prove that null set product $R^n$ is of measure zero












2












$begingroup$


Let $E$ be a set of measure zero. I want to prove that the Cartesian product of $E$ and $R^n$ is also of measure zero.



I thought of maybe using intervals to cover the product, as I know $E$ is of measure zero, but I couldn't really work it out.






calculus measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:37












  • $begingroup$
    It is in the space $R^m$. Could you show how do you use Fubini's theorem here
    $endgroup$
    – Gabi G
    Dec 30 '18 at 7:10
















2












$begingroup$


Let $E$ be a set of measure zero. I want to prove that the Cartesian product of $E$ and $R^n$ is also of measure zero.



I thought of maybe using intervals to cover the product, as I know $E$ is of measure zero, but I couldn't really work it out.






calculus measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:37












  • $begingroup$
    It is in the space $R^m$. Could you show how do you use Fubini's theorem here
    $endgroup$
    – Gabi G
    Dec 30 '18 at 7:10














2












2








2





$begingroup$


Let $E$ be a set of measure zero. I want to prove that the Cartesian product of $E$ and $R^n$ is also of measure zero.



I thought of maybe using intervals to cover the product, as I know $E$ is of measure zero, but I couldn't really work it out.






calculus measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$




Let $E$ be a set of measure zero. I want to prove that the Cartesian product of $E$ and $R^n$ is also of measure zero.



I thought of maybe using intervals to cover the product, as I know $E$ is of measure zero, but I couldn't really work it out.






calculus measure-theory multivariable-calculus




calculus measure-theory multivariable-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 22:21









Gabi GGabi G

408110




408110








  • 1




    $begingroup$
    You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:37












  • $begingroup$
    It is in the space $R^m$. Could you show how do you use Fubini's theorem here
    $endgroup$
    – Gabi G
    Dec 30 '18 at 7:10














  • 1




    $begingroup$
    You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:37












  • $begingroup$
    It is in the space $R^m$. Could you show how do you use Fubini's theorem here
    $endgroup$
    – Gabi G
    Dec 30 '18 at 7:10








1




1




$begingroup$
You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:37






$begingroup$
You have not said where $E$ is and w.r.t. what measure it has measure $0$. If it has measure $0$ w.r.t. Lebesgue measure on an Euclidean space then this is an immediate application of Fubini's Theorem
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:37














$begingroup$
It is in the space $R^m$. Could you show how do you use Fubini's theorem here
$endgroup$
– Gabi G
Dec 30 '18 at 7:10




$begingroup$
It is in the space $R^m$. Could you show how do you use Fubini's theorem here
$endgroup$
– Gabi G
Dec 30 '18 at 7:10










1 Answer
1






active

oldest

votes


















2












$begingroup$

I would use Tonellis theorem for non-negetive functions:
$$lambda^{m + n}(E times mathbb{R}^n) = int_{E times mathbb{R}^n} d lambda^{m+n}(x,y) = int_{mathbb{R}^n} int_E dlambda^m(x) dlambda^n(y) = 0.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks! It really helped me
    $endgroup$
    – Gabi G
    Dec 31 '18 at 8:00











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

I would use Tonellis theorem for non-negetive functions:
$$lambda^{m + n}(E times mathbb{R}^n) = int_{E times mathbb{R}^n} d lambda^{m+n}(x,y) = int_{mathbb{R}^n} int_E dlambda^m(x) dlambda^n(y) = 0.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks! It really helped me
    $endgroup$
    – Gabi G
    Dec 31 '18 at 8:00
















2












$begingroup$

I would use Tonellis theorem for non-negetive functions:
$$lambda^{m + n}(E times mathbb{R}^n) = int_{E times mathbb{R}^n} d lambda^{m+n}(x,y) = int_{mathbb{R}^n} int_E dlambda^m(x) dlambda^n(y) = 0.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks! It really helped me
    $endgroup$
    – Gabi G
    Dec 31 '18 at 8:00














2












2








2





$begingroup$

I would use Tonellis theorem for non-negetive functions:
$$lambda^{m + n}(E times mathbb{R}^n) = int_{E times mathbb{R}^n} d lambda^{m+n}(x,y) = int_{mathbb{R}^n} int_E dlambda^m(x) dlambda^n(y) = 0.$$






share|cite|improve this answer









$endgroup$



I would use Tonellis theorem for non-negetive functions:
$$lambda^{m + n}(E times mathbb{R}^n) = int_{E times mathbb{R}^n} d lambda^{m+n}(x,y) = int_{mathbb{R}^n} int_E dlambda^m(x) dlambda^n(y) = 0.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 9:49









eddieeddie

525110




525110








  • 1




    $begingroup$
    Thanks! It really helped me
    $endgroup$
    – Gabi G
    Dec 31 '18 at 8:00














  • 1




    $begingroup$
    Thanks! It really helped me
    $endgroup$
    – Gabi G
    Dec 31 '18 at 8:00








1




1




$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Dec 31 '18 at 8:00




$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Dec 31 '18 at 8:00


















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