# When expressing area of $f(D)$ using Jacobian, why exactly must $f$ be one-to-one?

I was working on a question very similar to this one: Expressing the area of the image of a holomorphic function by the coefficients of its expansion

Clearly the key lies in the formula $$\iint_{f(D)}dxdy=\iint_D\operatorname{Jac}(f)\,dx\,dy,$$ which turns out to be $$\iint_D|f'|^2\,dx\,dy$$ for holomorphic functions (using Cauchy-Riemann equations). But it has been pointed out that this is only true for one-to-one functions. Intuitively this makes sense to me since the integral would sum the same area more than once if it were not one-to-one. But then...

Is it correct to generalize this formula to functions which are not one-to-one by saying $$\iint_{f(D)}dxdy\leq \iint_D|f'|^2\,dx\,dy$$ with equality when $f$ is one-to-one? Would equality also hold if the set of points in the image which have more than one pre-image is non-empty but has measure zero? Or can stranger things happen that I have not considered?

If my question makes more sense with $\operatorname{Jac}(f)$ in place of $|f'|^2$, please let me know, and feel free to use the former instead.

Thanks.

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Yes I do. I've made the edit. Thanks. –  John Adamski Apr 26 '12 at 19:30
In the case of a non-constant holomorphic function, the set of points with more than one pre-image in an open set $D$ is open, so it can't be non-empty with measure $0$. –  Robert Israel Apr 26 '12 at 19:34
To motivate Christian's answer below without involving more general methods of analysis, it is useful to note that for many not-one-to-one maps $f$, it is possible to explicitly split the domain of $f$ into finitely many disjoint pieces on which $f$ is one-to-one. Applying the result for one-to-one functions on each piece and adding up, one finds that the integral over the domain of $J_f(x,y)$ is a sum of integrals of $1$ over pieces of the range of $f$, which can clearly be rewritten as an integral over the range of $f$ of the counting function $n(u,v)$ of Christian's answer. –  leslie townes Apr 26 '12 at 22:10

The map $f$, whether holomorphic or not, maps an infinitesimal rectangle $[x,x+dx]\times[y,y+dy]$ of area $dx\, dy$ onto an infinitesimal parallelogram of area $J_f(x,y)\cdot dx\,dy$. Therefore any "area element" of area ${\rm d}(x,y)$ near the point $(x,y)$ is mapped onto an area element of area $J_f(x,y)\cdot {\rm d}(x,y)$ near the point $f(x,y)$. When $f$ happens to be holomorphic then $J_f(z)=|f'(z)|^2$.
If $D$ is a "large" domain then the images of the area elements ${\rm d}(x,y)$ in $D$ cover $f(D)$, but some parts of $f(D)$ might be covered several times. Let $n(u,v)$ be the number of times the point $(u,v)\in f(D)$ is covered; in other words $n(u,v):=\#f^{-1}(u,v)$. Then our intuitive reasoning about infinitesimal parallelograms says that we should have a formula of the following form: $$\int_D J_f(x,y)\ {\rm d}(x,y)=\int_{f(D)} n(u,v)\ {\rm d}(u,v)\ .$$ This is Theorem 3.2.3(1) in Federer's "Geometric measure theory". When $f$ is holomorphic and essentially injective then $n(u,v)=1$ a.e.; whence $$A\bigl(f(D)\bigr)=\int_D \bigl|f'(z)\bigr|^2\ {\rm d}(x,y)$$ in this case.