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I am reading Gunning's Vol. I on holomorphic functions of several variables and am confused by his proof of the maximum modulus principle (Theorem A.4), which assumes the following fact.

Let $V$ be the volume measure on $\mathbb{R}^{2n}$ and hence on $\mathbb{C}^n$. Suppose $\Delta$ is centered at $a \in \mathbb{C}^n$, and $f$ is a function holomorphic on a neighborhood of $\overline{\Delta}$. Then $$ f (a) =\frac{1}{V(\Delta)} \int_{\Delta}{f({\zeta})dV(\zeta)}.$$ That is, $f (a)$ is an average of the values on a polydisc centered at $a$.

This question also shows up as an exercise in Jiri Lebl's book http://www.jirka.org/scv/scv.pdf as Exercise 1.2.7 (in fact, the above is quoted directly from Lebl).

I have found several responses to a similar question in one variable, but they only show that $$f(a) = \frac{1}{2\pi}\int_{|\zeta|=r}{f(\zeta)d\zeta}$$ which is clear to me as a direct application of Cauchy's integral formula. I am having trouble concluding from this that $$f(a) = \frac{1}{V(\Delta)}\int_{\Delta}{f(\zeta)dV(\zeta)}$$ even in the one dimensional case.

Any help?

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I was able to generalize the one-dimensional case. The multi-dimensional version of Cauchy's theorem is:

$$ f(a) = \frac{1}{(2\pi i)^n} \int_{|\zeta_j-a_j|=r_j} \frac{f(\zeta_1,\dots,\zeta_n)}{(\zeta_1-a_1)\dots(\zeta_n-a_n)}d\zeta_1\dots d\zeta_n. $$

Letting $\zeta_j=a_j+r_je^{i\theta_j}$, we get the iterated integral $$ f(a) = \frac{1}{(2\pi)^n} \int_{\zeta_n=0}^{2\pi} \dots \int_{\zeta_1=0}^{2\pi} f(a_1+r_1e^{i\theta_1},\dots,a_n+r_ne^{i\theta_n}) d\theta_1\dots d\theta_n. (1) $$

If the volume element $dV = dx_1dy_1 \dots dx_ndy_n$, we use the change-of-variables formulae $x_j=r_j\cos{\theta_j}$ and $y_j=r_j\sin{\theta_j}$ to get $dV = r_1 \dots r_n dr_1 d\theta_1 \dots dr_n d\theta_n$. This is because the resulting change-of-variables matrix is block diagonal with with $2\times2$ blocks, so its determinant is just the product of the determinants of the blocks.

Integrating both sides of $(1)$ with respect to $r_1 \dots r_n dr_1 \dots dr_n$ gives the result.

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