# Hausdorff dimension and non-empty intersections with lines

Let $$A\subseteq [0,1]^d$$, $$d\geq 2$$, a set with Hausdorff dimension $$\operatorname{dim}_{\mathcal{H}}A=s$$. What is the minimum $$s$$ (if any) which guarantee that $$A$$ has non-empty intersections with a positive fraction of lines passing through the origin? Thank you in advance for any suggestion.

• How do you define "a positive fraction of lines"? Commented Oct 21, 2022 at 17:15
• @IosifPinelis I am assuming Raffaele means that the set of $\omega \in \mathbf{R}P^{d-1}$ for which the corresponding line $l_\omega$ intersects $A$ has positive $d-1$-Hausdorff measure. Commented Oct 21, 2022 at 21:03
• Oops, I just realized that you were looking for a reference. Sorry about that... In any case, maybe the proof I gave in my answer still proves helpful. Commented Oct 21, 2022 at 21:59

This is true if the dimension of $$A$$ is strictly larger than $$d-1$$; on the other hand taking $$A = \{ x^d = 0 \}$$ shows that $$s = d-1$$ is not enough. To prove the first claim we use the co-area formula.

Remark. Generally, when working with Lipschitz functions for example, one has to be careful with the application of the co-area formula on irregular sets, which $$A$$ might well be. However, I believe that the formula as stated below is correct, because it 'conditions' on the values of a smooth function. I think the form below can be obtained from Fubini in spherical coordinates, integrating an indicator function.

A few preparations simplify the calculations.

First, pick $$\delta > 0$$ small enough that $$s - \delta > d-1$$. By definition of Hausdorff dimension, $$\mathcal{H}^{s-\delta}(A) = \infty$$, but we may consider a subset $$B \subset A$$ that has finite, but non-zero measure: $$0 < \mathcal{H}^{s-\delta}(B) < \infty$$. (If $$B$$ intersects a positive proportion of the lines through the origin, then $$A$$ does too.) To simplify the argument, we want to avoid the origin, so we may assume that $$B \subset [0,1]^d \setminus \mathbf{B}^{d}(\rho)$$ for some $$\rho > 0$$. Here $$\mathbf{B}^d(\rho)$$ is the closed ball of radius $$\rho$$ around the origin.

Let $$f: x \in \mathbf{R}^d \setminus \{ 0 \} \mapsto \frac{x}{\lvert x \rvert}$$. The level sets of $$f$$ are exactly the lines through the origin; we write $$l_\omega \subset \mathbf{R}^d$$ for the line directed by $$\omega \in \mathbf{R}P^{d-1}$$. The co-area formula essentially lets us 'condition' on these: $$$$\int_{B} Jf \, \mathrm{d} \mathcal{H}^{s-\delta} = \int_{\mathbf{R}P^{d-1}} \mathcal{H}^{s-\delta-(d-1)}(B \cap l_\omega) \, \mathrm{d} \mathcal{H}^{d-1},$$$$ where $$Jf = (\operatorname{det} Df \circ Df)^{1/2}$$ is the Jacobian area change factor of $$f$$.

The function $$f$$ in question is simple enough to calculate this explicitly if you like, but for us it's enough to note that—because $$f$$ is $$C^1$$ away from the origin for example—there are constants $$0 < c(d,\rho) < C(d,\rho)$$ so that $$$$c(d,\rho) < Jf < C(d,\rho) \quad \text{ on [0,1]^d \setminus \mathbf{B}^d(\rho)}.$$$$

We picked $$B$$ a subset of $$[0,1] \setminus \mathbf{B}^d(\rho)$$, so $$$$c(d,\rho) \mathcal{H}^{s-\delta}(B) \leq \int_{\mathbf{R}P^{d-1}} \mathcal{H}^{s-\delta-(d-1)}(B \cap l_\omega) \, \mathrm{d} \mathcal{H}^{d-1} \leq C(d,\rho) \mathcal{H}^{s-\delta}(B).$$$$ Therefore $$\mathcal{H}^{s- \delta - (d-1)}(B \cap l_\omega)$$ cannot vanish $$\mathcal{H}^{d-1}$$-a.e. everywhere on $$\mathbf{R}P^{d-1}$$, and in particular the set of $$\omega \in \mathbf{R}P^{d-1}$$ for which the intersection $$B \cap l_\omega$$ is non-empty has positive $$\mathcal{H}^{d-1}$$-measure.