Example of a set which is not in the product $sigma$-algebra
$begingroup$
Let $L_d$ be the $sigma$-algebra of Lebesgue measurable subsets of $mathbb{R}^d$.
By using Vitali's set $E subseteq [0,1]$, I am looking for an example of $A in L_2$ which is not in the product $sigma$-algebra $L_1 times L_1$.
I am also having trouble proving that $L_1 times L_1 subseteq L_2$. I can see that we can use $mathcal{B}(mathbb{R^2})=mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$ and that the Lebesgue measure $lambda_2$ on $(mathbb{R^2},mathcal{B}(mathbb{R^2}))$ is identical to the product measure $lambda_1 times lambda_1.$ Although I'm stuck afterwards.
measure-theory lebesgue-measure
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
$endgroup$
add a comment |
$begingroup$
Let $L_d$ be the $sigma$-algebra of Lebesgue measurable subsets of $mathbb{R}^d$.
By using Vitali's set $E subseteq [0,1]$, I am looking for an example of $A in L_2$ which is not in the product $sigma$-algebra $L_1 times L_1$.
I am also having trouble proving that $L_1 times L_1 subseteq L_2$. I can see that we can use $mathcal{B}(mathbb{R^2})=mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$ and that the Lebesgue measure $lambda_2$ on $(mathbb{R^2},mathcal{B}(mathbb{R^2}))$ is identical to the product measure $lambda_1 times lambda_1.$ Although I'm stuck afterwards.
measure-theory lebesgue-measure
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
$endgroup$
add a comment |
$begingroup$
Let $L_d$ be the $sigma$-algebra of Lebesgue measurable subsets of $mathbb{R}^d$.
By using Vitali's set $E subseteq [0,1]$, I am looking for an example of $A in L_2$ which is not in the product $sigma$-algebra $L_1 times L_1$.
I am also having trouble proving that $L_1 times L_1 subseteq L_2$. I can see that we can use $mathcal{B}(mathbb{R^2})=mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$ and that the Lebesgue measure $lambda_2$ on $(mathbb{R^2},mathcal{B}(mathbb{R^2}))$ is identical to the product measure $lambda_1 times lambda_1.$ Although I'm stuck afterwards.
measure-theory lebesgue-measure
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
$endgroup$
Let $L_d$ be the $sigma$-algebra of Lebesgue measurable subsets of $mathbb{R}^d$.
By using Vitali's set $E subseteq [0,1]$, I am looking for an example of $A in L_2$ which is not in the product $sigma$-algebra $L_1 times L_1$.
I am also having trouble proving that $L_1 times L_1 subseteq L_2$. I can see that we can use $mathcal{B}(mathbb{R^2})=mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$ and that the Lebesgue measure $lambda_2$ on $(mathbb{R^2},mathcal{B}(mathbb{R^2}))$ is identical to the product measure $lambda_1 times lambda_1.$ Although I'm stuck afterwards.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
asked Dec 5 '18 at 20:57
MilTomMilTom
1268
1268
add a comment |
add a comment |
1 Answer
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$begingroup$
$L_1otimes L_1$ is generated by $mathcal{C}={Atimes B:A,Bin L_1}$. Since $mathcal{C}subset L_2$, $L_1otimes L_1subseteq L_2$.
For a set $Nin L_1$ s.t. $Nne emptyset$ and $lambda_1(N)=0$, the set $Etimes Nin L_2$ ($because Etimes Nsubset [0,1]times N$ and $lambda_2([0,1]times N )=0$) but not in $L_1otimes L_1$ ($because$ for any $L_1otimes L_1$-measurable set $A$, the sections $A^y={xin mathbb{R}:(x,y)in A}$ belong to $L_1$).
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
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add a comment |
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1 Answer
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$begingroup$
$L_1otimes L_1$ is generated by $mathcal{C}={Atimes B:A,Bin L_1}$. Since $mathcal{C}subset L_2$, $L_1otimes L_1subseteq L_2$.
For a set $Nin L_1$ s.t. $Nne emptyset$ and $lambda_1(N)=0$, the set $Etimes Nin L_2$ ($because Etimes Nsubset [0,1]times N$ and $lambda_2([0,1]times N )=0$) but not in $L_1otimes L_1$ ($because$ for any $L_1otimes L_1$-measurable set $A$, the sections $A^y={xin mathbb{R}:(x,y)in A}$ belong to $L_1$).
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
$endgroup$
add a comment |
$begingroup$
$L_1otimes L_1$ is generated by $mathcal{C}={Atimes B:A,Bin L_1}$. Since $mathcal{C}subset L_2$, $L_1otimes L_1subseteq L_2$.
For a set $Nin L_1$ s.t. $Nne emptyset$ and $lambda_1(N)=0$, the set $Etimes Nin L_2$ ($because Etimes Nsubset [0,1]times N$ and $lambda_2([0,1]times N )=0$) but not in $L_1otimes L_1$ ($because$ for any $L_1otimes L_1$-measurable set $A$, the sections $A^y={xin mathbb{R}:(x,y)in A}$ belong to $L_1$).
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
$endgroup$
add a comment |
$begingroup$
$L_1otimes L_1$ is generated by $mathcal{C}={Atimes B:A,Bin L_1}$. Since $mathcal{C}subset L_2$, $L_1otimes L_1subseteq L_2$.
For a set $Nin L_1$ s.t. $Nne emptyset$ and $lambda_1(N)=0$, the set $Etimes Nin L_2$ ($because Etimes Nsubset [0,1]times N$ and $lambda_2([0,1]times N )=0$) but not in $L_1otimes L_1$ ($because$ for any $L_1otimes L_1$-measurable set $A$, the sections $A^y={xin mathbb{R}:(x,y)in A}$ belong to $L_1$).
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
$endgroup$
$L_1otimes L_1$ is generated by $mathcal{C}={Atimes B:A,Bin L_1}$. Since $mathcal{C}subset L_2$, $L_1otimes L_1subseteq L_2$.
For a set $Nin L_1$ s.t. $Nne emptyset$ and $lambda_1(N)=0$, the set $Etimes Nin L_2$ ($because Etimes Nsubset [0,1]times N$ and $lambda_2([0,1]times N )=0$) but not in $L_1otimes L_1$ ($because$ for any $L_1otimes L_1$-measurable set $A$, the sections $A^y={xin mathbb{R}:(x,y)in A}$ belong to $L_1$).
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
answered Dec 6 '18 at 6:37
d.k.o.d.k.o.
8,667528
8,667528
add a comment |
add a comment |
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