Tangent cone of null sets

Question

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 $$ T_S(x) = \mathbf{R}^2 \quad \textrm{for every $ x \in \{0\} \times \mathbf{R} $}.$$

Answer 1

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 $\{0\}\times \mathbb{R}$.

Answer 2

I think here is one example. For each rational in $[0, 1]$ of the form $k/2^{n}$, let

$$S_{k, n} := \{k/2^n\} \times \cup_{-2^n \leq j \leq 2^n} \{j2^{-2n}\}$$ and define

$$S = \bigcup_{k \in \mathbb Z, n \in \mathbb Z_+} S_{k, n} \cup ([0, 1] \times \{0\}).$$

It can be shown that the set of $x \in S$ with the liminf in question $0$ for all $v \in \mathbb R^2$ contains the set $E$ defined as follows:

For $z \in \mathbb R$, denote by $L_k (z)$ the length of the string of $0$’s or $1$’s beginning at the $k$’th decimal place (after the decimal point) of the binary expansion of $z$. Writing $(v_1, v_2)$ for $v \in \mathbb R^2$, and $\mathcal Q$ for the set of points in the circle $S^1$ with angles a dyadic rational multiple of $\pi$, define $E_q$ for $q \in \mathcal Q$ by

$$E_q := \{x \in [0, 1] \times \{0\}| \ \limsup_{k \to \infty} \min_{i = 1, 2} L_k ([x +2^{-k}q]_i) - k = +\infty\}.$$

Finally, set $E := \cup_{q \in \mathcal Q} E_q$.

I claim without proof that this set $E$ has full measure in $[0, 1]$, whence the set of $x \in S$ with $T_S (x) = \mathbb R^2$ is of $\mathcal H^1$ measure $1$.

Finally we note that $S$ is closed and has $\mathcal H^1$ measure $1$, and so $S$ satisfies the requirements of your problem.