If ${X_alpha}$ is a collection of mutually disjoint meas. subsets of $mathbb{R} rightarrow$ then at most...












3












$begingroup$


Let ${X_alpha}_{alpha in I}$ be a collection of mutually disjoint measurable subsets of $mathbb{R}. Show that at most countable of them has positive measure.



I want to see if my proof is correct.



Proof:



Let ${X_alpha}_{alpha in Gamma subset I}$ be the subcollection such that $m(X_alpha) > 0, forall alpha in Gamma$.



Since each set is measurable and has positive measure,



$$ forall X_{alpha}, alpha in Gamma, exists O_{alpha} = (a_alpha,b_alpha) mbox{ such that } O_alpha subset X_{alpha}$$



Now we must show that $|Gamma| leq |mathbb{N}|$



Let $f: {O_{alpha}}_{alpha in Gamma} to mathbb{Q}$ be the function that
$$f(O_alpha) = q_{alpha}, mbox{ where }q_alpha in (a_alpha,b_alpha)$$



This is possible because $mathbb{Q}$ is dense in $mathbb{R}$



This function is clearly injective since ${O_alpha}$ is a disjoint collection of open sets. Hence, the same rational can't be in two different open intervals from this collection.



Since $f$ is injective, ${O_alpha}_{alpha in Gamma}| = |Gamma| leq |mathbb{Q}| = |mathbb{N}| = aleph_0$



Q.E.D






real-analysis measure-theory proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
    $endgroup$
    – saz
    Jan 11 at 15:24








  • 1




    $begingroup$
    See this.
    $endgroup$
    – David Mitra
    Jan 11 at 15:33






  • 1




    $begingroup$
    No, a closed set need not contain a rational number.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:35






  • 2




    $begingroup$
    For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:38






  • 1




    $begingroup$
    FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
    $endgroup$
    – Dave L. Renfro
    Jan 11 at 16:12
















3












$begingroup$


Let ${X_alpha}_{alpha in I}$ be a collection of mutually disjoint measurable subsets of $mathbb{R}. Show that at most countable of them has positive measure.



I want to see if my proof is correct.



Proof:



Let ${X_alpha}_{alpha in Gamma subset I}$ be the subcollection such that $m(X_alpha) > 0, forall alpha in Gamma$.



Since each set is measurable and has positive measure,



$$ forall X_{alpha}, alpha in Gamma, exists O_{alpha} = (a_alpha,b_alpha) mbox{ such that } O_alpha subset X_{alpha}$$



Now we must show that $|Gamma| leq |mathbb{N}|$



Let $f: {O_{alpha}}_{alpha in Gamma} to mathbb{Q}$ be the function that
$$f(O_alpha) = q_{alpha}, mbox{ where }q_alpha in (a_alpha,b_alpha)$$



This is possible because $mathbb{Q}$ is dense in $mathbb{R}$



This function is clearly injective since ${O_alpha}$ is a disjoint collection of open sets. Hence, the same rational can't be in two different open intervals from this collection.



Since $f$ is injective, ${O_alpha}_{alpha in Gamma}| = |Gamma| leq |mathbb{Q}| = |mathbb{N}| = aleph_0$



Q.E.D






real-analysis measure-theory proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
    $endgroup$
    – saz
    Jan 11 at 15:24








  • 1




    $begingroup$
    See this.
    $endgroup$
    – David Mitra
    Jan 11 at 15:33






  • 1




    $begingroup$
    No, a closed set need not contain a rational number.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:35






  • 2




    $begingroup$
    For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:38






  • 1




    $begingroup$
    FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
    $endgroup$
    – Dave L. Renfro
    Jan 11 at 16:12














3












3








3





$begingroup$


Let ${X_alpha}_{alpha in I}$ be a collection of mutually disjoint measurable subsets of $mathbb{R}. Show that at most countable of them has positive measure.



I want to see if my proof is correct.



Proof:



Let ${X_alpha}_{alpha in Gamma subset I}$ be the subcollection such that $m(X_alpha) > 0, forall alpha in Gamma$.



Since each set is measurable and has positive measure,



$$ forall X_{alpha}, alpha in Gamma, exists O_{alpha} = (a_alpha,b_alpha) mbox{ such that } O_alpha subset X_{alpha}$$



Now we must show that $|Gamma| leq |mathbb{N}|$



Let $f: {O_{alpha}}_{alpha in Gamma} to mathbb{Q}$ be the function that
$$f(O_alpha) = q_{alpha}, mbox{ where }q_alpha in (a_alpha,b_alpha)$$



This is possible because $mathbb{Q}$ is dense in $mathbb{R}$



This function is clearly injective since ${O_alpha}$ is a disjoint collection of open sets. Hence, the same rational can't be in two different open intervals from this collection.



Since $f$ is injective, ${O_alpha}_{alpha in Gamma}| = |Gamma| leq |mathbb{Q}| = |mathbb{N}| = aleph_0$



Q.E.D






real-analysis measure-theory proof-verification







share|cite|improve this question











$endgroup$




Let ${X_alpha}_{alpha in I}$ be a collection of mutually disjoint measurable subsets of $mathbb{R}. Show that at most countable of them has positive measure.



I want to see if my proof is correct.



Proof:



Let ${X_alpha}_{alpha in Gamma subset I}$ be the subcollection such that $m(X_alpha) > 0, forall alpha in Gamma$.



Since each set is measurable and has positive measure,



$$ forall X_{alpha}, alpha in Gamma, exists O_{alpha} = (a_alpha,b_alpha) mbox{ such that } O_alpha subset X_{alpha}$$



Now we must show that $|Gamma| leq |mathbb{N}|$



Let $f: {O_{alpha}}_{alpha in Gamma} to mathbb{Q}$ be the function that
$$f(O_alpha) = q_{alpha}, mbox{ where }q_alpha in (a_alpha,b_alpha)$$



This is possible because $mathbb{Q}$ is dense in $mathbb{R}$



This function is clearly injective since ${O_alpha}$ is a disjoint collection of open sets. Hence, the same rational can't be in two different open intervals from this collection.



Since $f$ is injective, ${O_alpha}_{alpha in Gamma}| = |Gamma| leq |mathbb{Q}| = |mathbb{N}| = aleph_0$



Q.E.D






real-analysis measure-theory proof-verification




real-analysis measure-theory proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:31









Eric Wofsey

183k13211338




183k13211338










asked Jan 11 at 15:19









Richard ClareRichard Clare

1,066314




1,066314








  • 1




    $begingroup$
    No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
    $endgroup$
    – saz
    Jan 11 at 15:24








  • 1




    $begingroup$
    See this.
    $endgroup$
    – David Mitra
    Jan 11 at 15:33






  • 1




    $begingroup$
    No, a closed set need not contain a rational number.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:35






  • 2




    $begingroup$
    For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:38






  • 1




    $begingroup$
    FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
    $endgroup$
    – Dave L. Renfro
    Jan 11 at 16:12














  • 1




    $begingroup$
    No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
    $endgroup$
    – saz
    Jan 11 at 15:24








  • 1




    $begingroup$
    See this.
    $endgroup$
    – David Mitra
    Jan 11 at 15:33






  • 1




    $begingroup$
    No, a closed set need not contain a rational number.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:35






  • 2




    $begingroup$
    For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
    $endgroup$
    – David C. Ullrich
    Jan 11 at 15:38






  • 1




    $begingroup$
    FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
    $endgroup$
    – Dave L. Renfro
    Jan 11 at 16:12








1




1




$begingroup$
No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
$endgroup$
– saz
Jan 11 at 15:24






$begingroup$
No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] backslash mathbb{Q}$).
$endgroup$
– saz
Jan 11 at 15:24






1




1




$begingroup$
See this.
$endgroup$
– David Mitra
Jan 11 at 15:33




$begingroup$
See this.
$endgroup$
– David Mitra
Jan 11 at 15:33




1




1




$begingroup$
No, a closed set need not contain a rational number.
$endgroup$
– David C. Ullrich
Jan 11 at 15:35




$begingroup$
No, a closed set need not contain a rational number.
$endgroup$
– David C. Ullrich
Jan 11 at 15:35




2




2




$begingroup$
For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
$endgroup$
– David C. Ullrich
Jan 11 at 15:38




$begingroup$
For example, say the ratinoals are $q_1,dots$. Let $V=bigcup(q_n-1/n^2,q_n+1/n^2)$, $F=Bbb Rsetminus V$. Then $V$ is open so $F$ is closed, $V$ contains every rational so $F$ contains no rational, and $Fneemptyset$ since $m(V)<infty$.
$endgroup$
– David C. Ullrich
Jan 11 at 15:38




1




1




$begingroup$
FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
$endgroup$
– Dave L. Renfro
Jan 11 at 16:12




$begingroup$
FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero.
$endgroup$
– Dave L. Renfro
Jan 11 at 16:12










2 Answers
2






active

oldest

votes


















3












$begingroup$

No, that's totally wrong. Saying $X$ has positive measure does not imply $X$ contains an open interval.



Hint: For $n=1,2dots$ let $$Gamma_n={alpha:m([-n,n]cap X_alpha)>0}.$$Show that each $Gamma_n$ is countable and that $${alpha:m(X_alpha)>0}=bigcup_{n=1}^inftyGamma_n.$$



(To show $Gamma_n$ is countable: Show that ${alpha:m([-n,n]cap X_alpha)>1/k}$ is finite...)






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Hints: (I take it that you are considering $mathbb{R}$ equipped with the Lebesgue measure $lambda$.)




    1. Let $(Omega,mathcal{A},mu)$ be a probability space. Show that if $(A_{alpha})_{alpha in I}$ is a collection of measurable mutually disjoint sets, then $${alpha; mu(A_{alpha})>0}$$ is countable. Hint: Check that $${alpha; mu(A_{alpha}) > 1/k}$$ is a finite set for any $k in mathbb{N}$.

    2. Now let $(X_{alpha})_{alpha in I}$ be a collection of measurable mutually disjoints subsets of $mathbb{R}$. Define $$A_{ell,alpha} := X_{alpha} cap [ell,ell+1], qquad ell in mathbb{Z}.$$ Use Step 1 to prove that $${alpha; lambda(A_{ell,alpha})>0}$$ is countable for each $ell$. Conclude that $${alpha; lambda(X_{alpha})>0}$$ is countable.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      No, that's totally wrong. Saying $X$ has positive measure does not imply $X$ contains an open interval.



      Hint: For $n=1,2dots$ let $$Gamma_n={alpha:m([-n,n]cap X_alpha)>0}.$$Show that each $Gamma_n$ is countable and that $${alpha:m(X_alpha)>0}=bigcup_{n=1}^inftyGamma_n.$$



      (To show $Gamma_n$ is countable: Show that ${alpha:m([-n,n]cap X_alpha)>1/k}$ is finite...)






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        No, that's totally wrong. Saying $X$ has positive measure does not imply $X$ contains an open interval.



        Hint: For $n=1,2dots$ let $$Gamma_n={alpha:m([-n,n]cap X_alpha)>0}.$$Show that each $Gamma_n$ is countable and that $${alpha:m(X_alpha)>0}=bigcup_{n=1}^inftyGamma_n.$$



        (To show $Gamma_n$ is countable: Show that ${alpha:m([-n,n]cap X_alpha)>1/k}$ is finite...)






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          No, that's totally wrong. Saying $X$ has positive measure does not imply $X$ contains an open interval.



          Hint: For $n=1,2dots$ let $$Gamma_n={alpha:m([-n,n]cap X_alpha)>0}.$$Show that each $Gamma_n$ is countable and that $${alpha:m(X_alpha)>0}=bigcup_{n=1}^inftyGamma_n.$$



          (To show $Gamma_n$ is countable: Show that ${alpha:m([-n,n]cap X_alpha)>1/k}$ is finite...)






          share|cite|improve this answer











          $endgroup$



          No, that's totally wrong. Saying $X$ has positive measure does not imply $X$ contains an open interval.



          Hint: For $n=1,2dots$ let $$Gamma_n={alpha:m([-n,n]cap X_alpha)>0}.$$Show that each $Gamma_n$ is countable and that $${alpha:m(X_alpha)>0}=bigcup_{n=1}^inftyGamma_n.$$



          (To show $Gamma_n$ is countable: Show that ${alpha:m([-n,n]cap X_alpha)>1/k}$ is finite...)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 15:47

























          answered Jan 11 at 15:33









          David C. UllrichDavid C. Ullrich

          60k43994




          60k43994























              2












              $begingroup$

              Hints: (I take it that you are considering $mathbb{R}$ equipped with the Lebesgue measure $lambda$.)




              1. Let $(Omega,mathcal{A},mu)$ be a probability space. Show that if $(A_{alpha})_{alpha in I}$ is a collection of measurable mutually disjoint sets, then $${alpha; mu(A_{alpha})>0}$$ is countable. Hint: Check that $${alpha; mu(A_{alpha}) > 1/k}$$ is a finite set for any $k in mathbb{N}$.

              2. Now let $(X_{alpha})_{alpha in I}$ be a collection of measurable mutually disjoints subsets of $mathbb{R}$. Define $$A_{ell,alpha} := X_{alpha} cap [ell,ell+1], qquad ell in mathbb{Z}.$$ Use Step 1 to prove that $${alpha; lambda(A_{ell,alpha})>0}$$ is countable for each $ell$. Conclude that $${alpha; lambda(X_{alpha})>0}$$ is countable.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hints: (I take it that you are considering $mathbb{R}$ equipped with the Lebesgue measure $lambda$.)




                1. Let $(Omega,mathcal{A},mu)$ be a probability space. Show that if $(A_{alpha})_{alpha in I}$ is a collection of measurable mutually disjoint sets, then $${alpha; mu(A_{alpha})>0}$$ is countable. Hint: Check that $${alpha; mu(A_{alpha}) > 1/k}$$ is a finite set for any $k in mathbb{N}$.

                2. Now let $(X_{alpha})_{alpha in I}$ be a collection of measurable mutually disjoints subsets of $mathbb{R}$. Define $$A_{ell,alpha} := X_{alpha} cap [ell,ell+1], qquad ell in mathbb{Z}.$$ Use Step 1 to prove that $${alpha; lambda(A_{ell,alpha})>0}$$ is countable for each $ell$. Conclude that $${alpha; lambda(X_{alpha})>0}$$ is countable.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hints: (I take it that you are considering $mathbb{R}$ equipped with the Lebesgue measure $lambda$.)




                  1. Let $(Omega,mathcal{A},mu)$ be a probability space. Show that if $(A_{alpha})_{alpha in I}$ is a collection of measurable mutually disjoint sets, then $${alpha; mu(A_{alpha})>0}$$ is countable. Hint: Check that $${alpha; mu(A_{alpha}) > 1/k}$$ is a finite set for any $k in mathbb{N}$.

                  2. Now let $(X_{alpha})_{alpha in I}$ be a collection of measurable mutually disjoints subsets of $mathbb{R}$. Define $$A_{ell,alpha} := X_{alpha} cap [ell,ell+1], qquad ell in mathbb{Z}.$$ Use Step 1 to prove that $${alpha; lambda(A_{ell,alpha})>0}$$ is countable for each $ell$. Conclude that $${alpha; lambda(X_{alpha})>0}$$ is countable.






                  share|cite|improve this answer









                  $endgroup$



                  Hints: (I take it that you are considering $mathbb{R}$ equipped with the Lebesgue measure $lambda$.)




                  1. Let $(Omega,mathcal{A},mu)$ be a probability space. Show that if $(A_{alpha})_{alpha in I}$ is a collection of measurable mutually disjoint sets, then $${alpha; mu(A_{alpha})>0}$$ is countable. Hint: Check that $${alpha; mu(A_{alpha}) > 1/k}$$ is a finite set for any $k in mathbb{N}$.

                  2. Now let $(X_{alpha})_{alpha in I}$ be a collection of measurable mutually disjoints subsets of $mathbb{R}$. Define $$A_{ell,alpha} := X_{alpha} cap [ell,ell+1], qquad ell in mathbb{Z}.$$ Use Step 1 to prove that $${alpha; lambda(A_{ell,alpha})>0}$$ is countable for each $ell$. Conclude that $${alpha; lambda(X_{alpha})>0}$$ is countable.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 15:39









                  sazsaz

                  79.7k860124




                  79.7k860124






























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                      Antonio Litta Visconti Arese

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                      Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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