A clarification about Fubini's theorem 2
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Suppose $f(x,y)$ and $g(x,y)$ are measurable functions defined on a bounded open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$0leq int_{E} int_{F} f(x,y)dydxleq int_{E} int_{F} g(x,y)dydx $$
and $g$ is locally integrable (so we may apply Fubini's theorem to $g$). Can we apply also Fubini to $f$ and switch the order of the above integrals of $f$?
real-analysis probability integration
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
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add a comment |
$begingroup$
Suppose $f(x,y)$ and $g(x,y)$ are measurable functions defined on a bounded open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$0leq int_{E} int_{F} f(x,y)dydxleq int_{E} int_{F} g(x,y)dydx $$
and $g$ is locally integrable (so we may apply Fubini's theorem to $g$). Can we apply also Fubini to $f$ and switch the order of the above integrals of $f$?
real-analysis probability integration
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
$endgroup$
add a comment |
$begingroup$
Suppose $f(x,y)$ and $g(x,y)$ are measurable functions defined on a bounded open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$0leq int_{E} int_{F} f(x,y)dydxleq int_{E} int_{F} g(x,y)dydx $$
and $g$ is locally integrable (so we may apply Fubini's theorem to $g$). Can we apply also Fubini to $f$ and switch the order of the above integrals of $f$?
real-analysis probability integration
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
$endgroup$
Suppose $f(x,y)$ and $g(x,y)$ are measurable functions defined on a bounded open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$0leq int_{E} int_{F} f(x,y)dydxleq int_{E} int_{F} g(x,y)dydx $$
and $g$ is locally integrable (so we may apply Fubini's theorem to $g$). Can we apply also Fubini to $f$ and switch the order of the above integrals of $f$?
real-analysis probability integration
real-analysis probability integration
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
edited Dec 28 '18 at 5:25
M. Rahmat
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
asked Dec 28 '18 at 5:03
M. RahmatM. Rahmat
291212
291212
add a comment |
add a comment |
1 Answer
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Partial answer: I believe you're thinking of non-negative functions $f$ and $g$. Otherwise, it wouldn't be difficult to find an example in which this order of integration for $f$ gives $0$ as an answer but a positive value for the inverse order... and an apropriate characteristic function as $g$ would complete the counterexample.
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
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1 Answer
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1 Answer
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$begingroup$
Partial answer: I believe you're thinking of non-negative functions $f$ and $g$. Otherwise, it wouldn't be difficult to find an example in which this order of integration for $f$ gives $0$ as an answer but a positive value for the inverse order... and an apropriate characteristic function as $g$ would complete the counterexample.
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Partial answer: I believe you're thinking of non-negative functions $f$ and $g$. Otherwise, it wouldn't be difficult to find an example in which this order of integration for $f$ gives $0$ as an answer but a positive value for the inverse order... and an apropriate characteristic function as $g$ would complete the counterexample.
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Partial answer: I believe you're thinking of non-negative functions $f$ and $g$. Otherwise, it wouldn't be difficult to find an example in which this order of integration for $f$ gives $0$ as an answer but a positive value for the inverse order... and an apropriate characteristic function as $g$ would complete the counterexample.
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
Partial answer: I believe you're thinking of non-negative functions $f$ and $g$. Otherwise, it wouldn't be difficult to find an example in which this order of integration for $f$ gives $0$ as an answer but a positive value for the inverse order... and an apropriate characteristic function as $g$ would complete the counterexample.
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:30
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
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