Is the product operator lebesgue measurable?











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Define $T(x,y):=xy$ for $(x,y)in mathbb{R}^2$.



Since $T$ is continuous(moreover is of $C^infty$), it is Borel measurable. However, is $T:mathbb{R}^2rightarrow mathbb{R}$ Lebesgue measurable?



That is, if $A$ is an 1-dim Lebesgue measurable set, then is $T^{-1}(A)$ 2-dim Lebesgue measurable set?



EDIT



I wrote my proof for this as an answer below. I hope someone verifies this.. thank you in advance!






real-analysis measure-theory proof-verification lebesgue-measure







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  • Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
    – paul garrett
    Nov 28 at 2:28










  • Yes, I am sure. Would you verify my answer below?
    – Rubertos
    Nov 28 at 2:38















up vote
2
down vote

favorite












Define $T(x,y):=xy$ for $(x,y)in mathbb{R}^2$.



Since $T$ is continuous(moreover is of $C^infty$), it is Borel measurable. However, is $T:mathbb{R}^2rightarrow mathbb{R}$ Lebesgue measurable?



That is, if $A$ is an 1-dim Lebesgue measurable set, then is $T^{-1}(A)$ 2-dim Lebesgue measurable set?



EDIT



I wrote my proof for this as an answer below. I hope someone verifies this.. thank you in advance!






real-analysis measure-theory proof-verification lebesgue-measure







share|cite|improve this question
























  • Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
    – paul garrett
    Nov 28 at 2:28










  • Yes, I am sure. Would you verify my answer below?
    – Rubertos
    Nov 28 at 2:38













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Define $T(x,y):=xy$ for $(x,y)in mathbb{R}^2$.



Since $T$ is continuous(moreover is of $C^infty$), it is Borel measurable. However, is $T:mathbb{R}^2rightarrow mathbb{R}$ Lebesgue measurable?



That is, if $A$ is an 1-dim Lebesgue measurable set, then is $T^{-1}(A)$ 2-dim Lebesgue measurable set?



EDIT



I wrote my proof for this as an answer below. I hope someone verifies this.. thank you in advance!






real-analysis measure-theory proof-verification lebesgue-measure







share|cite|improve this question















Define $T(x,y):=xy$ for $(x,y)in mathbb{R}^2$.



Since $T$ is continuous(moreover is of $C^infty$), it is Borel measurable. However, is $T:mathbb{R}^2rightarrow mathbb{R}$ Lebesgue measurable?



That is, if $A$ is an 1-dim Lebesgue measurable set, then is $T^{-1}(A)$ 2-dim Lebesgue measurable set?



EDIT



I wrote my proof for this as an answer below. I hope someone verifies this.. thank you in advance!






real-analysis measure-theory proof-verification lebesgue-measure




real-analysis measure-theory proof-verification lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 2:46

























asked Nov 28 at 1:54









Rubertos

5,6632823




5,6632823












  • Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
    – paul garrett
    Nov 28 at 2:28










  • Yes, I am sure. Would you verify my answer below?
    – Rubertos
    Nov 28 at 2:38


















  • Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
    – paul garrett
    Nov 28 at 2:28










  • Yes, I am sure. Would you verify my answer below?
    – Rubertos
    Nov 28 at 2:38
















Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
– paul garrett
Nov 28 at 2:28




Are you sure that what you've said is what you want "Lebesgue measurable function" to mean?
– paul garrett
Nov 28 at 2:28












Yes, I am sure. Would you verify my answer below?
– Rubertos
Nov 28 at 2:38




Yes, I am sure. Would you verify my answer below?
– Rubertos
Nov 28 at 2:38










1 Answer
1






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0
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Let $E$ be an 1-dim Lebesgue measurable set such that $m(E)=0$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $T^{-1}(E)$ is measurable. Let us first fix $kinmathbb{Z}^+$. Define $B_k:={(x,y)in mathbb{R}^2|1/k leq |y| leq k}$.



Since $m(E)=0$, we can find a sequence $O_n$ of open sets such that $Esubset O_n$ and $lim_{nto infty} m(O_n)=0$.



Since $T$ is continuous, $T^{-1}(O_n)$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)cap B_k)=int int mathbb{1}_{O_n}(xy) mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= int frac{1}{|y|} mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.



Taking $nto infty$, the right hand side tends to $0$. Hence, $m(T^{-1}(O_n)cap B_k) to 0$ as $nto infty$. Thus, $m^*(T^{-1}(E)cap B_k)=0$ for all $k$. (Here, $m^*$ denotes the 2-dim Lebesgue outer measure)



By taking $kto infty$, we have $m^*(T^{-1}(E)cap {(x,y):|y|>0})=0$. Since $m^*({(x,y): |y|=0})=0$, we have $m^*(T^{-1}(E))=0$. Q.E.D.






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  • Looks OK to me.
    – Kavi Rama Murthy
    Nov 28 at 5:58











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Let $E$ be an 1-dim Lebesgue measurable set such that $m(E)=0$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $T^{-1}(E)$ is measurable. Let us first fix $kinmathbb{Z}^+$. Define $B_k:={(x,y)in mathbb{R}^2|1/k leq |y| leq k}$.



Since $m(E)=0$, we can find a sequence $O_n$ of open sets such that $Esubset O_n$ and $lim_{nto infty} m(O_n)=0$.



Since $T$ is continuous, $T^{-1}(O_n)$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)cap B_k)=int int mathbb{1}_{O_n}(xy) mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= int frac{1}{|y|} mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.



Taking $nto infty$, the right hand side tends to $0$. Hence, $m(T^{-1}(O_n)cap B_k) to 0$ as $nto infty$. Thus, $m^*(T^{-1}(E)cap B_k)=0$ for all $k$. (Here, $m^*$ denotes the 2-dim Lebesgue outer measure)



By taking $kto infty$, we have $m^*(T^{-1}(E)cap {(x,y):|y|>0})=0$. Since $m^*({(x,y): |y|=0})=0$, we have $m^*(T^{-1}(E))=0$. Q.E.D.






share|cite|improve this answer























  • Looks OK to me.
    – Kavi Rama Murthy
    Nov 28 at 5:58















up vote
0
down vote













Let $E$ be an 1-dim Lebesgue measurable set such that $m(E)=0$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $T^{-1}(E)$ is measurable. Let us first fix $kinmathbb{Z}^+$. Define $B_k:={(x,y)in mathbb{R}^2|1/k leq |y| leq k}$.



Since $m(E)=0$, we can find a sequence $O_n$ of open sets such that $Esubset O_n$ and $lim_{nto infty} m(O_n)=0$.



Since $T$ is continuous, $T^{-1}(O_n)$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)cap B_k)=int int mathbb{1}_{O_n}(xy) mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= int frac{1}{|y|} mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.



Taking $nto infty$, the right hand side tends to $0$. Hence, $m(T^{-1}(O_n)cap B_k) to 0$ as $nto infty$. Thus, $m^*(T^{-1}(E)cap B_k)=0$ for all $k$. (Here, $m^*$ denotes the 2-dim Lebesgue outer measure)



By taking $kto infty$, we have $m^*(T^{-1}(E)cap {(x,y):|y|>0})=0$. Since $m^*({(x,y): |y|=0})=0$, we have $m^*(T^{-1}(E))=0$. Q.E.D.






share|cite|improve this answer























  • Looks OK to me.
    – Kavi Rama Murthy
    Nov 28 at 5:58













up vote
0
down vote










up vote
0
down vote









Let $E$ be an 1-dim Lebesgue measurable set such that $m(E)=0$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $T^{-1}(E)$ is measurable. Let us first fix $kinmathbb{Z}^+$. Define $B_k:={(x,y)in mathbb{R}^2|1/k leq |y| leq k}$.



Since $m(E)=0$, we can find a sequence $O_n$ of open sets such that $Esubset O_n$ and $lim_{nto infty} m(O_n)=0$.



Since $T$ is continuous, $T^{-1}(O_n)$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)cap B_k)=int int mathbb{1}_{O_n}(xy) mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= int frac{1}{|y|} mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.



Taking $nto infty$, the right hand side tends to $0$. Hence, $m(T^{-1}(O_n)cap B_k) to 0$ as $nto infty$. Thus, $m^*(T^{-1}(E)cap B_k)=0$ for all $k$. (Here, $m^*$ denotes the 2-dim Lebesgue outer measure)



By taking $kto infty$, we have $m^*(T^{-1}(E)cap {(x,y):|y|>0})=0$. Since $m^*({(x,y): |y|=0})=0$, we have $m^*(T^{-1}(E))=0$. Q.E.D.






share|cite|improve this answer














Let $E$ be an 1-dim Lebesgue measurable set such that $m(E)=0$. Since the lebesgue measure is the completion of a Borel measure, it suffices to prove that $T^{-1}(E)$ is measurable. Let us first fix $kinmathbb{Z}^+$. Define $B_k:={(x,y)in mathbb{R}^2|1/k leq |y| leq k}$.



Since $m(E)=0$, we can find a sequence $O_n$ of open sets such that $Esubset O_n$ and $lim_{nto infty} m(O_n)=0$.



Since $T$ is continuous, $T^{-1}(O_n)$ is measurable. Applying Tonelli, $$m(T^{-1}(O_n)cap B_k)=int int mathbb{1}_{O_n}(xy) mathbb{1}_{[1/k,k]}(|y|) dm(x)dm(y)= int frac{1}{|y|} mathbb{1}_{[1/k,k]}(|y|) m(O_n)dm(y)$$.



Taking $nto infty$, the right hand side tends to $0$. Hence, $m(T^{-1}(O_n)cap B_k) to 0$ as $nto infty$. Thus, $m^*(T^{-1}(E)cap B_k)=0$ for all $k$. (Here, $m^*$ denotes the 2-dim Lebesgue outer measure)



By taking $kto infty$, we have $m^*(T^{-1}(E)cap {(x,y):|y|>0})=0$. Since $m^*({(x,y): |y|=0})=0$, we have $m^*(T^{-1}(E))=0$. Q.E.D.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 at 3:50

























answered Nov 28 at 2:37









Rubertos

5,6632823




5,6632823












  • Looks OK to me.
    – Kavi Rama Murthy
    Nov 28 at 5:58


















  • Looks OK to me.
    – Kavi Rama Murthy
    Nov 28 at 5:58
















Looks OK to me.
– Kavi Rama Murthy
Nov 28 at 5:58




Looks OK to me.
– Kavi Rama Murthy
Nov 28 at 5:58


















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