Binary expansion has a zero in each even position is Lebesgue measurable












1















Exercise: Show that the subset of points $A$ of the interval $[0,1]$ whose binary expansion has a zero in each even position of its expansion ($frac{2}{3}=0,101010...in A$ for example) is Lebesgue measurable and zero Lebesgue measure.




I know the following definition of measurability:
If $mu^*:mathscr{P}(Omega)tomathbb{R}^+$ is a outer measure. Then $E_i$ is measurable with respect to $mu^{*}$ iff $forall Ainmathscr{P}(Omega)::mu^*(A)=mu(Acap E)+mu(Asetminus E)$.



However despite knowing this definition I do not know if it supposed to be used.



I have no idea on how to solve this exercise.



Question:



Can someone provide an hint about how to prove measurability and that the measure is zero?



Thanks in advance!






real-analysis measure-theory







share|cite|improve this question






















  • It's a countable intersection of (at each stage) a finite family of dyadic intervals.
    – T. Bongers
    Nov 26 at 22:38










  • You can write the set as an intersection of borel sets.
    – Foobaz John
    Nov 26 at 22:38










  • @FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:45










  • @T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:46
















1















Exercise: Show that the subset of points $A$ of the interval $[0,1]$ whose binary expansion has a zero in each even position of its expansion ($frac{2}{3}=0,101010...in A$ for example) is Lebesgue measurable and zero Lebesgue measure.




I know the following definition of measurability:
If $mu^*:mathscr{P}(Omega)tomathbb{R}^+$ is a outer measure. Then $E_i$ is measurable with respect to $mu^{*}$ iff $forall Ainmathscr{P}(Omega)::mu^*(A)=mu(Acap E)+mu(Asetminus E)$.



However despite knowing this definition I do not know if it supposed to be used.



I have no idea on how to solve this exercise.



Question:



Can someone provide an hint about how to prove measurability and that the measure is zero?



Thanks in advance!






real-analysis measure-theory







share|cite|improve this question






















  • It's a countable intersection of (at each stage) a finite family of dyadic intervals.
    – T. Bongers
    Nov 26 at 22:38










  • You can write the set as an intersection of borel sets.
    – Foobaz John
    Nov 26 at 22:38










  • @FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:45










  • @T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:46














1












1








1








Exercise: Show that the subset of points $A$ of the interval $[0,1]$ whose binary expansion has a zero in each even position of its expansion ($frac{2}{3}=0,101010...in A$ for example) is Lebesgue measurable and zero Lebesgue measure.




I know the following definition of measurability:
If $mu^*:mathscr{P}(Omega)tomathbb{R}^+$ is a outer measure. Then $E_i$ is measurable with respect to $mu^{*}$ iff $forall Ainmathscr{P}(Omega)::mu^*(A)=mu(Acap E)+mu(Asetminus E)$.



However despite knowing this definition I do not know if it supposed to be used.



I have no idea on how to solve this exercise.



Question:



Can someone provide an hint about how to prove measurability and that the measure is zero?



Thanks in advance!






real-analysis measure-theory







share|cite|improve this question














Exercise: Show that the subset of points $A$ of the interval $[0,1]$ whose binary expansion has a zero in each even position of its expansion ($frac{2}{3}=0,101010...in A$ for example) is Lebesgue measurable and zero Lebesgue measure.




I know the following definition of measurability:
If $mu^*:mathscr{P}(Omega)tomathbb{R}^+$ is a outer measure. Then $E_i$ is measurable with respect to $mu^{*}$ iff $forall Ainmathscr{P}(Omega)::mu^*(A)=mu(Acap E)+mu(Asetminus E)$.



However despite knowing this definition I do not know if it supposed to be used.



I have no idea on how to solve this exercise.



Question:



Can someone provide an hint about how to prove measurability and that the measure is zero?



Thanks in advance!






real-analysis measure-theory




real-analysis measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 22:35









Pedro Gomes

1,6892720




1,6892720












  • It's a countable intersection of (at each stage) a finite family of dyadic intervals.
    – T. Bongers
    Nov 26 at 22:38










  • You can write the set as an intersection of borel sets.
    – Foobaz John
    Nov 26 at 22:38










  • @FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:45










  • @T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:46


















  • It's a countable intersection of (at each stage) a finite family of dyadic intervals.
    – T. Bongers
    Nov 26 at 22:38










  • You can write the set as an intersection of borel sets.
    – Foobaz John
    Nov 26 at 22:38










  • @FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:45










  • @T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
    – Pedro Gomes
    Nov 26 at 22:46
















It's a countable intersection of (at each stage) a finite family of dyadic intervals.
– T. Bongers
Nov 26 at 22:38




It's a countable intersection of (at each stage) a finite family of dyadic intervals.
– T. Bongers
Nov 26 at 22:38












You can write the set as an intersection of borel sets.
– Foobaz John
Nov 26 at 22:38




You can write the set as an intersection of borel sets.
– Foobaz John
Nov 26 at 22:38












@FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
– Pedro Gomes
Nov 26 at 22:45




@FoobazJohn Could you please give some more hints? It has been hard to understand what I am supposed to do.
– Pedro Gomes
Nov 26 at 22:45












@T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
– Pedro Gomes
Nov 26 at 22:46




@T.Bongers Could you please give some more hints? It has been hard to understand what I am supposed to do.
– Pedro Gomes
Nov 26 at 22:46










1 Answer
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Let $L_0=[0, 1[$ (${1}$ has null measure) for every $kinmathbb N$ we have
$$
L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[
$$
if $ninleft[frac{i}{2^k}, frac{i+1}{2^k}right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let
$$
L_k=bigcup_{i=0}^{2^{2k-1}-1}left[frac{2i}{2^{2k}}, frac{2i+1}{2^{2k}}right[\
L=bigcap_{kinmathbb N}L_k
$$
clearly $L$ is Lebesgue measurable



Prove that
$$
lvert L_ncap L_{n-1}capdotsbcap L_1rvert =frac 12lvert L_{n-1}capdotsbcap L_1rvert
$$
then you can easily continue the proof.






share|cite|improve this answer























  • I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
    – Pedro Gomes
    Nov 27 at 17:05










  • Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
    – P De Donato
    Nov 27 at 23:11













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1 Answer
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1 Answer
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1














Let $L_0=[0, 1[$ (${1}$ has null measure) for every $kinmathbb N$ we have
$$
L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[
$$
if $ninleft[frac{i}{2^k}, frac{i+1}{2^k}right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let
$$
L_k=bigcup_{i=0}^{2^{2k-1}-1}left[frac{2i}{2^{2k}}, frac{2i+1}{2^{2k}}right[\
L=bigcap_{kinmathbb N}L_k
$$
clearly $L$ is Lebesgue measurable



Prove that
$$
lvert L_ncap L_{n-1}capdotsbcap L_1rvert =frac 12lvert L_{n-1}capdotsbcap L_1rvert
$$
then you can easily continue the proof.






share|cite|improve this answer























  • I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
    – Pedro Gomes
    Nov 27 at 17:05










  • Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
    – P De Donato
    Nov 27 at 23:11


















1














Let $L_0=[0, 1[$ (${1}$ has null measure) for every $kinmathbb N$ we have
$$
L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[
$$
if $ninleft[frac{i}{2^k}, frac{i+1}{2^k}right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let
$$
L_k=bigcup_{i=0}^{2^{2k-1}-1}left[frac{2i}{2^{2k}}, frac{2i+1}{2^{2k}}right[\
L=bigcap_{kinmathbb N}L_k
$$
clearly $L$ is Lebesgue measurable



Prove that
$$
lvert L_ncap L_{n-1}capdotsbcap L_1rvert =frac 12lvert L_{n-1}capdotsbcap L_1rvert
$$
then you can easily continue the proof.






share|cite|improve this answer























  • I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
    – Pedro Gomes
    Nov 27 at 17:05










  • Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
    – P De Donato
    Nov 27 at 23:11
















1












1








1






Let $L_0=[0, 1[$ (${1}$ has null measure) for every $kinmathbb N$ we have
$$
L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[
$$
if $ninleft[frac{i}{2^k}, frac{i+1}{2^k}right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let
$$
L_k=bigcup_{i=0}^{2^{2k-1}-1}left[frac{2i}{2^{2k}}, frac{2i+1}{2^{2k}}right[\
L=bigcap_{kinmathbb N}L_k
$$
clearly $L$ is Lebesgue measurable



Prove that
$$
lvert L_ncap L_{n-1}capdotsbcap L_1rvert =frac 12lvert L_{n-1}capdotsbcap L_1rvert
$$
then you can easily continue the proof.






share|cite|improve this answer














Let $L_0=[0, 1[$ (${1}$ has null measure) for every $kinmathbb N$ we have
$$
L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[
$$
if $ninleft[frac{i}{2^k}, frac{i+1}{2^k}right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let
$$
L_k=bigcup_{i=0}^{2^{2k-1}-1}left[frac{2i}{2^{2k}}, frac{2i+1}{2^{2k}}right[\
L=bigcap_{kinmathbb N}L_k
$$
clearly $L$ is Lebesgue measurable



Prove that
$$
lvert L_ncap L_{n-1}capdotsbcap L_1rvert =frac 12lvert L_{n-1}capdotsbcap L_1rvert
$$
then you can easily continue the proof.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 23:07

























answered Nov 26 at 23:01









P De Donato

4047




4047












  • I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
    – Pedro Gomes
    Nov 27 at 17:05










  • Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
    – P De Donato
    Nov 27 at 23:11




















  • I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
    – Pedro Gomes
    Nov 27 at 17:05










  • Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
    – P De Donato
    Nov 27 at 23:11


















I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
– Pedro Gomes
Nov 27 at 17:05




I fail to see how you came to $L_0=bigcup_{i=0}^{2^k-1}left[frac{i}{2^k}, frac{i+1}{2^k}right[$ and the realized $i$ must be even. I need the intuition. Thanks for your answer!
– Pedro Gomes
Nov 27 at 17:05












Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
– P De Donato
Nov 27 at 23:11






Simply I separated the interval $[0, 1[$ in $2^k$ smaller intervals whose length is $frac 1{2^k}$. Let $k=3$ and $i=5=mathtt{101}$ in base $2$ then $5/2^3=mathtt{0.101}$ even in base $2$ (it's shifted to right by $3$ positions), so if $i$ is even then the least significant digit is $mathtt 0$ (otherwise is $mathtt 1$) that it'll be shifted to $k$ positions right when $i$ is divided by $2^k$.
– P De Donato
Nov 27 at 23:11




















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