Suppose $f \colon \mathbb{C} \to \mathbb{C}$ is holomorphic and $B = B(0,R)$ is the ball of radius $R > 0$ centered at $0$. Identifying $\mathbb{C}$ with $\mathbb{R}^2$ and using $\lambda$ to denote the Lebesgue measure on $\mathbb{R}^2$, we can integrate $f$ over $B$ in polar coordinates as $$\int_B f(z) \, d\lambda(z) = \int_0^R r\int_{S^1}f(r\xi) \, d\sigma(\xi) \, dr,$$ where $\sigma$ is the arclength measure on the unit circle $S^1$. My question is if the inner integral is the same as the integral below whose value is $2\pi f(0)$ by the mean value property for holomorphic functions: $$\int_0^{2\pi}f(re^{i\theta}) \, d\theta = 2\pi f(0).$$ Here I believe the integral is supposed to be taken with respect to the Lesbesgue measure on $[0,2\pi)$, which is equivalent to arclength measure on $S^1$ if we identify $[0,2\pi)$ with $S^1$ via $\theta \to e^{i\theta}$. So it seems to me that the only difference between the two integrals is notation.

  • $\begingroup$ The inner integral is $r$ times the integral you wrote because the arclength measure is $r d\theta$. $\endgroup$ – Ian Feb 13 at 10:44

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