Comparing measures in finite dimensional complex vector spaces I have a naive question that different people gave me different answers and I couldn't find any answer in dozens of textbooks.
In what follows $R^m$ denotes the $m$-dimensional euclidean space with the Lebesgue measure.

Do the following recipes give us different measures? 
A. the pushforward of $R^2 \to C$ 
B. area in the complex plane 
C. induced by $dz\wedge d\bar z$ 

Main question: What about in higher dimension? How different are 
1. the pushforward of $R^{2n} \to C^n$ 
2. the product measure using A, B, or C above 
3. a Haar measure 

Is there some other measure in $C^n$ that most people call "usual"?

Thanks in advance for explanations!

Comments.


*

*Feel free to edit the question. (For instance I don't know how to describe B analytically).

*I'm an undergraduate student starting in measure theory.
 A: $\def\R{\mathbb R}$
For (C), note that $i/2(dz \wedge d\bar z) = dx \wedge dy$, and this is the normalization that we want.  So I will just replace (C) with integration against the 2-form $dx \wedge dy$.  After that, the distinction between $\R^{2n}$ and $\mathbb C^n$ pretty much disappears from the discussion.
So for (A), (B), (C), you want to compare (A) "Lebesgue measure" on $\R^2$, (B) "area measure on $\R^2$" and (C) "$dx \wedge dy$" on $\R^2$.  The only two things you could possibly mean by (B) is either (A) or (C), so we are left to compare (A) and (C).    I think you might read about (A) and (C) in different places with different starting points, so what is a conclusion in one place might be a definition in another.  However one does it, the following things end up being the same:  translation invariant Borel measure $\mu$ on $\R^2$, satisfying $\mu(K) < \infty$ for $K$ compact, and  normalized so that  $\mu([0,1] \times [0,1]) = 1$;   product measure on $\R^2$ based on Lebesgue measure on $\R$; and the measure associated with the 2-form $dx \wedge dy$.
For (1), (2), (3), let's say we agree that "Lebesgue measure" on $\R^{2n}$ is the $2n$-fold product measure starting with Lebesgue measure on $\R$.  It is shown, in Rudin, Real and Complex Analysis, for example, that Lebesgue measure is the unique translation invariant Borel measure which is finite on compact sets and normalized so that the unit hypercube has measure 1.  So (1) and (3) agree.  
So you are only left to compare two product measure constructions:  $2n$-fold product measure starting with Lebesgue measure on $\R$ and $n$-fold product measure starting with Lebesgue measure on $\R^2$, which is itself $2$-fold product measure starting with Lebesgue measure on $\R$.  There is no way in the world that these could end up being different, so just muscle through the details one way or another, or more elegantly perhaps, mimic whatever argument Rudin used to show  $n$-fold product measure starting with Lebesgue measure on $\R$ ends up being translation invariant.
