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  "body": "\n## Proof\n\n1. Let $U$, $f_n$ and $f$ be as in the statement. Each $f_n$ is continuous, since a complex differentiable function is continuous at each point.\n2. Claim: $f$ is continuous. Let $z \\in U$ and pick $s \u003e 0$ with the closed disc $D = \\set{w : \\abs{w - z} \\le s} \\subseteq U$. It is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n3. Let $\\varepsilon \u003e 0$. By uniform convergence on $D$, choose $n$ with $\\abs{f_n - f} \u003c \\varepsilon/3$ on $D$. Choose $\\delta \\in (0, s)$ with $\\abs{f_n(w) - f_n(z)} \u003c \\varepsilon/3$ for $\\abs{w - z} \u003c \\delta$.\n4. For $\\abs{w - z} \u003c \\delta$, $\\abs{f(w) - f(z)} \\le \\abs{f(w) - f_n(w)} + \\abs{f_n(w) - f_n(z)} + \\abs{f_n(z) - f(z)} \u003c \\varepsilon$. This proves the claim.\n5. Let $B \\subseteq U$ be an open disc, and let $T \\subseteq B$ be a closed triangle with boundary path $\\partial T$ of length $L$.\n6. By [[holomorphic functions on a disc have primitives|02AC7JR6G6]], each $f_n$ has a primitive on $B$. By the [[fundamental theorem of contour integration|0YQ458D7PF]], $\\oint_{\\partial T} f_n(z) \\, dz = 0$.\n7. The triangle $T$ is closed and bounded, hence compact. Let $\\varepsilon_n = \\sup_{T} \\abs{f_n - f}$. Then $\\varepsilon_n \\to 0$.\n8. By step 6 and the [[estimation lemma|0DJMGPN8K9]], $\\abs{\\oint_{\\partial T} f} = \\abs{\\oint_{\\partial T} (f - f_n)} \\le \\varepsilon_n L$ for each $n$. So $\\oint_{\\partial T} f(z) \\, dz = 0$.\n9. By steps 4 and 8 and [[Morera's theorem|0VDD7W0EKZ]] on $B$, $f$ is [[holomorphic|0DG0Z72CQD]] on $B$.\n10. Each point of $U$ lies in an open disc $B \\subseteq U$. So $f$ is holomorphic on $U$.\n11. Now let $K \\subseteq U$ be compact. If $K$ is empty, there is nothing to prove, so let $K$ be nonempty. Claim: there is $\\delta \u003e 0$ such that $K' = \\set{w : \\abs{w - z} \\le \\delta \\text{ for some } z \\in K}$ is a compact subset of $U$.\n12. If $U = \\C$, take $\\delta = 1$. Otherwise let $d(z) = \\inf \\set{\\abs{z - w} : w \\notin U}$. Then $\\abs{d(z) - d(z')} \\le \\abs{z - z'}$, so $d$ is continuous.\n13. Since $U$ is open, $d \u003e 0$ on $K$. By [[attainment of bounds|0V4DET59NW]], $d$ has a positive minimum on $K$. Let $\\delta$ be half of it. Then $K' \\subseteq U$.\n14. The set $K'$ is bounded since $K$ is. It is closed: if $w_j \\in K'$ tends to $w$, pick $z_j \\in K$ with $\\abs{w_j - z_j} \\le \\delta$.\n15. By [[sequential compactness|0K8SGQSV7G]] of $K$, a subsequence of $(z_j)$ tends to some $z \\in K$. Then $\\abs{w - z} \\le \\delta$, so $w \\in K'$. By Heine–Borel, $K'$ is compact, which proves the claim.\n16. Let $\\eta_n = \\sup_{K'} \\abs{f_n - f}$. By uniform convergence on the compact set $K'$, $\\eta_n \\to 0$.\n17. Let $z \\in K$. The closed disc of radius $\\delta$ around $z$ lies in $K'$, and $f_n - f$ is holomorphic on $U$ by step 10.\n18. By [[Cauchy's estimates|0JJRENWMP6]] with $n = 1$ for $f_n - f$ on this disc, $\\abs{f_n'(z) - f'(z)} \\le \\eta_n / \\delta$.\n19. The bound in step 18 does not depend on $z \\in K$ and tends to $0$. So $f_n' \\to f'$ uniformly on $K$.\n\n## Idea\n\nIntegrals over triangles pass to the limit under uniform convergence. Each $f_n$ has zero integral over small triangles, so $f$ does too, and Morera's theorem makes $f$ holomorphic. For the derivatives, Cauchy's estimates bound $f_n' - f'$ at a point by the size of $f_n - f$ on a nearby circle.\n",
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