TITLE: Relation between circle mean value property and disk mean value property
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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}