Let $u$ be a continuous function on an open set $U$ of the complex plane. We say that $u$ satisfies the circle mean value property at a point $z_0\in U$ if $$ u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta$$
for all $r$ sufficiently small such that the disc centered at $z_0$ with radius $r$> is contained in $U$. We say that $u$ satisfies the disc mean value property at a point $z_0$ if $$u(z_0)=\frac{1}{\pi r^2}\iint_{D(z_0,r)}u dxdy$$
I think the two properties are related. In particular i'd like to show that the first implies the second. Is this an application of Green's Thm maybe?
The circle mean value property indeed implies the disc mean value property. To see this suppose without loss of generality that $z_0=0$ and introduce polar coordinates $$\begin{cases} x=\rho \cos \theta \\ y=\rho \sin \theta \end{cases} $$ We have for the area element the formula $$dxdy=\rho d\rho d\theta ,$$ meaning that we can rewrite an integral on the disc as a superposition of integrals over circles: \begin{equation} \begin{split} \frac{1}{\pi r^2} \iint_{D(r)}u(x+iy)\, dxdy&= \frac{1}{\pi r^2}\int_0^r\rho\, d\rho \int_0^{2\pi} u(\rho e^{i\theta})\, d\theta\\ &=\frac{2}{ r^2}\int_0^r \rho\,d\rho u(0)\\&=u(0). \end{split} \end{equation}