= Solution
For two moving vertices $p(s),q(s)$, the <integral> along their joining segment is
$$
J_{p,q}(s)=\bigl(q(s)-p(s)\bigr)\int_0^1 f\bigl((1-t)p(s)+tq(s)\bigr)\,dt.
$$
The <integral> factor is <continuous> in $s$ by the allowed fact, and the other factor is <continuous> because the vertices are <continuous>. Thus $J_{p,q}$ is <continuous>, including when its two endpoints coincide.
The <integral> around the <triangle> is exactly
$$
I(s)=J_{a,b}(s)+J_{b,c}(s)+J_{c,a}(s).
$$
Each term is <continuous>, so their sum is <continuous> at zero and
$$
\boxed{\lim_{s\to0}\int_{\Delta_s}f(z)\,dz=\int_{\Delta_0}f(z)\,dz.}
$$
No holomorphicity is used here. <Compactness> of the fixed containing <triangle> also gives a direct proof of the allowed fact through <uniform continuity> of $f$.
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