Let $f:U\rightarrow V$ be some quasiconformal map in the plane. I want to show that the formula $$\mathrm{area}(f(E)) = \int_{E}|f_z|^2-|f_{\bar{z}}|^2\mathrm{d}x\, \mathrm{d}y$$ is valid.
There are several proofs on this based on Green's formula, for instance in Ahlfors book but I am interested in proving this without the use of line integrals. There are some notes by Chris Bishop which can be found here where he proves this result without the use of Green's formula.
The following is from Corollary 5.6 on p. 101 in Bishops notes. There are some parts of the proofs that I don't understand and would like help unraveling.
The procedure is the following. We fix a rectangle $Q$ and take a sequence $f_n$ obtained from $f$ via convolution with bump functions. The $f_n$ then converges uniformly to $f$. In the end he has the following statement:
This is what I have trouble understanding. Just because $f_n$ converges uniformly to $f$ is it necessarily so that if $K$ is a compact subset of the interior of $Q$ that $f_n(Q)$ eventually contains $f(K)$? I just don't see why this should be true. My thought goes to space filling curves where we can find sequences $f_n:[0,1]\rightarrow \mathbb{R}^2$ such that $\mathrm{area}(f_n[0,1]) = 0$ however $f([0,1]) = [0,1]\times [0,1]$. Of course in this example the domain is not in $\mathbb{R}^2$ and the map $f$ is not 1-1 however I think this requires some additional argument?Why does he need that the Jacobian is $L^p$ for some $p>1$? He only seem to use that $f_z$ and $f_{\bar{z}}$ are in $L^2$ when he applies the Hardy little maximal theorem.
I will be thankful for any help on this!
Suppose that $\Omega$ is a relatively compact domain in ${\mathbb R}^n$ and $f: cl\Omega\to {\mathbb R}^n$ is a continuous map. Then for $y\notin f(\partial \Omega)$ one defines the degree of $f$ at $y$, $deg(f; y)$, see for instance, here. The degree is stable under proper homotopy of maps. In particular, suppose that you have a sequence of maps $f_k: cl\Omega\to {\mathbb R}^n$ converging uniformly to $f: cl\Omega\to {\mathbb R}^n$. Then, assuming that $deg(f; y)$ is defined, we have $$ deg(f; y)= deg(f_k; y) $$ for all large $k$. In particular, if $f$ is an orientation-preserving homeomorphism to its image, then for $y\in f(\Omega)$, $deg(f; y)=+1$. Thus, for all large $k$, $deg(f_k; y)=1$, hence, $y\in f_k(\Omega)$.