A Sierpinski gasket computed by hand

Every length-$n$ word gives a lower-left corner $p={\sum }_{j\le n}{2}^{-j}{d}_j$ and a containing closed square $Q=p+[0,2^{-n}{]}^2$. There are $3^n$ words, giving distinct grid squares because each coordinate prefix has a unique length-$n$ binary code. Their diameters are $\sqrt{2}{2}^{-n}$; hence their $s$-cost is $3^n(\sqrt{2}{2}^{-n}{)}^s=2^{s\mathrm{/}2}$. The series converge coordinatewise by geometric tails, and these covers at arbitrarily small scales give ${\mathcal{H}}^s(K)\le 2^{s\mathrm{/}2}$.

For $u\in [0,1)$ define $e_j(u)=\lfloor 3^{j}u\rfloor -3\lfloor 3^{j-1}u\rfloor \in \{0,1,2\}$, identify these three values with the listed members of $D$, and put $T(u)={\sum }_{j\ge 1}{2}^{-j}d(e_j(u))$. Each coordinate is a limit of Borel step functions, hence Borel measurable. The vector map is Borel since preimages of open rational rectangles are Borel and those rectangles form a countable basis. Define $\mu (B)={\lambda }_1(T^{-1}(B))$ for Borel $B\subseteq {\mathbb{R}}^2$, with preimages taken in $[0,1)$. Disjoint Borel preimages prove countable additivity; thus $\mu$ is a Borel probability.

Every ternary prefix event is a half-open interval of length $3^{-n}$, hence has probability $3^{-n}$. Its image lies in the corresponding square. Also $T([0,1))\subseteq K$, so every Borel superset of $K$ has $\mu$-measure one and ${\mu }^{\ast }(K)=1$; no measurability claim about an arbitrary image is needed.

For nonempty $U$ of diameter $r$ with $2^{-n}\le r<2^{1-n}$, each coordinate projection lies in an interval of length at most $r<2\cdot 2^{-n}$. Such an interval meets at most four closed grid intervals of side $2^{-n}$, allowing all boundary contacts. Thus $U$ meets at most sixteen level-$n$ grid squares. Let $R$ be the closed coordinate bounding rectangle of $U$; its coordinate side lengths are at most $r$, so it meets at most sixteen squares. Every $u$ with $T(u)\in R$ has its own prefix square meeting $R$, so ${\mu }^{\ast }(U)\le \mu (R)\le 163^{-n}\le 16r^s$.

For a singleton use its coordinate point rectangle at arbitrarily fine levels; at most four squares contain the point, so its mass is at most $43^{-n}\to 0$. Empty sets have zero mass. The diameter estimate therefore holds also at zero. Apply mass distribution with constant sixteen and outer mass one to obtain ${\mathcal{H}}^s(K)\ge 1\mathrm{/}16$. Combined with the finite upper bound, this gives ${\dim }_{H}K=s$.