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.
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1$\begingroup$ How do you define "a positive fraction of lines"? $\endgroup$– Iosif PinelisCommented Oct 21, 2022 at 17:15
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$\begingroup$ @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. $\endgroup$– Leo MoosCommented Oct 21, 2022 at 21:03
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$\begingroup$ 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
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$\begingroup$ 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 $d-1$; 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: \begin{equation} \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}, \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^{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). \end{equation} 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.