Lebesgue measure (unit square/integral)
$begingroup$
Let $f:[0,1]times [0,1]to mathbb{R^2}, (x,y) mapsto (y,x^2y+x)$.
How to determine the Lebesgue measure of the image $f([0,1]times [0,1])$?
Since it's surjective and a rectangle the Lebesgue measure should be $2 cdot 1=2$, but it also ranges between $0$ and $y+1$ so the Lebesgue measure could be also $frac{3}{2}$.
How to find out which of these solutions is correct?
measure-theory lebesgue-integral
asked Dec 4 '18 at 23:06
TartulopTartulop
656
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]times [0,1]to mathbb{R^2}, (x,y) mapsto (y,x^2y+x)$.
How to determine the Lebesgue measure of the image $f([0,1]times [0,1])$?
Since it's surjective and a rectangle the Lebesgue measure should be $2 cdot 1=2$, but it also ranges between $0$ and $y+1$ so the Lebesgue measure could be also $frac{3}{2}$.
How to find out which of these solutions is correct?
measure-theory lebesgue-integral
asked Dec 4 '18 at 23:06
TartulopTartulop
656
$endgroup$
1
$begingroup$
The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17
add a comment |
$begingroup$
Let $f:[0,1]times [0,1]to mathbb{R^2}, (x,y) mapsto (y,x^2y+x)$.
How to determine the Lebesgue measure of the image $f([0,1]times [0,1])$?
Since it's surjective and a rectangle the Lebesgue measure should be $2 cdot 1=2$, but it also ranges between $0$ and $y+1$ so the Lebesgue measure could be also $frac{3}{2}$.
How to find out which of these solutions is correct?
measure-theory lebesgue-integral
asked Dec 4 '18 at 23:06
TartulopTartulop
656
$endgroup$
Let $f:[0,1]times [0,1]to mathbb{R^2}, (x,y) mapsto (y,x^2y+x)$.
How to determine the Lebesgue measure of the image $f([0,1]times [0,1])$?
Since it's surjective and a rectangle the Lebesgue measure should be $2 cdot 1=2$, but it also ranges between $0$ and $y+1$ so the Lebesgue measure could be also $frac{3}{2}$.
How to find out which of these solutions is correct?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Dec 4 '18 at 23:06
TartulopTartulop
656
asked Dec 4 '18 at 23:06
TartulopTartulop
656
asked Dec 4 '18 at 23:06
TartulopTartulop
656
asked Dec 4 '18 at 23:06
TartulopTartulop
656
asked Dec 4 '18 at 23:06
TartulopTartulop
656
656
1
$begingroup$
The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17
add a comment |
1
$begingroup$
The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17
1
1
$begingroup$
The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17
$begingroup$
The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17
add a comment |
1 Answer
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You can verify that the range is exactly ${(x,y): 0leq x leq 1,0leq y leq x+1}$. The measure of this set is 1.5. If you draw a picture you can see that the set is made up of a rectangle and a triangle. So it is easy to compute the area.
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
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add a comment |
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$begingroup$
You can verify that the range is exactly ${(x,y): 0leq x leq 1,0leq y leq x+1}$. The measure of this set is 1.5. If you draw a picture you can see that the set is made up of a rectangle and a triangle. So it is easy to compute the area.
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
You can verify that the range is exactly ${(x,y): 0leq x leq 1,0leq y leq x+1}$. The measure of this set is 1.5. If you draw a picture you can see that the set is made up of a rectangle and a triangle. So it is easy to compute the area.
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
You can verify that the range is exactly ${(x,y): 0leq x leq 1,0leq y leq x+1}$. The measure of this set is 1.5. If you draw a picture you can see that the set is made up of a rectangle and a triangle. So it is easy to compute the area.
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
You can verify that the range is exactly ${(x,y): 0leq x leq 1,0leq y leq x+1}$. The measure of this set is 1.5. If you draw a picture you can see that the set is made up of a rectangle and a triangle. So it is easy to compute the area.
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 4 '18 at 23:31
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
52.8k32055
add a comment |
add a comment |
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The map is not surjective - certainly the image of $f$ is not all of $mathbb{R}^{2}$, since the first coordinate can only be between $0$ and $1$
$endgroup$
– pwerth
Dec 4 '18 at 23:17