Folland Exercise 3.17











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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.










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  • Relate math.stackexchange.com/questions/1003666
    – Nosrati
    2 hours ago















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4
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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.










share|cite|improve this question
























  • Relate math.stackexchange.com/questions/1003666
    – Nosrati
    2 hours ago













up vote
4
down vote

favorite









up vote
4
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.










share|cite|improve this question
















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






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share|cite|improve this question













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edited 2 hours ago

























asked 2 hours ago









Sank

13811




13811












  • Relate math.stackexchange.com/questions/1003666
    – Nosrati
    2 hours ago


















  • Relate math.stackexchange.com/questions/1003666
    – Nosrati
    2 hours ago
















Relate math.stackexchange.com/questions/1003666
– Nosrati
2 hours ago




Relate math.stackexchange.com/questions/1003666
– Nosrati
2 hours ago










1 Answer
1






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up vote
4
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).






share|cite|improve this answer























  • 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
    2 hours ago












  • 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
    2 hours ago










  • 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
    2 hours ago










  • 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
    2 hours ago










  • Thank you, this was so helpful!
    – Sank
    2 hours ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
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).






share|cite|improve this answer























  • 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
    2 hours ago












  • 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
    2 hours ago










  • 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
    2 hours ago










  • 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
    2 hours ago










  • Thank you, this was so helpful!
    – Sank
    2 hours ago















up vote
4
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).






share|cite|improve this answer























  • 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
    2 hours ago












  • 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
    2 hours ago










  • 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
    2 hours ago










  • 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
    2 hours ago










  • Thank you, this was so helpful!
    – Sank
    2 hours ago













up vote
4
down vote



accepted







up vote
4
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).






share|cite|improve this answer














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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours ago









user25959

1,071714




1,071714












  • 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
    2 hours ago












  • 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
    2 hours ago










  • 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
    2 hours ago










  • 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
    2 hours ago










  • Thank you, this was so helpful!
    – Sank
    2 hours ago


















  • 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
    2 hours ago












  • 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
    2 hours ago










  • 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
    2 hours ago










  • 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
    2 hours ago










  • Thank you, this was so helpful!
    – Sank
    2 hours ago
















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
2 hours ago






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
2 hours ago














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
2 hours ago




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
2 hours ago












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
2 hours ago




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
2 hours ago












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
2 hours ago




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
2 hours ago












Thank you, this was so helpful!
– Sank
2 hours ago




Thank you, this was so helpful!
– Sank
2 hours ago


















 

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