2-dimensional Lebesgue measure of certain sets in $R^3$ Let $\theta >0$ and $E \subseteq \mathbb{R}^3$ be a closed set (I've added closedness as a new requirement) which satisfies the following condition:
For any $x \in E$, there exist at least two lines $L_1,L_2 \subseteq E$ passing through $x$ with $\angle(L_1,L_2) \in (\theta, \frac{\pi}{2})$.
I am trying to figure out whether $\mathscr{L}^2(E)>0$ (excluding non-measurable sets). In fact, I can't even come up with 2-dimensional (Hausdorff) sets satisfying the above condition other than 2-planes or a union of them.
Would appreciate some suggestions.
 A: You can reduce from $\mathbb{R}^3$ to $\mathbb{R}^2$ by projecting onto the plane $Z:=\{z=0\}$, at the cost of a non-uniform $\theta$. Let $P \subset E$ be the set of points $p \in E$ where $E$ contains a line orthogonal to $Z$ through $p$, and let $\Pi$ denote projection onto $Z$. If the projection $\Pi(P)$ has positive $\mathcal{H}^1$ measure, then $E$ has positive $\mathcal{H}^2$ measure, so we can assume $\Pi(P)$ has zero $\mathcal{H}^1$ measure. Thus, there are two lines in $\Pi(E)\subset Z$ through almost every point in $\Pi(E)$.
Furthermore, each point in $\Pi(P)$ lies on a line $\ell(p)$ contained in $\Pi(E)$, by construction. For $p \in \Pi(P)$, let $\nu(p)$ denote the line through $p$ orthogonal to $\ell(p)$, and let $N:=\bigcup_{p \in \Pi(P)} \nu(p)$ denote the union of these normals. Then in any bounded set $\Omega \subset \mathbb{R}^2$, Fubini's theorem implies $N \cap \Omega$ has zero measure: $\mathcal{H}^2(N \cap \Omega) \le \int_{\Pi(P)} \mathcal{H}(\nu(p) \cap \Omega)\,d\mathcal{H}^1(p)=0$.
Then we can run @Shorty's argument from the comments on $Z \backslash N$ (which is fine with non-uniform $\theta$): choose some point $q \in Z \backslash (\Pi(E) \cup N)$ with $dist(q,\Pi(E))>0$, and let $B(q,\epsilon)$ be a minimal closed ball centred at $q$ which meets $\Pi(E)$, and $y \in \Pi(E)$ be any point in $B(q,\epsilon) \cap \Pi(E)$. Then $y\not\in \Pi(P)$ since $q \not\in N$, so there are two distinct lines in $\Pi(E)$ through $y$, contradicting $y$ being the closest point in $\Pi(E)$ to $q$ (notice this doesn't require $\theta$ to be uniform, just positive at $y$). Hence $\Pi(E)$ is dense in $Z \backslash N$ (and hence $Z$), and the fact $E$ is closed means that $\Pi(E)$ contains all of $Z$ in the $\mathcal{H}^1(\Pi(P))=0$ case (assuming $E$ is non-empty).
