# Lebesgue integral and contour integration

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.

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