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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                      What does “Dominus providebit” mean?

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                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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