Write $[0,1]^mathbb{N}$ as a disjoint union of two measurable subsets such that non of which contain a...












0












$begingroup$


Consider the space $X=[0,1]^mathbb{N}$ endowed with the Borel $sigma$-algebra (with respect to the product topology). I'd like to think of $X$ as the set of all sequences with values in $[0,1]$.



My Question: Is it possible to split $X$ into a disjoint union of two measurable subsets $X=mathcal{A}_1sqcupmathcal{A}_2$ such that no cylinder of $X$ lies in any of the $mathcal{A}_i$?



By a cylinder I mean a set of all sequences whose values at some finitely many indices $n_1,...,n_k$ are fixed. (i.e. sets of the form ${a_nin X : a_{n_1}=b_1,...,a_{n_k}=b_k}$ where $b_1,...,b_kin[0,1]$ are given, for example the set of all the sequences which begin with $frac{1}{2}$ is a cylinder).



Edit: It turns out I recieved a positive answer in which one of the sets $mathcal{A}_1$ is of measure zero (with respect to the standard Borel measure on $X$). I wonder if one could generalize the given construction for the case where both $mathcal{A}_1$ and $mathcal{A}_2$ are of positive measure.






measure-theory







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Consider the space $X=[0,1]^mathbb{N}$ endowed with the Borel $sigma$-algebra (with respect to the product topology). I'd like to think of $X$ as the set of all sequences with values in $[0,1]$.



    My Question: Is it possible to split $X$ into a disjoint union of two measurable subsets $X=mathcal{A}_1sqcupmathcal{A}_2$ such that no cylinder of $X$ lies in any of the $mathcal{A}_i$?



    By a cylinder I mean a set of all sequences whose values at some finitely many indices $n_1,...,n_k$ are fixed. (i.e. sets of the form ${a_nin X : a_{n_1}=b_1,...,a_{n_k}=b_k}$ where $b_1,...,b_kin[0,1]$ are given, for example the set of all the sequences which begin with $frac{1}{2}$ is a cylinder).



    Edit: It turns out I recieved a positive answer in which one of the sets $mathcal{A}_1$ is of measure zero (with respect to the standard Borel measure on $X$). I wonder if one could generalize the given construction for the case where both $mathcal{A}_1$ and $mathcal{A}_2$ are of positive measure.






    measure-theory







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the space $X=[0,1]^mathbb{N}$ endowed with the Borel $sigma$-algebra (with respect to the product topology). I'd like to think of $X$ as the set of all sequences with values in $[0,1]$.



      My Question: Is it possible to split $X$ into a disjoint union of two measurable subsets $X=mathcal{A}_1sqcupmathcal{A}_2$ such that no cylinder of $X$ lies in any of the $mathcal{A}_i$?



      By a cylinder I mean a set of all sequences whose values at some finitely many indices $n_1,...,n_k$ are fixed. (i.e. sets of the form ${a_nin X : a_{n_1}=b_1,...,a_{n_k}=b_k}$ where $b_1,...,b_kin[0,1]$ are given, for example the set of all the sequences which begin with $frac{1}{2}$ is a cylinder).



      Edit: It turns out I recieved a positive answer in which one of the sets $mathcal{A}_1$ is of measure zero (with respect to the standard Borel measure on $X$). I wonder if one could generalize the given construction for the case where both $mathcal{A}_1$ and $mathcal{A}_2$ are of positive measure.






      measure-theory







      share|cite|improve this question











      $endgroup$




      Consider the space $X=[0,1]^mathbb{N}$ endowed with the Borel $sigma$-algebra (with respect to the product topology). I'd like to think of $X$ as the set of all sequences with values in $[0,1]$.



      My Question: Is it possible to split $X$ into a disjoint union of two measurable subsets $X=mathcal{A}_1sqcupmathcal{A}_2$ such that no cylinder of $X$ lies in any of the $mathcal{A}_i$?



      By a cylinder I mean a set of all sequences whose values at some finitely many indices $n_1,...,n_k$ are fixed. (i.e. sets of the form ${a_nin X : a_{n_1}=b_1,...,a_{n_k}=b_k}$ where $b_1,...,b_kin[0,1]$ are given, for example the set of all the sequences which begin with $frac{1}{2}$ is a cylinder).



      Edit: It turns out I recieved a positive answer in which one of the sets $mathcal{A}_1$ is of measure zero (with respect to the standard Borel measure on $X$). I wonder if one could generalize the given construction for the case where both $mathcal{A}_1$ and $mathcal{A}_2$ are of positive measure.






      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 Dec 13 '18 at 16:38







      Yanko

















      asked Dec 13 '18 at 14:54









      YankoYanko

      6,8931629




      6,8931629






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Let $$A_1 = {x: x_n = 0 text{eventually}}$$
          and $A_2$ its complement.



          EDIT:
          For an example with the additional requirement that both $A_i$ are of positive measure, let
          $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
            $endgroup$
            – Yanko
            Dec 13 '18 at 16:36












          • $begingroup$
            After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:09






          • 1




            $begingroup$
            Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:11












          • $begingroup$
            Of course. I think it works now, is it necessary that $n$ is odd?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:15










          • $begingroup$
            My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:26



















          2












          $begingroup$

          Let $a_m,b_m in [0,1]^{mathbb N}$ be such that $a_m(n)=delta_{m,n}$ and $b_m(n)=1-delta_{m,n}=1-a_m(n)$. Then
          $$
          A_1 = {xin [0,1]^{mathbb N}: x_0 leq 1/2} cup {b_m:minmathbb N} setminus {a_m:minmathbb N}
          $$
          and
          $$
          A_2 = {xin [0,1]^{mathbb N}: x_0 > 1/2} cup {a_m:minmathbb N} setminus {b_m:minmathbb N}
          $$
          should do the job.



          Moreover the measures are $|A_1|=|A_2|=1/2$.



          Edit: I had previously read ${0,1}^{mathbb N}$ instead of $[0,1]^{mathbb N}$. Now it is fixed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:05










          • $begingroup$
            There is an easy fix for that. I'm adding it
            $endgroup$
            – Federico
            Dec 13 '18 at 19:10










          • $begingroup$
            No I was wrong, I cannot fix that. sorry :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:13












          • $begingroup$
            It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:14










          • $begingroup$
            Yes I've read that. Very nice solution. Thank you for the problem :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:15











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let $$A_1 = {x: x_n = 0 text{eventually}}$$
          and $A_2$ its complement.



          EDIT:
          For an example with the additional requirement that both $A_i$ are of positive measure, let
          $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
            $endgroup$
            – Yanko
            Dec 13 '18 at 16:36












          • $begingroup$
            After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:09






          • 1




            $begingroup$
            Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:11












          • $begingroup$
            Of course. I think it works now, is it necessary that $n$ is odd?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:15










          • $begingroup$
            My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:26
















          2












          $begingroup$

          Let $$A_1 = {x: x_n = 0 text{eventually}}$$
          and $A_2$ its complement.



          EDIT:
          For an example with the additional requirement that both $A_i$ are of positive measure, let
          $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
            $endgroup$
            – Yanko
            Dec 13 '18 at 16:36












          • $begingroup$
            After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:09






          • 1




            $begingroup$
            Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:11












          • $begingroup$
            Of course. I think it works now, is it necessary that $n$ is odd?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:15










          • $begingroup$
            My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:26














          2












          2








          2





          $begingroup$

          Let $$A_1 = {x: x_n = 0 text{eventually}}$$
          and $A_2$ its complement.



          EDIT:
          For an example with the additional requirement that both $A_i$ are of positive measure, let
          $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.






          share|cite|improve this answer











          $endgroup$



          Let $$A_1 = {x: x_n = 0 text{eventually}}$$
          and $A_2$ its complement.



          EDIT:
          For an example with the additional requirement that both $A_i$ are of positive measure, let
          $ A_1 $ be the set of sequences $x$ such that there is an odd $n$ with $x_n < 2^{-n}$ but $x_j ge 2^{-j}$ for all $j > n$, $A_2$ its complement. Note that almost every $x$ has $x_n < 2^{-n}$ for only finitely many $n$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 19:07

























          answered Dec 13 '18 at 16:01









          Robert IsraelRobert Israel

          323k23213467




          323k23213467












          • $begingroup$
            Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
            $endgroup$
            – Yanko
            Dec 13 '18 at 16:36












          • $begingroup$
            After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:09






          • 1




            $begingroup$
            Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:11












          • $begingroup$
            Of course. I think it works now, is it necessary that $n$ is odd?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:15










          • $begingroup$
            My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:26


















          • $begingroup$
            Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
            $endgroup$
            – Yanko
            Dec 13 '18 at 16:36












          • $begingroup$
            After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:09






          • 1




            $begingroup$
            Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:11












          • $begingroup$
            Of course. I think it works now, is it necessary that $n$ is odd?
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:15










          • $begingroup$
            My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 19:26
















          $begingroup$
          Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
          $endgroup$
          – Yanko
          Dec 13 '18 at 16:36






          $begingroup$
          Oh looks like I forgot an important assumption that I need $A_1,A_2$ to be of positive measure... I will wait to see if there are any results with the additional assumption. If not I'll just accept yours. Thanks for this answer too.
          $endgroup$
          – Yanko
          Dec 13 '18 at 16:36














          $begingroup$
          After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:09




          $begingroup$
          After the edit: Why is $A_1$ of positive measure? Even if you just take $x_j>2^{-j}$ for infinitely many $j$ you already get a set of measure zero. Am I missing something?
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:09




          1




          1




          $begingroup$
          Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 19:11






          $begingroup$
          Yes, you are. The measure of $x$ for which $x_j > 2^{-j}$ for all $jge 1$ is $prod_{j=1}^infty (1 - 2^{-j})$, an infinite product which converges to a nonzero limit.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 19:11














          $begingroup$
          Of course. I think it works now, is it necessary that $n$ is odd?
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:15




          $begingroup$
          Of course. I think it works now, is it necessary that $n$ is odd?
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:15












          $begingroup$
          My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 19:26




          $begingroup$
          My example has the additional property that any set of positive measure depending on only finitely many coordinates has intersections of positive measure with $A_1$ and $A_2$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 19:26











          2












          $begingroup$

          Let $a_m,b_m in [0,1]^{mathbb N}$ be such that $a_m(n)=delta_{m,n}$ and $b_m(n)=1-delta_{m,n}=1-a_m(n)$. Then
          $$
          A_1 = {xin [0,1]^{mathbb N}: x_0 leq 1/2} cup {b_m:minmathbb N} setminus {a_m:minmathbb N}
          $$
          and
          $$
          A_2 = {xin [0,1]^{mathbb N}: x_0 > 1/2} cup {a_m:minmathbb N} setminus {b_m:minmathbb N}
          $$
          should do the job.



          Moreover the measures are $|A_1|=|A_2|=1/2$.



          Edit: I had previously read ${0,1}^{mathbb N}$ instead of $[0,1]^{mathbb N}$. Now it is fixed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:05










          • $begingroup$
            There is an easy fix for that. I'm adding it
            $endgroup$
            – Federico
            Dec 13 '18 at 19:10










          • $begingroup$
            No I was wrong, I cannot fix that. sorry :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:13












          • $begingroup$
            It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:14










          • $begingroup$
            Yes I've read that. Very nice solution. Thank you for the problem :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:15
















          2












          $begingroup$

          Let $a_m,b_m in [0,1]^{mathbb N}$ be such that $a_m(n)=delta_{m,n}$ and $b_m(n)=1-delta_{m,n}=1-a_m(n)$. Then
          $$
          A_1 = {xin [0,1]^{mathbb N}: x_0 leq 1/2} cup {b_m:minmathbb N} setminus {a_m:minmathbb N}
          $$
          and
          $$
          A_2 = {xin [0,1]^{mathbb N}: x_0 > 1/2} cup {a_m:minmathbb N} setminus {b_m:minmathbb N}
          $$
          should do the job.



          Moreover the measures are $|A_1|=|A_2|=1/2$.



          Edit: I had previously read ${0,1}^{mathbb N}$ instead of $[0,1]^{mathbb N}$. Now it is fixed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:05










          • $begingroup$
            There is an easy fix for that. I'm adding it
            $endgroup$
            – Federico
            Dec 13 '18 at 19:10










          • $begingroup$
            No I was wrong, I cannot fix that. sorry :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:13












          • $begingroup$
            It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:14










          • $begingroup$
            Yes I've read that. Very nice solution. Thank you for the problem :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:15














          2












          2








          2





          $begingroup$

          Let $a_m,b_m in [0,1]^{mathbb N}$ be such that $a_m(n)=delta_{m,n}$ and $b_m(n)=1-delta_{m,n}=1-a_m(n)$. Then
          $$
          A_1 = {xin [0,1]^{mathbb N}: x_0 leq 1/2} cup {b_m:minmathbb N} setminus {a_m:minmathbb N}
          $$
          and
          $$
          A_2 = {xin [0,1]^{mathbb N}: x_0 > 1/2} cup {a_m:minmathbb N} setminus {b_m:minmathbb N}
          $$
          should do the job.



          Moreover the measures are $|A_1|=|A_2|=1/2$.



          Edit: I had previously read ${0,1}^{mathbb N}$ instead of $[0,1]^{mathbb N}$. Now it is fixed.






          share|cite|improve this answer











          $endgroup$



          Let $a_m,b_m in [0,1]^{mathbb N}$ be such that $a_m(n)=delta_{m,n}$ and $b_m(n)=1-delta_{m,n}=1-a_m(n)$. Then
          $$
          A_1 = {xin [0,1]^{mathbb N}: x_0 leq 1/2} cup {b_m:minmathbb N} setminus {a_m:minmathbb N}
          $$
          and
          $$
          A_2 = {xin [0,1]^{mathbb N}: x_0 > 1/2} cup {a_m:minmathbb N} setminus {b_m:minmathbb N}
          $$
          should do the job.



          Moreover the measures are $|A_1|=|A_2|=1/2$.



          Edit: I had previously read ${0,1}^{mathbb N}$ instead of $[0,1]^{mathbb N}$. Now it is fixed.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 17:40

























          answered Dec 13 '18 at 17:09









          FedericoFederico

          5,074514




          5,074514












          • $begingroup$
            Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:05










          • $begingroup$
            There is an easy fix for that. I'm adding it
            $endgroup$
            – Federico
            Dec 13 '18 at 19:10










          • $begingroup$
            No I was wrong, I cannot fix that. sorry :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:13












          • $begingroup$
            It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:14










          • $begingroup$
            Yes I've read that. Very nice solution. Thank you for the problem :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:15


















          • $begingroup$
            Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:05










          • $begingroup$
            There is an easy fix for that. I'm adding it
            $endgroup$
            – Federico
            Dec 13 '18 at 19:10










          • $begingroup$
            No I was wrong, I cannot fix that. sorry :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:13












          • $begingroup$
            It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
            $endgroup$
            – Yanko
            Dec 13 '18 at 19:14










          • $begingroup$
            Yes I've read that. Very nice solution. Thank you for the problem :)
            $endgroup$
            – Federico
            Dec 13 '18 at 19:15
















          $begingroup$
          Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:05




          $begingroup$
          Thank a lot. However this result doesn't help me much more than the original . One can change $A_1$ or $A_2$ by a set of measure zero and then it would contain a cylinder so this is not so much different than the previous answer. However you helped me discover that I need to think much better about how to formalize my question to avoid all sort of "measure zero counterexamples".
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:05












          $begingroup$
          There is an easy fix for that. I'm adding it
          $endgroup$
          – Federico
          Dec 13 '18 at 19:10




          $begingroup$
          There is an easy fix for that. I'm adding it
          $endgroup$
          – Federico
          Dec 13 '18 at 19:10












          $begingroup$
          No I was wrong, I cannot fix that. sorry :)
          $endgroup$
          – Federico
          Dec 13 '18 at 19:13






          $begingroup$
          No I was wrong, I cannot fix that. sorry :)
          $endgroup$
          – Federico
          Dec 13 '18 at 19:13














          $begingroup$
          It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:14




          $begingroup$
          It's ok looks like Robert changed his answer and it looks right to me. Anyway your solution was interesting and I managed to learn from it. Thank you.
          $endgroup$
          – Yanko
          Dec 13 '18 at 19:14












          $begingroup$
          Yes I've read that. Very nice solution. Thank you for the problem :)
          $endgroup$
          – Federico
          Dec 13 '18 at 19:15




          $begingroup$
          Yes I've read that. Very nice solution. Thank you for the problem :)
          $endgroup$
          – Federico
          Dec 13 '18 at 19:15


















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