⋂ A n ) □ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square □ Prop. : {\color{Orange} \text{Prop. :} } Prop. :
A ∈ M , λ ∈ R ⟹ λ + A ∈ M \begin{array}{l}
\text{$A\in\mathcal{M},\lambda \in \mathbb{R}\implies\lambda+A\in\mathcal{M}$ }
\end{array} A ∈ M , λ ∈ R ⟹ λ + A ∈ M Dem: {\color{Orange} \text{Dem:} } Dem: f : R → R ∣ f ( x ) = λ + x f:\mathbb{R}\to \mathbb{R}\:|\:f(x)=\lambda+x f : R → R ∣ f ( x ) = λ + x f ( M ) : = { λ + A : A ⊆ M } ⊆ P ( R ) f(\mathcal{M}):=\{ \lambda+A:A\subseteq\mathcal{M} \}\subseteq\mathcal{P}(\mathbb{R}) f ( M ) := { λ + A : A ⊆ M } ⊆ P ( R )
Ejercicio: f ( M ) f(\mathcal{M}) f ( M ) σ − a ˊ l g e b r a \sigma-álgebra σ − a ˊ l g e b r a f f f f ( M ) = M f(\mathcal{M})=\mathcal{M} f ( M ) = M
I I I ⟺ f ( I ) \iff f(I) ⟺ f ( I ) ⟺ f − 1 ( I ) \iff f^{-1}(I) ⟺ f − 1 ( I ) z z z ⟺ E j e r c i c i o f ( z ) \underset{Ejercicio}{\iff}f(z) E j er c i c i o ⟺ f ( z ) ⟺ f − 1 ( z ) \iff f^{-1}(z) ⟺ f − 1 ( z ) ∀ A \forall A ∀ A ∀ A \forall A ∀ A
f ( M ) ∋ f ( f − 1 ( A ) ⏟ ∈ M ) = A f(\mathcal{M})\ni f(\underset{\in \mathcal{M}}{\underbrace{f^{-1}(A)}} )=A f ( M ) ∋ f ( ∈ M f − 1 ( A ) ) = A ∴ μ es la mas chica f ( M ) ⊇ M \underset{\text{$\mu$ es la mas chica}}{\therefore\: }f(\mathcal{M})\supseteq \mathcal{M} μ es la mas chica ∴ f ( M ) ⊇ M Análogamente, f − 1 ( M ) ⊇ M f^{-1}(\mathcal{M})\supseteq \mathcal{M} f − 1 ( M ) ⊇ M f , M ⊇ f ( M ) f,\mathcal{M} \supseteq f(\mathcal{M}) f , M ⊇ f ( M ) = = =
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Invariante por traslacion A ∈ M , λ ∈ R . Entonces μ ( A + λ ) = μ ( A ) \begin{array}{l}
\textbf{Invariante por traslacion} \\
\text{$A\in\mathcal{M},\lambda \in \mathbb{R}$. Entonces $\mu(A+\lambda)=\mu(A)$ }
\end{array} Invariante por traslacion A ∈ M , λ ∈ R . Entonces μ ( A + λ ) = μ ( A ) Dem: {\color{violet} \text{Dem:} } Dem: ν : M → [ 0 , + ∞ ] \nu:\mathcal{M}\to[0,+\infty] ν : M → [ 0 , + ∞ ]
ν ( A ) = μ ( A + λ ) \nu(A)=\mu(A+\lambda) ν ( A ) = μ ( A + λ ) Quiero ver que μ = ν \mu=\nu μ = ν μ \mu μ
ν ( ( a , b ) ) = μ ( ( a , b ) + λ ⏟ ( a + λ , b + λ ) ) = ( b + λ ) − ( a + λ ) = b − a \nu((a,b))=\mu(\underset{(a+\lambda,b+\lambda)}{\underbrace{(a,b)+\lambda}})=(b+\lambda)-(a+\lambda)=b-a ν (( a , b )) = μ ( ( a + λ , b + λ ) ( a , b ) + λ ) = ( b + λ ) − ( a + λ ) = b − a σ − \sigma- σ − σ − a d i t i v i d a d \sigma-aditividad σ − a d i t i v i d a d G − r e g u l a r i d a d G-regularidad G − r e g u l a r i d a d E > 0 \mathcal{E}>0 E > 0 μ \mu μ G − r e g G-reg G − r e g
∃ V ⊇ A + λ c o n μ ( V ∖ A + λ ) < E \:\exists\:V\supseteq A+\lambda\:con\:\mu(V\setminus A+\lambda)<\mathcal{E} ∃ V ⊇ A + λ co n μ ( V ∖ A + λ ) < E Pongo U = V + ( − λ ) U=V+(-\lambda) U = V + ( − λ ) U ⊇ A U\supseteq A U ⊇ A
ν ( U ∖ A ) = μ ( λ + ( V + ( − λ ) ∖ A ) ) = μ ( V ∖ A + λ ) \nu(U\setminus A)=\mu(\lambda+(V+(-\lambda)\setminus A))=\mu(V\setminus A+\lambda) ν ( U ∖ A ) = μ ( λ + ( V + ( − λ ) ∖ A )) = μ ( V ∖ A + λ ) El conjunto de Cantor
-> 00:00
Dado un intervalo I = a , b I=a,b I = a , b L ( I ) = [ a , a + b − a 3 ] L(I)=\left[ a,a+\frac{b-a}{3} \right] L ( I ) = [ a , a + 3 b − a ] R ( I ) = [ b − b − a 3 , b ] R(I)=\left[ b-\frac{b-a}{3},b \right] R ( I ) = [ b − 3 b − a , b ] F n \mathcal{F}_{n} F n
F 1 = { [ 0 , 1 ] } \mathcal{F}_{1}=\{ [0,1] \} F 1 = {[ 0 , 1 ]} F n + 1 = ⋃ I ∈ I n { L ( I ) , R ( I ) } \mathcal{F}_{n+1}=\bigcup_{I\in I_{n}}\{ L(I),R(I) \} F n + 1 = ⋃ I ∈ I n { L ( I ) , R ( I )}
F 2 = { [ 0 , 1 3 ] , [ 2 3 , 1 ] } \mathcal{F}_{2}=\left\{ \left[ 0, \frac{1}{3} \right],\left[ \frac{2}{3},1 \right] \right\} F 2 = { [ 0 , 3 1 ] , [ 3 2 , 1 ] } F 3 = { [ 0 , 1 9 ] , [ 2 9 , 3 9 ] , [ 6 9 , 7 9 ] , [ 8 9 , 1 ] } \mathcal{F}_{3}=\left\{ \left[ 0,\frac{1}{9} \right],\left[ \frac{2}{9},\frac{3}{9} \right],\left[ \frac{6}{9},\frac{7}{9} \right],\left[ \frac{8}{9},1 \right] \right\} F 3 = { [ 0 , 9 1 ] , [ 9 2 , 9 3 ] , [ 9 6 , 9 7 ] , [ 9 8 , 1 ] }
Sea J n = ⋃ I ∈ F n I n J_{n}=\bigcup_{I\in\mathcal{F}_{n}}I_{n} J n = ⋃ I ∈ F n I n
Def. : {\color{Cyan} \text{Def. :} } Def. :
El conjunto de Cantor es C = ⋂ n ≥ 1 J n ⊆ [ 0 , 1 ] \begin{array}{l}
\text{El conjunto de Cantor es } \\
C=\bigcap_{n\geq 1}J_{n}\subseteq[0,1]
\end{array} El conjunto de Cantor es C = ⋂ n ≥ 1 J n ⊆ [ 0 , 1 ] O b s : Obs: O b s :
J n ⊇ J n + 1 J_{n}\supseteq J_{n+1} J n ⊇ J n + 1 Así:
C C C C ∈ M C\in\mathcal{M} C ∈ M C 0 = ∅ C^{0}=\emptyset C 0 = ∅ C ′ = C C'=C C ′ = C C C C
μ ( C ) ≤ μ ( J n ) ≤ 2 n − 1 ⏟ # J n . 1 3 n − 1 ⏟ l o n g ( I n ) → n → + ∞ 0 \mu(C)\leq \mu(J_{n})\leq \underset{\#J_{n}}{\underbrace{2^{n-1}}} .\underset{long(I_{n})}{\underbrace{\frac{1}{3^{n-1} }}} \underset{n\to +\infty}{\to } 0 μ ( C ) ≤ μ ( J n ) ≤ # J n 2 n − 1 . l o n g ( I n ) 3 n − 1 1 n → + ∞ → 0 Prop. : {\color{Orange} \text{Prop. :} } Prop. :
# C = c \#C=c # C = c Más aún, φ : C → { 0 , 2 } N ∣ φ ( x ) = ( x n ) n ∈ N \varphi:C\to \{ 0,2 \}^{\mathbb{N}}\:|\:\varphi(x)=( x_{n} )_{n \in \mathbb{N}} φ : C → { 0 , 2 } N ∣ φ ( x ) = ( x n ) n ∈ N x ∈ I ∈ F n − 1 x \in I\in\mathcal{F}_{n-1} x ∈ I ∈ F n − 1
x n = { 0 x ∈ L ( I ) 2 x ∈ R ( I ) x_{n}=\begin{cases}
0 & x \in L(I) \\
2 & x \in R(I)
\end{cases} x n = { 0 2 x ∈ L ( I ) x ∈ R ( I ) es biyectiva.
-->21:20
Dem: {\color{Orange} \text{Dem:} } Dem: Ψ : { 0 , 2 } N → C \varPsi:\{ 0,2 \}^{\mathbb{N}}\to C Ψ : { 0 , 2 } N → C ( x n ) n ⊆ { 0 , 2 } N (x_{n})_{n}\subseteq \{ 0,2 \}^{\mathbb{N}} ( x n ) n ⊆ { 0 , 2 } N
I n + 1 = { L ( I n ) x n = 0 R ( I n ) x n = 2 I_{n+1}=\begin{cases}
L(I_{n}) & x_{n}=0 \\
R(I_{n}) & x_{n}=2
\end{cases} I n + 1 = { L ( I n ) R ( I n ) x n = 0 x n = 2 Así, I n ∈ F n ∀ n I_{n}\in\mathcal{F}_{n}\quad\forall n I n ∈ F n ∀ n Ψ ( ( x n ) n ) = x , \varPsi((x_{n})_{n})=x, Ψ (( x n ) n ) = x , { x } = ⋂ n I n \{ x \}=\bigcap_{n}I_{n} { x } = ⋂ n I n x x x
Def. : {\color{Cyan} \text{Def. :} } Def. :
E ∈ M , Sea p : E → { V , F } Decimos que p se cumple en casi todo punto de E si μ ( { x ∈ E : p ( x ) = F } ) = 0 \begin{array}{l}
\text{$E\in\mathcal{M}$, Sea $p:E\to \{ V,F \}$ }\\
\text{Decimos que $p$ se cumple en casi todo punto de $E$ si}\\
\mu(\{ x \in E:p(x)=F \})=0
\end{array} E ∈ M , Sea p : E → { V , F } Decimos que p se cumple en casi todo punto de E si μ ({ x ∈ E : p ( x ) = F }) = 0 N o t : Not: N o t :
p ( x ) para casi todo x ∈ E p(x) \text{ para casi todo } x\in E p ( x ) para casi todo x ∈ E p ( x ) c . t . p . x ∈ E p(x)\:c.t.p.\:x \in E p ( x ) c . t . p . x ∈ E p ( x ) a . e . p(x)\:a.e. p ( x ) a . e . x ∈ E , p ( x ) p . p . x ∈ E x \in E,p(x)\:p.p.x \in E x ∈ E , p ( x ) p . p . x ∈ E
Def. : {\color{Cyan} \text{Def. :} } Def. :
E ⊆ R su funci o ˊ n caracter ı ˊ stica es X E : R → { 0 , 1 } X E ( x ) = { 1 x ∈ E 0 x ∉ E \begin{array}{l}
\text{$E\subseteq \mathbb{R}$ su función característica es $\mathcal{X}_{E}:\mathbb{R}\to \{ 0,1 \}$ }\\
\mathcal{X}_{E}(x)=\begin{cases}
1 & x \in E \\
0 & x \not\in E
\end{cases}
\end{array} E ⊆ R su funci o ˊ n caracter ı ˊ stica es X E : R → { 0 , 1 } X E ( x ) = { 1 0 x ∈ E x ∈ E E j e m p l o : Ejemplo: E j e m pl o : X C ( x ) = 0 p . p . x ∈ [ 0 , 1 ] \mathcal{X}_{C}(x)=0\quad p.p.\:x \in[0,1] X C ( x ) = 0 p . p . x ∈ [ 0 , 1 ] { x ∈ [ 0 , 1 ] : X C ( x ) ≠ 0 } = C \{ x \in[0,1]:\mathcal{X}_{C}(x)\neq0 \}=C { x ∈ [ 0 , 1 ] : X C ( x ) = 0 } = C
No es porque se cumple para todo salvo finitos o numerable, sino que son algo de un conjunto de medida 0. El ejemplo es este que se acaba de dar: el conjunto de Cantor no es numerable y y tiene medida nula
Citas y Comentarios
Franks, 3.1