Integrating holomorphic function with respect to one component yields holomorphic function

The following statement and proof is taken from Complex Geometry\textit{Complex Geometry} (Huybrechts).

Let U⊂CnU \subset \mathbb{C}^{n} be an open subset and let V⊂CV \subset \mathbb{C} be an open neighbourhood of the boundary of Bε(0)⊂CB_{\varepsilon}(0)\subset \mathbb{C}. Assume that f:V×U→Cf\colon V\times U \to \mathbb{C} is a holomorphic function. Then
g(z):=g(z_1,\ldots, z_n):= \int_{\vert \xi \vert = \varepsilon} f(\xi, z_1,\ldots,z_n) d\xi

is a holomorphic function on UU.

Proof.\textit{Proof}. Let z∈Uz\in U. If |ξ|=ε\vert \xi \vert = \varepsilon then there exists a polydisc Bδ(ξ)(ξ)×Bδ′(ξ)(z)⊂V×UB_{\delta(\xi)}(\xi) \times B_{\delta'(\xi)}(z) \subset V\times U on which ff has a power series expansion.
Since ∂Bε(0)\partial{B_{\varepsilon}}(0) is compact we can find a finite number of points ξ1,…,ξk∈∂Bε(0)\xi_1,\ldots,\xi_k\in \partial{B_{\varepsilon}(0)} and positive real numbers δ(ξ1),…,δ(ξk)\delta(\xi_1),\ldots, \delta(\xi_k) such that
⋃∂Bε(0)∩Bδ(ξi)/2(ξi) is a disjoint union
\bigcup \partial{B_{\varepsilon}}(0) \cap B_{\delta(\xi_i)/2}(\xi_i) \text{ is a disjoint union}

\partial{B_{\varepsilon}(0)} = \bigcup \partial{B_{\varepsilon}(0)} \cap \overline{B_{\delta(\xi_i)/2}(\xi_i)}.

Hence, g(z)=∫|ξ|=εf(ξ,z1,…,zn)dξ=∑ki=1∫|ξ|=ε,|ξi−ξ|<δ(ξi)/2fdξg(z)=\int_{\vert \xi \vert = \varepsilon}f(\xi,z_1,\ldots,z_n)d\xi = \sum_{i=1}^{k} \int_{\vert \xi \vert = \varepsilon, \vert \xi_i-\xi \vert < \delta(\xi_i)/2} f d\xi. It clear to me that the balls Bδ(ξ)(ξ)B_{\delta(\xi)}(\xi) cover the boundary of Bε(0)B_{\varepsilon}(0), and so finitely many suffice to cover it. However, I do not see how we can choose them such that the two written properties hold. Moreover, assuming we found a finite number of points having those two properties, I am wondering why we have ∫|ξ|=εf(ξ,z1,…,zn)dξ=k∑i=1∫|ξ|=ε,|ξi−ξ|<δ(ξi)/2fdξ$ \int_{\vert \xi \vert = \varepsilon}f(\xi,z_1,\ldots,z_n)d\xi = \sum_{i=1}^{k} \int_{\vert \xi \vert = \varepsilon, \vert \xi_i-\xi \vert < \delta(\xi_i)/2} f d\xi$ Set D:=⋃∂Bε(0)∩Bδ(ξi)/2(ξi)D:=\bigcup \partial{B_{\varepsilon}}(0) \cap B_{\delta(\xi_i)/2}(\xi_i) and let ¯D\overline{D} denote the closure of DD in ∂Bε(0)\partial{B_{\varepsilon}(0)}, then we have ¯D=∂Bε(0)\overline{D}= \partial{B_{\varepsilon}(0)} by assumption. Then ∑ki=1∫|ξ|=ε,|ξi−ξ|<δ(ξi)/2fdξ=∫Dfdξ\sum_{i=1}^{k} \int_{\vert \xi \vert = \varepsilon, \vert \xi_i-\xi \vert < \delta(\xi_i)/2} f d\xi = \int_{D}fd\xi because the union is disjoint. How is this equal to the ∫¯Dfdξ\int_{\overline{D}}f d\xi? ================= ================= =================