The exercise in general says:
Find $k$ if the joint probability density of $X$, $Y$, and $Z$ is given by
$$f(x,y,z)=kxy(1−z) \text{ for } 0<x<1, 0<y<1, x+y+z<1,$$
then the answers define the following integral
$$\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-y-z} kxy(1−z) \, dx \, dy \, dz.$$
I understand $0 <z <1$, and $0<x <1-y-z$, but I don't understand where it comes from the interval $0<y<1-z$ could someone explain to me :(.
$y <1-x-z$ because $x+y+z <1$ Since $x >0$ this gives $y <1-z$.