Product measure : difference between $mathcal Mtimes mathcal M$ and ${Atimes Bmid A,Bin mathcal M}$
$begingroup$
Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$
Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :
1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)
2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?
measure-theory
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$
Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :
1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)
2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?
measure-theory
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
$endgroup$
$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51
add a comment |
$begingroup$
Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$
Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :
1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)
2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?
measure-theory
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
$endgroup$
Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$
Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :
1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)
2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?
measure-theory
measure-theory
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
asked Dec 4 '18 at 9:11
NewMathNewMath
4059
4059
$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51
add a comment |
$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51
$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51
$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
add a comment |
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$begingroup$
If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
add a comment |
$begingroup$
If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
add a comment |
$begingroup$
If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
edited Dec 4 '18 at 10:09
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Dec 4 '18 at 9:18
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
add a comment |
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12
1
1
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
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– Kavi Rama Murthy
Dec 4 '18 at 10:17
$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17
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$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
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– Vera
Dec 4 '18 at 11:51