Q:

Changing order of integration, bound is a function

Consider the following double integral:
$$\int_{g(v)}^v \int_{g(z)}^z f(z,y)\mathrm{d} y \mathrm{d} z$$
I need to change the order of integration to $\mathrm{d}z \mathrm{d}y$.
$g$ is some increasing function that is below the $45°$-line.
1) Is it correct that, if $g(x) = 0 \space\forall \space x$, then the the double integral above could be rewritten as
$$\int_{0}^v \int_{0}^z f(z,y) \mathrm{d} y \mathrm{d} z = \int_{0}^v \int_{y}^v f(z,y) \mathrm{d} z \mathrm{d} y\: ?$$
2) If this was correct, i don't know how to do the trick for another $g$ function. Say $g(x)= \cfrac{x}{2}$. Then the integration area should be this 
If I now want to integrate with respect to z first, the bounds should be 
$$g(v) < z < v.$$
But then the upper and lower level in the y dimension depend on $z$.
Is it necessary to slice the area into two (in this particular case $g(x)= \frac{x}{2}$) triangles?

A:

$g(z)\leq y\leq z$, $g(v)\leq z\leq v$. $g$ is increasing so you can certainly use its inverse $g^{-1}$. $g(z)\leq y$ means $z\leq g^{-1}(y)$. Combining with $y\leq z$, we get:
$$y\leq z\leq g^{-1}(y).$$
We know $y\leq z\leq v$, so $y\leq v$. We know $g(v)\leq z$, so $v\leq g^{-1}(z)$. But $z\leq g^{-1}(y)$ so $g^{-1}(z)\leq g^{-2}(y)$, so $v\leq g^{-2}(y)$ or $g(g(v))\leq y$. In other words:
$$g(g(v))\leq y\leq v.$$
That should give the bounds for the inversion.
Edit
It seems that, at least in some cases (e.g. that presented in comments), the integration domain obtained with these new inequalities is strictly bigger than the one of the old ones. I think this might be because those new inequalities do not imply $g(v)\leq z\leq v$ directly. At least, in the case of comments this was the problem. So integrating in $z$ from $\max\{g(v),y\}$ to $\min\{v,g^{-1}(y)\}$ should fix the problem in at least some cases (e.g. that of comments).

