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 nonempty intersections with a positive fraction of lines passing through the origin? Thank you in advance for any suggestion.

1$\begingroup$ How do you define "a positive fraction of lines"? $\endgroup$– Iosif PinelisCommented Oct 21, 2022 at 17:15

$\begingroup$ @IosifPinelis I am assuming Raffaele means that the set of $\omega \in \mathbf{R}P^{d1}$ for which the corresponding line $l_\omega$ intersects $A$ has positive $d1$Hausdorff measure. $\endgroup$– Leo MoosCommented Oct 21, 2022 at 21:03

$\begingroup$ 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. $\endgroup$– Leo MoosCommented Oct 21, 2022 at 21:59

$\begingroup$ Thank you, Leo, your answer is really helpful. $\endgroup$– RaffaeleScandoneCommented Oct 22, 2022 at 13:29
1 Answer
This is true if the dimension of $A$ is strictly larger than $d1$; on the other hand taking $A = \{ x^d = 0 \}$ shows that $s = d1$ is not enough. To prove the first claim we use the coarea formula.
Remark. Generally, when working with Lipschitz functions for example, one has to be careful with the application of the coarea 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 $s  \delta > d1$. By definition of Hausdorff dimension, $\mathcal{H}^{s\delta}(A) = \infty$, but we may consider a subset $B \subset A$ that has finite, but nonzero 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 $\omega \in \mathbf{R}P^{d1}$. The coarea formula essentially lets us 'condition' on these: \begin{equation} \int_{B} Jf \, \mathrm{d} \mathcal{H}^{s\delta} = \int_{\mathbf{R}P^{d1}} \mathcal{H}^{s\delta(d1)}(B \cap l_\omega) \, \mathrm{d} \mathcal{H}^{d1}, \end{equation} 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 \begin{equation} c(d,\rho) < Jf < C(d,\rho) \quad \text{ on $[0,1]^d \setminus \mathbf{B}^d(\rho)$}. \end{equation}
We picked $B$ a subset of $[0,1] \setminus \mathbf{B}^d(\rho)$, so \begin{equation} c(d,\rho) \mathcal{H}^{s\delta}(B) \leq \int_{\mathbf{R}P^{d1}} \mathcal{H}^{s\delta(d1)}(B \cap l_\omega) \, \mathrm{d} \mathcal{H}^{d1} \leq C(d,\rho) \mathcal{H}^{s\delta}(B). \end{equation} Therefore $\mathcal{H}^{s \delta  (d1)}(B \cap l_\omega)$ cannot vanish $\mathcal{H}^{d1}$a.e. everywhere on $\mathbf{R}P^{d1}$, and in particular the set of $\omega \in \mathbf{R}P^{d1}$ for which the intersection $B \cap l_\omega$ is nonempty has positive $\mathcal{H}^{d1}$measure.