Let f,g be bounded measurable functions on a set E of finite measure. Show that: If f=a.e.g then ∫f=∫g...
This question already has an answer here:
Lebesgue integration: $f = g$ a.e. $ Rightarrow int_Omega f = int_Omega g$
2 answers
Let f,g be bounded measurable functions on a set E of finite measure. Show that:
If f=a.e.g then ∫f=∫g on E
I have this proof from Cupta book, but I can't understand how this step done, depends on what?
measure-theory
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
marked as duplicate by KReiser, José Carlos Santos, Paul Frost, Davide Giraudo measure-theory
Users with the measure-theory badge can single-handedly close measure-theory questions as duplicates and reopen them as needed.
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Nov 26 '18 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Lebesgue integration: $f = g$ a.e. $ Rightarrow int_Omega f = int_Omega g$
2 answers
Let f,g be bounded measurable functions on a set E of finite measure. Show that:
If f=a.e.g then ∫f=∫g on E
I have this proof from Cupta book, but I can't understand how this step done, depends on what?
measure-theory
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
marked as duplicate by KReiser, José Carlos Santos, Paul Frost, Davide Giraudo measure-theory
Users with the measure-theory badge can single-handedly close measure-theory questions as duplicates and reopen them as needed.
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Nov 26 '18 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02
|
show 2 more comments
This question already has an answer here:
Lebesgue integration: $f = g$ a.e. $ Rightarrow int_Omega f = int_Omega g$
2 answers
Let f,g be bounded measurable functions on a set E of finite measure. Show that:
If f=a.e.g then ∫f=∫g on E
I have this proof from Cupta book, but I can't understand how this step done, depends on what?
measure-theory
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
This question already has an answer here:
Lebesgue integration: $f = g$ a.e. $ Rightarrow int_Omega f = int_Omega g$
2 answers
Let f,g be bounded measurable functions on a set E of finite measure. Show that:
If f=a.e.g then ∫f=∫g on E
I have this proof from Cupta book, but I can't understand how this step done, depends on what?
This question already has an answer here:
Lebesgue integration: $f = g$ a.e. $ Rightarrow int_Omega f = int_Omega g$
2 answers
measure-theory
measure-theory
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
asked Nov 26 '18 at 5:46
Duaa Hamzeh
614
614
marked as duplicate by KReiser, José Carlos Santos, Paul Frost, Davide Giraudo measure-theory
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Nov 26 '18 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by KReiser, José Carlos Santos, Paul Frost, Davide Giraudo measure-theory
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Nov 26 '18 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02
|
show 2 more comments
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02
|
show 2 more comments
1 Answer
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Let's prove this: Suppose $f geq 0$ on a measurable set $E.$ Then $int_E f geq 0.$
Choose a simple function $u$ supported on $E$ so $u leq f$ and $u geq 0.$ Then
$$ int_E u = sum u_j m(E_j) $$
where $m$ is Lebesgue measure, $u_j geq 0$, and the $E_j$ is a partition of $E$. Then the above is nonnegative. It follows that the supremum over all simple functions $u$ that approximate $f$ from below is nonnegative, and hence $int_E f geq 0.$
The result that $f leq g$ a.e. on $E$ implies $int_E f leq int_E g$ follows by replacing $f$ with $g-f$ in the above.
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let's prove this: Suppose $f geq 0$ on a measurable set $E.$ Then $int_E f geq 0.$
Choose a simple function $u$ supported on $E$ so $u leq f$ and $u geq 0.$ Then
$$ int_E u = sum u_j m(E_j) $$
where $m$ is Lebesgue measure, $u_j geq 0$, and the $E_j$ is a partition of $E$. Then the above is nonnegative. It follows that the supremum over all simple functions $u$ that approximate $f$ from below is nonnegative, and hence $int_E f geq 0.$
The result that $f leq g$ a.e. on $E$ implies $int_E f leq int_E g$ follows by replacing $f$ with $g-f$ in the above.
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
add a comment |
Let's prove this: Suppose $f geq 0$ on a measurable set $E.$ Then $int_E f geq 0.$
Choose a simple function $u$ supported on $E$ so $u leq f$ and $u geq 0.$ Then
$$ int_E u = sum u_j m(E_j) $$
where $m$ is Lebesgue measure, $u_j geq 0$, and the $E_j$ is a partition of $E$. Then the above is nonnegative. It follows that the supremum over all simple functions $u$ that approximate $f$ from below is nonnegative, and hence $int_E f geq 0.$
The result that $f leq g$ a.e. on $E$ implies $int_E f leq int_E g$ follows by replacing $f$ with $g-f$ in the above.
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
add a comment |
Let's prove this: Suppose $f geq 0$ on a measurable set $E.$ Then $int_E f geq 0.$
Choose a simple function $u$ supported on $E$ so $u leq f$ and $u geq 0.$ Then
$$ int_E u = sum u_j m(E_j) $$
where $m$ is Lebesgue measure, $u_j geq 0$, and the $E_j$ is a partition of $E$. Then the above is nonnegative. It follows that the supremum over all simple functions $u$ that approximate $f$ from below is nonnegative, and hence $int_E f geq 0.$
The result that $f leq g$ a.e. on $E$ implies $int_E f leq int_E g$ follows by replacing $f$ with $g-f$ in the above.
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
Let's prove this: Suppose $f geq 0$ on a measurable set $E.$ Then $int_E f geq 0.$
Choose a simple function $u$ supported on $E$ so $u leq f$ and $u geq 0.$ Then
$$ int_E u = sum u_j m(E_j) $$
where $m$ is Lebesgue measure, $u_j geq 0$, and the $E_j$ is a partition of $E$. Then the above is nonnegative. It follows that the supremum over all simple functions $u$ that approximate $f$ from below is nonnegative, and hence $int_E f geq 0.$
The result that $f leq g$ a.e. on $E$ implies $int_E f leq int_E g$ follows by replacing $f$ with $g-f$ in the above.
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
edited Nov 26 '18 at 6:36
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
answered Nov 26 '18 at 6:24
Sean Roberson
6,40531327
6,40531327
add a comment |
add a comment |
The integral preserves order. If $phi$ dominates $f-g$ and the latter is a.e. zero, then $phi$ is a.e. nonnegative, and hence the integral is nonnegative.
– Sean Roberson
Nov 26 '18 at 5:55
Thanks alot. But I confused about why integral of (f-g) must be nonnegative?
– Duaa Hamzeh
Nov 26 '18 at 5:58
@DuaaHamzeh $f-g=0$ a.e., so $f-g$ is non-negative a.e. I dont like the proof of this book, it is complicated without any sense
– Masacroso
Nov 26 '18 at 5:59
Again. The domination property. If $u leq v$ a.e. then the intergal of $u$ is less than or equal to that of $v$. This follows from saying if $v-u leq 0$ a.e. then $int v-u leq 0.$ Prove this with an approximation argument.
– Sean Roberson
Nov 26 '18 at 6:01
Perhaps also share what Theorem 2.2 (b) is?
– AlkaKadri
Nov 26 '18 at 6:02