Sets that project to zero measure on all lines except one

Question

It is a (difficult) exercise to show that there exists a measurable set $E \subset [0,1]^2$ (necessarily with zero 2-dimensional Lebesgue measure) such that the projection on every line passing through the origin has zero $\mathscr H^1$-measure (Hausdorff measure) except for the two coordinate axis (see for example https://math.stackexchange.com/questions/1546045/projection-of-f-onto-any-line-nonparallel-to-coordinate-axes-has-measure-zero).

Question: does there exist such a set that projects to zero on all lines passing through the origin but except one (say the $x$-axis)?

EDIT: at the moment it seems that we do not even have a proof for the two lines case.

Answer 1

Consider a sequence of rational numbers $s_i$ with the following property: For each rational number $s$ and nonnegative integer $n$, there exists $i$ such that $ s=s_i =s_{i+1} = \dots = s_{i+n}$, and $|s_{i+j}| \leq (3/2)^j$ for $j>n$.

Such a sequence certainly exists: The set of pairs $(s,n)$ is countable, so we can order all the pairs and satisfy them in sequence. After satisfying some of the pairs, we have fixed some initial segment $s_1,\dots, s_m$. If we didn't have the second condition we could simply set $s_{m+1} = \dots = s_{m+1+n} =s $. With the second condity, we may have to choose $s_{m+1},\dots, s_{m+k}$ small for some $k$ and then choose $s_{m+k+1} = \dots = s_{m+k+1+n}=s$, but we can always do this.

Now consider the set of vectors in $\mathbb R^2$ of the form $$ \left\{\sum_{i=1}^\infty e_i 2^{-i} (1,s_i) \mid e_1,e_2,\dots \in \{0,1\}\right\}$$. In other words, this is the image of the set of infinite binary sequences under the map which sends each binary sequence $e_1,e_2,\dots$ to the sum $\sum_{i=1}^\infty e_i 2^{-i} (1,s_i)$.

The projection of this set onto the $x$ axis is $$\left \{\sum_{i=1}^\infty e_i 2^{-i} \mid e_1,e_2,\dots \in \{0,1\}\right\}=[0,1]$$ which has positive measure.

Let's show the projection onto every line not parallel to the $x$ axis has measure $0$. It suffices to check for the projection onto the line $x= -ty$ for every real number $t$. The projection of the set onto that line is given by

$$\left \{\sum_{i=1}^\infty e_i 2^{-i} (t-s_i) \mid e_1,e_2,\dots \in \{0,1\}\right\}.$$

Observe that for any $j$, this set is contained in the union of $2^{j-1}$ intervals of length

$$\sum_{i=j}^\infty 2^{-i} |t-s_i|$$

where each interval consists of points lying between the maximum and minimum possible value of $\sum_{i=1}^\infty e_i 2^{-i} (t-s_i)$ for $e_1,\dots,e_{j-1}$ taking some fixed $j-1$-tuple of values. Hence the measure is at most

$$\sum_{i=j}^\infty 2^{j-1-i} |t-s_i|.$$

If we choose $j$ so that $s_j= s_{j+1} = \dots = s_{j+n}=s$ for some rational approximation $s$ of $t$ and $s_{j+k} \leq (3/2)^k$ for $k>n$ then we have

$$\sum_{i=j}^\infty 2^{j-1-i} |t-s_i| = \sum_{i=j}^{j+n}2^{j-1-i} |t-s_i| + \sum_{i=j+n+1}^\infty |t-s_i|$$ $$ \leq \sum_{i=j}^{j+n}2^{j-1-i} |t-s| + \sum_{i=j+n+1}^{\infty}2^{j-1-i} ( |t| + (3/2)^{i-j} )$$

$$ \leq |t-s| + 2^{-n} |t| + 2 (3/4)^{n+1}$$

which we can make arbitrarily small by choosing $n$ large and $s$ a good rational approximation of $t$.

Answer 2

This can be done with an iterated Venetian blind construction.

The idea is you start with a horizontal line segment, replace it with many small rotated parallel line segments, and repeat. In each step, you ensure that the union of the segments has the same projection onto the x-axis as the original line segment. By rotating the segments in each step, the size of projections in other directions decrease. Taking an appropriate limit of this constuction gives the desired result.

For the specific question you ask, I don't know if the details are written down anywhere. But you can see, e.g., Lemma 11.8 of Mattila's Fourier analysis and Hausdorff dimension for something related.

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An alternative proof is to use Baire category:

Let $\mathcal A$ be the set of all compact subsets $E \subset [0,1] \times \mathbb R$ with full projection onto the $x$-axis. Let $\mathcal B \subset \mathcal A$ be the set of all $E \in A$ such that the projection of $E$ onto every 1-dimensional subspace except the $x$-axis has 1-dimensional Hausdorff measure zero. Then by a Venetian blind-like argument, one can show that $\mathcal B$ is residual/co-meager in $\mathcal A$. (This is essentially the point-line dual of the construction in Körner's article Besicovitch via Baire.)

The details of this are in Section 4 of some unpublished notes of mine. See Corollary 4.8.