Lebesgue measure (unit square/integral)












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$begingroup$


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







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  • 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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














  • 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










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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.






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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.






    share|cite|improve this answer









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      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















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        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 23:31









        Kavi Rama MurthyKavi Rama Murthy

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