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# 8.5.E: Problemas en la Integración de Funciones Reales Extendidas

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## Ejercicio$$\PageIndex{1}$$

Usando las fórmulas en (1) y nuestras convenciones, verificar que
(i)$$\overline{\int}_{A} f=+\infty$$ iff$$\overline{\int}_{A} f^{+}=\infty$$;
(ii)$$\underline{\int}_{A} f=\infty$$ iff$$\underline{\int}_{A} f^{+}=\infty ;$$ y
(iii)$$\overline{f}_{A} f=-\infty$$ iff$$\underline{\int}_{A} f^{-}=\infty$$ y$$\overline{\int}_{A} f^{+}<\infty$$.
(iv) Derivar una condición similar a (iii) para$$\underline{\int}_{A} f=-\infty$$.
v) Revisión Problema 6 del Capítulo 4, §4.

## Ejercicio$$\PageIndex{2}$$

Rellene los datos de prueba faltantes en Teoremas 1 a 3 y Lemmas 1 y 2.

## Ejercicio$$\PageIndex{3}$$

Demostrar que si$$\underline{\int_{A}} f=\infty,$$ hay un mapa elemental y (extendido) real$$g \leq f$$ en$$A,$$ con$$\int_{A} g=\infty$$.
[Esquema: Por Problema$$1,$$ tenemos
\ [
\ subrayado {\ int_ {A}} f^ {+} =\ infty.
\]
Como Lemmas 1 y 2 seguramente se mantienen para funciones no negativas, arregla una medible$$F \leq f^{+}$$$$(F \geq 0),$$ con

\ [\ int_ {A} F=\ subrayado {\ int_ {A}} f^ {+} =\ infty.
\]
Argumentando como en Teorema$$3,$$ encontramos un mapa elemental y no negativo$$g \leq F,$$ con
\ [
(1+\ varepsilon)\ int_ {A} g=\ int_ {A} F=\ infty;
\]
así$$\int_{A} g=\infty$$ y$$0 \leq g \leq F \leq f^{+}$$ sucesivamente$$A$$.
Vamos
\ [
A_ {+} =A (F>0)\ in\ mathcal {M}
\]
y
\ [
A_ {0} =A (F=0)\ in\ mathcal {M}
\]
(Teorema 1 en §2). On$$A_{+},$$
\ [
g\ leq F\ leq f^ {+} =f (\ text {¿por qué? }),
\]
mientras está encendido$$A_{0}, g=F=0 ;$$ así
\ [
\ int_ {A_ {+}} g=\ int_ {A} g=\ infty (\ mathrm {por qué}?) .
\]
Ahora redefine$$g=-\infty$$ on$$A_{0}$$ (solo). Mostrar que$$g$$ es entonces la función requerida.]

## Ejercicio$$\PageIndex{4}$$

Para cualquier$$f: S \rightarrow E^{*},$$ prueba lo siguiente.
(a) Si$$\overline{\int}_{A} f<\infty,$$ entonces$$f<\infty$$ a.e. on$$A$$.
(b) Si$$\underline{\int_{A}} f$$ es ortodoxo y$$>-\infty,$$ luego$$f>-\infty$$ a.e. on$$A$$.
[Pista: Utilice el Problema 1 y aplique el Corolario 1 para$$f^{+} ;$$ así probar (a). Entonces para (b), use el Teorema 1 (e').]

## Ejercicio$$\PageIndex{5}$$

$$\Rightarrow 5$$. Para cualquier$$f, g: S \rightarrow E^{*},$$ prueba que
(i)$$\overline{\int}_{A} f+\overline{\int}_{A} g \geq \overline{\int}_{A}(f+g),$$ y
(ii)$$\underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f+\underline{\int}_{A} g \quad$$ si$$\left|\underline{\int}_{A} g\right|<\infty$$.
[Pista: Supongamos que

\ [\ overline {\ int} _ {A} f+\ overline {\ int} _ {A} g<\ overline {\ int} _ {A} (f+g).
\]
Luego están los números
\ [
u>\ overline {\ int} _ {A} f\ text {y} v>\ overline {\ int} _ {A} g,
\]
con
\ [
u+v\ leq
overline {\ int} _ {A} (f+g).
\]
(¿Por qué?) Así Lema 1 produce mapas elementales y (extendidos) reales$$F \geq f$$ y$$G \geq g$$ tales que
\ [
u>\ overline {\ int} _ {A} F\ text {y} v>\ overline {\ int} _ {A} G
\]
Como$$f+g \leq F+G$$ en$$A,$$ Teorema$$1(\mathrm{c})$$ de §5 y Problema 6 de §4 mostrar que
\ [
\ overline {\ int} _ {A} (f+g)\ leq\ int_ {A} (F+G) =\ int_ {A} F+\ int_ {A} g<u+v,
\]
contrario a
\ [
u+v\ leq\ overline {\ int} _ _ {A} (f+g).
\]
Similarmente probar la cláusula (ii).]

## Ejercicio$$\PageIndex{6}$$

Problema continuo$$5,$$ probar que
\ [
\ overline {\ int} _ {A} (f+g)\ geq\ overline {\ int} _ {A} f+\ subrayado {\ int} _ {A} _ {A} g\ geq\ subrayado {\ int} _ {A} (f+g)\ geq\ subrayado {\ int} _ _ {A} f+\ subrayado {\ int} _ A} g
\]
siempre$$\left|\underline{\int}_{A} g\right|<\infty$$.
[Pista para la segunda desigualdad: Podemos suponer que
\ [
\ overline {\ int} _ {A} (f+g) <\ infty\ text {and}\ overline {\ int} _ {A} f>-\ infty.
\]
(¿Por qué?) Aplicar Problemas 5 y$$4(\mathrm{a})$$ a
\ [
\ overline {\ int} _ {A} ((f+g) + (-g)).
\]
Usar teorema$$\left.1\left(\mathrm{e}^{\prime}\right) .\right]$$

## Ejercicio$$\PageIndex{7}$$

Demostrar lo siguiente.
(i)\ [
\ overline {\ int} _ {A} |f|<\ infty\ text {iff} -\ infty<\ subrayado {\ int} _ {A} f\ leq\ overline {\ int} _ {A} f<\ infty.
\]
(ii) Si$$\overline{f}_{A}|f|<\infty$$ y$$\overline{\int}_{A}|g|<\infty,$$ luego
\ [
\ izquierda|\ overline {\ int} _ {A} f-\ overline {\ int} _ {A} g\ derecha|\ leq\ overline {\ int} _ {A} |f-g|
\]
y
\ [
\ izquierda|\ subrayado {\ int} _ {A} f-\ subrayado {\ int} _ {A} g\ derecha|\ leq\ overline {\ int} _ {A} |f-g|.
\]
[Pista: Problemas de uso$$5 \text { and } 6 .]$$

## Ejercicio$$\PageIndex{8}$$

Demostrar que cualquier medida firmada$$\left.\bar{s}_{f} \text { (Note } 4\right)$$ es la diferencia de dos medidas:$$\bar{s}_{f}=\bar{s}_{f+}-\bar{s}_{f-}$$.

8.5.E: Problemas en la Integración de Funciones Reales Extendidas is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.