[proofplan]
We derive the mean value property directly from the Cauchy Integral Formula. We parametrise the circle $\partial B(z_0, r)$ by $w = z_0 + re^{i\theta}$ for $\theta \in [0, 2\pi]$, substitute into the Cauchy formula evaluated at $z = z_0$, and simplify the resulting integral to obtain the claimed average.
[/proofplan]
[step:Apply the Cauchy Integral Formula at $z = z_0$]
Since $f \in \mathcal{O}(\Omega)$ and $\overline{B}(z_0, r) \subset \Omega$, the [Cauchy Integral Formula](/theorems/345) applies with centre $z_0$ and radius $r$. Evaluating at $z = z_0$:
\begin{align*}
f(z_0) = \frac{1}{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{w - z_0}\, dw,
\end{align*}
where $\partial B(z_0, r)$ is traversed counter-clockwise.
[guided]
The Cauchy Integral Formula states that if $f$ is holomorphic on an open set containing $\overline{B}(z_0, r)$, then for every $z \in B(z_0, r)$,
\begin{align*}
f(z) = \frac{1}{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{w - z}\, dw.
\end{align*}
We verify the hypothesis: $f \in \mathcal{O}(\Omega)$ and $\overline{B}(z_0, r) \subset \Omega$, so $f$ is holomorphic on the open set $\Omega$ containing $\overline{B}(z_0, r)$. In particular, $z_0 \in B(z_0, r)$, so we may set $z = z_0$ to obtain
\begin{align*}
f(z_0) = \frac{1}{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{w - z_0}\, dw.
\end{align*}
Why evaluate at the centre? Because the kernel $1/(w - z_0)$ simplifies dramatically when $w$ lies on the circle centred at $z_0$ -- the denominator has constant modulus $r$, which will allow us to extract a clean average in the next step.
[/guided]
[/step]
[step:Parametrise the contour and simplify to obtain the mean value]
Parametrise $\partial B(z_0, r)$ counter-clockwise by
\begin{align*}
\gamma: [0, 2\pi] &\to \mathbb{C} \\
\theta &\mapsto z_0 + re^{i\theta}.
\end{align*}
Then $\gamma'(\theta) = ire^{i\theta}$ and $\gamma(\theta) - z_0 = re^{i\theta}$, so
\begin{align*}
\frac{\gamma'(\theta)}{\gamma(\theta) - z_0} = \frac{ire^{i\theta}}{re^{i\theta}} = i.
\end{align*}
Substituting into the Cauchy integral:
\begin{align*}
f(z_0) &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z_0 + re^{i\theta})}{re^{i\theta}} \cdot ire^{i\theta}\, d\mathcal{L}^1(\theta) \\
&= \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + re^{i\theta}) \cdot i\, d\mathcal{L}^1(\theta) \\
&= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})\, d\mathcal{L}^1(\theta).
\end{align*}
This is the claimed mean value property.
[guided]
We parametrise the circle of integration. The standard counter-clockwise parametrisation of $\partial B(z_0, r)$ is
\begin{align*}
\gamma: [0, 2\pi] &\to \mathbb{C} \\
\theta &\mapsto z_0 + re^{i\theta},
\end{align*}
with derivative $\gamma'(\theta) = ire^{i\theta}$. The contour integral becomes a Lebesgue integral over $[0, 2\pi]$ via the substitution $w = \gamma(\theta)$, $dw = \gamma'(\theta)\, d\mathcal{L}^1(\theta)$:
\begin{align*}
f(z_0) &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(\gamma(\theta))}{\gamma(\theta) - z_0} \cdot \gamma'(\theta)\, d\mathcal{L}^1(\theta).
\end{align*}
Now we compute the ratio $\gamma'(\theta) / (\gamma(\theta) - z_0)$. Since $\gamma(\theta) - z_0 = re^{i\theta}$ and $\gamma'(\theta) = ire^{i\theta}$, the factor $re^{i\theta}$ cancels:
\begin{align*}
\frac{\gamma'(\theta)}{\gamma(\theta) - z_0} = \frac{ire^{i\theta}}{re^{i\theta}} = i.
\end{align*}
This cancellation is the key simplification that makes the mean value property work: because we evaluate the Cauchy formula at the centre $z_0$ of the circle, the kernel $1/(w - z_0)$ has constant modulus on the contour, and the phase cancels exactly with the phase of $dw$.
Substituting:
\begin{align*}
f(z_0) &= \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + re^{i\theta}) \cdot i\, d\mathcal{L}^1(\theta) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})\, d\mathcal{L}^1(\theta).
\end{align*}
This completes the proof: the value of $f$ at the centre $z_0$ equals the average of $f$ over the circle $\partial B(z_0, r)$.
[/guided]
[/step]
