$f$ integrable on $A times B$, show subset of $A$ is negligible
$begingroup$
$A subset mathbb R^m$ and $B subset mathbb R^n$ are boxes, and $f$ is integrable in the riemann sense on $(Atimes B)$.
We define $$A_1 =left{a in A left|exists int_{B}f(a,b)dbright.right} subset A.$$
Show that $A setminus A_1$ has zero volume.
integration measure-theory
edited Dec 23 '18 at 20:37
gt6989b
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
$endgroup$
add a comment |
$begingroup$
$A subset mathbb R^m$ and $B subset mathbb R^n$ are boxes, and $f$ is integrable in the riemann sense on $(Atimes B)$.
We define $$A_1 =left{a in A left|exists int_{B}f(a,b)dbright.right} subset A.$$
Show that $A setminus A_1$ has zero volume.
integration measure-theory
edited Dec 23 '18 at 20:37
gt6989b
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
$endgroup$
$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50
add a comment |
$begingroup$
$A subset mathbb R^m$ and $B subset mathbb R^n$ are boxes, and $f$ is integrable in the riemann sense on $(Atimes B)$.
We define $$A_1 =left{a in A left|exists int_{B}f(a,b)dbright.right} subset A.$$
Show that $A setminus A_1$ has zero volume.
integration measure-theory
edited Dec 23 '18 at 20:37
gt6989b
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
$endgroup$
$A subset mathbb R^m$ and $B subset mathbb R^n$ are boxes, and $f$ is integrable in the riemann sense on $(Atimes B)$.
We define $$A_1 =left{a in A left|exists int_{B}f(a,b)dbright.right} subset A.$$
Show that $A setminus A_1$ has zero volume.
integration measure-theory
integration measure-theory
edited Dec 23 '18 at 20:37
gt6989b
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
edited Dec 23 '18 at 20:37
gt6989b
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
edited Dec 23 '18 at 20:37
gt6989b
36k22557
edited Dec 23 '18 at 20:37
gt6989b
36k22557
edited Dec 23 '18 at 20:37
gt6989b
36k22557
36k22557
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
asked Dec 23 '18 at 20:18
Oria GruberOria Gruber
6,54232464
6,54232464
$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50
add a comment |
$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50
$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50
$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50
add a comment |
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$begingroup$
I have a strong feeling this is related to Fubini's theorem. Something along the lines of $int_{A times B} f = int_{A} int_B f(a,b)dbda = int_{A_1}int_B f(a,b)dbda + int_{A setminus A_1}int_B f(a,b)dbda$
$endgroup$
– Oria Gruber
Dec 23 '18 at 20:50