Help with integrals in Gunning's book In Gunning's book Introduction to Holomorphic Functions of Severals Variables, Vol I, on pages 44 and 45 have some statements about integrals that I can not understand.
First, on page 44, equation $(3)$ we have the integral
$$\iint_{D_{r}} \frac{\partial f(\zeta)}{\partial \overline{\zeta}} \frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z},$$
where f is a $\mathcal{C}^{\infty}$ function in an open neighborhood of $\overline{D}$, $D$ an open subset of the complex plane bounded by a rectifiable simple curve $\gamma$, $z \in D$ and $D_{r} = D - \overline{\Delta}(z; r)$ for $r$ such that the disc is in $D$.
Gunning says that since $(\zeta - z)^{-1} d \overline{\zeta} \wedge d \zeta$ is a bounded measure in the plane, the integral above converges to the integral over $D$ as $r$ tends to zero. I can not understand $(\zeta - z)^{-1} d \overline{\zeta} \wedge d \zeta$ as a measure, much less bounded in the plane.
On page 45, we have the integral
$$g(z) = \frac{1}{2\pi i} \iint_{\mathbb{C}^{1}} \frac{f(\zeta)}{\zeta - z} d \zeta \wedge d \overline{\zeta},$$
where $f$ is a $\mathcal{C}^{\infty}$ function that vanishes outside a compact subset of $\mathbb{C}$.
Gunning says that this function is $\mathcal{C}^{\infty}$ in $\mathbb{C}^{1}$ (if $f$ is holomorphic, then $g$ is holomorphic). Why, if we have $(\zeta - z)$ in the denominator? I think this is related to the first question.
Perhaps I just need references to study more about integrals. Thanks.
 A: For fixed $z$, we write $\zeta-z=x+iy$. Then $d\zeta=dx+idy$ and $d\overline{\zeta}=dx-idy$. This implies that 
$$\frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z}=\frac{2id x \wedge d y}{x+iy}=\frac{2i(x-iy)d x \wedge d y}{x^2+y^2}.$$
Hence, in polar coordinates, we have $(x,y)=(\rho\cos\theta,\rho\sin\theta)$, and 
$$\frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z}=\frac{2i(\rho\cos\theta-i\rho\sin\theta)\rho\, d\rho \wedge d \theta}{\rho^2}=2i(\cos\theta-i\sin\theta)d\rho \wedge d \theta,$$
which is bounded. Since $f$ is a $C^\infty$ function in an open neighborhood of $\overline{D}$, $f$ is bounded on $\overline{D}$. Hence, 
 we can by conclude that (by dominated convergence theorem for example)
$$\int_{D_{r}} \frac{\partial f(\zeta)}{\partial \overline{\zeta}} \frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z}\rightarrow \int_{D} \frac{\partial f(\zeta)}{\partial \overline{\zeta}} \frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z}$$
since $D_{r} = D - \overline{\Delta}(z; r)\rightarrow D-\{z\}$ as $r\rightarrow 0$. 
Since $\displaystyle\frac{d \overline{\zeta} \wedge d \zeta}{\zeta - z}$ is bounded by the proof above, 
$$g(z) = \frac{1}{2\pi i} \iint_{\mathbb{C}^{1}} \frac{f(\zeta)}{\zeta - z} d \zeta \wedge d \overline{\zeta},$$
is well-defined since $f$ is a $C^\infty$ function that vanishes outside a compact subset of $\mathbb{C}$. By change of variable $\zeta\rightarrow\zeta+z$, we have
$$g(z) = \frac{1}{2\pi i} \iint_{\mathbb{C}^{1}} \frac{f(\zeta+z)}{\zeta} d \zeta \wedge d \overline{\zeta}.$$
Therefore, if $f$ is holomorphic, then $g$ is holomorphic.
