Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions

For the vanishing-integrals-to-holomorphy direction, fix $a\in \mathrm{\Omega }$ and choose $r>0$ with $\bar{D}(a,r)\subseteq \mathrm{\Omega }$; the disc $D(a,r)$ is open and contains $a$. For $z$ in it and $0\le t\le 1$, $\mathrm{\mid }a+t(z-a)-a\mathrm{\mid }=t\mathrm{\mid }z-a\mathrm{\mid }<r$; openness and segment transport in The complex coordinate isometry transports topology and affine segments show its coordinate image is star-shaped at $\mathrm{\Phi }(a)$ in Star-shaped open subsets of Euclidean space, and every filled triangle in it is among the triangles covered by the assumed condition and [L1].

For the holomorphy-to-vanishing-integrals direction, if $f$ is holomorphic on $\mathrm{\Omega }$, [L4] gives zero integral around every filled triangle of [L1] contained in $\mathrm{\Omega }$, including those with repeated or collinear vertices.

For the vanishing-integrals-to-holomorphy direction, [L2] applied on $D(a,r)$ supplies a holomorphic function $F$ there with $F^{\mathrm{\prime }}=f$.

For the vanishing-integrals-to-holomorphy direction, [L3] makes the derivative $F^{\mathrm{\prime }}$ holomorphic, so $f=F^{\mathrm{\prime }}$ is holomorphic on $D(a,r)$.

If $\mathrm{\Omega }$ is nonempty, the point in step 1.1 was arbitrary, so step 3.1 proves holomorphy throughout $\mathrm{\Omega }$ under the integral condition, while step 1.2 proves the converse; if $\mathrm{\Omega }$ is empty, both directions are vacuous.