I know the question may seem foolish to you but I am not quite sure how to show it in a decent way. My problem is to show that for bounded Borel measurable $f:\mathbb{D}^2\to\mathbb{R}$, (D1) is equivalent to (D2).
(D1) for all sphere $S(x,r)\subset \mathbb{D}^2$, where $S(x,r):=\{y \in \mathbb{D}^2|d(y,x)=r\}$, we have $$f(x)={1 \over {2\pi r}}\int_{S(x,r)}f(y)\ d\sigma(y)$$ Here $d\sigma$ is the surface measure on $S(x,r)$.
(D2) for all ball $B(x,r)\subset \mathbb{D}^2$, where $B(x,r):=\{y \in \mathbb{D}^2|d(y,x)\leq r\}$, we have $$f(x)={1 \over {\pi r^2}}\int_{B(x,r)}f(y)\ dy$$
I recognize that (D2) is equaivalent to the fact that $f$ is harmonic in $\mathbb{D}^2$, but I just have no idea of how to show the seemingly easier part as stated.
If (D1) holds for any $(x,r)$, we have \begin{align*} \frac 1{\pi r^2} \int_{B(x,r)} f(y) \, dy &= \frac 1{\pi r^2} \int_0^r \int_{S(x,s)} f(y) \, d\sigma(y)\, ds\\ &= \frac 1{\pi r^2} \int_0^r 2\pi s f(x)\, ds\\ &= \frac 2{r^2} f(x) \cdot \int_0^r s\, ds\\ &= f(x). \end{align*} On the other hand, if (D2) holds, we have \begin{align*} f(x) &= \frac 1{\pi r^2}\int_{B(x,r)} f(y)\, dy\\ &= \frac 1{\pi r^2} \int_0^r \int_{S(x,s)} f(y) \, d\sigma(y)\, ds\\ \end{align*} Differentiating with respect to $r$, we get $$ 0 = -\frac 2{\pi r^3} \int_0^r\int_{S(x,s)} f(y)\, d\sigma(y)\, ds + \frac 1{\pi r^2} \int_{S(x,r)} f(y)\, d\sigma(y). $$ Or $$ \frac 2{\pi r^3} \int_{B(x,r)} f(y)\, dy = \frac 1{\pi r^2}\int_{S(x,r)} f(y)\, d\sigma(y) \iff \frac 1{2\pi r}\int_{S(x,r)} f(y) \, d\sigma(y) = f(x). $$