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Hahn Decomposition Theorem In Folland
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Equivalent Measures via Hahn-Kolmogorov and $sigma-$finitenesswhat is the meaning of this statement (a use of Hahn-Banach theorem)?Uniqueness of decomposition in the Radon-Nikodym theoremAbout the proof of the Hahn decomposition theorem on WikipediaOn the example of the The Hahn decompositionA non-constructive proof of Hahn decomposition theoremCan't follow Proof in Folland: Outer Measures and Pre-measures.Prove that every signed measure $mu$ has a decomposition into difference $mu=mu^+ - mu^-$ , where $mu^+ , mu^-$ are measures .How to show sequnce of function converges uniformly on any set on which function is bounded?Doubt in understading Proof of Matrix lie group and Lie algebra locally homemorphic
$begingroup$
I was reading the proof of Hahn Decomposition theorem from the textbook of Folland: precisely I was looking at the following text
I have the following question:
- As Highlighted in the text above, why $m$ is finite? It may be infinite as there is no restriction on $X$. Why does the author consider it finite?
- Again why $nu(A)<infty$ ? I do not understand also this.
- I understand that even for $A$ also we get some $B$ with the property that $nu(B)>nu(A)+1/n$ but I do not understand how this leads to a contradiction.
I would be really thankful if someone could help me. Any help will be appreciated.
real-analysis measure-theory proof-explanation lebesgue-measure
edited Mar 31 at 9:50
dmtri
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
$endgroup$
add a comment |
$begingroup$
I was reading the proof of Hahn Decomposition theorem from the textbook of Folland: precisely I was looking at the following text
I have the following question:
- As Highlighted in the text above, why $m$ is finite? It may be infinite as there is no restriction on $X$. Why does the author consider it finite?
- Again why $nu(A)<infty$ ? I do not understand also this.
- I understand that even for $A$ also we get some $B$ with the property that $nu(B)>nu(A)+1/n$ but I do not understand how this leads to a contradiction.
I would be really thankful if someone could help me. Any help will be appreciated.
real-analysis measure-theory proof-explanation lebesgue-measure
edited Mar 31 at 9:50
dmtri
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
$endgroup$
add a comment |
$begingroup$
I was reading the proof of Hahn Decomposition theorem from the textbook of Folland: precisely I was looking at the following text
I have the following question:
- As Highlighted in the text above, why $m$ is finite? It may be infinite as there is no restriction on $X$. Why does the author consider it finite?
- Again why $nu(A)<infty$ ? I do not understand also this.
- I understand that even for $A$ also we get some $B$ with the property that $nu(B)>nu(A)+1/n$ but I do not understand how this leads to a contradiction.
I would be really thankful if someone could help me. Any help will be appreciated.
real-analysis measure-theory proof-explanation lebesgue-measure
edited Mar 31 at 9:50
dmtri
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
$endgroup$
I was reading the proof of Hahn Decomposition theorem from the textbook of Folland: precisely I was looking at the following text
I have the following question:
- As Highlighted in the text above, why $m$ is finite? It may be infinite as there is no restriction on $X$. Why does the author consider it finite?
- Again why $nu(A)<infty$ ? I do not understand also this.
- I understand that even for $A$ also we get some $B$ with the property that $nu(B)>nu(A)+1/n$ but I do not understand how this leads to a contradiction.
I would be really thankful if someone could help me. Any help will be appreciated.
real-analysis measure-theory proof-explanation lebesgue-measure
real-analysis measure-theory proof-explanation lebesgue-measure
edited Mar 31 at 9:50
dmtri
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
edited Mar 31 at 9:50
dmtri
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
edited Mar 31 at 9:50
dmtri
1,7612521
edited Mar 31 at 9:50
dmtri
1,7612521
edited Mar 31 at 9:50
dmtri
1,7612521
1,7612521
asked Mar 31 at 8:28
SRJSRJ
1,8981620
asked Mar 31 at 8:28
SRJSRJ
1,8981620
asked Mar 31 at 8:28
SRJSRJ
1,8981620
1,8981620
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $nu : mathcal X to (-infty,+infty)$.
According to Wikipedia:
There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.
Suppose we allow infinite values. Then Lebesgue measure on $mathbb R$ would be a finite measure, $E = mathbb R$ would be a positive set, and $nu(E)=infty$ on that case.
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
$endgroup$
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $nu : mathcal X to (-infty,+infty)$.
According to Wikipedia:
There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.
Suppose we allow infinite values. Then Lebesgue measure on $mathbb R$ would be a finite measure, $E = mathbb R$ would be a positive set, and $nu(E)=infty$ on that case.
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
$endgroup$
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
add a comment |
$begingroup$
Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $nu : mathcal X to (-infty,+infty)$.
According to Wikipedia:
There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.
Suppose we allow infinite values. Then Lebesgue measure on $mathbb R$ would be a finite measure, $E = mathbb R$ would be a positive set, and $nu(E)=infty$ on that case.
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
$endgroup$
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
add a comment |
$begingroup$
Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $nu : mathcal X to (-infty,+infty)$.
According to Wikipedia:
There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.
Suppose we allow infinite values. Then Lebesgue measure on $mathbb R$ would be a finite measure, $E = mathbb R$ would be a positive set, and $nu(E)=infty$ on that case.
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
$endgroup$
Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $nu : mathcal X to (-infty,+infty)$.
According to Wikipedia:
There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.
Suppose we allow infinite values. Then Lebesgue measure on $mathbb R$ would be a finite measure, $E = mathbb R$ would be a positive set, and $nu(E)=infty$ on that case.
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
edited Mar 31 at 10:59
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
answered Mar 31 at 10:37
GEdgarGEdgar
63.5k269175
63.5k269175
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
add a comment |
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
1
1
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
Did you look this up? Signed measures can take infinite values as well.
$endgroup$
– user370967
Mar 31 at 10:45
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
$begingroup$
@GEdgar Dear Sir , Signed measure can take at most on infinite value either positive or negative.
$endgroup$
– SRJ
Mar 31 at 10:57
add a comment |
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