Folland Exercise 3.17
up vote
7
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Let $(X, mathcal M, mu)$ be a finite measure space, $mathcal N$ a sub-$sigma$-algebra of $mathcal M$, and $nu = mu|mathcal N$. If $f in L^1(mu)$, there exists $g in L^1(nu)$ (thus $g$ is $mathcal N$-measurable) such that $int_E f dmu = int_E g dnu$ for all $E in mathcal N$; if $g'$ is another such function then $g = g'$ $nu$-a.e. (In probability theory, $g$ is called the conditional expectation of $f$ on $scr{N}$.)
I have managed to prove this statement to be true by defining a measure $lambda$ such that $dlambda = gdnu$ and then using Lebesgue-Radon-Nikodym theorem. Now as an extension of the problem, I want to characterize $g$ in terms of $f$ when $mathcal N = {emptyset, X}$, and when $mathcal N={emptyset, X, E, E^c}$ for some $Einmathcal M$. Now I'm not sure how to do the last bit, and completely stuck here.
I would like to get some help on how to tackle the last part.
real-analysis measure-theory
asked Nov 23 at 2:20
Sank
16511
add a comment |
up vote
7
down vote
favorite
Let $(X, mathcal M, mu)$ be a finite measure space, $mathcal N$ a sub-$sigma$-algebra of $mathcal M$, and $nu = mu|mathcal N$. If $f in L^1(mu)$, there exists $g in L^1(nu)$ (thus $g$ is $mathcal N$-measurable) such that $int_E f dmu = int_E g dnu$ for all $E in mathcal N$; if $g'$ is another such function then $g = g'$ $nu$-a.e. (In probability theory, $g$ is called the conditional expectation of $f$ on $scr{N}$.)
I have managed to prove this statement to be true by defining a measure $lambda$ such that $dlambda = gdnu$ and then using Lebesgue-Radon-Nikodym theorem. Now as an extension of the problem, I want to characterize $g$ in terms of $f$ when $mathcal N = {emptyset, X}$, and when $mathcal N={emptyset, X, E, E^c}$ for some $Einmathcal M$. Now I'm not sure how to do the last bit, and completely stuck here.
I would like to get some help on how to tackle the last part.
real-analysis measure-theory
asked Nov 23 at 2:20
Sank
16511
Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $(X, mathcal M, mu)$ be a finite measure space, $mathcal N$ a sub-$sigma$-algebra of $mathcal M$, and $nu = mu|mathcal N$. If $f in L^1(mu)$, there exists $g in L^1(nu)$ (thus $g$ is $mathcal N$-measurable) such that $int_E f dmu = int_E g dnu$ for all $E in mathcal N$; if $g'$ is another such function then $g = g'$ $nu$-a.e. (In probability theory, $g$ is called the conditional expectation of $f$ on $scr{N}$.)
I have managed to prove this statement to be true by defining a measure $lambda$ such that $dlambda = gdnu$ and then using Lebesgue-Radon-Nikodym theorem. Now as an extension of the problem, I want to characterize $g$ in terms of $f$ when $mathcal N = {emptyset, X}$, and when $mathcal N={emptyset, X, E, E^c}$ for some $Einmathcal M$. Now I'm not sure how to do the last bit, and completely stuck here.
I would like to get some help on how to tackle the last part.
real-analysis measure-theory
asked Nov 23 at 2:20
Sank
16511
Let $(X, mathcal M, mu)$ be a finite measure space, $mathcal N$ a sub-$sigma$-algebra of $mathcal M$, and $nu = mu|mathcal N$. If $f in L^1(mu)$, there exists $g in L^1(nu)$ (thus $g$ is $mathcal N$-measurable) such that $int_E f dmu = int_E g dnu$ for all $E in mathcal N$; if $g'$ is another such function then $g = g'$ $nu$-a.e. (In probability theory, $g$ is called the conditional expectation of $f$ on $scr{N}$.)
I have managed to prove this statement to be true by defining a measure $lambda$ such that $dlambda = gdnu$ and then using Lebesgue-Radon-Nikodym theorem. Now as an extension of the problem, I want to characterize $g$ in terms of $f$ when $mathcal N = {emptyset, X}$, and when $mathcal N={emptyset, X, E, E^c}$ for some $Einmathcal M$. Now I'm not sure how to do the last bit, and completely stuck here.
I would like to get some help on how to tackle the last part.
real-analysis measure-theory
real-analysis measure-theory
asked Nov 23 at 2:20
Sank
16511
asked Nov 23 at 2:20
Sank
16511
edited Nov 23 at 2:33
asked Nov 23 at 2:20
Sank
16511
asked Nov 23 at 2:20
Sank
16511
asked Nov 23 at 2:20
Sank
16511
16511
Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34
add a comment |
Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34
Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34
Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34
add a comment |
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
When $mathcal{N} = {emptyset,X}$, you can check that
$$g = left( frac{1}{mu(X)}int_X f , dmu right) mathbb{I}_X$$ (i.e., a constant function) does the job.
When $mathcal{N} = {emptyset,E,E^c,X}$,
$$g = left( frac{1}{mu(E)}int_E f , dmu right) mathbb{I}_E + left( frac{1}{mu(E^c)}int_{E^c} f , dmu right) mathbb{I}_{E^c}$$
does the same thing.
To understand the construction we can think of a concrete example which parallels your problem: let's say you are a video game designer building a virtual world with 7 billion humans. You want them to approximate real-world humans in terms of their physical heights. The absolute best you could do is to match every human in the world to a video game human and make their heights correspond (this corresponds to taking $mathcal{N}=mathcal{M})$. The worst you could do is to make every video game human the same height, with that height being the average real human height (what else would it be?) A slight improvement is to divide the virtual humans into male and female, then make all females have the average real female height, and all males have the average real male height. These last two cases correspond to your question.
By the way, to make sure you understand the construction you should try it for the case when $mathcal{N}$ is generated by a countable partition of $X$ (i.e. when $X = sqcup E_i$ for $E_i$ measurable).
answered Nov 23 at 2:35
user25959
1,559816
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
When $mathcal{N} = {emptyset,X}$, you can check that
$$g = left( frac{1}{mu(X)}int_X f , dmu right) mathbb{I}_X$$ (i.e., a constant function) does the job.
When $mathcal{N} = {emptyset,E,E^c,X}$,
$$g = left( frac{1}{mu(E)}int_E f , dmu right) mathbb{I}_E + left( frac{1}{mu(E^c)}int_{E^c} f , dmu right) mathbb{I}_{E^c}$$
does the same thing.
To understand the construction we can think of a concrete example which parallels your problem: let's say you are a video game designer building a virtual world with 7 billion humans. You want them to approximate real-world humans in terms of their physical heights. The absolute best you could do is to match every human in the world to a video game human and make their heights correspond (this corresponds to taking $mathcal{N}=mathcal{M})$. The worst you could do is to make every video game human the same height, with that height being the average real human height (what else would it be?) A slight improvement is to divide the virtual humans into male and female, then make all females have the average real female height, and all males have the average real male height. These last two cases correspond to your question.
By the way, to make sure you understand the construction you should try it for the case when $mathcal{N}$ is generated by a countable partition of $X$ (i.e. when $X = sqcup E_i$ for $E_i$ measurable).
answered Nov 23 at 2:35
user25959
1,559816
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
add a comment |
up vote
7
down vote
accepted
When $mathcal{N} = {emptyset,X}$, you can check that
$$g = left( frac{1}{mu(X)}int_X f , dmu right) mathbb{I}_X$$ (i.e., a constant function) does the job.
When $mathcal{N} = {emptyset,E,E^c,X}$,
$$g = left( frac{1}{mu(E)}int_E f , dmu right) mathbb{I}_E + left( frac{1}{mu(E^c)}int_{E^c} f , dmu right) mathbb{I}_{E^c}$$
does the same thing.
To understand the construction we can think of a concrete example which parallels your problem: let's say you are a video game designer building a virtual world with 7 billion humans. You want them to approximate real-world humans in terms of their physical heights. The absolute best you could do is to match every human in the world to a video game human and make their heights correspond (this corresponds to taking $mathcal{N}=mathcal{M})$. The worst you could do is to make every video game human the same height, with that height being the average real human height (what else would it be?) A slight improvement is to divide the virtual humans into male and female, then make all females have the average real female height, and all males have the average real male height. These last two cases correspond to your question.
By the way, to make sure you understand the construction you should try it for the case when $mathcal{N}$ is generated by a countable partition of $X$ (i.e. when $X = sqcup E_i$ for $E_i$ measurable).
answered Nov 23 at 2:35
user25959
1,559816
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
When $mathcal{N} = {emptyset,X}$, you can check that
$$g = left( frac{1}{mu(X)}int_X f , dmu right) mathbb{I}_X$$ (i.e., a constant function) does the job.
When $mathcal{N} = {emptyset,E,E^c,X}$,
$$g = left( frac{1}{mu(E)}int_E f , dmu right) mathbb{I}_E + left( frac{1}{mu(E^c)}int_{E^c} f , dmu right) mathbb{I}_{E^c}$$
does the same thing.
To understand the construction we can think of a concrete example which parallels your problem: let's say you are a video game designer building a virtual world with 7 billion humans. You want them to approximate real-world humans in terms of their physical heights. The absolute best you could do is to match every human in the world to a video game human and make their heights correspond (this corresponds to taking $mathcal{N}=mathcal{M})$. The worst you could do is to make every video game human the same height, with that height being the average real human height (what else would it be?) A slight improvement is to divide the virtual humans into male and female, then make all females have the average real female height, and all males have the average real male height. These last two cases correspond to your question.
By the way, to make sure you understand the construction you should try it for the case when $mathcal{N}$ is generated by a countable partition of $X$ (i.e. when $X = sqcup E_i$ for $E_i$ measurable).
answered Nov 23 at 2:35
user25959
1,559816
When $mathcal{N} = {emptyset,X}$, you can check that
$$g = left( frac{1}{mu(X)}int_X f , dmu right) mathbb{I}_X$$ (i.e., a constant function) does the job.
When $mathcal{N} = {emptyset,E,E^c,X}$,
$$g = left( frac{1}{mu(E)}int_E f , dmu right) mathbb{I}_E + left( frac{1}{mu(E^c)}int_{E^c} f , dmu right) mathbb{I}_{E^c}$$
does the same thing.
To understand the construction we can think of a concrete example which parallels your problem: let's say you are a video game designer building a virtual world with 7 billion humans. You want them to approximate real-world humans in terms of their physical heights. The absolute best you could do is to match every human in the world to a video game human and make their heights correspond (this corresponds to taking $mathcal{N}=mathcal{M})$. The worst you could do is to make every video game human the same height, with that height being the average real human height (what else would it be?) A slight improvement is to divide the virtual humans into male and female, then make all females have the average real female height, and all males have the average real male height. These last two cases correspond to your question.
By the way, to make sure you understand the construction you should try it for the case when $mathcal{N}$ is generated by a countable partition of $X$ (i.e. when $X = sqcup E_i$ for $E_i$ measurable).
answered Nov 23 at 2:35
user25959
1,559816
edited Nov 23 at 2:55
answered Nov 23 at 2:35
user25959
1,559816
answered Nov 23 at 2:35
user25959
1,559816
answered Nov 23 at 2:35
user25959
1,559816
1,559816
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
add a comment |
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
thank you for your answer. $mathbb{I}_{X}$ what does this notation mean? Also, could you elaborate just a bit more on what you mean by"coarsening"? I'm having a hard time seeing how these constructions arose :/
– Sank
Nov 23 at 2:44
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
That is the indicator function on $X$. It equals 1 everywhere. The indicator function on $E$ equals 1 for points in $E$, and 0 elsewhere.
– user25959
Nov 23 at 2:46
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
If I may paraphrase, since we know the area under $g$ and $f$ are equal on certain sets, we are constructing $g$ from that information right - letting $g$ equal to the average value over the set that it equals $f$, right? Also, do we need to worry about measurability at all here?
– Sank
Nov 23 at 2:51
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Right. and measurability of these functions is a given - the indicator functions over measurable sets are basically by definition measurable functions, and scalar multiples and sums of measurable functions are measurable.
– user25959
Nov 23 at 2:53
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
Thank you, this was so helpful!
– Sank
Nov 23 at 2:54
add a comment |
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Relate math.stackexchange.com/questions/1003666
– Nosrati
Nov 23 at 2:34