Goursat's triangle theorem remains valid for a continuous function holomorphic away from one point

If $p\mathrm{\notin }T$, then $T\subseteq U\setminus \{p\}$; this punctured set is open because every $z\mathrm{\ne }p$ has the ball $B(z,\mathrm{\mid }z-p\mathrm{\mid }\mathrm{/}2)$ inside it, so [L1] applies. If $T$ is degenerate, order its distinct collinear vertices along their common line and split each directed edge at the intervening vertices; every resulting subsegment occurs equally often in both orientations, so [L2] gives zero.

It remains to treat a nondegenerate triangle $\mathrm{\Delta }[p,u,v]$ having $p$ as a vertex. For $0<t<1$, put $u_t=p+t(u-p)$ and $v_t=p+t(v-p)$. The triangles $\mathrm{\Delta }[u_t,u,v]$ and $\mathrm{\Delta }[u_t,v,v_t]$ lie in $\mathrm{\Delta }[p,u,v]$ and avoid $p$, so [L1] makes their boundary integrals zero; adding their boundaries and cancelling opposite internal edges by [L2] gives $I_f[p,u,v]=I_f[p,u_t,v_t]$.

Continuity at $p$ gives a neighborhood on which $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }f(p)\mathrm{\mid }+1$. For all sufficiently small $t>0$, the small triangle lies in that neighborhood, has perimeter $tP(\mathrm{\Delta }[p,u,v])$, and [L3] gives $\mathrm{\mid }{I}_f[p,u_t,v_t]\mathrm{\mid }\le (\mathrm{\mid }f(p)\mathrm{\mid }+1)tP(\mathrm{\Delta }[p,u,v])$.

Letting $t$ be arbitrarily small in steps 1.2 and 2.1 forces $I_f[p,u,v]=0$, so every triangle having $p$ as a vertex has zero boundary integral.

If $p\in T$, the three filled triangles $\mathrm{\Delta }[p,a,b]$, $\mathrm{\Delta }[p,b,c]$, and $\mathrm{\Delta }[p,c,a]$ cover $T$ with compatible orientations; their internal edges cancel by [L2], leaving $\mathrm{\partial }T$. Their integrals vanish by steps 1.1 and 3.1, so $I_f(T)=0$. Together with the case $p\mathrm{\notin }T$ from step 1.1, this proves the claim for every position of $p$.