The Koch snowflake is a non-rectifiable quasicircle

The maps in [F2] send $T$ into itself; direct substitution of its three vertices verifies this. Their triangles meet only at consecutive retained vertices, and nonconsecutive triangles are separated by at least $1\mathrm{/}6$: their real projections lie respectively in $[-1\mathrm{/}2,-1\mathrm{/}6]$, $[-1\mathrm{/}6,0]$, $[0,1\mathrm{/}6]$, $[1\mathrm{/}6,1\mathrm{/}2]$. For adjacent triangles, their cones at the common vertex have angular separation at least $\pi \mathrm{/}3$, so their intersection is just that vertex. The same maps send $V$ into $V$, as substitution of its four vertices verifies, and their interiors lie in the corresponding disjoint open real-coordinate strips. Iteration gives level-$n$ rhombi of diameter $3^{-n}$ with disjoint interiors. Define $\rho (t)$ by the nested triangles prescribed by the base-four digits of $t\in [0,1]$, interpreting $1$ by the all-three digits. Nested diameters tend to zero, so completeness gives a unique point. At a double expansion, the two addresses end in all-three and all-zero digits; their triangles shrink to the same consecutive vertex, so $\rho$ is well-defined. Parameters within $4^{-n}$ lie in the same or adjacent level-$n$ parameter intervals, whose triangles have union diameter at most $2\cdot 3^{-n}$; hence $\rho$ is continuous. The polygonal parametrizations differ uniformly from it by at most $3^{-n}$ and have the retained vertices as interval endpoints, proving surjectivity onto the Hausdorff limit. If two parameters have different first child addresses, triangle separation forces any common image to be their shared endpoint; the endpoint's only addresses are the corresponding all-three/all-zero tails. Thus the parameters agree, proving injectivity. The three outward copies of $T$ on the initial equilateral triangle meet only at the initial vertices: their endpoint cones again have separation at least $\pi \mathrm{/}3$, and away from the vertices they lie on different exterior sides. The concatenation of the three side parametrizations is therefore continuous, with equal endpoints and injective on $[0,1)$. It induces a continuous bijection from the compact circle to $S$, a homeomorphism by A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Hence $S$ is Jordan. Every level-$n$ rhombus has inradius $3^{-n}\mathrm{/}4$, so it contains an open axis-parallel square of side $3^{-n}\mathrm{/}4$; this is valid after any rotation, since that square's circumradius is less than the inradius.

The closed parametrisation traverses the three side parametrisations in cyclic order. By [F2], subdividing each side parameter interval into $4^n$ equal pieces inscribes exactly the $4^n$ retained level-$n$ edges on that side. Each has length $3^{-n}$, so the concatenated partition has polygonal sum $3\cdot 4^n{3}^{-n}=3(4\mathrm{/}3{)}^n$, which tends to $+\mathrm{\infty }$ by [F10]. By [F1], the limiting boundary is not rectifiable. A chord-arc curve is rectifiable by definition, so its chord-arc condition fails here.

Put $\alpha =\log 4\mathrm{/}\log 3$. For each $n$, the $3\cdot 4^n$ level-$n$ rhombi covering the three Koch sides have diameter $3^{-n}$. Since $4^n(3^{-n}{)}^{\alpha }=1$, these covers have total $\alpha$-cost $3$ and diameters tending to zero. Thus ${\mathcal{H}}^{\alpha }(S)\le 3<\mathrm{\infty }$.

Let $K$ be one side and $(U_j)$ any countable cover of $K$ with $\mathrm{diam}{U}_j\le \delta <1$. For a nonempty $U$ with $d=\mathrm{diam}U>0$, choose $n\ge 1$ with $3^{-n}\le d<3^{-n+1}$. At most $1024$ level-$n$ rhombi meet $U$: each contains an open axis-parallel square of side $3^{-n}\mathrm{/}4$, these squares are pairwise disjoint, and all such squares lie in one axis-parallel square of side $8\cdot 3^{-n}$; finite additivity and monotonicity of planar area give $N(3^{-n}{)}^2\mathrm{/}16\le 64(3^{-n}{)}^2$. The base-four parameter intervals of the cells meeting $U$ cover ${\rho }^{-1}(U)$, so ${\lambda }_1^{\ast }({\rho }^{-1}(U))\le 1024\cdot 4^{-n}\le 1024d^{\alpha }$. If $d=0$, then $U$ is empty or a singleton, and injectivity of $\rho$ makes its preimage empty or a singleton, of outer measure zero. Countable subadditivity and ${\lambda }_1^{\ast }([0,1])=1$ now give $1\le 1024{\sum }_j(\mathrm{diam}{U}_j{)}^{\alpha }$. Taking the infimum over every such cover and then the small-scale limit yields ${\mathcal{H}}^{\alpha }(K)\ge 1\mathrm{/}1024$; monotonicity gives ${\mathcal{H}}^{\alpha }(S)\ge 1\mathrm{/}1024>0$.

First consider the side arc from a point $x$ to an endpoint $v$. If $x=v$, its diameter is zero. Otherwise let $T_w$ be the deepest nested endpoint triangle containing $x$, of diameter $a=3^{-\mathrm{\mid }w\mathrm{\mid }}$. The endpoint children are ${\psi }_0$ and ${\psi }_3$; their repeated triangles shrink to the endpoint, so this depth is finite. The other three children of $T_w$ are at distance at least $a\mathrm{/}3$ from $v$, by the real-coordinate strips in step 1.1. Thus $\mathrm{\mid }x-v\mathrm{\mid }\ge a\mathrm{/}3$, while the entire endpoint subarc lies in $T_w$ and has diameter at most $a\le 3\mathrm{\mid }x-v\mathrm{\mid }$. Now take distinct $x,y$ on one side and their deepest common triangle, of diameter $a$. If their first distinct children are nonconsecutive, their distance is at least $a\mathrm{/}6$ and the intervening subarc has diameter at most $a$, giving ratio at most $6$. If the children are consecutive with shared vertex $v$, their endpoint cones have separation at least $\pi \mathrm{/}3$ by step 1.1. Writing $r=\mathrm{\mid }x-v\mathrm{\mid }$, $s=\mathrm{\mid }y-v\mathrm{\mid }$, the cosine law gives $\mathrm{\mid }x-y{\mathrm{\mid }}^2\ge r^2+s^2-rs\ge (r+s{)}^2\mathrm{/}4$. The two endpoint tails have combined diameter at most $3(r+s)\le 6\mathrm{\mid }x-y\mathrm{\mid }$. This also includes a zero tail when one point is the vertex. Points on different initial sides admit the subarc through their shared initial vertex, with the identical cone and endpoint-tail estimate. These cases prove bounded turning with bound $6$, hence with the advertised bound $12$; no sharpness is asserted.

By [F4], the finite positive $\alpha$-measure gives ${\dim }_{H}S=\alpha$; since $\alpha >1$, the same theorem gives ${\mathcal{H}}^1(S)=+\mathrm{\infty }$.

The Jordan curve $S$ has bounded turning by step 2.4, so [F7] makes it a quasicircle. Applying [F8] then proves that $S$ is conformally removable.