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







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    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 21 at 19:31









      Michał Zapała

      11812




      11812






















          1 Answer
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          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
          $$
          sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
          $$






          share|cite|improve this answer





















          • 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











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          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
          $$
          sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
          $$






          share|cite|improve this answer





















          • 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















          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
          $$
          sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
          $$






          share|cite|improve this answer





















          • 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













          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
          $$
          sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
          $$






          share|cite|improve this answer












          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
          $$
          sup_{Ainmathcal A} sigma(A) = 0 neq infty = sigmaleft(bigcupmathcal Aright) .
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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