Mixing
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.
measure-theory
asked 22 hours ago
KnobbyWan
177110
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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
asked 22 hours ago
KnobbyWan
177110
add a comment |
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
asked 22 hours ago
KnobbyWan
177110
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
asked 22 hours ago
KnobbyWan
177110
asked 22 hours ago
KnobbyWan
177110
asked 22 hours ago
KnobbyWan
177110
asked 22 hours ago
KnobbyWan
177110
asked 22 hours ago
KnobbyWan
177110
177110
add a comment |
add a comment |
1 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.
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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.
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
add a comment |
up vote
0
down vote
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.
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
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.
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
answered 21 hours ago
Kavi Rama Murthy
40.1k31750
40.1k31750
add a comment |
add a comment |
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