Why L is not a σ-algebra?












1












$begingroup$


Why $L$ is not a $σ$-algebra?



Let $L$ be the collection of all finite disjoint unions of all intervals of the form:



$(−∞, a], (a, b], (b, ∞), ∅, R.$



Then L is an algebra over R, but not a σ-algebra because



union of sets $$bigcup _{i=1}^{infty} {(0,(i − 1)/i]} = (0,1)$$ which doesn't belong to L



Why (0, 1) doesn't belong to L ?



Any help?






measure-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Why $L$ is not a $σ$-algebra?



    Let $L$ be the collection of all finite disjoint unions of all intervals of the form:



    $(−∞, a], (a, b], (b, ∞), ∅, R.$



    Then L is an algebra over R, but not a σ-algebra because



    union of sets $$bigcup _{i=1}^{infty} {(0,(i − 1)/i]} = (0,1)$$ which doesn't belong to L



    Why (0, 1) doesn't belong to L ?



    Any help?






    measure-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Why $L$ is not a $σ$-algebra?



      Let $L$ be the collection of all finite disjoint unions of all intervals of the form:



      $(−∞, a], (a, b], (b, ∞), ∅, R.$



      Then L is an algebra over R, but not a σ-algebra because



      union of sets $$bigcup _{i=1}^{infty} {(0,(i − 1)/i]} = (0,1)$$ which doesn't belong to L



      Why (0, 1) doesn't belong to L ?



      Any help?






      measure-theory







      share|cite|improve this question











      $endgroup$




      Why $L$ is not a $σ$-algebra?



      Let $L$ be the collection of all finite disjoint unions of all intervals of the form:



      $(−∞, a], (a, b], (b, ∞), ∅, R.$



      Then L is an algebra over R, but not a σ-algebra because



      union of sets $$bigcup _{i=1}^{infty} {(0,(i − 1)/i]} = (0,1)$$ which doesn't belong to L



      Why (0, 1) doesn't belong to L ?



      Any help?






      measure-theory




      measure-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 15:19









      greedoid

      38.7k114797




      38.7k114797










      asked Jan 8 at 14:53









      LauraLaura

      1196




      1196






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+infty)$ or $(a,b]$, neither of which fits $(0,1)$.



          If you want a more rigorous proof, suppose $(0,1)in L$, then there is a finite set $mathcal{A} = {I_1,dots, I_n}$ of intervals of the given form such that $I := bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + frac{1-b}{2}in (0,1)$ but $b+ frac{1-b}{2} notin I$; or $bgeq 1$, so $1in I$ but $1notin (0,1)$. Both contradicts $(0,1) = I$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Amazing @Atudo!
            $endgroup$
            – Laura
            Jan 8 at 15:18



















          2












          $begingroup$

          Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
            $endgroup$
            – Laura
            Jan 8 at 15:07












          • $begingroup$
            Not sure what are you asking?
            $endgroup$
            – greedoid
            Jan 8 at 15:09










          • $begingroup$
            If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
            $endgroup$
            – Laura
            Jan 8 at 15:12











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+infty)$ or $(a,b]$, neither of which fits $(0,1)$.



          If you want a more rigorous proof, suppose $(0,1)in L$, then there is a finite set $mathcal{A} = {I_1,dots, I_n}$ of intervals of the given form such that $I := bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + frac{1-b}{2}in (0,1)$ but $b+ frac{1-b}{2} notin I$; or $bgeq 1$, so $1in I$ but $1notin (0,1)$. Both contradicts $(0,1) = I$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Amazing @Atudo!
            $endgroup$
            – Laura
            Jan 8 at 15:18
















          1












          $begingroup$

          Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+infty)$ or $(a,b]$, neither of which fits $(0,1)$.



          If you want a more rigorous proof, suppose $(0,1)in L$, then there is a finite set $mathcal{A} = {I_1,dots, I_n}$ of intervals of the given form such that $I := bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + frac{1-b}{2}in (0,1)$ but $b+ frac{1-b}{2} notin I$; or $bgeq 1$, so $1in I$ but $1notin (0,1)$. Both contradicts $(0,1) = I$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Amazing @Atudo!
            $endgroup$
            – Laura
            Jan 8 at 15:18














          1












          1








          1





          $begingroup$

          Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+infty)$ or $(a,b]$, neither of which fits $(0,1)$.



          If you want a more rigorous proof, suppose $(0,1)in L$, then there is a finite set $mathcal{A} = {I_1,dots, I_n}$ of intervals of the given form such that $I := bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + frac{1-b}{2}in (0,1)$ but $b+ frac{1-b}{2} notin I$; or $bgeq 1$, so $1in I$ but $1notin (0,1)$. Both contradicts $(0,1) = I$.






          share|cite|improve this answer









          $endgroup$



          Basically this is because elements in $L$ are constituted as finite unions of those intervals. And a non-trivial such union can only have the form $(a,+infty)$ or $(a,b]$, neither of which fits $(0,1)$.



          If you want a more rigorous proof, suppose $(0,1)in L$, then there is a finite set $mathcal{A} = {I_1,dots, I_n}$ of intervals of the given form such that $I := bigcup_{i=1}^{n} =(0,1)$. Obviously each nonempty $I_i$ can only be of the form $(a_i, b_i]$ for some finite reals $a,b$. By finiteness, there must be some largest $b_k$ among $b_1, dots, b_n$. Then we are in a dilemma: either $b< 1$ , so $b + frac{1-b}{2}in (0,1)$ but $b+ frac{1-b}{2} notin I$; or $bgeq 1$, so $1in I$ but $1notin (0,1)$. Both contradicts $(0,1) = I$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 15:16









          AtugoAtugo

          363




          363












          • $begingroup$
            Amazing @Atudo!
            $endgroup$
            – Laura
            Jan 8 at 15:18


















          • $begingroup$
            Amazing @Atudo!
            $endgroup$
            – Laura
            Jan 8 at 15:18
















          $begingroup$
          Amazing @Atudo!
          $endgroup$
          – Laura
          Jan 8 at 15:18




          $begingroup$
          Amazing @Atudo!
          $endgroup$
          – Laura
          Jan 8 at 15:18











          2












          $begingroup$

          Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
            $endgroup$
            – Laura
            Jan 8 at 15:07












          • $begingroup$
            Not sure what are you asking?
            $endgroup$
            – greedoid
            Jan 8 at 15:09










          • $begingroup$
            If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
            $endgroup$
            – Laura
            Jan 8 at 15:12
















          2












          $begingroup$

          Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
            $endgroup$
            – Laura
            Jan 8 at 15:07












          • $begingroup$
            Not sure what are you asking?
            $endgroup$
            – greedoid
            Jan 8 at 15:09










          • $begingroup$
            If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
            $endgroup$
            – Laura
            Jan 8 at 15:12














          2












          2








          2





          $begingroup$

          Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.






          share|cite|improve this answer











          $endgroup$



          Say interval $(0,1)$ would be union of disjunct interval of form $(a,b]$. But such union contains most rigt bound, so $(0,1)$ is not in that family.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 15:17

























          answered Jan 8 at 15:04









          greedoidgreedoid

          38.7k114797




          38.7k114797












          • $begingroup$
            But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
            $endgroup$
            – Laura
            Jan 8 at 15:07












          • $begingroup$
            Not sure what are you asking?
            $endgroup$
            – greedoid
            Jan 8 at 15:09










          • $begingroup$
            If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
            $endgroup$
            – Laura
            Jan 8 at 15:12


















          • $begingroup$
            But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
            $endgroup$
            – Laura
            Jan 8 at 15:07












          • $begingroup$
            Not sure what are you asking?
            $endgroup$
            – greedoid
            Jan 8 at 15:09










          • $begingroup$
            If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
            $endgroup$
            – Laura
            Jan 8 at 15:12
















          $begingroup$
          But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
          $endgroup$
          – Laura
          Jan 8 at 15:07






          $begingroup$
          But if contains most right bound , (a,b] contains (0,1) ? Right? If it was something like this: (0,1] would work?
          $endgroup$
          – Laura
          Jan 8 at 15:07














          $begingroup$
          Not sure what are you asking?
          $endgroup$
          – greedoid
          Jan 8 at 15:09




          $begingroup$
          Not sure what are you asking?
          $endgroup$
          – greedoid
          Jan 8 at 15:09












          $begingroup$
          If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
          $endgroup$
          – Laura
          Jan 8 at 15:12




          $begingroup$
          If it was something like this: (0,1] would work? Or (−∞, 1]. It would be a sigma algebra?
          $endgroup$
          – Laura
          Jan 8 at 15:12


















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