almost everywhere zero in product measure space











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Let $M_{1}$ and $M_2$ be measure spaces with measures $mu_{1}$ and $mu_{2}$ respectively, and $f:M_{1}times M_{2}rightarrow mathbb{R}$ be a measurable function (I assume $M_{1}times M_{2}$ is equipped with product measure).



Suppose there exists a measurable set $Nsubset M_{2}$ with $mu_{2}(N)=0$ such that for any fixed $yin M_{2}-N$, the function $g(x)=f(x,y)$ is almost everywhere zero with respect to measure $mu_{1}$. Is it necessarily true that $f(x,y)=0$ almost everywhere with respect to the product measure?



I think it is true but I kind of have difficulty proving this statement as it may involve uncountable union. Thanks.






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    Let $M_{1}$ and $M_2$ be measure spaces with measures $mu_{1}$ and $mu_{2}$ respectively, and $f:M_{1}times M_{2}rightarrow mathbb{R}$ be a measurable function (I assume $M_{1}times M_{2}$ is equipped with product measure).



    Suppose there exists a measurable set $Nsubset M_{2}$ with $mu_{2}(N)=0$ such that for any fixed $yin M_{2}-N$, the function $g(x)=f(x,y)$ is almost everywhere zero with respect to measure $mu_{1}$. Is it necessarily true that $f(x,y)=0$ almost everywhere with respect to the product measure?



    I think it is true but I kind of have difficulty proving this statement as it may involve uncountable union. Thanks.






    measure-theory







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $M_{1}$ and $M_2$ be measure spaces with measures $mu_{1}$ and $mu_{2}$ respectively, and $f:M_{1}times M_{2}rightarrow mathbb{R}$ be a measurable function (I assume $M_{1}times M_{2}$ is equipped with product measure).



      Suppose there exists a measurable set $Nsubset M_{2}$ with $mu_{2}(N)=0$ such that for any fixed $yin M_{2}-N$, the function $g(x)=f(x,y)$ is almost everywhere zero with respect to measure $mu_{1}$. Is it necessarily true that $f(x,y)=0$ almost everywhere with respect to the product measure?



      I think it is true but I kind of have difficulty proving this statement as it may involve uncountable union. Thanks.






      measure-theory







      share|cite|improve this question













      Let $M_{1}$ and $M_2$ be measure spaces with measures $mu_{1}$ and $mu_{2}$ respectively, and $f:M_{1}times M_{2}rightarrow mathbb{R}$ be a measurable function (I assume $M_{1}times M_{2}$ is equipped with product measure).



      Suppose there exists a measurable set $Nsubset M_{2}$ with $mu_{2}(N)=0$ such that for any fixed $yin M_{2}-N$, the function $g(x)=f(x,y)$ is almost everywhere zero with respect to measure $mu_{1}$. Is it necessarily true that $f(x,y)=0$ almost everywhere with respect to the product measure?



      I think it is true but I kind of have difficulty proving this statement as it may involve uncountable union. Thanks.






      measure-theory




      measure-theory






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      asked 22 hours ago









      KnobbyWan

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          Let $E={(x,y): f(x,y) neq 0}$. Fro any $yin M_2$ the section $E^{y}$ is ${x:f(x,y)=0}$ and its $mu_1$ measure is $0$ by hypothesis. By Fubini's Theorem this implies $(mu_1times mu_2) (E)=0$ which is what we have to prove.






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            Let $E={(x,y): f(x,y) neq 0}$. Fro any $yin M_2$ the section $E^{y}$ is ${x:f(x,y)=0}$ and its $mu_1$ measure is $0$ by hypothesis. By Fubini's Theorem this implies $(mu_1times mu_2) (E)=0$ which is what we have to prove.






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              Let $E={(x,y): f(x,y) neq 0}$. Fro any $yin M_2$ the section $E^{y}$ is ${x:f(x,y)=0}$ and its $mu_1$ measure is $0$ by hypothesis. By Fubini's Theorem this implies $(mu_1times mu_2) (E)=0$ which is what we have to prove.






              share|cite|improve this answer























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                Let $E={(x,y): f(x,y) neq 0}$. Fro any $yin M_2$ the section $E^{y}$ is ${x:f(x,y)=0}$ and its $mu_1$ measure is $0$ by hypothesis. By Fubini's Theorem this implies $(mu_1times mu_2) (E)=0$ which is what we have to prove.






                share|cite|improve this answer












                Let $E={(x,y): f(x,y) neq 0}$. Fro any $yin M_2$ the section $E^{y}$ is ${x:f(x,y)=0}$ and its $mu_1$ measure is $0$ by hypothesis. By Fubini's Theorem this implies $(mu_1times mu_2) (E)=0$ which is what we have to prove.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered 21 hours ago









                Kavi Rama Murthy

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