Mixing
How do I show that $mu wedge nu leq mu, nu$?
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I'm using Royden & Fitzpatrick for real analysis, and one of the problems defines (for signed measures $mu$ and $nu$) $mu wedge nu = frac{1}{2}(mu + nu - |mu - nu|)$ and $mu vee nu = mu + nu - mu wedge nu$, and requires us to show that $mu wedge nu leq mu, nu$ and that if $eta$ is any other signed measure smaller than $mu, nu$, then $eta leq mu wedge nu$. I've tried everything I can think of, but the best I've been able to come up with is $mu wedge nu leq |nu|, |mu|$ with the reverse triangle inequality. Can anyone point me in the right direction?
real-analysis measure-theory
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
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add a comment |
$begingroup$
I'm using Royden & Fitzpatrick for real analysis, and one of the problems defines (for signed measures $mu$ and $nu$) $mu wedge nu = frac{1}{2}(mu + nu - |mu - nu|)$ and $mu vee nu = mu + nu - mu wedge nu$, and requires us to show that $mu wedge nu leq mu, nu$ and that if $eta$ is any other signed measure smaller than $mu, nu$, then $eta leq mu wedge nu$. I've tried everything I can think of, but the best I've been able to come up with is $mu wedge nu leq |nu|, |mu|$ with the reverse triangle inequality. Can anyone point me in the right direction?
real-analysis measure-theory
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
$endgroup$
2
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
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What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12
add a comment |
$begingroup$
I'm using Royden & Fitzpatrick for real analysis, and one of the problems defines (for signed measures $mu$ and $nu$) $mu wedge nu = frac{1}{2}(mu + nu - |mu - nu|)$ and $mu vee nu = mu + nu - mu wedge nu$, and requires us to show that $mu wedge nu leq mu, nu$ and that if $eta$ is any other signed measure smaller than $mu, nu$, then $eta leq mu wedge nu$. I've tried everything I can think of, but the best I've been able to come up with is $mu wedge nu leq |nu|, |mu|$ with the reverse triangle inequality. Can anyone point me in the right direction?
real-analysis measure-theory
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
$endgroup$
I'm using Royden & Fitzpatrick for real analysis, and one of the problems defines (for signed measures $mu$ and $nu$) $mu wedge nu = frac{1}{2}(mu + nu - |mu - nu|)$ and $mu vee nu = mu + nu - mu wedge nu$, and requires us to show that $mu wedge nu leq mu, nu$ and that if $eta$ is any other signed measure smaller than $mu, nu$, then $eta leq mu wedge nu$. I've tried everything I can think of, but the best I've been able to come up with is $mu wedge nu leq |nu|, |mu|$ with the reverse triangle inequality. Can anyone point me in the right direction?
real-analysis measure-theory
real-analysis measure-theory
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
asked Jan 26 at 20:06
IsomorphismIsomorphism
1339
1339
2
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
$begingroup$
What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12
add a comment |
2
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
$begingroup$
What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12
2
2
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
$begingroup$
What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12
$begingroup$
What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $A$ be an element of the corresponding $sigma$-algebra $Sigma$. Then for any $varepsilon > 0$ we find disjoint $A_1,ldots,A_nsubset A$ in $Sigma$ such that
$$
|mu-nu|(A) = sum_{j=1}^n|mu(A_j)-nu(A_j)| + varepsilon',
$$
where $varepsilon'levarepsilon$. Hence,
begin{align*}
(muwedgenu)(A)
&= sum_{j=1}^nbig[mu(A_j) + nu(A_j) - |mu(A_j)-nu(A_j)|big] - varepsilon'\
&= sum_{j=1}^nmin{mu(A_j),nu(A_j)} - varepsilon'.
end{align*}
From this everything follows.
answered Jan 26 at 20:55
amsmathamsmath
2,842417
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A$ be an element of the corresponding $sigma$-algebra $Sigma$. Then for any $varepsilon > 0$ we find disjoint $A_1,ldots,A_nsubset A$ in $Sigma$ such that
$$
|mu-nu|(A) = sum_{j=1}^n|mu(A_j)-nu(A_j)| + varepsilon',
$$
where $varepsilon'levarepsilon$. Hence,
begin{align*}
(muwedgenu)(A)
&= sum_{j=1}^nbig[mu(A_j) + nu(A_j) - |mu(A_j)-nu(A_j)|big] - varepsilon'\
&= sum_{j=1}^nmin{mu(A_j),nu(A_j)} - varepsilon'.
end{align*}
From this everything follows.
answered Jan 26 at 20:55
amsmathamsmath
2,842417
$endgroup$
add a comment |
$begingroup$
Let $A$ be an element of the corresponding $sigma$-algebra $Sigma$. Then for any $varepsilon > 0$ we find disjoint $A_1,ldots,A_nsubset A$ in $Sigma$ such that
$$
|mu-nu|(A) = sum_{j=1}^n|mu(A_j)-nu(A_j)| + varepsilon',
$$
where $varepsilon'levarepsilon$. Hence,
begin{align*}
(muwedgenu)(A)
&= sum_{j=1}^nbig[mu(A_j) + nu(A_j) - |mu(A_j)-nu(A_j)|big] - varepsilon'\
&= sum_{j=1}^nmin{mu(A_j),nu(A_j)} - varepsilon'.
end{align*}
From this everything follows.
answered Jan 26 at 20:55
amsmathamsmath
2,842417
$endgroup$
add a comment |
$begingroup$
Let $A$ be an element of the corresponding $sigma$-algebra $Sigma$. Then for any $varepsilon > 0$ we find disjoint $A_1,ldots,A_nsubset A$ in $Sigma$ such that
$$
|mu-nu|(A) = sum_{j=1}^n|mu(A_j)-nu(A_j)| + varepsilon',
$$
where $varepsilon'levarepsilon$. Hence,
begin{align*}
(muwedgenu)(A)
&= sum_{j=1}^nbig[mu(A_j) + nu(A_j) - |mu(A_j)-nu(A_j)|big] - varepsilon'\
&= sum_{j=1}^nmin{mu(A_j),nu(A_j)} - varepsilon'.
end{align*}
From this everything follows.
answered Jan 26 at 20:55
amsmathamsmath
2,842417
$endgroup$
Let $A$ be an element of the corresponding $sigma$-algebra $Sigma$. Then for any $varepsilon > 0$ we find disjoint $A_1,ldots,A_nsubset A$ in $Sigma$ such that
$$
|mu-nu|(A) = sum_{j=1}^n|mu(A_j)-nu(A_j)| + varepsilon',
$$
where $varepsilon'levarepsilon$. Hence,
begin{align*}
(muwedgenu)(A)
&= sum_{j=1}^nbig[mu(A_j) + nu(A_j) - |mu(A_j)-nu(A_j)|big] - varepsilon'\
&= sum_{j=1}^nmin{mu(A_j),nu(A_j)} - varepsilon'.
end{align*}
From this everything follows.
answered Jan 26 at 20:55
amsmathamsmath
2,842417
answered Jan 26 at 20:55
amsmathamsmath
2,842417
answered Jan 26 at 20:55
amsmathamsmath
2,842417
answered Jan 26 at 20:55
amsmathamsmath
2,842417
2,842417
add a comment |
add a comment |
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2
$begingroup$
For $a,binmathbb R$ you have $tfrac 12(a+b-|a-b|) = min(a,b)$.
$endgroup$
– amsmath
Jan 26 at 20:09
$begingroup$
What concerns me about that is that $mu wedge nu$ is a signed measure, but in general, $min{mu, nu}$ is not, because it fails to be countably additive. So I don't think we can force that equality.
$endgroup$
– Isomorphism
Jan 26 at 20:12