Let $U \subset \Bbb C$ be a star domain and $f:U \to \Bbb C$ continuous, s.t. $$\int_{\partial \bigtriangleup}f(z)dz=0$$ for all triangles $\bigtriangleup$ which are in $U$. It is to be shown that $f$ has a primitive.
Let $z_0 \in U$ be a center of the star domain, so for $z \in U$ the segment between $z_0$ and $z$ is in $U$. We denote the segment between $z_0$ and $z$ with $[z_0,z]$. Let $$F(z):=\int_{[z_0,z]}f(w)dw$$ Now $U$ is open and so it exists $r>0$ s.t. $B(z,r) \subset U$. If we take $\vert u \vert <r$, then the segment $[z,z+u]$ lies in $U$. Because the integral over a triangle is $0$ it follows for the triangle $[z_0,z,z+u]$ that $$F(z+u)-F(z)=\int_{[z,z+u]}f(w)dw.$$ But now how can I conclude the proof? I should show that $$\lim_{u \to 0}\frac{F(z+u)-F(z)}{u}=f(z)$$ But I got stuck...
With the parametrization $[0, 1] \ni t \mapsto z +tu$ of the segment $[z, z+u]$ we get $$ F(z+u)-F(z)= u \int_0^1 f(z +t u) \, dt $$ and therefore $$ \frac{F(z+u)-F(z)}{u} - f(z) = \int_0^1 \bigl(f(z +t u) -f(z) \bigr) \, dt \, . $$ $f$ is continuous at $z$, so given $\epsilon > 0$ there is a $\delta > 0$ such that $|f(w) - f(z)| < \epsilon $ if $|w-z| < \delta$. It follows that if $|u| < \delta$ then $$ \left|\frac{F(z+u)-F(z)}{u} - f(z) \right| \le \int_0^1 \underbrace{|f(z +t u) -f(z) |}_{< \epsilon} \, dt < \epsilon $$ and that proves the desired limit.