Thu-19-06-2025 03:57
profe: Natalia Accomazzo Scotti
status:
tags:
Def. : {\color{Cyan} \text{Def. :} } Def. :
Sea E ⊆ R , E = ⋃ i = 1 d n E i , E i medible, φ ( x ) = ∑ i = 1 n α i ⋅ X E i ( x ) . Defino \begin{array}{l}
\text{Sea $E\subseteq \mathbb{R},E=\displaystyle \overset{ d }{ \bigcup_{i=1} }^{n}E_{i},E_{i}$ medible, ${\Large\varphi}(x)=\displaystyle \sum_{i=1}^{n}\alpha_{i}\cdot \mathcal{X}_{E_{i}}(x)$. Defino }
\end{array} Sea E ⊆ R , E = i = 1 ⋃ d n E i , E i medible, φ ( x ) = i = 1 ∑ n α i ⋅ X E i ( x ) . Defino ∫ E φ ( x ) d μ ( x ) = ∫ E φ d μ = ∑ i = 1 n α i ⋅ μ ( E i ) \int_{E}{\Large\varphi}(x)\:d\mu(x)=\int_{E}{\Large\varphi}\:d\mu=\sum_{i=1}^{n} \alpha_{i}\cdot \mu(E_{i}) ∫ E φ ( x ) d μ ( x ) = ∫ E φ d μ = i = 1 ∑ n α i ⋅ μ ( E i ) E j e m p l o : Ejemplo: E j e m pl o : φ ( x ) = { 1 x ∈ Q 0 c c {\Large\varphi}(x)=\begin{cases}
1 & x \in \mathbb{Q} \\
0 & cc
\end{cases} φ ( x ) = { 1 0 x ∈ Q cc En el [ 0 , 1 ] , [0,1], [ 0 , 1 ] ,
φ ( x ) = 1 ⋅ X Q ∩ [ 0 , 1 ] ( x ) ⟹ ∫ [ 0 , 1 ] φ d μ = μ ( Q ∩ [ 0 , 1 ] ) = 0 {\Large\varphi}(x)=1\cdot \mathcal{X}_{\mathbb{Q}\cap [0,1]}(x)\implies\int_{[0,1]}{\Large\varphi}\:d\mu =\mu(\mathbb{Q}\cap[0,1])=0 φ ( x ) = 1 ⋅ X Q ∩ [ 0 , 1 ] ( x ) ⟹ ∫ [ 0 , 1 ] φ d μ = μ ( Q ∩ [ 0 , 1 ]) = 0 Def. : {\color{Cyan} \text{Def. :} } Def. :
Teorema \begin{array}{l}
\text{Teorema}
\end{array} Teorema Sea f : E → [ 0 , + ∞ ] f:E\to[0,+\infty] f : E → [ 0 , + ∞ ]
∫ E f d μ = sup { ∫ E φ d μ ∣ 0 ≤ φ ≤ f , φ simple } \int_{E}f\:d\mu=\sup\:\left\{ \int_{E}{\Large\varphi}\:d\mu\bigm| 0\leq {\Large\varphi}\leq f,{\Large\varphi}\text{ simple} \right\} ∫ E f d μ = sup { ∫ E φ d μ 0 ≤ φ ≤ f , φ simple }
El supremo podría valer + ∞ . +\infty. + ∞.
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Sean f , g : E → [ 0 , + ∞ ] medibles. Entonces: 1. 0 ≤ f ≤ g ⟹ ∫ E f d μ ≤ ∫ E g d μ 2. Si ∫ E f d μ < ∞ ⟹ f < ∞ ctp. 3. Si A ⊆ E , A ∈ M ⟹ ∫ A f d μ ≤ ∫ E f d μ \begin{array}{l}
\text{Sean $f,g:E\to[0,+\infty]$ medibles. Entonces:}\\
\text{1. $0\leq f\leq g\implies \displaystyle \int_{E}f\:d\mu\leq \int_{E}g\:d\mu$}\\
\text{2. Si $\displaystyle \int_{E}f\:d\mu<\infty\implies f<\infty$ ctp.}\\
\text{3. Si $A\subseteq E,A \in\mathcal{M}\implies \displaystyle \int_{A}f\:d\mu\leq \int_{E}f\:d\mu$ }\\
\end{array} Sean f , g : E → [ 0 , + ∞ ] medibles. Entonces: 1. 0 ≤ f ≤ g ⟹ ∫ E f d μ ≤ ∫ E g d μ 2. Si ∫ E f d μ < ∞ ⟹ f < ∞ ctp. 3. Si A ⊆ E , A ∈ M ⟹ ∫ A f d μ ≤ ∫ E f d μ Dem: {\color{violet} \text{Dem:} } Dem: ( 1 ) : (1): ( 1 ) :
Si 0 ≤ φ ≤ f ⟹ φ ≤ g . 0\leq {\Large\varphi}\leq f\implies {\Large\varphi}\leq g. 0 ≤ φ ≤ f ⟹ φ ≤ g . ∫ E f d μ ≤ ∫ E g d μ \displaystyle \int_{E}f\:d\mu\leq \int_{E}g\:d\mu ∫ E f d μ ≤ ∫ E g d μ
( 2 ) : (2): ( 2 ) : A = { f = + ∞ } A=\{ f=+\infty \} A = { f = + ∞ } μ ( A ) > 0 \mu(A)>0 μ ( A ) > 0
φ n ( x ) = n ⋅ X A ( x ) + 0 ⋅ X A c ( x ) {\Large\varphi}_{n}(x)=n\cdot\mathcal{X}_{A} (x)+0\cdot\mathcal{X}_{A^{c} }(x) φ n ( x ) = n ⋅ X A ( x ) + 0 ⋅ X A c ( x ) ∫ E φ n ( x ) d μ ( x ) = n ⋅ μ ( A ) ⟶ n → ∞ + ∞ \int_{E}{\Large\varphi}_{n}(x)\:d\mu(x)=n\cdot \mu(A)\underset{ n\to \infty }{ \longrightarrow } +\infty ∫ E φ n ( x ) d μ ( x ) = n ⋅ μ ( A ) n → ∞ ⟶ + ∞ ⟹ ∫ E f d μ = + ∞ . \implies \int_{E}f\:d\mu=+\infty. ⟹ ∫ E f d μ = + ∞. Absurdo.
( 3 ) : (3): ( 3 ) :
A ⊆ E , A ∈ M A\subseteq E,A \in\mathcal{M} A ⊆ E , A ∈ M φ {\Large\varphi} φ φ ( x ) = ∑ i = 1 n α i ⋅ X E i ( x ) {\Large\varphi}(x)=\displaystyle \sum_{i=1}^{n}\alpha_{i}\cdot\mathcal{X}_{E_{i}}(x) φ ( x ) = i = 1 ∑ n α i ⋅ X E i ( x ) x ∈ A x \in A x ∈ A
φ ( x ) = ∑ i = 1 n α i ⋅ X E i ∩ A ( x ) , A = ⋃ i = 1 d E i ∩ A {\Large\varphi}(x)=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}\cap A}(x),\quad A=\overset{ d }{ \bigcup_{ i=1 }} E_{i}\cap A φ ( x ) = i = 1 ∑ n α i ⋅ X E i ∩ A ( x ) , A = i = 1 ⋃ d E i ∩ A
E i ∩ A \displaystyle E_{i}\cap A E i ∩ A
∫ A φ ( x ) d μ ( x ) = ∑ i = 1 n α i ⋅ μ ( E i ∩ A ) ≤ ∑ i = 1 n α i ⋅ μ ( E i ) = ∫ E φ d μ \int_{A}{\Large\varphi}(x)\:d\mu(x)=\sum_{i=1}^{n} \alpha_{i}\cdot \mu(E_{i}\cap A)\leq \sum_{i=1}^{n} \alpha_{i}\cdot \mu(E_{i})=\int_{E}{\Large\varphi}\:d\mu ∫ A φ ( x ) d μ ( x ) = i = 1 ∑ n α i ⋅ μ ( E i ∩ A ) ≤ i = 1 ∑ n α i ⋅ μ ( E i ) = ∫ E φ d μ Tomando supremos
∫ A f d μ ≤ ∫ E f d μ \int_{A}f\:d\mu\leq \int_{E}f\:d\mu ∫ A f d μ ≤ ∫ E f d μ O b s : Obs: O b s : φ : E → R {\Large\varphi}:E\to \mathbb{R} φ : E → R φ ( x ) = ∑ i = 1 n α i ⋅ X E i ( x ) {\Large\varphi}(x)=\displaystyle \sum_{i=1}^{n}\alpha_{i}\cdot\mathcal{X}_{E_{i}}(x) φ ( x ) = i = 1 ∑ n α i ⋅ X E i ( x ) x ∈ A x \in A x ∈ A
φ ( x ) ⋅ X A ( x ) = ∑ i = 1 n α i ⋅ X E i ∩ A ( x ) {\Large\varphi}(x)\cdot\mathcal{X}_{A} (x)=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}\cap A} (x) φ ( x ) ⋅ X A ( x ) = i = 1 ∑ n α i ⋅ X E i ∩ A ( x ) ∫ E φ ( x ) ⋅ X A ( x ) d μ ( x ) = ∫ A φ ( x ) d μ ( x ) \int_{E}{\Large\varphi}(x)\cdot\mathcal{X}_{A} (x)\:d\mu(x)=\int_{A}{\Large\varphi}(x)\:d\mu(x) ∫ E φ ( x ) ⋅ X A ( x ) d μ ( x ) = ∫ A φ ( x ) d μ ( x ) También vale para f : E → [ 0 , + ∞ ] f:E\to[0,+\infty] f : E → [ 0 , + ∞ ]
∫ E f ⋅ X A d μ = ∫ A f d μ \int_{E}f\cdot\mathcal{X}_{A}\:d\mu=\int_{A}f\:d\mu ∫ E f ⋅ X A d μ = ∫ A f d μ A ⊆ E , A ∈ M A\subseteq E,A \in\mathcal{M} A ⊆ E , A ∈ M
Multiplicar por característicos es como intersecar conjuntos.
Si E = A ⋃ d B E=A\overset{ d }{ \bigcup} B E = A ⋃ d B φ : E → R , X E = X A + X B {\Large\varphi}:E\to \mathbb{R},\mathcal{X}_{E}=\mathcal{X}_{A}+\mathcal{X}_{B} φ : E → R , X E = X A + X B
∫ E φ d μ = ∫ E φ ( X A + X B ) d μ = ∫ A φ d μ + ∫ B φ d μ \int_{E}{\Large\varphi}\:d\mu=\int_{E}{\Large\varphi}(\mathcal{X}_{A} +\mathcal{X}_{B} )\:d\mu=\int _{A}{\Large\varphi}\:d\mu+\int_{B}{\Large\varphi}\:d\mu ∫ E φ d μ = ∫ E φ ( X A + X B ) d μ = ∫ A φ d μ + ∫ B φ d μ
Esto por linealidad, que todavía no vimos.
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Sea ( f n ) : E → [ 0 , + ∞ ] medibles tal que 0 ≤ f n ≤ f n + 1 ∀ n ∈ N Sea f ( x ) = lim n → ∞ f n ( x ) Entonces ∫ E f d μ = lim n → ∞ ∫ E f n d μ \begin{array}{l}
\text{Sea $(f_{n}):E\to[0,+\infty]$ medibles tal que $0\leq f_{n}\leq f_{n+1}\quad\forall n\in \mathbb{N}$}\\
\text{Sea $f(x)=\underset{ n\to \infty }{ \lim }f_{n}(x)$ }\\
\text{Entonces $\displaystyle \int_{E}f\:d\mu=\underset{ n\to \infty }{ \lim }\int_{E} f_{n} \, d\mu$ }
\end{array} Sea ( f n ) : E → [ 0 , + ∞ ] medibles tal que 0 ≤ f n ≤ f n + 1 ∀ n ∈ N Sea f ( x ) = n → ∞ lim f n ( x ) Entonces ∫ E f d μ = n → ∞ lim ∫ E f n d μ
f ( x ) f(x) f ( x ) ∀ x ( f n ) \forall x(f_{n}) ∀ x ( f n ) f ( x ) f(x) f ( x ) + ∞ . +\infty. + ∞.
Dem: {\color{violet} \text{Dem:} } Dem:
Como f n ≤ f ⟹ ∫ E f n d μ ≤ ∫ E f d μ ∀ n f_{n}\leq f\implies \displaystyle \int_{E} f_{n} \, d\mu\leq \int_{E} f \, d\mu \quad\forall n f n ≤ f ⟹ ∫ E f n d μ ≤ ∫ E f d μ ∀ n
⟹ lim n → ∞ ∫ E f n d μ ≤ ∫ E f d μ \implies \underset{ n\to \infty }{ \lim } \int_{E} f_{n} \, d\mu \leq \int_{E} f \, d\mu ⟹ n → ∞ lim ∫ E f n d μ ≤ ∫ E f d μ Por el lema, 0 ≤ φ ≤ f , φ 0\leq {\Large\varphi}\leq f,{\Large\varphi} 0 ≤ φ ≤ f , φ
⟹ ∫ E φ d μ ≤ lim n → ∞ ∫ E f n d μ \implies \int_{E} {\Large\varphi} \, d\mu \leq \underset{ n\to \infty }{ \lim } \int_{E} f_{n} \, d\mu ⟹ ∫ E φ d μ ≤ n → ∞ lim ∫ E f n d μ ∫ E f d μ = sup { ∫ E φ d μ ∣ 0 ≤ φ ≤ f φ simple } \int_{E} f \, d\mu =\sup \left\{ \int_{E} {\Large\varphi} \, d\mu \bigm| 0\leq {\Large\varphi}\leq f\quad {\Large\varphi}\text{ simple} \right\} ∫ E f d μ = sup { ∫ E φ d μ 0 ≤ φ ≤ f φ simple } Tomando supremo,
∫ E f d μ ≤ lim n → ∞ ∫ E f n d μ \int_{E} f \, d\mu \leq \underset{ n\to \infty }{ \lim } \int_{E} f_{n} \, d\mu ∫ E f d μ ≤ n → ∞ lim ∫ E f n d μ Q.E.D. □ \quad \quad \quad \quad \quad \quad \quad \quad \boxed{\text{Q.E.D.}} \quad \square Q.E.D. □ Teorema : {\color{violet} \text{Teorema :} } Teorema :
Si f , g ≥ 0 medibles ⟹ ∫ E f + g d μ = ∫ E f d μ + ∫ E g d μ Adem a ˊ s, E = A ⋃ d B , A , B ∈ M ⟹ ∫ E f d μ = ∫ A f d μ + ∫ B f d μ \begin{array}{l}
\text{Si $f,g\geq 0$ medibles $\implies \displaystyle \int_{E} f+g \, d\mu=\int_{E} f \, d\mu +\int_{E} g \, d\mu$ }\\
\text{Además, $E=A\overset{ d }{ \bigcup}B$, $A,B \in\mathcal{M}$ $\displaystyle\implies \int_{E} f \, d\mu=\int_{A} f \, d\mu+\int_B f \, d\mu$ }
\end{array} Si f , g ≥ 0 medibles ⟹ ∫ E f + g d μ = ∫ E f d μ + ∫ E g d μ Adem a ˊ s, E = A ⋃ d B , A , B ∈ M ⟹ ∫ E f d μ = ∫ A f d μ + ∫ B f d μ Dem: {\color{violet} \text{Dem:} } Dem: f f f g g g
f ( x ) = ∑ i = 1 n α i ⋅ X E i ( x ) E = ⋃ i = 1 d n E i f(x)=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}}(x)\quad E=\overset{ n }{ \overset{ d }{ \bigcup_{ i=1 }} } E_{i} f ( x ) = i = 1 ∑ n α i ⋅ X E i ( x ) E = i = 1 ⋃ d n E i g ( x ) = ∑ j = 1 m β i ⋅ X F i ( x ) E = ⋃ j = 1 d m F j g(x)=\sum_{j=1}^{m} \beta_{i}\cdot\mathcal{X}_{F_{i}}(x)\quad E=\overset{ m }{ \overset{ d }{ \bigcup_{ j=1 }} } F_{j} g ( x ) = j = 1 ∑ m β i ⋅ X F i ( x ) E = j = 1 ⋃ d m F j La suma de simples da simple:
( f + g ) ( x ) = ∑ i = 1 n ∑ j = 1 m ( α i + β j ) ⋅ X E i ∩ F i ( x ) (f+g)(x)=\sum_{i=1}^{n} \sum_{j=1}^{m} (\alpha_{i}+\beta_{j})\cdot\mathcal{X}_{E_{i}\cap F_{i}} (x) ( f + g ) ( x ) = i = 1 ∑ n j = 1 ∑ m ( α i + β j ) ⋅ X E i ∩ F i ( x ) ∫ E f + g d μ = ∑ i = 1 n ∑ j = 1 m ( α i + β j ) ⋅ μ ( E i ∩ F i ) \int_{E} f+g \, d\mu =\sum_{i=1}^{n} \sum_{j=1}^{m} (\alpha_{i}+\beta_{j})\cdot \mu(E_{i}\cap F_{i}) ∫ E f + g d μ = i = 1 ∑ n j = 1 ∑ m ( α i + β j ) ⋅ μ ( E i ∩ F i ) = ∑ i = 1 n ∑ j = 1 m α i ⋅ μ ( E i ∩ F i ) + ∑ i = 1 n ∑ j = 1 m β j ⋅ μ ( E i ∩ F i ) = ∑ i = 1 n α i ⋅ μ ( E i ) + ∑ j = 1 m β j ⋅ μ ( F j ) =\sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{i}\cdot \mu(E_{i}\cap F_{i})+\sum_{i=1}^{n} \sum_{j=1}^{m} \beta_{j}\cdot \mu(E_{i}\cap F_{i})=\sum_{i=1}^{n} \alpha_{i}\cdot \mu(E_{i})+\sum_{j=1}^{m} \beta_{j}\cdot \mu(F_{j}) = i = 1 ∑ n j = 1 ∑ m α i ⋅ μ ( E i ∩ F i ) + i = 1 ∑ n j = 1 ∑ m β j ⋅ μ ( E i ∩ F i ) = i = 1 ∑ n α i ⋅ μ ( E i ) + j = 1 ∑ m β j ⋅ μ ( F j ) = ∫ E f d μ + ∫ E g d μ =\int_{E} f \, d\mu +\int_{E} g \, d\mu = ∫ E f d μ + ∫ E g d μ
E i = ⋃ j = 1 d m E i ∩ F j ⟹ μ ( E i ) = ∑ j = 1 m μ ( E i ∩ F j ) E_{i}=\displaystyle \overset{ m }{ \overset{ d }{ \bigcup_{j=1}} } E_{i}\cap F_{j}\implies \mu(E_{i})=\sum_{j=1}^{m}\mu(E_{i}\cap F_{j}) E i = j = 1 ⋃ d m E i ∩ F j ⟹ μ ( E i ) = j = 1 ∑ m μ ( E i ∩ F j )
Si f , g f,g f , g ∃ ( φ n ) , ( ψ n ) \:\exists\:({\Large\varphi}_{n}),({\Large\psi}_{n}) ∃ ( φ n ) , ( ψ n ) 0 ≤ φ n ≤ φ n + 1 , 0 ≤ ψ n ≤ ψ n + 1 0\leq {\Large\varphi}_{n}\leq {\Large\varphi}_{n+1},0\leq {\Large\psi}_{n}\leq {\Large\psi}_{n+1} 0 ≤ φ n ≤ φ n + 1 , 0 ≤ ψ n ≤ ψ n + 1 φ n ⟶ f {\Large\varphi}_{n}\longrightarrow f φ n ⟶ f ψ n ⟶ g {\Large\psi}_{n}\longrightarrow g ψ n ⟶ g h n = ( φ n + ψ n ) h_{n}=({\Large\varphi}_{n}+{\Large\psi}_{n}) h n = ( φ n + ψ n ) 0 ≤ h n ≤ h n + 1 0\leq h_{n}\leq h_{n+1} 0 ≤ h n ≤ h n + 1 h n ⟶ f + g h_{n}\longrightarrow f+g h n ⟶ f + g
∫ E f + g d μ = c o n v . m o n . lim n → ∞ ∫ E φ n + ψ n d μ \displaystyle \int_{E} f+g \, d\mu\underset{ conv. mon. }{ = }\underset{ n\to \infty }{ \lim }\int_{E} {\Large\varphi}_{n}+{\Large\psi}_{n} \, d\mu ∫ E f + g d μ co n v . m o n . = n → ∞ lim ∫ E φ n + ψ n d μ = lim n → ∞ ∫ E φ n d μ + ∫ E ψ n d μ = Linealidad del lim. y conv. mon. ∫ E f d μ + ∫ E g d μ =\underset{ n\to \infty }{ \lim } \int_{E} {\Large\varphi}_{n} \, d\mu +\int_{E} {\Large\psi}_{n} \, d\mu \underset{ \text{Linealidad del lim. y conv. mon.} }{ = }\int_{E} f \, d\mu +\int_{E} g \, d\mu = n → ∞ lim ∫ E φ n d μ + ∫ E ψ n d μ Linealidad del lim. y conv. mon. = ∫ E f d μ + ∫ E g d μ Q.E.D. □ \quad \quad \quad \quad \quad \quad \quad \quad \boxed{\text{Q.E.D.}} \quad \square Q.E.D. □ Def. : {\color{Cyan} \text{Def. :} } Def. :
Teorema \begin{array}{l}
\text{Teorema}
\end{array} Teorema Sea f : E → R ∪ { − ∞ , + ∞ } f:E\to \mathbb{R}\cup \{ -\infty,+\infty \} f : E → R ∪ { − ∞ , + ∞ }
f + ( x ) = { f ( x ) f ( x ) ≥ 0 0 f ( x ) ≤ 0 f − ( x ) = { − f ( x ) f ( x ) ≤ 0 0 f ( x ) ≥ 0 f^{+} (x)=\begin{cases}
f(x) & f(x)\geq 0 \\
0 & f(x)\leq 0
\end{cases}
\quad f^{-} (x)=\begin{cases}
-f(x) & f(x)\leq 0 \\
0 & f(x)\geq 0
\end{cases} f + ( x ) = { f ( x ) 0 f ( x ) ≥ 0 f ( x ) ≤ 0 f − ( x ) = { − f ( x ) 0 f ( x ) ≤ 0 f ( x ) ≥ 0 Decimos que ∫ E f d μ \displaystyle \int_{E} f \, d\mu ∫ E f d μ ∫ E f + d μ < ∞ \displaystyle \int_{E} f^{+} \, d\mu<\infty ∫ E f + d μ < ∞ ∫ E f − d μ < ∞ \displaystyle \int_{E} f^{-} \, d\mu<\infty ∫ E f − d μ < ∞
∫ E f d μ = ∫ E f + d μ − ∫ E f − d μ \int_{E} f \, d\mu =\int_{E} f^{+} \, d\mu -\int_{E} f^{-} \, d\mu ∫ E f d μ = ∫ E f + d μ − ∫ E f − d μ
Puede dar + ∞ +\infty + ∞ − ∞ -\infty − ∞ ∫ E f + d μ \int_{E} f^{+} \, d\mu ∫ E f + d μ ∫ E f − d μ \int_{E} f^{-}\, d\mu ∫ E f − d μ ∞ \infty ∞
Decimos que f f f ∣ ∫ E f d μ ∣ < ∞ \left| \displaystyle \int_{E} f \, d\mu \right|<\infty ∫ E f d μ < ∞ ∫ E f + d μ < ∞ \int_{E} f^{+} \, d\mu<\infty ∫ E f + d μ < ∞ ∫ E f − d μ < ∞ \int_{E} f^{-} \, d\mu<\infty ∫ E f − d μ < ∞
Integrable ≠ ∃ \neq \:\exists\: = ∃
Prop. : {\color{Orange} \text{Prop. :} } Prop. :
Sean f , g : E → R ∪ { − ∞ , + ∞ } medibles. \begin{array}{l}
\text{Sean $f,g:E\to \mathbb{R}\cup \{ -\infty,+\infty \}$ medibles.}
\end{array} Sean f , g : E → R ∪ { − ∞ , + ∞ } medibles.
Si ∃ ∫ E f d μ ⟹ ∣ ∫ E f d μ ∣ ≤ ∫ E ∣ f ∣ d μ \:\exists\:\displaystyle \int_{E} f \, d\mu\implies \left| \int_{E} f \, d\mu \right|\leq \int_{E} |f| \, d\mu ∃ ∫ E f d μ ⟹ ∫ E f d μ ≤ ∫ E ∣ f ∣ d μ
f f f ⟺ ∣ f ∣ \iff |f| ⟺ ∣ f ∣ f ≤ g ( f + ≤ g + g − ≤ f − ) ⟹ ∫ E f d μ ≤ ∫ E g d μ \underset{ (f^{+}\leq g^{+}\quad g^{-}\leq f^{-}) }{ f\leq g }\implies \displaystyle \int_{E}f \, d\mu\leq \int_{E} g \, d\mu ( f + ≤ g + g − ≤ f − ) f ≤ g ⟹ ∫ E f d μ ≤ ∫ E g d μ Si ∫ E f d μ \displaystyle \int_{E} f \, d\mu ∫ E f d μ λ ∈ R ⟹ ∫ E λ ⋅ f d μ = λ ⋅ ∫ E f d μ \lambda \in \mathbb{R}\implies \displaystyle \int_{E} \lambda\cdot f \, d\mu=\lambda\cdot \int_{E} f \, d\mu λ ∈ R ⟹ ∫ E λ ⋅ f d μ = λ ⋅ ∫ E f d μ
∫ E f + g d μ = ∫ E f d μ + ∫ E g d μ \displaystyle \int_{E} f+g \, d\mu=\int_{E} f \, d\mu+\int_{E} g \, d\mu ∫ E f + g d μ = ∫ E f d μ + ∫ E g d μ E = A ⋃ d B ⟹ ∫ E f d μ = ∫ A f d μ + ∫ B f d μ E=A\overset{ d }{ \bigcup} B\implies \displaystyle \int_{E} f \, d\mu=\int_{A} f \, d\mu+\int_{B}f \, d\mu E = A ⋃ d B ⟹ ∫ E f d μ = ∫ A f d μ + ∫ B f d μ
Dem: {\color{Orange} \text{Dem:} } Dem: ( 1 ) : (1): ( 1 ) :
∣ ∫ E f d μ ∣ = ∣ ∫ E f + d μ − ∫ E f − d μ ∣ ≤ ∫ E f + d μ + ∫ E f − d μ = ∫ E f + + f − d μ = \left| \int_{E} f \, d\mu \right| =\left| \int_{E} f^{+} \, d\mu -\int_{E} f^{-} \, d\mu \right| \leq \int_{E} f^{+} \, d\mu +\int_{E} f^{-} \, d\mu =\int_{E} f^{+} +f^{-} \, d\mu = ∫ E f d μ = ∫ E f + d μ − ∫ E f − d μ ≤ ∫ E f + d μ + ∫ E f − d μ = ∫ E f + + f − d μ = = ∫ E ∣ f ∣ d μ =\int_{E} |f| \, d\mu = ∫ E ∣ f ∣ d μ
La (2) sale de (1). El resto son tarea, el (5) es más complicado.
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Si f : [ a , b ] → R integrable Riemann ⟹ f es integrable y ∫ a b f ( x ) d x = ∫ [ a , b ] f d μ \begin{array}{l}
\text{Si $f:[a,b]\to \mathbb{R}$ integrable Riemann $\implies f$ es integrable }\\
\text{y $\displaystyle \int_{a}^{b} f(x) \, dx=\int_{[a,b]} f \, d\mu$ }
\end{array} Si f : [ a , b ] → R integrable Riemann ⟹ f es integrable y ∫ a b f ( x ) d x = ∫ [ a , b ] f d μ Dem: {\color{violet} \text{Dem:} } Dem:
Teorema : {\color{violet} \text{Teorema :} } Teorema :
Sea ( f n ) : E → R ∪ { − ∞ , + ∞ } medibles tal que f n → f (limite puntual.) Si ∃ ϕ integrable tal que ∣ f n ∣ ≤ ϕ ∀ n ∈ N ⟹ ∫ E f d μ = lim n → ∞ ∫ E f n d μ \begin{array}{l}
\text{Sea $(f_{n}):E\to \mathbb{R}\cup \{ -\infty,+\infty \}$ medibles tal que $f_{n}\to f$ (limite puntual.)}\\
\text{Si $\:\exists\:\phi$ integrable tal que $|f_{n}|\leq \phi \quad\forall n\in \mathbb{N}$ }\\
\implies \int_{E} f \, d\mu =\underset{ n\to \infty }{ \lim } \int_{E} f_{n} \, d\mu
\end{array} Sea ( f n ) : E → R ∪ { − ∞ , + ∞ } medibles tal que f n → f (limite puntual.) Si ∃ ϕ integrable tal que ∣ f n ∣ ≤ ϕ ∀ n ∈ N ⟹ ∫ E f d μ = n → ∞ lim ∫ E f n d μ Dem: {\color{violet} \text{Dem:} } Dem:
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