If $A times B$ is Lebesgue measurable in $mathbb{R}^2$ and $B$ is Lebesgue measurable in $mathbb{R}$ then $A$...
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Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial:
Write $A= A times {0}= cup_i(A_{i} times B_{i})$,
$m_1(A)=m_2(A times {0})=cup_i m_1(A_i)m_1(B_i)$
I don't know what to do next! Is my approach correct?
where $m_1,m_2$ are 1 and 2 dimensional measures in $mathbb{R},mathbb{R^2}$ respectively.
real-analysis measure-theory lebesgue-measure
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
$endgroup$
add a comment |
$begingroup$
Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial:
Write $A= A times {0}= cup_i(A_{i} times B_{i})$,
$m_1(A)=m_2(A times {0})=cup_i m_1(A_i)m_1(B_i)$
I don't know what to do next! Is my approach correct?
where $m_1,m_2$ are 1 and 2 dimensional measures in $mathbb{R},mathbb{R^2}$ respectively.
real-analysis measure-theory lebesgue-measure
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
$endgroup$
add a comment |
$begingroup$
Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial:
Write $A= A times {0}= cup_i(A_{i} times B_{i})$,
$m_1(A)=m_2(A times {0})=cup_i m_1(A_i)m_1(B_i)$
I don't know what to do next! Is my approach correct?
where $m_1,m_2$ are 1 and 2 dimensional measures in $mathbb{R},mathbb{R^2}$ respectively.
real-analysis measure-theory lebesgue-measure
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
$endgroup$
Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial:
Write $A= A times {0}= cup_i(A_{i} times B_{i})$,
$m_1(A)=m_2(A times {0})=cup_i m_1(A_i)m_1(B_i)$
I don't know what to do next! Is my approach correct?
where $m_1,m_2$ are 1 and 2 dimensional measures in $mathbb{R},mathbb{R^2}$ respectively.
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
asked Dec 20 '18 at 0:39
InfinityInfinity
321112
321112
add a comment |
add a comment |
3 Answers
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The statement is not true.
Hint: Consider a set $B$ with measure 0.
The statement is true if $B$ has positive measure. To prove this, show that $mu(A times B) = mu^*(A)mu^*(B) = mu_*(A) mu_*(B)$ where $mu^*$ is outer Lebesgue measure and $mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
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add a comment |
$begingroup$
This is false. For a counterexample, take $B=mathbb Q $ (or even $B=emptyset)$ and $A=$ a Vitali set. Then, $lambda^*_2(Atimes B)=0$ so $Atimes Bin mathscr M(mathbb R^2)$ but $Anotin mathscr M(mathbb R).$
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
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add a comment |
$begingroup$
As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.
Consider the characteristic function $f(x,y) = chi_{A times B}(x, y) = chi_A(x) chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : mathbb{R} to mathbb{R}, x mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y in B$ such that this condition holds. But then, $f_y(x) = chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is not true.
Hint: Consider a set $B$ with measure 0.
The statement is true if $B$ has positive measure. To prove this, show that $mu(A times B) = mu^*(A)mu^*(B) = mu_*(A) mu_*(B)$ where $mu^*$ is outer Lebesgue measure and $mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
$endgroup$
add a comment |
$begingroup$
The statement is not true.
Hint: Consider a set $B$ with measure 0.
The statement is true if $B$ has positive measure. To prove this, show that $mu(A times B) = mu^*(A)mu^*(B) = mu_*(A) mu_*(B)$ where $mu^*$ is outer Lebesgue measure and $mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
$endgroup$
add a comment |
$begingroup$
The statement is not true.
Hint: Consider a set $B$ with measure 0.
The statement is true if $B$ has positive measure. To prove this, show that $mu(A times B) = mu^*(A)mu^*(B) = mu_*(A) mu_*(B)$ where $mu^*$ is outer Lebesgue measure and $mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
$endgroup$
The statement is not true.
Hint: Consider a set $B$ with measure 0.
The statement is true if $B$ has positive measure. To prove this, show that $mu(A times B) = mu^*(A)mu^*(B) = mu_*(A) mu_*(B)$ where $mu^*$ is outer Lebesgue measure and $mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
edited Dec 20 '18 at 1:22
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
answered Dec 20 '18 at 0:56
zoidbergzoidberg
1,065113
1,065113
add a comment |
add a comment |
$begingroup$
This is false. For a counterexample, take $B=mathbb Q $ (or even $B=emptyset)$ and $A=$ a Vitali set. Then, $lambda^*_2(Atimes B)=0$ so $Atimes Bin mathscr M(mathbb R^2)$ but $Anotin mathscr M(mathbb R).$
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
$endgroup$
add a comment |
$begingroup$
This is false. For a counterexample, take $B=mathbb Q $ (or even $B=emptyset)$ and $A=$ a Vitali set. Then, $lambda^*_2(Atimes B)=0$ so $Atimes Bin mathscr M(mathbb R^2)$ but $Anotin mathscr M(mathbb R).$
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
$endgroup$
add a comment |
$begingroup$
This is false. For a counterexample, take $B=mathbb Q $ (or even $B=emptyset)$ and $A=$ a Vitali set. Then, $lambda^*_2(Atimes B)=0$ so $Atimes Bin mathscr M(mathbb R^2)$ but $Anotin mathscr M(mathbb R).$
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
$endgroup$
This is false. For a counterexample, take $B=mathbb Q $ (or even $B=emptyset)$ and $A=$ a Vitali set. Then, $lambda^*_2(Atimes B)=0$ so $Atimes Bin mathscr M(mathbb R^2)$ but $Anotin mathscr M(mathbb R).$
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
answered Dec 20 '18 at 0:59
MatematletaMatematleta
10.6k2918
10.6k2918
add a comment |
add a comment |
$begingroup$
As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.
Consider the characteristic function $f(x,y) = chi_{A times B}(x, y) = chi_A(x) chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : mathbb{R} to mathbb{R}, x mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y in B$ such that this condition holds. But then, $f_y(x) = chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
$endgroup$
add a comment |
$begingroup$
As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.
Consider the characteristic function $f(x,y) = chi_{A times B}(x, y) = chi_A(x) chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : mathbb{R} to mathbb{R}, x mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y in B$ such that this condition holds. But then, $f_y(x) = chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
$endgroup$
add a comment |
$begingroup$
As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.
Consider the characteristic function $f(x,y) = chi_{A times B}(x, y) = chi_A(x) chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : mathbb{R} to mathbb{R}, x mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y in B$ such that this condition holds. But then, $f_y(x) = chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
$endgroup$
As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.
Consider the characteristic function $f(x,y) = chi_{A times B}(x, y) = chi_A(x) chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : mathbb{R} to mathbb{R}, x mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y in B$ such that this condition holds. But then, $f_y(x) = chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
answered Dec 20 '18 at 1:19
Daniel ScheplerDaniel Schepler
8,4941618
8,4941618
add a comment |
add a comment |
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