Proof of $fracdleftlvert nurightrvertd mu=leftlvertfracdnudmurightrvert$
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My current task at hand is to prove that on a measurable space $(X,mathcal A)$, for a complex measure $nu$ and a $sigma$-finite measure $mu$ with $nullmu$, there holds
$$
fracdleftlvert nurightrvertd mu=leftlvertfracdnudmurightrvert quad mutext-a.e.,
$$
where $leftlvert nurightrvert$ denotes the total variation of $nu$. The inequality $fracdleftlvert nurightrvertd muleqleftlvertfracdnudmurightrvert$ is quite straightforward for it is enough to prove that
$$
int_Xfracdleftlvert nurightrvertd mu,mathrm dmu leq int_X leftlvertfracdnudmurightrvert,mathrm dmu
$$
and since LHS is equal to $lvertnurvert(X)$, we use the definition of the total variation, i.e. pick an arbitrary measurable finite partition $(A_i)_i=1^n$ of $X$ and show that
$$sum_ileftlvertnu(A_i)rightrvert=sum_i leftlvertint_A_i fracdnudmu,mathrm dmurightrvertleqsum_iint_A_ileftlvertfracdnudmurightrvert,mathrm dmu =int_X leftlvertfracdnudmurightrvert,mathrm dmu.$$
The only remaining question I have up to this point is whether it's enough to consider $X$ or does this need to hold $forall Einmathcal A$.
However, the inverse inequality is more tricky and I've problems with even beginning to tackle the problem. I've looked in some textbooks (e.g. Folland), but they usually state this as an exercise.
measure-theory
asked Aug 14 at 11:47
suhogrozdje
164111
add a comment |Â
up vote
1
down vote
favorite
My current task at hand is to prove that on a measurable space $(X,mathcal A)$, for a complex measure $nu$ and a $sigma$-finite measure $mu$ with $nullmu$, there holds
$$
fracdleftlvert nurightrvertd mu=leftlvertfracdnudmurightrvert quad mutext-a.e.,
$$
where $leftlvert nurightrvert$ denotes the total variation of $nu$. The inequality $fracdleftlvert nurightrvertd muleqleftlvertfracdnudmurightrvert$ is quite straightforward for it is enough to prove that
$$
int_Xfracdleftlvert nurightrvertd mu,mathrm dmu leq int_X leftlvertfracdnudmurightrvert,mathrm dmu
$$
and since LHS is equal to $lvertnurvert(X)$, we use the definition of the total variation, i.e. pick an arbitrary measurable finite partition $(A_i)_i=1^n$ of $X$ and show that
$$sum_ileftlvertnu(A_i)rightrvert=sum_i leftlvertint_A_i fracdnudmu,mathrm dmurightrvertleqsum_iint_A_ileftlvertfracdnudmurightrvert,mathrm dmu =int_X leftlvertfracdnudmurightrvert,mathrm dmu.$$
The only remaining question I have up to this point is whether it's enough to consider $X$ or does this need to hold $forall Einmathcal A$.
However, the inverse inequality is more tricky and I've problems with even beginning to tackle the problem. I've looked in some textbooks (e.g. Folland), but they usually state this as an exercise.
measure-theory
asked Aug 14 at 11:47
suhogrozdje
164111
I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My current task at hand is to prove that on a measurable space $(X,mathcal A)$, for a complex measure $nu$ and a $sigma$-finite measure $mu$ with $nullmu$, there holds
$$
fracdleftlvert nurightrvertd mu=leftlvertfracdnudmurightrvert quad mutext-a.e.,
$$
where $leftlvert nurightrvert$ denotes the total variation of $nu$. The inequality $fracdleftlvert nurightrvertd muleqleftlvertfracdnudmurightrvert$ is quite straightforward for it is enough to prove that
$$
int_Xfracdleftlvert nurightrvertd mu,mathrm dmu leq int_X leftlvertfracdnudmurightrvert,mathrm dmu
$$
and since LHS is equal to $lvertnurvert(X)$, we use the definition of the total variation, i.e. pick an arbitrary measurable finite partition $(A_i)_i=1^n$ of $X$ and show that
$$sum_ileftlvertnu(A_i)rightrvert=sum_i leftlvertint_A_i fracdnudmu,mathrm dmurightrvertleqsum_iint_A_ileftlvertfracdnudmurightrvert,mathrm dmu =int_X leftlvertfracdnudmurightrvert,mathrm dmu.$$
The only remaining question I have up to this point is whether it's enough to consider $X$ or does this need to hold $forall Einmathcal A$.
However, the inverse inequality is more tricky and I've problems with even beginning to tackle the problem. I've looked in some textbooks (e.g. Folland), but they usually state this as an exercise.
measure-theory
asked Aug 14 at 11:47
suhogrozdje
164111
My current task at hand is to prove that on a measurable space $(X,mathcal A)$, for a complex measure $nu$ and a $sigma$-finite measure $mu$ with $nullmu$, there holds
$$
fracdleftlvert nurightrvertd mu=leftlvertfracdnudmurightrvert quad mutext-a.e.,
$$
where $leftlvert nurightrvert$ denotes the total variation of $nu$. The inequality $fracdleftlvert nurightrvertd muleqleftlvertfracdnudmurightrvert$ is quite straightforward for it is enough to prove that
$$
int_Xfracdleftlvert nurightrvertd mu,mathrm dmu leq int_X leftlvertfracdnudmurightrvert,mathrm dmu
$$
and since LHS is equal to $lvertnurvert(X)$, we use the definition of the total variation, i.e. pick an arbitrary measurable finite partition $(A_i)_i=1^n$ of $X$ and show that
$$sum_ileftlvertnu(A_i)rightrvert=sum_i leftlvertint_A_i fracdnudmu,mathrm dmurightrvertleqsum_iint_A_ileftlvertfracdnudmurightrvert,mathrm dmu =int_X leftlvertfracdnudmurightrvert,mathrm dmu.$$
The only remaining question I have up to this point is whether it's enough to consider $X$ or does this need to hold $forall Einmathcal A$.
However, the inverse inequality is more tricky and I've problems with even beginning to tackle the problem. I've looked in some textbooks (e.g. Folland), but they usually state this as an exercise.
measure-theory
asked Aug 14 at 11:47
suhogrozdje
164111
edited Aug 14 at 12:50
asked Aug 14 at 11:47
suhogrozdje
164111
asked Aug 14 at 11:47
suhogrozdje
164111
asked Aug 14 at 11:47
suhogrozdje
164111
164111
I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18
add a comment |Â
I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18
I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18
add a comment |Â
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I didn't see this in Folland, what is the number of the exercise?
– Math1000
Aug 14 at 12:36
Sorry, I wrote that last sentence ambiguously - I checked Folland and didn't see it proven there (nor as an exercise); I did see this problem as an exercise in a couple of textbooks, though.
– suhogrozdje
Aug 14 at 12:42
hint: there is a measurable function $h$ s.t $|h|=1$ and $dnu=hd|nu|.$
– Driver 8
Aug 14 at 13:18