Problem about some corollaries of the Moreraˊs theorem As I was studying complex analysis,I met a problem about some corollaries of the Cauchy integral formula,including the Moreraˊs theorem.
Here is the problem:
Let $D$ be the open connected region of $\mathbb{C}$,$f$ is continuous within $D$,if for any arched region $G \cup \partial G \subset D$,we have $\int_{\partial G} f(z) dz =0$,then $f \in H(D)$,which means that $f$ is holomorphic in$D$.Is the conclusion still hold if we replace the arched region by disc?
For the first part,we first consider a triangle $\Delta ABC$ that is small enough to allow its circumscribed circle $B(z_0,\epsilon) \subset D$,since $\int_{\overline{AC}}f(z)dz + \int_{\widehat{CBA}}f(z)dz = 0$ and $\int_{\widehat{CBA}}f(z)dz = \int_{\widehat{CB}}f(z)dz + \int_{\widehat{BA}}f(z)dz = \int_{\overline{CB}}f(z)dz + \int_{\overline{BA}}f(z)dz=0$ , we can easily deduce that $\int_{\partial \Delta ABC }f(z)dz=0$.For any triangle in $D$,use line paralleled to sides of the triangle to divide it into many small triangles such that their circumscribed circles in $D$.Hence the integral of $f$ on the boundary of the triangle is zero.Consequently $f$ is holomorphic in $D$.
But I couldn´t go any further when the arched region is replaced by disc.    
 A: First assume that $f$ is real differentiable,then for any $z_0 \in D$ there exists $B(z_0,\epsilon) \subset D$.Note that
$$f(z)=f(z_0) + \frac{\partial f}{\partial z} (z_0)(z-z_0)+\frac{\partial f}{\partial \overline{z_0}}(z_0) \overline{(z-z_0)} + o(|z-z_0|)$$
Moreover
$$\begin{align*}
0 &= \int_{\partial {B(z_0,\epsilon)}}f(z)dz\\
&= \int_{0}^{2\pi}(f(z_0) + \frac{\partial f}{\partial z}(z_0)\epsilon e^{i \theta} + \frac{\partial f}{\partial \overline{z_0}}(z_0)\epsilon e^{-i \theta} + o(\epsilon))i\epsilon e^{i \theta}d\theta\\
&= \int_{0}^{2\pi}(\frac{\partial f}{\partial \overline{z}}(z_0) + \frac{o(\epsilon)}{\epsilon}e^{i\theta})d\theta
\end{align*}
$$
Let $\epsilon \to 0$ and $\frac{\partial f}{\partial \overline{z}}(z_0)=0$, hence $f$ holomorphic.
Generally,define $g(w)=g(x,y)(w=x+iy)(supp g \subset \subset D)$ a smooth function on $\mathbb{R^2}$ such that $0 \le g \le 1$ and $\int_{\mathbb{R^2}}g(w)dV(w)=1$,where $dV(w)=dxdy$.
For any $\epsilon \gt 0$,let $g_{\epsilon}(z)=\epsilon^2 g(\frac{z}{\epsilon})$ and $f_{\epsilon}(z)=g_{\epsilon} * f(z)$.Note that $f_{\epsilon}$ is real differential and
$$
\begin{align*}
\frac{\partial{f_{\epsilon}}}{\partial z}(z_0)&=g_{\epsilon}*\frac{\partial f}{\partial z}(z_0),\\\frac{\partial{f_{\epsilon}}}{\partial{\overline{z}}}(z_0)&=g_{\epsilon}*\frac{\partial f}{\partial \overline{z}}(z_0)
\end{align*}
$$
On one hand we have $$\int_{\partial B(z_0,\epsilon)}f_{\epsilon}(z)dz=i\epsilon^2 \int_{0}^{2\pi}(\frac{\partial{f_{\epsilon}}}{\partial \overline{z}}(z_0)+\frac{o(\epsilon)}{\epsilon}e^{i\theta})d\theta$$.
On the other hand we have
$$\int_{\partial B(z_0,\epsilon)}f_{\epsilon}(z)dz = \int_{\mathbb{R^2}}g_{\epsilon}(w)(\int_{\partial B(z_0,\epsilon)}(f(z_0) + \frac{\partial f}{\partial z}(z_0)(z-w-z_0) + \frac{\partial f}{\partial \overline{z}}(z_0)\overline{(z-w-z_0)}+o(|z-w-z_0|))dV(w)=0$$.
Combine the two equations we have $\frac{\partial f_{\epsilon}}{\partial \overline{z}}(z_0)=0$,hence $f_{\epsilon} \in H(D)$.
Next we prove that $f_{\epsilon}$ uniformly converges to $f$ on any compact set $\Omega \subset D$.Actually for any $z \in \Omega \subset \subset D$,$$|f_{\epsilon}(z)-f(z)|
=|\int_{supp g_{\epsilon}}\epsilon^2g(\frac{x}{\epsilon},\frac{y}{\epsilon})(f(z-w)-f(z))dV(w)|=|\int_{supp g}g(w)(f(z-w\epsilon)-f(z)dV(w)| \le sup_{w \in suppg}|f(z-w\epsilon)-f(z)| \int_{suppg}g(w)dV(w) = sup_{w \in suppg}|f(z-w\epsilon)-f(z)| \to 0(\epsilon \to 0)$$.
Since a sequence of holomorphic functions $\{ f_{\epsilon} \}_{\epsilon \gt 0}$ is a internally closed uniform convergence and $f_{\epsilon} \to f(\epsilon \to 0)$,consequently $f \in H(D)$.
