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?

  • 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

You can verify that the range is exactly $\{(x,y): 0\leq x \leq 1,0\leq y \leq x+1\}$. 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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