Suppose a collection of unit vectors has measure zero on the sphere. Can ${\mathbb R}^d$ be the union of the subspaces perpendicular to the vectors? So if a union of proper subspaces has measure zero (e.g. countably many subspacees), then ${\mathbb R}^{d}$ is not the union of these proper subspaces. But what if we have a union of $d-1$ dimensional subspaces such that the set of normal vectors to these subspaces has measure $0$ on the sphere? Can we still say ${\mathbb R}^d$ is not the union of these proper sub-spaces? If so, how? I'm not sure how an arbitrary small cover of the normal vectors easily leads to an arbitrary small cover of the union of subspaces.
 A: No, it's possible to cover $\mathbb{R}^d$ by subspaces with normal vector of measure zero whenever $d>2$, since this is when the sphere is higher than 1-dimensional: intuitively, you need more than one dimension of normal vectors to get positive measure, but it only takes a one-parameter family of $d-1$-dimensional objects to cover a $d$-dimensional object. Note that the restriction is necessary: every point of $\mathbb{R}^2$ lies in a unique $1$-dimensional subspace, and $\mathbb{R}$ can't be written as a union of $0$-dimensional subspaces at all.
For a specific example, take the orthocomplements of $(\cos\theta,\sin\theta,...,0)$ as $\theta$ runs over $[0,\pi)$. For example in $\mathbb{R}^2$ this is just all the vertical planes. One gets the span of $e_2$ through $e_d$ in the orthocomplement of $(1,0,...,0)$ and more generally the span of a rotation of $e_2$ in the $(x_1,x)2)$ plane along with $e_3$ through $e_d$ in the other hyperplanes. And indeed every vector in $\mathbb{R}^d$ may be expressed as a multiple of a rotation of $e_2$ in the $(x_1,x_2)$ plane plus an element of the span of $e_3$ through $e_d$.
But these normal vectors only cover a 1-dimensional equator of the sphere, which is of measure zero since the sphere is $d-1$-dimensional and $d-1>1$.  
