Stronger continuity of measure from below?
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If $(X,mathcal F, sigma)$ is any space with a measure and $mathcal{A subseteq F}$ is a countable chain then $sigma(bigcup mathcal A)=sup_{A in mathcal A} sigma(A)$; this is well-known. Can there be a space $(X,mathcal F, sigma)$ where an uncountable chain $mathcal {A subseteq F}$ such that $bigcup mathcal A in mathcal F$ and $sigma(bigcup mathcal A)neqsup_{A in mathcal A}sigma(A)$ can be found?
I generally accept Axiom of Choice.
measure-theory
asked Nov 21 at 19:31
Michał Zapała
11812
add a comment |
up vote
0
down vote
favorite
If $(X,mathcal F, sigma)$ is any space with a measure and $mathcal{A subseteq F}$ is a countable chain then $sigma(bigcup mathcal A)=sup_{A in mathcal A} sigma(A)$; this is well-known. Can there be a space $(X,mathcal F, sigma)$ where an uncountable chain $mathcal {A subseteq F}$ such that $bigcup mathcal A in mathcal F$ and $sigma(bigcup mathcal A)neqsup_{A in mathcal A}sigma(A)$ can be found?
I generally accept Axiom of Choice.
measure-theory
asked Nov 21 at 19:31
Michał Zapała
11812
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $(X,mathcal F, sigma)$ is any space with a measure and $mathcal{A subseteq F}$ is a countable chain then $sigma(bigcup mathcal A)=sup_{A in mathcal A} sigma(A)$; this is well-known. Can there be a space $(X,mathcal F, sigma)$ where an uncountable chain $mathcal {A subseteq F}$ such that $bigcup mathcal A in mathcal F$ and $sigma(bigcup mathcal A)neqsup_{A in mathcal A}sigma(A)$ can be found?
I generally accept Axiom of Choice.
measure-theory
asked Nov 21 at 19:31
Michał Zapała
11812
If $(X,mathcal F, sigma)$ is any space with a measure and $mathcal{A subseteq F}$ is a countable chain then $sigma(bigcup mathcal A)=sup_{A in mathcal A} sigma(A)$; this is well-known. Can there be a space $(X,mathcal F, sigma)$ where an uncountable chain $mathcal {A subseteq F}$ such that $bigcup mathcal A in mathcal F$ and $sigma(bigcup mathcal A)neqsup_{A in mathcal A}sigma(A)$ can be found?
I generally accept Axiom of Choice.
measure-theory
measure-theory
asked Nov 21 at 19:31
Michał Zapała
11812
asked Nov 21 at 19:31
Michał Zapała
11812
asked Nov 21 at 19:31
Michał Zapała
11812
asked Nov 21 at 19:31
Michał Zapała
11812
asked Nov 21 at 19:31
Michał Zapała
11812
11812
add a comment |
add a comment |
1 Answer
1
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oldest
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up vote
2
down vote
accepted
Consider $(aleph_1, mathcal P(aleph_1), sigma)$ where $aleph_1$ is the smallest uncountable ordinal and $sigma(E)=0$ if $E$ is countable, $sigma(E)=infty$ if $E$ is uncountable. Consider
$$
mathcal A = bigl{ A_x={yinaleph_1:yleq x}: xinaleph_1 bigr}.
$$
Then each $A_x$ is countable therefore
$$
sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
$$
answered Nov 21 at 20:01
Federico
4,107512
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Consider $(aleph_1, mathcal P(aleph_1), sigma)$ where $aleph_1$ is the smallest uncountable ordinal and $sigma(E)=0$ if $E$ is countable, $sigma(E)=infty$ if $E$ is uncountable. Consider
$$
mathcal A = bigl{ A_x={yinaleph_1:yleq x}: xinaleph_1 bigr}.
$$
Then each $A_x$ is countable therefore
$$
sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
$$
answered Nov 21 at 20:01
Federico
4,107512
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
add a comment |
up vote
2
down vote
accepted
Consider $(aleph_1, mathcal P(aleph_1), sigma)$ where $aleph_1$ is the smallest uncountable ordinal and $sigma(E)=0$ if $E$ is countable, $sigma(E)=infty$ if $E$ is uncountable. Consider
$$
mathcal A = bigl{ A_x={yinaleph_1:yleq x}: xinaleph_1 bigr}.
$$
Then each $A_x$ is countable therefore
$$
sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
$$
answered Nov 21 at 20:01
Federico
4,107512
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Consider $(aleph_1, mathcal P(aleph_1), sigma)$ where $aleph_1$ is the smallest uncountable ordinal and $sigma(E)=0$ if $E$ is countable, $sigma(E)=infty$ if $E$ is uncountable. Consider
$$
mathcal A = bigl{ A_x={yinaleph_1:yleq x}: xinaleph_1 bigr}.
$$
Then each $A_x$ is countable therefore
$$
sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
$$
answered Nov 21 at 20:01
Federico
4,107512
Consider $(aleph_1, mathcal P(aleph_1), sigma)$ where $aleph_1$ is the smallest uncountable ordinal and $sigma(E)=0$ if $E$ is countable, $sigma(E)=infty$ if $E$ is uncountable. Consider
$$
mathcal A = bigl{ A_x={yinaleph_1:yleq x}: xinaleph_1 bigr}.
$$
Then each $A_x$ is countable therefore
$$
sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
$$
answered Nov 21 at 20:01
Federico
4,107512
answered Nov 21 at 20:01
Federico
4,107512
answered Nov 21 at 20:01
Federico
4,107512
answered Nov 21 at 20:01
Federico
4,107512
4,107512
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
add a comment |
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
Notice also that $mathcal A=aleph_1$
– Federico
Nov 21 at 20:05
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
This construction does not require the axiom of choice
– Federico
Nov 21 at 20:07
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Wait, can you actually prove that countable family of countable sets has countable union without axiom of choice?
– Michał Zapała
Nov 21 at 20:47
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
Well, you are right, I can prove the $sigma$-additivity of my measure $sigma$ assuming at least the axiom of countable choice.
– Federico
Nov 22 at 17:25
add a comment |
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