Why the function f must be measurable? [on hold]
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In this theorem:
Let f be a bounded measurable function on a set of finite measure E. Then f is integrable over E.
Why the condition "f is measurable" is necessary?
What if it doesn't?
I know why f need to be bounded and why E need to have finite measure, but I can't determine what will be happened if f not a measurable.
measure-theory
asked Nov 15 at 10:13
Duaa Hamzeh
604
put on hold as off-topic by TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel Nov 17 at 8:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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In this theorem:
Let f be a bounded measurable function on a set of finite measure E. Then f is integrable over E.
Why the condition "f is measurable" is necessary?
What if it doesn't?
I know why f need to be bounded and why E need to have finite measure, but I can't determine what will be happened if f not a measurable.
measure-theory
asked Nov 15 at 10:13
Duaa Hamzeh
604
put on hold as off-topic by TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel Nov 17 at 8:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
1
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
1
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
1
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
1
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In this theorem:
Let f be a bounded measurable function on a set of finite measure E. Then f is integrable over E.
Why the condition "f is measurable" is necessary?
What if it doesn't?
I know why f need to be bounded and why E need to have finite measure, but I can't determine what will be happened if f not a measurable.
measure-theory
asked Nov 15 at 10:13
Duaa Hamzeh
604
In this theorem:
Let f be a bounded measurable function on a set of finite measure E. Then f is integrable over E.
Why the condition "f is measurable" is necessary?
What if it doesn't?
I know why f need to be bounded and why E need to have finite measure, but I can't determine what will be happened if f not a measurable.
measure-theory
measure-theory
asked Nov 15 at 10:13
Duaa Hamzeh
604
asked Nov 15 at 10:13
Duaa Hamzeh
604
asked Nov 15 at 10:13
Duaa Hamzeh
604
asked Nov 15 at 10:13
Duaa Hamzeh
604
asked Nov 15 at 10:13
Duaa Hamzeh
604
604
put on hold as off-topic by TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel Nov 17 at 8:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel Nov 17 at 8:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, amWhy, José Carlos Santos, Rushabh Mehta, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
1
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
1
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
1
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
1
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41
|
show 3 more comments
1
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
1
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
1
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
1
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
1
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41
1
1
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
1
1
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
1
1
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
1
1
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
1
1
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41
|
show 3 more comments
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1
Because you cannot define the integral of a non-measurable function.
– TheGeekGreek
Nov 15 at 10:18
1
This is a question in measure theory and in measure theory integrals are defined only for measurable functions. What type of integrals are you thinking of and how do you define integration?.
– Kavi Rama Murthy
Nov 15 at 10:19
1
@KaviRamaMurthy there is nothing in the definition of the integral that actually requires the function to be measurable, except maybe the phrase "let f be measurable...". The definition as the sup of the integrals of measurable step functions not exceeding f still goes through. It seems like a good exercise to see what breaks in the theory when you do this.
– Pliny the ill
Nov 15 at 10:23
1
@KaviRamaMurthy There is no such theory to my knowledge. But if the OP is asking because he or she doesn't understand why measurability is important in the Lebesgue theory of integration, the right way to answer him or her is to point to some kind of motivation, e.g. the breakdown of a particular fact or theorem that we find desirable. It's not so obvious, especially to students.
– Pliny the ill
Nov 15 at 10:33
1
@Plinytheill i think an example is $f = chi_N$, $g = 1-chi_N$ where $N subseteq [0,1]$ is non-measurable. $int (f+g) = 1$ while $int f = int g = 0$.
– mathworker21
Nov 15 at 12:41