If $|mu(E)|le csqrtnu(E)$ for all $E$ belonging to a generator, then the total variation of $mu$ is bounded by $csqrtnu$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does the signed measure based on a Jordan decomposition of a function with bounded variation depend on the decomposition?Exercise: signed measures, total variation.If the variation of a measure $μ$ is smaller than the variation of a measure $ν$ on any compact interval, then $μ$ is absolutely continuous wrt $ν$If $mu$ is a finite measure and $nu$ is a signed measure, can we extend the inequality $|nu|le Cmu$ from a generator to the whole σ-algebra?If $mu$ is a finite measure and $ν$ is a signed measure with $|nu|le Cmu$, are we able to show $left|fracrm dνrm dmuright|le C$?If $mu$ is a vector measure and $nu$ is a measure with $left|muright|lenu$ on a generating semiring, are we able to conclude $|mu|lenu$?Characteristic property of the total variation of a vector measureEquality of finite signed measures by showing that the integrals of every bounded continuous function coincideCalculate product measure of off-diagonal setShow that any nondecreasing $f:[0,infty)tomathbb R$ of locally bounded variation admits a unique signed measure $mu$ with $mu([a,b])=f(b)-f(a)$
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If $|mu(E)|le csqrtnu(E)$ for all $E$ belonging to a generator, then the total variation of $mu$ is bounded by $csqrtnu$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does the signed measure based on a Jordan decomposition of a function with bounded variation depend on the decomposition?Exercise: signed measures, total variation.If the variation of a measure $μ$ is smaller than the variation of a measure $ν$ on any compact interval, then $μ$ is absolutely continuous wrt $ν$If $mu$ is a finite measure and $nu$ is a signed measure, can we extend the inequality $|nu|le Cmu$ from a generator to the whole σ-algebra?If $mu$ is a finite measure and $ν$ is a signed measure with $|nu|le Cmu$, are we able to show $left|fracrm dνrm dmuright|le C$?If $mu$ is a vector measure and $nu$ is a measure with $left|muright|lenu$ on a generating semiring, are we able to conclude $|mu|lenu$?Characteristic property of the total variation of a vector measureEquality of finite signed measures by showing that the integrals of every bounded continuous function coincideCalculate product measure of off-diagonal setShow that any nondecreasing $f:[0,infty)tomathbb R$ of locally bounded variation admits a unique signed measure $mu$ with $mu([a,b])=f(b)-f(a)$
$begingroup$
Let $(Omega,mathcal A)$ be a measurable space, $mu$ be a signed measure on $(Omega,mathcal A)$, $nu$ be a measure on $(E,mathcal E)$ and $cge0$ with $$left|mu(E)right|le csqrtnu(E);;;textfor all Einmathcal Etag1$$ for some $cap$-stable $mathcal Esubseteqmathcal A$ with $sigma(mathcal E)=mathcal A$.
Now, let $|mu|$ denote the variation of $|mu|$, i.e. $$|mu|(A):=supleft:kinmathbb Ntext and A_1,ldots,A_kinmathcal Atext are mutually disjoint with biguplus_i=1^kA_isubseteq Aright$$ for $AsubseteqOmega$.
Are we able to show $$|mu|(A)le csqrtnu(A)tag2$$ for all $Ainmathcal A$?
Remark: I know how I can conclude $|mu|(A)lesqrtnu_1(A)sqrtnu_2(A)$ for all $Ainmathcal A$ from $|mu(E)|lesqrtnu_1(E)sqrtnu_2(E)$ for all $Einmathcal E$ (where $nu_i$ is a measure on $(E,mathcal E)$), but in the case of $(1)$ and $(2)$ we cannot apply the Cauchy-Schwarz inequality.
real-analysis measure-theory bounded-variation signed-measures
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal A)$ be a measurable space, $mu$ be a signed measure on $(Omega,mathcal A)$, $nu$ be a measure on $(E,mathcal E)$ and $cge0$ with $$left|mu(E)right|le csqrtnu(E);;;textfor all Einmathcal Etag1$$ for some $cap$-stable $mathcal Esubseteqmathcal A$ with $sigma(mathcal E)=mathcal A$.
Now, let $|mu|$ denote the variation of $|mu|$, i.e. $$|mu|(A):=supleft:kinmathbb Ntext and A_1,ldots,A_kinmathcal Atext are mutually disjoint with biguplus_i=1^kA_isubseteq Aright$$ for $AsubseteqOmega$.
Are we able to show $$|mu|(A)le csqrtnu(A)tag2$$ for all $Ainmathcal A$?
Remark: I know how I can conclude $|mu|(A)lesqrtnu_1(A)sqrtnu_2(A)$ for all $Ainmathcal A$ from $|mu(E)|lesqrtnu_1(E)sqrtnu_2(E)$ for all $Einmathcal E$ (where $nu_i$ is a measure on $(E,mathcal E)$), but in the case of $(1)$ and $(2)$ we cannot apply the Cauchy-Schwarz inequality.
real-analysis measure-theory bounded-variation signed-measures
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal A)$ be a measurable space, $mu$ be a signed measure on $(Omega,mathcal A)$, $nu$ be a measure on $(E,mathcal E)$ and $cge0$ with $$left|mu(E)right|le csqrtnu(E);;;textfor all Einmathcal Etag1$$ for some $cap$-stable $mathcal Esubseteqmathcal A$ with $sigma(mathcal E)=mathcal A$.
Now, let $|mu|$ denote the variation of $|mu|$, i.e. $$|mu|(A):=supleft:kinmathbb Ntext and A_1,ldots,A_kinmathcal Atext are mutually disjoint with biguplus_i=1^kA_isubseteq Aright$$ for $AsubseteqOmega$.
Are we able to show $$|mu|(A)le csqrtnu(A)tag2$$ for all $Ainmathcal A$?
Remark: I know how I can conclude $|mu|(A)lesqrtnu_1(A)sqrtnu_2(A)$ for all $Ainmathcal A$ from $|mu(E)|lesqrtnu_1(E)sqrtnu_2(E)$ for all $Einmathcal E$ (where $nu_i$ is a measure on $(E,mathcal E)$), but in the case of $(1)$ and $(2)$ we cannot apply the Cauchy-Schwarz inequality.
real-analysis measure-theory bounded-variation signed-measures
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
$endgroup$
Let $(Omega,mathcal A)$ be a measurable space, $mu$ be a signed measure on $(Omega,mathcal A)$, $nu$ be a measure on $(E,mathcal E)$ and $cge0$ with $$left|mu(E)right|le csqrtnu(E);;;textfor all Einmathcal Etag1$$ for some $cap$-stable $mathcal Esubseteqmathcal A$ with $sigma(mathcal E)=mathcal A$.
Now, let $|mu|$ denote the variation of $|mu|$, i.e. $$|mu|(A):=supleft:kinmathbb Ntext and A_1,ldots,A_kinmathcal Atext are mutually disjoint with biguplus_i=1^kA_isubseteq Aright$$ for $AsubseteqOmega$.
Are we able to show $$|mu|(A)le csqrtnu(A)tag2$$ for all $Ainmathcal A$?
Remark: I know how I can conclude $|mu|(A)lesqrtnu_1(A)sqrtnu_2(A)$ for all $Ainmathcal A$ from $|mu(E)|lesqrtnu_1(E)sqrtnu_2(E)$ for all $Einmathcal E$ (where $nu_i$ is a measure on $(E,mathcal E)$), but in the case of $(1)$ and $(2)$ we cannot apply the Cauchy-Schwarz inequality.
real-analysis measure-theory bounded-variation signed-measures
real-analysis measure-theory bounded-variation signed-measures
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
edited Apr 3 at 11:54
0xbadf00d
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
asked Apr 3 at 11:44
0xbadf00d0xbadf00d
1,65341534
1,65341534
add a comment |
add a comment |
1 Answer
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$begingroup$
This is not true, even when $mu = nu.$ For concreteness let $mu = nu = mathcal L^n$ be the Lebesgue measure on $mathbb R^n$ equipped with the Borel $sigma$-algebra, and consider,
$$ mathcalE = E subset mathbb R^n : Q text Borel with mathcal L^n(Q) leq 1 . $$
This family is certainly stable under finite unions, and for any $E in mathcalE$ we have,
$$ mathcal L^n(E) leq sqrtmathcal L^n(E), $$
since $sqrtt - t geq 0$ for $0 leq t leq 1.$ However this inequality does not hold for any $A subset mathbb R^n$ Borel with $1 < mathcal L^n(A) < infty,$ since $sqrtt - t < 0$ whenever $t>1.$
answered Apr 8 at 17:26
ktoiktoi
2,5411618
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not true, even when $mu = nu.$ For concreteness let $mu = nu = mathcal L^n$ be the Lebesgue measure on $mathbb R^n$ equipped with the Borel $sigma$-algebra, and consider,
$$ mathcalE = E subset mathbb R^n : Q text Borel with mathcal L^n(Q) leq 1 . $$
This family is certainly stable under finite unions, and for any $E in mathcalE$ we have,
$$ mathcal L^n(E) leq sqrtmathcal L^n(E), $$
since $sqrtt - t geq 0$ for $0 leq t leq 1.$ However this inequality does not hold for any $A subset mathbb R^n$ Borel with $1 < mathcal L^n(A) < infty,$ since $sqrtt - t < 0$ whenever $t>1.$
answered Apr 8 at 17:26
ktoiktoi
2,5411618
$endgroup$
add a comment |
$begingroup$
This is not true, even when $mu = nu.$ For concreteness let $mu = nu = mathcal L^n$ be the Lebesgue measure on $mathbb R^n$ equipped with the Borel $sigma$-algebra, and consider,
$$ mathcalE = E subset mathbb R^n : Q text Borel with mathcal L^n(Q) leq 1 . $$
This family is certainly stable under finite unions, and for any $E in mathcalE$ we have,
$$ mathcal L^n(E) leq sqrtmathcal L^n(E), $$
since $sqrtt - t geq 0$ for $0 leq t leq 1.$ However this inequality does not hold for any $A subset mathbb R^n$ Borel with $1 < mathcal L^n(A) < infty,$ since $sqrtt - t < 0$ whenever $t>1.$
answered Apr 8 at 17:26
ktoiktoi
2,5411618
$endgroup$
add a comment |
$begingroup$
This is not true, even when $mu = nu.$ For concreteness let $mu = nu = mathcal L^n$ be the Lebesgue measure on $mathbb R^n$ equipped with the Borel $sigma$-algebra, and consider,
$$ mathcalE = E subset mathbb R^n : Q text Borel with mathcal L^n(Q) leq 1 . $$
This family is certainly stable under finite unions, and for any $E in mathcalE$ we have,
$$ mathcal L^n(E) leq sqrtmathcal L^n(E), $$
since $sqrtt - t geq 0$ for $0 leq t leq 1.$ However this inequality does not hold for any $A subset mathbb R^n$ Borel with $1 < mathcal L^n(A) < infty,$ since $sqrtt - t < 0$ whenever $t>1.$
answered Apr 8 at 17:26
ktoiktoi
2,5411618
$endgroup$
This is not true, even when $mu = nu.$ For concreteness let $mu = nu = mathcal L^n$ be the Lebesgue measure on $mathbb R^n$ equipped with the Borel $sigma$-algebra, and consider,
$$ mathcalE = E subset mathbb R^n : Q text Borel with mathcal L^n(Q) leq 1 . $$
This family is certainly stable under finite unions, and for any $E in mathcalE$ we have,
$$ mathcal L^n(E) leq sqrtmathcal L^n(E), $$
since $sqrtt - t geq 0$ for $0 leq t leq 1.$ However this inequality does not hold for any $A subset mathbb R^n$ Borel with $1 < mathcal L^n(A) < infty,$ since $sqrtt - t < 0$ whenever $t>1.$
answered Apr 8 at 17:26
ktoiktoi
2,5411618
answered Apr 8 at 17:26
ktoiktoi
2,5411618
answered Apr 8 at 17:26
ktoiktoi
2,5411618
answered Apr 8 at 17:26
ktoiktoi
2,5411618
2,5411618
add a comment |
add a comment |
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