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I would like to evaluate the Lebesgue integral $int_{Omega}x-y^2 ,dx,dy$ using the level set definition of the Lebesgue integral. Here $Omega = -sqrt{x} < y < x, 0 < x < 1$.

It seems easy enough to handle the $y$ term if it's over the real line. Just realize that you're dealing with $mu {y: -y^2 > t} = mu {y: y^2 < -t} = mu {y: -y^2 > t} = mu {y: |y| < sqrt{-t}}$, where $mu$ is Lebesgue measure and $t$ is a real number. We can then use the definition of the Lebesgue measure of a ball to see that this equals $pi t^2$. However, this is not the case, so I'm not quite sure what to do.

I think that the next step involves Fubini's theorem. However, I'm having difficulty getting the measure of $x$, since it doesn't look like we can use the definition of the Lebesgue measure of a ball.

Some hints would be appreciated here!

Level set definition works for a nonnegative function. Thus, you cannot apply it to $int xdxdy$ and $int -y^2 dxdy$ separately. If correctly done,
$$
int_Omega (x-y^2) dxdy = int_0^1 dxleft(int_{-sqrt{x}}^{x} (x-y^2)dyright)=int_0^1 dxleft(int_0^infty mu(yin(-sqrt{x},x);|;x-y^2>t)dtright).
$$
We can see that $$mu(yin(-sqrt{x},x);|;x-y^2>t) = sqrt{x-t} + [sqrt{x-t} wedge x] $$is equal to $sqrt{x-t}+x$ for $0<t<x-x^2$ and $2sqrt{x-t}$ for $x-x^2 <t<x$. By integrating separately, we get
$$
int_0^{x-x^2} sqrt{x-t}+x;dt + int_{x-x^2}^x 2sqrt{x-t};dt = int^x_{x^2}sqrt{t} dt + x^2(1-x) + 2int_0^{x^2}sqrt{t} dt = frac{2}{3}x^{1.5} + x^2 -frac{1}{3}x^3.
$$
Finally, the answer is $$
int_0^1 frac{2}{3}x^{1.5} + x^2 -frac{1}{3}x^3dx = frac{31}{60}.
$$
(Apply Fubini's theorem first, then 1-dimensional integral about $y$ is calculated using your method.)

If you want to apply level set method in 2-dimension directly, you should calculate
$$
mu_2(xin [0,1], yin [-sqrt{x},x];|; x-y^2 >t ),
$$ which also requires Fubini's theorem to calculate 2-dimensional Lebesgue meausure.

(Note: We can see that $int_{-sqrt{x}}^x (x-y^2)dy = frac{2}{3}x^{1.5} + x^2 -frac{1}{3}x^3$ as expected.)