In mathematics, a measure on a real vector space is said to be **transverse** to a given set if it assigns measure zero to every translate of that set, while assigning finite and positive (i.e. non-zero) measure to some compact set.

##
Definition

Let *V* be a real vector space together with a metric space structure with respect to which it is complete. A Borel measure *μ* is said to be **transverse** to a Borel-measurable subset *S* of *V* if

- there exists a compact subset
*K* of *V* with 0 < *μ*(*K*) < +∞; and
*μ*(*v* + *S*) = 0 for all *v* ∈ *V*, where

- $v+S=\{v+s\in V|s\in S\))$

- is the translate of
*S* by *v*.

The first requirement ensures that, for example, the trivial measure is not considered to be a transverse measure.

##
Example

As an example, take *V* to be the Euclidean plane **R**^{2} with its usual Euclidean norm/metric structure. Define a measure *μ* on **R**^{2} by setting *μ*(*E*) to be the one-dimensional Lebesgue measure of the intersection of *E* with the first coordinate axis:

- $\mu (E)=\lambda ^{1}{\big (}\{x\in \mathbf {R} |(x,0)\in E\subseteq \mathbf {R} ^{2}\}{\big )}.$

An example of a compact set *K* with positive and finite *μ*-measure is *K* = *B*_{1}(0), the closed unit ball about the origin, which has *μ*(*K*) = 2. Now take the set *S* to be the second coordinate axis. Any translate (*v*_{1}, *v*_{2}) + *S* of *S* will meet the first coordinate axis in precisely one point, (*v*_{1}, 0). Since a single point has Lebesgue measure zero, *μ*((*v*_{1}, *v*_{2}) + *S*) = 0, and so *μ* is transverse to *S*.