Given a set $S\subseteq \mathbf{R}^n $ and $ x \in \overline{S} $ we define the tangent cone $ T\_S(x) $ to be the collection of all vectors $ v \in \mathbf{R}^n $ such that
$$ \liminf\_{r \to 0+} r^{-1} \mathrm{dist}(x + r v, S) =0. $$


It seems reasonable that there exists an example of a *closed* set $ S \subseteq \mathbf{R}^2 $ such that $ \mathcal{H}^1(S) < \infty $ and $ T\_S(x) = \mathbf{R}^2 $ for every $ x $ in a subset of $ S $ with positive $ \mathcal{H}^1 $ measure. Can anyone provide (or point out a reference for) a construction of such example?


Of course, if we replace the hypothesis of (locally) finiteness of the $ \mathcal{H}^1 $-measure with $ \sigma $-finiteness of the $ \mathcal{H}^1 $-measure then the construction is trivial. We can simply take $ S = (\{0\} \times \mathbf{R}) \cup \bigcup\_{i=1}^\infty\big[(\{i^{-1}\} \times \mathbf{R}) \cup (\{-i^{-1}\} \times \mathbf{R})\big] $, for which evidently
$$ T\_S(x) = \mathbf{R}^2 \quad \textrm{for every $ x \in \{0\} \times \mathbf{R} $}.$$



Tangent cone of null sets
One can use your infinite-density example, but replace the outer lines with very sparse dotted lines:


$$S = (\{0\} \times \mathbb{R}) \cup \bigcup\_{i=1}^\infty \{i^{-1},-i^{-1}\} \times \left[\bigcup\_{j \in \mathbb{Z}} [i^{-2}j,i^{-2}j+i^{-4}]\right] $$


By symmetry it is enough to consider the intersection with $[0,1]^2$ for local finiteness. There, what is left of the $i$-th line consists of $i^2$ pieces of length $i^{-4}$, i.e. contributes a total of $i^{-2}$ to the Hausdorff measure, so their sum is finite.


Additionally, each of the line-pieces is closed and isolated, their only accumulation points are on the line $\{0\}\times \mathbb{R}$, which is already included in $S$, so $S$ is closed.


Finally, take $x=(0,x\_2)$ and $v=(v\_1,v\_2) \in \mathbb{R}^2$. For $v\_1=0$ it is clear that $v$ is in the tangent cone. For $v\_1 \neq 0$ and any $i \in \mathbb{N}$ pick $r = \frac{1}{i|v\_1|}$. Then $x+rv = (\frac{\pm1}{i},x\_2+rv\_2)$, which no matter the second component is of distance less than $i^{-2}=r^2|v\_1|^2$ to a point of the form $(\pm i^{-1},i^{-2}j)$ and thus in $S$. This gives you a sequence for the liminf and shows that $v$ is in the tangent cone of $x$.


Also as an additional sanity check, note that this does not violate the Besicovitch-Federer structure theorem: While the tangent cones at $x\in \{0\}\times \mathbb{R}$ are degenerate and the set has positive integralgeometric measure, we have that due to the low density of $S$, the approximate tangent cone at each $x\in \{0\}\times \mathbb{R}$ is still $\{0\}\times \mathbb{R}$.


