Product of a null set
Let $E subset mathbb{R}^m$, $F subset mathbb{R}^n$ and $mu$ the Lebesgue measure.
How to prove:
If $E$ is a Lebesgue null set of $mathbb{R}^m Rightarrow E times F$ is a Lebesgue null set of $mathbb{R}^{m+n}$.
I know there is a way to show that implication with the theorems of Fubini, Tonelli and Cavalieri, but how can it be proved without them?
I used:
Since $E times F$ is a rectangle, by definition: $mu_{m+n}(E times F)=mu_m(E) cdot mu_n(F)$.
So if $mu_m(E)=0Rightarrow mu_{m+n}(E times F)=0$
Is this way correct? And how can it be shown if $mu_n(F)=infty$?
measure-theory lebesgue-measure
edited Nov 30 at 9:29
Henrik
5,98392030
asked Nov 30 at 9:21
Tartulop
656
add a comment |
Let $E subset mathbb{R}^m$, $F subset mathbb{R}^n$ and $mu$ the Lebesgue measure.
How to prove:
If $E$ is a Lebesgue null set of $mathbb{R}^m Rightarrow E times F$ is a Lebesgue null set of $mathbb{R}^{m+n}$.
I know there is a way to show that implication with the theorems of Fubini, Tonelli and Cavalieri, but how can it be proved without them?
I used:
Since $E times F$ is a rectangle, by definition: $mu_{m+n}(E times F)=mu_m(E) cdot mu_n(F)$.
So if $mu_m(E)=0Rightarrow mu_{m+n}(E times F)=0$
Is this way correct? And how can it be shown if $mu_n(F)=infty$?
measure-theory lebesgue-measure
edited Nov 30 at 9:29
Henrik
5,98392030
asked Nov 30 at 9:21
Tartulop
656
Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14
add a comment |
Let $E subset mathbb{R}^m$, $F subset mathbb{R}^n$ and $mu$ the Lebesgue measure.
How to prove:
If $E$ is a Lebesgue null set of $mathbb{R}^m Rightarrow E times F$ is a Lebesgue null set of $mathbb{R}^{m+n}$.
I know there is a way to show that implication with the theorems of Fubini, Tonelli and Cavalieri, but how can it be proved without them?
I used:
Since $E times F$ is a rectangle, by definition: $mu_{m+n}(E times F)=mu_m(E) cdot mu_n(F)$.
So if $mu_m(E)=0Rightarrow mu_{m+n}(E times F)=0$
Is this way correct? And how can it be shown if $mu_n(F)=infty$?
measure-theory lebesgue-measure
edited Nov 30 at 9:29
Henrik
5,98392030
asked Nov 30 at 9:21
Tartulop
656
Let $E subset mathbb{R}^m$, $F subset mathbb{R}^n$ and $mu$ the Lebesgue measure.
How to prove:
If $E$ is a Lebesgue null set of $mathbb{R}^m Rightarrow E times F$ is a Lebesgue null set of $mathbb{R}^{m+n}$.
I know there is a way to show that implication with the theorems of Fubini, Tonelli and Cavalieri, but how can it be proved without them?
I used:
Since $E times F$ is a rectangle, by definition: $mu_{m+n}(E times F)=mu_m(E) cdot mu_n(F)$.
So if $mu_m(E)=0Rightarrow mu_{m+n}(E times F)=0$
Is this way correct? And how can it be shown if $mu_n(F)=infty$?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Nov 30 at 9:29
Henrik
5,98392030
asked Nov 30 at 9:21
Tartulop
656
edited Nov 30 at 9:29
Henrik
5,98392030
asked Nov 30 at 9:21
Tartulop
656
edited Nov 30 at 9:29
Henrik
5,98392030
edited Nov 30 at 9:29
Henrik
5,98392030
edited Nov 30 at 9:29
Henrik
5,98392030
5,98392030
asked Nov 30 at 9:21
Tartulop
656
asked Nov 30 at 9:21
Tartulop
656
asked Nov 30 at 9:21
Tartulop
656
656
Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14
add a comment |
Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14
Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14
Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14
add a comment |
1 Answer
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Hint: consider $F_k = F cap prod_{i=1}^n [-k,k]$. From you definition of product measure, $mu(Etimes F_k)=0$. Apply the regularity of measure and let $kto+infty$ to conclude.
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
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1 Answer
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votes
Hint: consider $F_k = F cap prod_{i=1}^n [-k,k]$. From you definition of product measure, $mu(Etimes F_k)=0$. Apply the regularity of measure and let $kto+infty$ to conclude.
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
add a comment |
Hint: consider $F_k = F cap prod_{i=1}^n [-k,k]$. From you definition of product measure, $mu(Etimes F_k)=0$. Apply the regularity of measure and let $kto+infty$ to conclude.
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
add a comment |
Hint: consider $F_k = F cap prod_{i=1}^n [-k,k]$. From you definition of product measure, $mu(Etimes F_k)=0$. Apply the regularity of measure and let $kto+infty$ to conclude.
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
Hint: consider $F_k = F cap prod_{i=1}^n [-k,k]$. From you definition of product measure, $mu(Etimes F_k)=0$. Apply the regularity of measure and let $kto+infty$ to conclude.
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 30 at 9:33
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
add a comment |
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Yes, that's right. Doesn't matter whether $mu_n(F)=infty$, because by definition in this context $0cdotinfty=0$.
– David C. Ullrich
Nov 30 at 15:14