Context: Suppose $w(z,\bar{z})\in C^1$ and $D$ a simply connected region bounded by a sufficiently smooth curve $\Gamma$. Note that $w$ is complex but is not required to be analytic. Then,
$$
\boxed{\oint_{\Gamma}dz\,w = i\int_Dd^2z\,\partial_{\bar{z}}w}\qquad(1)
$$
which can be demonstrated, e.g., by writing everything in real coordinates, $z=x+iy$, $w(z,\bar{z})=u(x,y)+iv(x,y)$, $d^2z=2dxdy$ and $\partial_{\bar{z}}=\frac{1}{2}(\partial_x+i\partial_y)$, and then equating real and imaginary parts. The resulting equations are precisely Green's Theorem,
$$
\oint_{\Gamma}\big(dx \,v+dy\,u\big) = \int_Ddxdy\big(\partial_xu-\partial_yv\big)
$$
which we take for granted. Eqn (1) makes the correspondence with Cauchy's theorem manifest (to which it reduces when $w=w(z)$ is analytic).
Show under the same hypotheses leading to (1) and with $z$ a fixed point in $D$ that:
$$
\boxed{w(z,\bar{z}) = \oint_{\Gamma}\frac{d\zeta}{2\pi i}\frac{w(\zeta,\bar{\zeta})}{\zeta-z}-\frac{1}{2\pi}\int_Dd^2\zeta\,\frac{\partial_{\bar{\zeta}}w(\zeta,\bar{\zeta})}{\zeta-z}}\qquad(2)
$$
Following L. Bers, Riemann Surfaces, 1957-1958, Lect. 3, to prove this apply (1) to the function
$$
W(\zeta,\bar{\zeta}):= \frac{1}{2\pi i}\frac{w(\zeta,\bar{\zeta})}{\zeta-z},
$$
choose to integrate over a region $D-D'$ bounded by $\Gamma+\Gamma'$ as shown in the figure (the standard keyhole),
and take $D'$ centred at $z\in D'$. The resulting relation is:
$$
-\oint\limits_{\Gamma'}\frac{d\zeta}{2\pi i}\frac{w(\zeta,\bar{\zeta})}{\zeta-z}=\oint\limits_{\Gamma}\frac{d\zeta}{2\pi i}\frac{w(\zeta,\bar{\zeta})}{\zeta-z}-\frac{1}{2\pi}\int\limits_{D-D'}d^2\zeta\,\frac{\partial_{\bar{\zeta}}w(\zeta,\bar{\zeta})}{\zeta-z}.\qquad(3)
$$
Bers states that the integral on the LHS in (3) reduces to $w(z,\bar{z})$ when we take the radius of $D'$ (call it $r=|\zeta-z|$) to zero, and that this can be shown by estimating the difference between this integral and $w(z,\bar{z})$ and taking $r\rightarrow 0$. The result (2) follows.
My questions:
(a) Given $w(\zeta,\bar{\zeta})$ is only $C^1$ (and not infinitely differentiable) how can we show that the LHS of (3) tends to $w(z,\bar{z})$ as $r\rightarrow 0$? E.g., we cannot use Taylor series, so what is the estimation procedure in these cases?
(update: I suppose that as $r\rightarrow 0$, $w(\zeta,\bar{\zeta})\rightarrow w(z,\bar{z})$ by definition, following from the fact that it is continuous in the domain containing $D$. Is this reasoning correct?)
(b) How to show that the integral over $D-D'$ in (3) reduces to the integral over $D$ shown in (2) when we take $r\rightarrow 0$? For example, is it clear there is no obstruction due to the apparent pole at $\zeta=z$?