2C2025 - Teo Integral 1

4 de noviembre de 202028 min de lectura

Thu-20-11-2025 08:56 profe: Pablo Turco - Cecilia De Vita | Leonard Ehrhorn - Tomer Damián Cohen Mizrahi - Iván Moyano Tassara
status: tags:

Lo que quedó pendiente de la teo anterior

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{Si $f,g:E\to \mathbb{R}$ medibles $\implies f+g$ medible.} \end{array}

${\color{Orange} \text{Dem:} }$ Supongamos que $\{ f=+\infty \}=\emptyset=\{ g=\pm \infty \}$

Si $a \in \mathbb{R}$ $\{ f+g<a \}$ es medible Veamos que

\{ f+g<a \}=\bigcup_{q \in \mathbb{Q}}\underbrace{ \{ f<g \} }_{ \text{medible} }\cap \underbrace{ \{ g<a-q \} }_{ \text{medible} }

-- 00:00

$\supseteq)$ Si $x \in \{ f<q \}\cap \{ q<a-g \}$

\implies f(x)+g(x)<q+g(x)<a-g(x)+g(x)=a

$\subseteq)$ Usamos esto: $A\subseteq B\implies X \setminus B \subseteq X \setminus A$ Tomemos $x \in E.$ $\:\exists\:q \in \mathbb{Q}\bigm|f(x)<q$

\:\exists\:( q_{n} )_{n \in \mathbb{N}}\subset \mathbb{Q}\bigm| f(x)<q_{n}\land q_{n}\searrow f(x)

Si $x\neq \bigcup_{q \in \mathbb{Q}}\{ f<q \}\cap \{ g<a-q \}$ $\implies f(x)<q_{n}\implies g(x)\geq a- q_{n}$

\implies f(x)+g(x)\geq f(x)+a-q_{n}\quad\forall n\in \mathbb{N}

\implies f(x)+g(x)\geq f(x)+a-f(x)=a

\implies x \not\in \{ f+g<a \}

Sea ${\Large\varphi}:E\to \mathbb{R}$ simple medible y ${\Large\varphi}\geq 0$ Existe partición finita $E_{1},\dots,E_{n}$ medibles $\bigm|\displaystyle E=\dot{\bigcup_{i=1}}E_{i}$ y $\alpha_{1},\dots,\alpha_{n}\geq 0\bigm|\mathscr{l}=\sum \alpha_{i}\cdot\mathcal{X}_{E_{i}}\implies I(\mathscr{l})=\displaystyle \sum_{i=1}^{n}\alpha_{i}\cdot \mu(E_{i})$

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{Sean ${\Large\varphi},\phi:E\to \mathbb{R}$ con ${\Large\varphi},\phi\geq 0$ simples medibles. }\\ \text{1. $I({\Large\varphi}+\phi)=I({\Large\varphi})+I(\phi )$ } \\ \text{2. $I(\alpha\cdot {\Large\varphi})=\alpha\cdot I({\Large\varphi})$ } \end{array}

${\color{Orange} \text{Dem 1):} }$ Sean $E_{1},\dots,E_{n}$ medibles disjuntos, $\alpha_{1},\dots,\alpha_{n}\geq 0$ $D_{1},...,D_{m}$ medibles disjuntos $\beta_{1},\cdots,\beta _m\geq 0$ tal que

E=\dot{\bigcup^n} E_{i} =\dot{\bigcup^m}D_{j}

{\Large\varphi}=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}} ,\quad \phi=\sum_{j=1}^{m} \beta_{j}\cdot\mathcal{X}_{D_{j}}

Como $\displaystyle E=\dot{\bigcup_{i}}\dot{\bigcup_{j}}E_{i}\cap D_{j}$ y ${\Large\varphi}+\phi(x)=\alpha_{i}+\beta_{j}\quad\forall x \in E_{i}\cap D_{j}$

\implies {\Large\varphi}+\phi=\sum_{i=1}^{n} \sum_{j=1}^{m} (\alpha _{i}+\beta_{j})\cdot \mathcal{X}_{E_{i}\cap D_{j}}

I({\Large\varphi}+\phi)=\sum_{i=1}^{n} \sum_{j=1}^{m} (\alpha_{i}+\beta_{j})\cdot\mu(E_{i}\cap D_{j})

=\sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{i}\cdot \mu(E_{i}\cap D_{j})+\beta_{j}\cdot \mu(E_{i}\cap D_{j})

=\sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{i}\cdot \mu(E_{i}\cap D_{j})+\sum_{i=1}^{n} \sum_{j=1}^{m}\beta_{j}\cdot \mu(E_{i}\cap D_{j})

=\sum_{i=1}^{n} \alpha_{i}\cdot \sum_{j=1}^{m} \mu(E_{i}\cap D_{j})+\sum_{j=1}^{m} \beta_{j}\sum_{i=1}^{n} \mu(E_{i}\cap D_{j})

=\sum_{i=1}^{n} \alpha_{i}\cdot \mu(E_{i})+\sum_{j=1}^{m} \beta_{j}\cdot \mu(D_{j})

=I({\Large\varphi})+I(\phi)

${\color{Cyan} \text{Def. :} }$

\begin{array}{l} \text{Teorema} \end{array}

Sea $f:E\to \mathbb{R}$ medible.

\int_{E} f =\int_{E} f \, d\mu =\int_{E}f(x) \,\, d\mu(x)

=\underset{ }{ \sup }\{ I({\Large\varphi}):0\leq {\Large\varphi}\leq f,{\Large\varphi} \text{ simple medible} \}

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{Si $f:E\to \mathbb{R}$ simple medible con $f\geq 0$}\\ \displaystyle \int_{E} f=I(f) \end{array}

${\color{Orange} \text{Dem:} }$

Ver que si $0\leq {\Large\varphi}\leq f$ simple medible $\implies I({\Large\varphi})\leq I(f)$

Queda como ejercicio.

--46:35, Al final si dió la demo

Sean ${\Large\varphi},f:E\to \mathbb{R}$ simples medibles.

f=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}} \quad \quad {\Large\varphi}=\sum_{j=1}^{m} \beta _{j}\cdot\mathcal{X}_{D_{j}}

f=\sum_{i=1}^{n} \alpha_{i}\cdot \mathcal{X}_{E_{i}\cap\displaystyle \dot{\cup}_{j}D_{j}} =\sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{i}\cdot\mathcal{X}_{E_{i}\cap D_{j}}

De la misma manera

{\Large\varphi}=\sum_{j=1}^{m} \sum_{i=1}^{n} \beta_{j}\cdot\mathcal{X}_{E_{i}\cap D_{j}}

Si $x \in E_{i}\cap D_{j}$ $f(x)=\alpha_{i}$ y ${\Large\varphi}(x)=\beta_{j}\implies\alpha_{i}\geq\beta_{j}$

Esto ultimo porque suponemos que $f\geq {\Large\varphi}$

I({\Large\varphi})=\sum_{i=1}^{n} \sum_{j=1}^{m} B_{j}\cdot \mu(E_{i}\cap D_{j})\leq \sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{i}\cdot \mu(E_{i}\cap D_{j})=I(f)

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{$f:E \to \mathbb{R}$. Si $E$ es nulo, $\displaystyle \int_{E} f \, d\mu=0$ } \end{array}

${\color{Orange} \text{Dem:} }$

Si ${\Large\varphi}:E\to \mathbb{R}$ simple tal que ${\Large\varphi}\leq f$

{\Large\varphi}=\sum_{i=1}^{n} \alpha_{i}\cdot\mathcal{X}_{E_{i}}\quad \quad E=\dot{\bigcup}_{i=1}^n E_{i}

I({\Large\varphi})=\sum_{i=1}^{n} \alpha_{i}\cdot \underbrace{ \mu(E_{i}) }_{ =0 }=0

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{Sean $f,g:E\to \mathbb{R}$ medibles con $0\leq f\leq g$ }\\ \displaystyle \int_{E}f \, d\mu =\int_{E} g \, d\mu \end{array}

${\color{Orange} \text{Dem:} }$ Si $0\leq {\Large\varphi}\leq f$ es simple medible $\implies {\Large\varphi}\leq g$ --59:30

${\color{Orange} \text{Prop. :} }$

\begin{array}{l} \text{proposicion} \end{array}

Sea $f:E\to \mathbb{R}$ medible con $f\geq 0$ y $A\subseteq E$ medible.

\implies \int_{A} f \, d\mu =\int_{E} f\cdot \mathcal{X}_{A} \, d\mu

Luego

\int_{A} f \, d\mu \leq \int_{E} f \, d\mu

pues $f\cdot\mathcal{X}_{A}\leq f$

$Obs:$ Vimos que puede ocurrir que

f_{n} \longrightarrow f\text{ pero } \int f_{n}\cancel{ \longrightarrow } \int f

Ejemplos:

f_{n}(x)=\begin{cases} n^{2}x & 0\leq x\leq \frac{1}{2n} \\ n^{2}\left( \frac{1}{n}-x \right) & \frac{1}{2n}\leq x \leq \frac{1}{n} \\ 0 & x> \frac{1}{n} \end{cases}

Teorema de convergencia monótona

Sean $( f_{n} )_{n \in \mathbb{N}}$ sucesión de funciones medibles positivas y sea $f$ medible tal que $f_{n}\to f$
Si $f_{n}\leq f_{n+1} \quad\forall n\in \mathbb{N}$ (monótonas crecientes, $f_{n}\nearrow f$ ) entonces

\underset{ n\to \infty }{ \lim } \int_{E} f_{n} \, d\mu =\int_{E} \underset{ n\to \infty }{ \lim } f_{n} \, d\mu =\int_{E}f \, d\mu

$Dem:$

Se deja como ejercicio.

${\color{violet} \text{Teorema :} }$

\begin{array}{l} \text{Teorema} \end{array}

Sean $f,g:E\to \mathbb{R}$ medibles y positivas y sea $\lambda>0$ . Entonces

\int_{E} f+\lambda \cdot g \, d\mu =\int_{E} f \, d\mu +\lambda\cdot \int_{E} g \, d\mu

${\color{violet} \text{Dem:} }$

\int_{E} f+\lambda g \, d\mu =I(f+\lambda g)=I(f)+\lambda \cdot I(g)=\int_{E} f \, d\mu +\lambda\int_{E} g \, d\mu

Por algo anterior existen $( {\Large\varphi}_{n} )_{n \in \mathbb{N}}$ medibles simples positivas con ${\Large\varphi}_{n}\nearrow f$ Existen $( \phi_{n} )_{n \in \mathbb{N}}$ medibles simples positivas tal que $\phi_{n}\nearrow g$ Luego

{\Large\varphi}_{n}+\lambda \cdot \phi_{n}\nearrow f + \lambda \cdot g

\int_{E} f+\lambda \cdot g \, d\mu =\int_{E} \underset{ n\to \infty }{ \lim } {\Large\varphi}_{n}+\lambda \cdot \phi_{n} \, d\mu \underbrace{ = }_{ \text{teor conv mon} } \underset{ n\to \infty }{ \lim } \int_{E} {\Large\varphi}_{n} + \lambda \cdot \phi_{n} \, d\mu

\underset{ {\Large\varphi}_{n}+\lambda \phi_{n }\text{ son simples} }{ = }\underset{ n\to \infty }{ \lim } \int_{E} {\Large\varphi}_{n} \, d\mu +\lambda \int_{E} \phi_{n} \, d\mu =\underset{ n\to \infty }{ \lim } \int_{E} {\Large\varphi}_{n} \, d\mu + \lambda\underset{ n\to \infty }{ \lim } \int_{E} \phi_{n} \, d\mu

\underset{ \text{teor conv mon} }{ = } \int_{E} \underset{ n\to \infty }{ \lim } {\Large\varphi}_{n} \, d\mu +\lambda \cdot \int_{E} \underset{ n\to \infty }{ \lim } \phi_{n} \, d\mu =\int_{E} f \, d\mu +\lambda \cdot \int_{E}g \, d\mu

${\color{Red}\text{Corolario}:}$

\begin{array}{l} \text{Corolario} \end{array}

Sea $f:E\to \mathbb{R}$ medible positiva Sean $E_{1},E_{2}\subseteq E$ medibles con $E_{1}\cap E_{2}=\emptyset$

\int_{E_{1}\dot{\bigcup}E_{2}} f \, d\mu =\int_{E_{1}} f \, d\mu +\int_{E_{2}} f \, d\mu

${\color{Red} \text{Dem}:}$

\int_{E_{1}\dot{\cup}E_{2}}f=\int_{E}f \cdot\mathcal{X}_{E_{1}\dot{\cup}E_{2}} =\int_{E}f\cdot (\mathcal{X}_{E_{1}} +\mathcal{X}_{E_{2}} )

=\int_{E}f\cdot\mathcal{X}_{E_{1}} +f \cdot \mathcal{X}_{E_{2}} =\int_{E}f \cdot \mathcal{X}_{E_{1}} + \int_{E}f \cdot \mathcal{X}_{E_{2}}=\int_{E_{1}}f + \int_{E_{2}}f

${\color{Red}\text{Corolario}:}$

\begin{array}{l} \text{Corolario} \end{array}

Sean $f,g:E\longrightarrow \mathbb{R}$ medibles positivas. Si $f=g$ en casi todo punto

\int_{E} f = \int_{E}g

${\color{Red} \text{Dem}:}$ --01:36:00

\int_{E}f=\int_{E\cap \{ f=g \}} f+ \int_{E\cap \{ f\neq g \}}f=\int_{E\cap \{ f=g \}}g+0 \leq \int_{E}g

\implies \int_{E}f\leq \int_{E}g

Análogamente empiezo con $g$ y obtenga la otra desigualdad.

${\color{Red}\text{Corolario}:}$

\begin{array}{l} \text{Corolario} \end{array}

\int_{\mathbb{R}} \mathcal{X}_{\mathbb{Q}} = 0

${\color{Red} \text{Dem}:}$ Como $\mu(\mathbb{Q})= 0\implies\mathcal{X}_{\mathbb{Q}}\equiv0$ en casi todo punto. Además $\displaystyle \int_{E}0=0$

${\color{Red}\text{Corolario}:}$

\begin{array}{l} \text{Corolario} \end{array}

Sea $f:E\longrightarrow \mathbb{R}$ medible y positiva. $\displaystyle \int_{E}f=0$ si y solo si $f=0$ en casi todo punto.

${\color{Red} \text{Dem}:}$ $\implies)$ --01:44:12 Sea $E_{n}=\left\{ f> \frac{1}{n} \right\}$ medible.

Z=\{ f=0 \}

0=\int_{E}f=\int_{Z\dot{\cup}\left( \bigcup_{n}E_{n} \right)} f=\int_{Z}f+\int_{\bigcup_{n}E_{n}}f=0+\int_{\bigcup E_{n}}f

\geq \int_{E_{n}}f> \int_{E_{n}} \frac{1}{n}=\int_{E} \frac{1}{n}\cdot\mathcal{X}_{E_{n}} = \frac{1}{n} \cdot \mu(E_{n})

\implies_{0}\geq \frac{1}{n}\cdot \mu(E_{n})\geq 0\implies \mu(E_{n})= 0

\implies \mu(\{ f\neq 0 \})=\mu\left( \bigcup_{n\in \mathbb{N}}E_{n} \right)\leq \underset{ n\to \infty }{ \lim } \mu(E_{n})= 0

$Obs:$ Si $f,g\geq 0$ Me gustaría que

\int f-g=\int f - \int g

${\color{Cyan} \text{Def. :} }$

\begin{array}{l} \text{Teorema} \end{array}

Sea $f:E\to \mathbb{R}$

f^{+} =\begin{cases} f(x) & f(x)\geq 0 \\ 0 & f(x)< 0 \end{cases},\quad f^{-} =\begin{cases} -f(x) & f(x)< 0 \\ 0 & f(x)\geq 0 \end{cases}

$f^{+}\geq 0,f^{-}\geq 0$ Además

f=f^{+} -f^{-}

f^{+} =f \cdot \mathcal{X}_{\{ f\geq 0 \}} \quad \quad f^{-} =-f \cdot\mathcal{X}_{\{ f\leq 0 \}}

${\color{Cyan} \text{Def. :} }$

\begin{array}{l} \text{Teorema} \end{array}

$f:E\to \mathbb{R}$ medible.

\int_{E}f=\int_{E}f^{+} - \int_{E}f^{-}

Ejercicio: Todo lo que hicimos para funciones positivas ( $f\geq 0$ ) vale para $f$ en general

Lo que quedó pendiente de la teo anterior

Teorema de convergencia monótona

Citas y Comentarios