Help understanding the proof of Morera's Theorem Theorem:  If $f$ is defined and continuous on all of $\mathbb{C}$ and for any smooth closed curve $\mathbb{C}$ we have $$\int_C f(z)dz=0,$$ then $f$ is entire.
Proof (attempted): Let $a\in\mathbb{C}$.  Define $f$ by $$F(z)=\int_a^z f(z)dz.$$  This function is well defined since for any two paths $A$ and $B$ connecting $a$ to $z$ we have $\int_{A+B} f(z)dz=0$ and thus $\int_A f(z)dz=\int_B f(z)dz$.
Now I am stuck.  I want to show that $F'(z)=f(z)$.  The only way I can think to go is by definition of derivative.  But I just can't get it.
 A: $$ \lim_{w \to 0} \dfrac{F(z+w) - F(z)}{w} - f(z) = \lim_{w \to 0} \dfrac{1}{w} \int_z^{z+w} (f(\zeta) - f(z))\; d\zeta$$
Now by continuity of $f$ at $z$, for any $\epsilon > 0$ there is $\delta > 0$ such that $|f(\zeta) - f(z)| < \epsilon$ when $|\zeta - z| < \delta$.  Thus when
$|w| < \delta$ we have 
$$\left|\int_{z}^{z+w} (f(\zeta) - f(z))\; d\zeta \right| \le |w| \epsilon$$
Since this is true for every $\epsilon$, 
$$\lim_{w \to 0} \dfrac{1}{w} \int_z^{z+w} (f(\zeta) - f(z))\; d\zeta = 0$$
A: If you take $\gamma(t)$ to end in a vertical segment, an explicit parameterization yields ${\partial_y F(x + iy)} = i f(x + iy)$. If you take $\gamma(t)$ to end in a horizontal segment, then an explicit parameterization yields ${\partial_x F(x + iy)} = f(x + iy)$. 
Hence one has ${\partial_y F(x + iy)} = i{\partial_x F(x + iy)}$.
Thus writing $F(x + iy) = U(x + iy) + i V(x + iy)$  you have
$$ {\partial U \over \partial y} (x + iy) + i {\partial V \over \partial y}(x + iy) = 
 i {\partial U \over \partial x} (x + iy) - {\partial V \over \partial x}(x + iy) $$
The real and imaginary parts of this are exactly the Cauchy-Riemann equations, so $F$ is analytic. And the formula ${\partial_x F(x + iy)} = f(x + iy)$ is equivalent to saying $F'(z) = f(z)$.
A: We have $F(z+h)-F(z) = \int_\gamma f(w) dw $, where $\gamma$ is any path
from $z$ to $z+h$.
Pick the path $\gamma(t) = z + th$, $t \in [0,1]$.
Then ${F(z+h)-F(z) \over h}-f(z) = \int_0^1 (f(z+th)-f(z)) dt$.
Let $\epsilon>0$, then there is some $\delta>0$ such that if $|h|< \delta$ then
$|f(z+h)-f(z)| < \epsilon$.
Hence if $|h| < \delta$ then
$|{F(z+h)-F(z) \over h}-f(z)| \le \int_0^1 |f(z+th)-f(z)| dt < \epsilon$, and
so we see that $F$ is differentiable and $F'(z) = f(z)$.
