Cartesian product of Lebesgue measurable sets is Lebesgue measurable












0












$begingroup$


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











$endgroup$












  • $begingroup$
    You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 17:27










  • $begingroup$
    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.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 18:54










  • $begingroup$
    What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:15








  • 1




    $begingroup$
    By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:22










  • $begingroup$
    I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 19:54


















0












$begingroup$


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











$endgroup$












  • $begingroup$
    You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 17:27










  • $begingroup$
    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.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 18:54










  • $begingroup$
    What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:15








  • 1




    $begingroup$
    By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:22










  • $begingroup$
    I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 19:54
















0












0








0





$begingroup$


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











$endgroup$




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 '18 at 18:56







Olsgur

















asked Nov 26 '18 at 17:13









OlsgurOlsgur

544




544












  • $begingroup$
    You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 17:27










  • $begingroup$
    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.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 18:54










  • $begingroup$
    What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:15








  • 1




    $begingroup$
    By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:22










  • $begingroup$
    I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 19:54




















  • $begingroup$
    You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 17:27










  • $begingroup$
    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.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 18:54










  • $begingroup$
    What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:15








  • 1




    $begingroup$
    By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
    $endgroup$
    – Keen-ameteur
    Nov 26 '18 at 19:22










  • $begingroup$
    I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
    $endgroup$
    – Olsgur
    Nov 26 '18 at 19:54


















$begingroup$
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
$endgroup$
– Keen-ameteur
Nov 26 '18 at 17:27




$begingroup$
You are probably discussing measurable with respect to $mathcal{L}{n+m}$ measure on $mathbb{R}^{n+m}$, or am I wrong?
$endgroup$
– Keen-ameteur
Nov 26 '18 at 17:27












$begingroup$
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.
$endgroup$
– Olsgur
Nov 26 '18 at 18:54




$begingroup$
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.
$endgroup$
– Olsgur
Nov 26 '18 at 18:54












$begingroup$
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 19:15






$begingroup$
What does $mathcal{L}^{n+m}(Usetminus V)< epsilon$ then? It must be the Lebesgue measure on $mathbb{R}^{n+m}$.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 19:15






1




1




$begingroup$
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
$endgroup$
– Keen-ameteur
Nov 26 '18 at 19:22




$begingroup$
By the way are you familiar with the terms $sigma$ algebra and product $sigma$ algebra?
$endgroup$
– Keen-ameteur
Nov 26 '18 at 19:22












$begingroup$
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
$endgroup$
– Olsgur
Nov 26 '18 at 19:54






$begingroup$
I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra.
$endgroup$
– Olsgur
Nov 26 '18 at 19:54












1 Answer
1






active

oldest

votes


















0












$begingroup$

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









$endgroup$













  • $begingroup$
    Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
    $endgroup$
    – Olsgur
    Nov 27 '18 at 13:56










  • $begingroup$
    I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
    $endgroup$
    – Keen-ameteur
    Nov 27 '18 at 18:34










  • $begingroup$
    Yes, I mean the standard argument with the $sigma$-finite case.
    $endgroup$
    – Olsgur
    Nov 27 '18 at 23:23











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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









$endgroup$













  • $begingroup$
    Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
    $endgroup$
    – Olsgur
    Nov 27 '18 at 13:56










  • $begingroup$
    I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
    $endgroup$
    – Keen-ameteur
    Nov 27 '18 at 18:34










  • $begingroup$
    Yes, I mean the standard argument with the $sigma$-finite case.
    $endgroup$
    – Olsgur
    Nov 27 '18 at 23:23
















0












$begingroup$

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









$endgroup$













  • $begingroup$
    Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
    $endgroup$
    – Olsgur
    Nov 27 '18 at 13:56










  • $begingroup$
    I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
    $endgroup$
    – Keen-ameteur
    Nov 27 '18 at 18:34










  • $begingroup$
    Yes, I mean the standard argument with the $sigma$-finite case.
    $endgroup$
    – Olsgur
    Nov 27 '18 at 23:23














0












0








0





$begingroup$

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









$endgroup$



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 '18 at 10:45









Keen-ameteurKeen-ameteur

1,365316




1,365316












  • $begingroup$
    Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
    $endgroup$
    – Olsgur
    Nov 27 '18 at 13:56










  • $begingroup$
    I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
    $endgroup$
    – Keen-ameteur
    Nov 27 '18 at 18:34










  • $begingroup$
    Yes, I mean the standard argument with the $sigma$-finite case.
    $endgroup$
    – Olsgur
    Nov 27 '18 at 23:23


















  • $begingroup$
    Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
    $endgroup$
    – Olsgur
    Nov 27 '18 at 13:56










  • $begingroup$
    I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
    $endgroup$
    – Keen-ameteur
    Nov 27 '18 at 18:34










  • $begingroup$
    Yes, I mean the standard argument with the $sigma$-finite case.
    $endgroup$
    – Olsgur
    Nov 27 '18 at 23:23
















$begingroup$
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
$endgroup$
– Olsgur
Nov 27 '18 at 13:56




$begingroup$
Thanks for the explanation. If there is a standard way to show it, how would it look like in this case?
$endgroup$
– Olsgur
Nov 27 '18 at 13:56












$begingroup$
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
$endgroup$
– Keen-ameteur
Nov 27 '18 at 18:34




$begingroup$
I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'?
$endgroup$
– Keen-ameteur
Nov 27 '18 at 18:34












$begingroup$
Yes, I mean the standard argument with the $sigma$-finite case.
$endgroup$
– Olsgur
Nov 27 '18 at 23:23




$begingroup$
Yes, I mean the standard argument with the $sigma$-finite case.
$endgroup$
– Olsgur
Nov 27 '18 at 23:23


















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