Vanishing integrals around triangles construct a primitive for a continuous function on a star-shaped domain

Fix $z\in U$. The segment ${\mathrm{\ell }}_{az}$ is rectifiable and $f$ is continuous on it, so $F(z)$ exists by [L5]. Since $U$ is open, choose $\rho >0$ with $B(z,\rho )\subseteq U$. If $0<\mathrm{\mid }h\mathrm{\mid }<\rho$, the short segment from $z$ to $z+h$ lies in that ball, and every segment from $a$ to a point of the short segment lies in $U$ by star-shapedness; hence $\mathrm{\Delta }[a,z,z+h]\subseteq U$.

Parametrizing the short edge by $\zeta =z+th$ and using [L2] gives ${\int }_{{\mathrm{\ell }}_{z,z+h}}f(z)d\zeta =f(z)h$.

The zero boundary integral of that triangle reads $F(z)+{\int }_{{\mathrm{\ell }}_{z,z+h}}f-F(z+h)=0$ by [L1], and therefore $F(z+h)-F(z)={\int }_{{\mathrm{\ell }}_{z,z+h}}f(\zeta )d\zeta$.

By [L3], steps 2.1 and 1.2 imply $\mid (F(z+h)-F(z))\mathrm{/}h-f(z)\mid \le {\sup }_{0\le t\le 1}\mathrm{\mid }f(z+th)-f(z)\mathrm{\mid }$.

Continuity of $f$ at $z$ makes the right side of step 3.1 tend to zero as $h\to 0$. Thus $F^{\mathrm{\prime }}(z)=f(z)$ for arbitrary $z\in U$, including $z=a$; [L4] says exactly that $F$ is a primitive, and $h=0$ was only the excluded difference-quotient value.