Cartesian product of Lebesgue measurable sets is Lebesgue measurable












0














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







share|cite|improve this question
























  • 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


















0














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







share|cite|improve this question
























  • 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
















0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 18:56

























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




















  • 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












1 Answer
1






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oldest

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0














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.






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
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0














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.






share|cite|improve this answer





















  • 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
















0














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.






share|cite|improve this answer





















  • 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














0












0








0






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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Bundesstraße 106

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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