Cartesian product of Lebesgue measurable sets is Lebesgue measurable
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A times B$ is Lebesgue measurable.
For showing that the cartesian product is Lebesgue measurable, I tried to use:
$A times B$ is Lebesgue measurable if $varepsilon>0$, and an open set $U$ and a closed set $V$ exist such that $V subset A times B subset U$ and $mathcal{L^{n+m}}(U$ $V)<varepsilon$.
I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.
measure-theory lebesgue-measure
asked Nov 26 at 17:13
Olsgur
444
add a comment |
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A times B$ is Lebesgue measurable.
For showing that the cartesian product is Lebesgue measurable, I tried to use:
$A times B$ is Lebesgue measurable if $varepsilon>0$, and an open set $U$ and a closed set $V$ exist such that $V subset A times B subset U$ and $mathcal{L^{n+m}}(U$ $V)<varepsilon$.
I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.
measure-theory lebesgue-measure
asked Nov 26 at 17:13
Olsgur
444
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
1
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54
add a comment |
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A times B$ is Lebesgue measurable.
For showing that the cartesian product is Lebesgue measurable, I tried to use:
$A times B$ is Lebesgue measurable if $varepsilon>0$, and an open set $U$ and a closed set $V$ exist such that $V subset A times B subset U$ and $mathcal{L^{n+m}}(U$ $V)<varepsilon$.
I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.
measure-theory lebesgue-measure
asked Nov 26 at 17:13
Olsgur
444
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A times B$ is Lebesgue measurable.
For showing that the cartesian product is Lebesgue measurable, I tried to use:
$A times B$ is Lebesgue measurable if $varepsilon>0$, and an open set $U$ and a closed set $V$ exist such that $V subset A times B subset U$ and $mathcal{L^{n+m}}(U$ $V)<varepsilon$.
I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 26 at 17:13
Olsgur
444
asked Nov 26 at 17:13
Olsgur
444
edited Nov 26 at 18:56
asked Nov 26 at 17:13
Olsgur
444
asked Nov 26 at 17:13
Olsgur
444
asked Nov 26 at 17:13
Olsgur
444
444
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
1
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54
add a comment |
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
1
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
1
1
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54
add a comment |
1 Answer
1
active
oldest
votes
As a prelude, let me first say that $Atimes B$ is clearly in the product $sigma$ algebra, since the product sigma algebra $mathcal{F}otimes mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$mathcal{F}otimes mathcal{G}= sigmaBig( { Etimes F: Ein mathcal{F}, Fin mathcal{G} } Big)$
Going by this definition, your set $Atimes B$ is Lebesgue measurable in $mathbb{R}^{n+m}$. However going by your criterion, recall first that for $Ain mathcal{L}^n$ and $Bin mathcal{L}^m$, we have:
$lambda_n otimes lambda_m (Atimes B)= lambda_n(A)times lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1supseteq A supseteq V_1$ , $U_2supseteq B supseteq V_2$ while $lambda_n(U_1setminus V_1)leq frac{1}{2} tilde{epsilon}$ and $lambda_m(U_2setminus V_2)leq frac{1}{2} tilde{epsilon}$
Notice that $U_1times U_2supseteq Atimes B supseteq V_1times V_2$ while $U_1times U_2$ is open and $V_1 times V_2$ is closed. Since you can write:
$U_1times U_2= Big( U_1times V_2 Big) sqcup Big( U_1 times (U_2 setminus V_2) Big)= Big( V_1times V_2 Big) sqcup Big( (U_1setminus V_1)times V_2 Big) sqcup Big( U_1 times (U_2setminus V_2) Big) $
Then:
$lambda_{n+m}Big( (U_1times U_2) setminus (V_1 times V_2) Big)= lambda_{n+m}Big( (U_1setminus V_1)times V_2
Big) + lambda_{n+m}Big( U_1 times (U_2setminus V_2)
Big)= $
$= lambda_n(U_1 setminus V_1)cdot lambda_m(V_2)+ lambda_n(U)cdot lambda_m(U_2 setminus V_2)$
If you assume $lambda_n(U_1), lambda_m(U_2)leq M$, then for $tilde{epsilon}= dfrac{epsilon}{2M}$, you've shown that the difference is of measure less than $epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $sigma$-finite case.
answered Nov 27 at 10:45
Keen-ameteur
1,265316
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
add a comment |
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As a prelude, let me first say that $Atimes B$ is clearly in the product $sigma$ algebra, since the product sigma algebra $mathcal{F}otimes mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$mathcal{F}otimes mathcal{G}= sigmaBig( { Etimes F: Ein mathcal{F}, Fin mathcal{G} } Big)$
Going by this definition, your set $Atimes B$ is Lebesgue measurable in $mathbb{R}^{n+m}$. However going by your criterion, recall first that for $Ain mathcal{L}^n$ and $Bin mathcal{L}^m$, we have:
$lambda_n otimes lambda_m (Atimes B)= lambda_n(A)times lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1supseteq A supseteq V_1$ , $U_2supseteq B supseteq V_2$ while $lambda_n(U_1setminus V_1)leq frac{1}{2} tilde{epsilon}$ and $lambda_m(U_2setminus V_2)leq frac{1}{2} tilde{epsilon}$
Notice that $U_1times U_2supseteq Atimes B supseteq V_1times V_2$ while $U_1times U_2$ is open and $V_1 times V_2$ is closed. Since you can write:
$U_1times U_2= Big( U_1times V_2 Big) sqcup Big( U_1 times (U_2 setminus V_2) Big)= Big( V_1times V_2 Big) sqcup Big( (U_1setminus V_1)times V_2 Big) sqcup Big( U_1 times (U_2setminus V_2) Big) $
Then:
$lambda_{n+m}Big( (U_1times U_2) setminus (V_1 times V_2) Big)= lambda_{n+m}Big( (U_1setminus V_1)times V_2
Big) + lambda_{n+m}Big( U_1 times (U_2setminus V_2)
Big)= $
$= lambda_n(U_1 setminus V_1)cdot lambda_m(V_2)+ lambda_n(U)cdot lambda_m(U_2 setminus V_2)$
If you assume $lambda_n(U_1), lambda_m(U_2)leq M$, then for $tilde{epsilon}= dfrac{epsilon}{2M}$, you've shown that the difference is of measure less than $epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $sigma$-finite case.
answered Nov 27 at 10:45
Keen-ameteur
1,265316
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
add a comment |
As a prelude, let me first say that $Atimes B$ is clearly in the product $sigma$ algebra, since the product sigma algebra $mathcal{F}otimes mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$mathcal{F}otimes mathcal{G}= sigmaBig( { Etimes F: Ein mathcal{F}, Fin mathcal{G} } Big)$
Going by this definition, your set $Atimes B$ is Lebesgue measurable in $mathbb{R}^{n+m}$. However going by your criterion, recall first that for $Ain mathcal{L}^n$ and $Bin mathcal{L}^m$, we have:
$lambda_n otimes lambda_m (Atimes B)= lambda_n(A)times lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1supseteq A supseteq V_1$ , $U_2supseteq B supseteq V_2$ while $lambda_n(U_1setminus V_1)leq frac{1}{2} tilde{epsilon}$ and $lambda_m(U_2setminus V_2)leq frac{1}{2} tilde{epsilon}$
Notice that $U_1times U_2supseteq Atimes B supseteq V_1times V_2$ while $U_1times U_2$ is open and $V_1 times V_2$ is closed. Since you can write:
$U_1times U_2= Big( U_1times V_2 Big) sqcup Big( U_1 times (U_2 setminus V_2) Big)= Big( V_1times V_2 Big) sqcup Big( (U_1setminus V_1)times V_2 Big) sqcup Big( U_1 times (U_2setminus V_2) Big) $
Then:
$lambda_{n+m}Big( (U_1times U_2) setminus (V_1 times V_2) Big)= lambda_{n+m}Big( (U_1setminus V_1)times V_2
Big) + lambda_{n+m}Big( U_1 times (U_2setminus V_2)
Big)= $
$= lambda_n(U_1 setminus V_1)cdot lambda_m(V_2)+ lambda_n(U)cdot lambda_m(U_2 setminus V_2)$
If you assume $lambda_n(U_1), lambda_m(U_2)leq M$, then for $tilde{epsilon}= dfrac{epsilon}{2M}$, you've shown that the difference is of measure less than $epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $sigma$-finite case.
answered Nov 27 at 10:45
Keen-ameteur
1,265316
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
add a comment |
As a prelude, let me first say that $Atimes B$ is clearly in the product $sigma$ algebra, since the product sigma algebra $mathcal{F}otimes mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$mathcal{F}otimes mathcal{G}= sigmaBig( { Etimes F: Ein mathcal{F}, Fin mathcal{G} } Big)$
Going by this definition, your set $Atimes B$ is Lebesgue measurable in $mathbb{R}^{n+m}$. However going by your criterion, recall first that for $Ain mathcal{L}^n$ and $Bin mathcal{L}^m$, we have:
$lambda_n otimes lambda_m (Atimes B)= lambda_n(A)times lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1supseteq A supseteq V_1$ , $U_2supseteq B supseteq V_2$ while $lambda_n(U_1setminus V_1)leq frac{1}{2} tilde{epsilon}$ and $lambda_m(U_2setminus V_2)leq frac{1}{2} tilde{epsilon}$
Notice that $U_1times U_2supseteq Atimes B supseteq V_1times V_2$ while $U_1times U_2$ is open and $V_1 times V_2$ is closed. Since you can write:
$U_1times U_2= Big( U_1times V_2 Big) sqcup Big( U_1 times (U_2 setminus V_2) Big)= Big( V_1times V_2 Big) sqcup Big( (U_1setminus V_1)times V_2 Big) sqcup Big( U_1 times (U_2setminus V_2) Big) $
Then:
$lambda_{n+m}Big( (U_1times U_2) setminus (V_1 times V_2) Big)= lambda_{n+m}Big( (U_1setminus V_1)times V_2
Big) + lambda_{n+m}Big( U_1 times (U_2setminus V_2)
Big)= $
$= lambda_n(U_1 setminus V_1)cdot lambda_m(V_2)+ lambda_n(U)cdot lambda_m(U_2 setminus V_2)$
If you assume $lambda_n(U_1), lambda_m(U_2)leq M$, then for $tilde{epsilon}= dfrac{epsilon}{2M}$, you've shown that the difference is of measure less than $epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $sigma$-finite case.
answered Nov 27 at 10:45
Keen-ameteur
1,265316
As a prelude, let me first say that $Atimes B$ is clearly in the product $sigma$ algebra, since the product sigma algebra $mathcal{F}otimes mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$mathcal{F}otimes mathcal{G}= sigmaBig( { Etimes F: Ein mathcal{F}, Fin mathcal{G} } Big)$
Going by this definition, your set $Atimes B$ is Lebesgue measurable in $mathbb{R}^{n+m}$. However going by your criterion, recall first that for $Ain mathcal{L}^n$ and $Bin mathcal{L}^m$, we have:
$lambda_n otimes lambda_m (Atimes B)= lambda_n(A)times lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1supseteq A supseteq V_1$ , $U_2supseteq B supseteq V_2$ while $lambda_n(U_1setminus V_1)leq frac{1}{2} tilde{epsilon}$ and $lambda_m(U_2setminus V_2)leq frac{1}{2} tilde{epsilon}$
Notice that $U_1times U_2supseteq Atimes B supseteq V_1times V_2$ while $U_1times U_2$ is open and $V_1 times V_2$ is closed. Since you can write:
$U_1times U_2= Big( U_1times V_2 Big) sqcup Big( U_1 times (U_2 setminus V_2) Big)= Big( V_1times V_2 Big) sqcup Big( (U_1setminus V_1)times V_2 Big) sqcup Big( U_1 times (U_2setminus V_2) Big) $
Then:
$lambda_{n+m}Big( (U_1times U_2) setminus (V_1 times V_2) Big)= lambda_{n+m}Big( (U_1setminus V_1)times V_2
Big) + lambda_{n+m}Big( U_1 times (U_2setminus V_2)
Big)= $
$= lambda_n(U_1 setminus V_1)cdot lambda_m(V_2)+ lambda_n(U)cdot lambda_m(U_2 setminus V_2)$
If you assume $lambda_n(U_1), lambda_m(U_2)leq M$, then for $tilde{epsilon}= dfrac{epsilon}{2M}$, you've shown that the difference is of measure less than $epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $sigma$-finite case.
answered Nov 27 at 10:45
Keen-ameteur
1,265316
answered Nov 27 at 10:45
Keen-ameteur
1,265316
answered Nov 27 at 10:45
Keen-ameteur
1,265316
answered Nov 27 at 10:45
Keen-ameteur
1,265316
1,265316
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
add a comment |
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
– Olsgur
Nov 27 at 13:56
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
– Keen-ameteur
Nov 27 at 18:34
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
Yes, I mean the standard argument with the $sigma$-finite case.
– Olsgur
Nov 27 at 23:23
add a comment |
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You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
– Keen-ameteur
Nov 26 at 17:27
Let $A subset mathbb{R^m}$, $B subset mathbb{R^n}$ and $lambda_n$ , thats all what is given, so I suppose it could be right.
– Olsgur
Nov 26 at 18:54
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
– Keen-ameteur
Nov 26 at 19:15
1
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
– Keen-ameteur
Nov 26 at 19:22
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
– Olsgur
Nov 26 at 19:54