On the meaning of the complex measure $\int_{\mathbb{C}} d z d \bar{z}$ I have a problem to understand the meaning of a complex measure; i.e., when someone writes ($i \equiv \sqrt{-1}$)
$$
\int_{\mathbb{R}^2} d\mathrm{Re}z \, d\mathrm{Im}z  \equiv \int_{\mathbb{C}} \frac{dz d\bar{z}}{2i} \quad (\ast)
$$
The lefthand-side will yield a real number (after performing the integration over a real-valued function), while it is not obvious that the righthand-side yields a real number. 
Furthermore, how can one obtain the equivalence relation? Is the factor $\frac{1}{2i}$ the Jacobian of some transformation like
$$
z = \mathrm{Re} z + i \, \mathrm{Im} z ,\\
\bar{z} = \mathrm{Re} z - i \, \mathrm{Im} z ,\\
$$
So, the complex measure $dz d\bar{z}$ does not have the same meaning as a ‘simple’ complex integration which in complex calculus (integration over a path in the complex plane).
Please provide an explanation for the complex measure and the equivalence relation ($\ast$) above.


Notes


1. A similar question is asked here; yet no clear proof or justification is provided.


2. An example of the relation appears, for instance, in Altland, A. and B. D. Simons. “Condensed Matter Field Theory” (2nd ed., 2010), p. 102:


 A: Probably the easiest way to see this (up to a sign that has me confused --- are you sure the order of differentials is right in the formula you quoted) is in terms of differential forms.  With $z=x+iy$ (where $x$ and $y$ are real variables), we have $\bar z=x-iy$ and so
$$
dz\land d\bar z=(dx+i\,dy)\land(dx-i\,dy)=(dx\land dx)+i(dy\land dx)
-i(dx\land dy)-(dy\land dy).
$$
Since the wedge product is skew-symmetric, the first and last terms here vanish, and the other two terms combine to give $-2i\,dx\land dy$.
A: You have to  be aware that there are intagration of measures and integration of differiential forms. You cab see the difference when using the Change of variables formula.
For example if $V,W$ are vector spaces of same dimension, $\lambda,\mu$ Lebesgue measures on $V$ and $W$ (or rather a Haar measure), and $f : V \rightarrow W$ smooth and compactely supported, then :
$$\int_{x \in V} d\mu(f(x)) = \int_{x \in V} |\mathrm{Jac}(f(x))| ~ d\lambda(x).$$
In your context it is integration of differential form, and you need the following formula.
If $V,W$ are oriented vector spaces not necessarily of the same dimension, $f : V \rightarrow W$ smooth, $v$ an invertible differential form on $V$ (i.e. volume form), $w$ a differential form on $W$ of degree $\dim V$ then
$$\int_{x \in V} f^*v~(x) = \int_{x \in V} \mathrm{Jac}(f(x)) ~ v(x),$$
where $\mathrm{Jac}(f(x))$ is the unique number such that $f^*v~(x) = \mathrm{Jac}(f(x)) ~ v(x)$ (recall that $v$ is invertible, and it coincides with the determinant when $V$ and $W$ are of the same dimension).
In your situation $V=\mathbb{C}$, $W = \mathbb{C} \times \mathbb{C}$, $f : z=x+iy \mapsto (z,\bar{z})$, $v= dx \wedge dy$, and $w = dz_1 \wedge dz_2$, which yields :
$$\int_{z \in \mathbb{C}} dz \wedge d\bar{z} = \int_{z \in \mathbb{C}} -2i ~ dx \wedge dy.$$
Note : $dz$, $d\bar{z}$, $dz.d\bar{z}$ are not measures nor complex measures. They are complex differential forms of degree 1 or 2.
