Lusin Areal Integral Formula

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be an analytic and $1-1$ function and let $\mathbb{D}$ be the open unitary disk. Prove that:

$$\iint \limits_{\mathbb{D}} \left |f'(z) \right| \, {\rm d}z = f \left(\mathbb{D} \right)$$

Solution

 Let $f=u+ iv$. Then the Jacobian of $f$ is the real mapping from $\mathbb{R}^2$ to $\mathbb{R}^2$ that is defined to be:

$$\mathcal{J}= \det \begin{pmatrix}
u_x & u_y\\
 v_x& v_y
\end{pmatrix}$$

Since $f$ is analytic then invoking the Cauchy - Riemann Equations we have that:

$$\det \begin{pmatrix}
u_x &u_y \\
 v_x& v_y
\end{pmatrix} = \det \begin{pmatrix}
u_x &-v_x \\
 v_x& u_x
\end{pmatrix}= u_x^2 +v_x^2 = \left | f' \right |^2$$

Hence:

$$\iint \limits_{\Omega_1} \left | f'(z) \right |^2 \, {\rm d}z = \iint \limits_{\Omega_1} \mathcal{J} \, {\rm d}A = \iint \limits _{\Omega_2} 1 \, {\rm d}A = f \left ( \mathbb{D} \right )$$