The Koch curve is a uniform limit of polygonal paths of lengths $(4\mathrm{/}3{)}^n$ but is not rectifiable

Define $R(x,y):=(x\mathrm{/}2-\sqrt{3}y\mathrm{/}2,\sqrt{3}x\mathrm{/}2+y\mathrm{/}2)$. Since $(\sqrt{3}{)}^2=3$, direct expansion gives $\parallel Rv{\parallel }_2=\parallel v{\parallel }_2$ and $\parallel v-Rv{\parallel }_2=\parallel v{\parallel }_2$ for every $v\in {\mathbb{R}}^2$.

Put ${\kappa }_0(t)=(t,0)$. Recursively, suppose ${\kappa }_n$ is affine between consecutive points of ${\sigma }_n:=\{j\mathrm{/}{4}^n:0\le j\le 4^n\}$. For an old edge from $A={\kappa }_n(j\mathrm{/}{4}^n)$ to $A+v={\kappa }_n((j+1)\mathrm{/}{4}^n)$, prescribe the five successive values of ${\kappa }_{n+1}$ at parameters $(4j+r)\mathrm{/}{4}^{n+1}$, $0\le r\le 4$, to be $A$, $A+v\mathrm{/}3$, $A+v\mathrm{/}3+Rv\mathrm{/}3$, $A+2v\mathrm{/}3$, and $A+v$, and make ${\kappa }_{n+1}$ affine between them. The endpoint prescriptions agree on adjacent old edges, so [L1] gives the sequence.

Every old vertex is retained. By step 1.1, each old edge of length $\mathrm{\ell }$ is replaced by four edges of length $\mathrm{\ell }\mathrm{/}3$. Induction therefore gives $4^n$ edges of length $3^{-n}$ in ${\kappa }_n$, and [L5] gives $L({\kappa }_n)=4^n{3}^{-n}=(4\mathrm{/}3{)}^n$.

On an old edge of length $3^{-n}$, compare ${\kappa }_{n+1}$ with the affine chord ${\kappa }_n$. At the five subdivision parameters their differences have norms at most $0,3^{-n}\mathrm{/}12,3^{-n}\mathrm{/}2,3^{-n}\mathrm{/}12,0$. On each intervening interval the difference is affine, so the triangle inequality gives ${\sup }_t\parallel {\kappa }_{n+1}(t)-{\kappa }_n(t){\parallel }_2\le 1\mathrm{/}(2\cdot 3^n)$. Consequently, for $m>n$, telescoping and the finite geometric-sum identity give ${\sup }_t\parallel {\kappa }_m(t)-{\kappa }_n(t){\parallel }_2<3\mathrm{/}(4\cdot 3^n)$.

Each coordinate sequence is uniformly Cauchy by step 4.1, so [L3] supplies uniform coordinate limits. Let $\kappa$ be the resulting vector-valued limit. The inequality $\parallel z{\parallel }_2\le \mathrm{\mid }{z}_1\mathrm{\mid }+\mathrm{\mid }{z}_2\mathrm{\mid }$, obtained from the coordinate decomposition and norm axioms in [L2], makes the convergence ${\kappa }_n\to \kappa$ uniform. Each ${\kappa }_n$ is continuous, and [L3] makes $\kappa$ continuous, hence a path.

Fix $n$. Every vertex of ${\sigma }_n$ remains unchanged in every later path, so uniform convergence gives $\kappa (j\mathrm{/}{4}^n)={\kappa }_n(j\mathrm{/}{4}^n)$ for all $0\le j\le 4^n$. Thus the inscribed polygonal sum of $\kappa$ on ${\sigma }_n$ is ${\mathrm{\ell }}_{{\sigma }_n}(\kappa )=4^n{3}^{-n}=(4\mathrm{/}3{)}^n$. These sums are unbounded by [L4]. The defining supremum in [L5] is therefore $+\mathrm{\infty }$, so $\kappa$ is not rectifiable.

The classical Koch snowflake boundary is the concatenation of three isometric copies of $\kappa$. By [L7], each copy is nonrectifiable, and a rectifiable concatenation would have rectifiable restrictions. Hence the snowflake boundary is nonrectifiable.

Lower semicontinuity is consistent with the result but cannot prove it: [L6] yields only $L(\kappa )\le {\mathrm{lim\hspace{0.17em}inf}}_n(4\mathrm{/}3{)}^n=+\mathrm{\infty }$, a vacuous upper bound. The retained-vertex partitions in step 6.1 supply the necessary lower bounds.