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# Halfspace depth for general measures: The ray basis theorem and its
consequences

Petra Laketa  and  Stanislav Nagy<EMAIL_ADDRESS>Charles University,
Faculty of Mathematics and Physics, Prague, Czech Republic

###### Abstract.

The halfspace depth is a prominent tool of nonparametric multivariate
analysis. The upper level sets of the depth, termed the trimmed regions of a
measure, serve as a natural generalization of the quantiles and inter-quantile
regions to higher-dimensional spaces. The smallest non-empty trimmed region,
coined the halfspace median of a measure, generalizes the median. We focus on
the (inverse) ray basis theorem for the halfspace depth, a crucial theoretical
result that characterizes the halfspace median by a covering property. First,
a novel elementary proof of that statement is provided, under minimal
assumptions on the underlying measure. The proof applies not only to the
median, but also to other trimmed regions. Motivated by the technical
development of the amended ray basis theorem, we specify connections between
the trimmed regions, floating bodies, and additional equi-affine convex sets
related to the depth. As a consequence, minimal conditions for the strict
monotonicity of the depth are obtained. Applications to the computation of the
depth and robust estimation are outlined.

##  1\. Introduction: Characterization of the halfspace median

The lack of natural ordering of the points in multidimensional spaces makes
the nonparametric analysis of multivariate data challenging. The depth
introduces a data-dependent ordering of the sample points, in the direction
from the most central observations being those that attain high depth values,
to the peripheral ones with low depth [13, 32, 34, 4]. A definition of depth-
based central regions of the data, which are the regions where the depth
exceeds given thresholds, ensues naturally. The smallest non-empty depth
region is often termed the depth median set of the data. The depth medians
provide convenient robust location estimators, well studied in the literature.

We consider the seminal halfspace (or Tukey) depth, and the general setup of
finite Borel (not necessarily probability) measures. Write $\mathbb{R}^{d}$
for the $d$-dimensional Euclidean space equipped with the inner product
$\left\langle\cdot,\cdot\right\rangle$, and $\mathbb{S}^{d-1}$ for the unit
sphere in $\mathbb{R}^{d}$. The set of all closed halfspaces in
$\mathbb{R}^{d}$ is denoted by $\mathcal{H}$. Elements of $\mathcal{H}$ can be
represented as

$H_{x,u}=\left\\{y\in\mathbb{R}^{d}\colon\left\langle
x,u\right\rangle\leq\left\langle y,u\right\rangle\right\\}$

for $x\in\mathbb{R}^{d}$ a point in the boundary hyperplane and
$u\in\mathbb{S}^{d-1}$ the inner unit normal of $H_{x,u}$. We write
$\mathcal{H}(x)=\left\\{H_{x,u}\colon u\in\mathbb{S}^{d-1}\right\\}$ for all
halfspaces whose boundary passes through $x\in\mathbb{R}^{d}$. The collection
of all finite Borel measures on $\mathbb{R}^{d}$ is denoted by
$\mathcal{M}\left(\mathbb{R}^{d}\right)$. The halfspace depth was first
considered by Tukey, [28] and later substantially elaborated on by Donoho and
Gasko, [5]. The _halfspace depth_ of a point $x\in\mathbb{R}^{d}$ with respect
to (w.r.t.) a measure $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ is
defined as

(1) $D(x;\mu)=\inf_{H\in\mathcal{H}(x)}\mu(H)=\inf_{H\in\mathcal{H},x\in
H}\mu(H).$

The two expressions for the depth are easily seen to be equivalent. Our
principal interest lies in the concept of the _depth-trimmed regions_ (also
called _central regions_) of $\mu$, defined for $\alpha\in\mathbb{R}$ by
$D_{\alpha}(\mu)=\left\\{x\in\mathbb{R}^{d}\colon
D\left(x;\mu\right)\geq\alpha\right\\}$. It is a simple observation that the
regions $D_{\alpha}(\mu)$ are always closed convex sets, non-increasing in
$\alpha\in\mathbb{R}$ in the sense of set inclusion. For $\alpha\leq 0$ we
have $D_{\alpha}(\mu)=\mathbb{R}^{d}$; for all $\alpha$ large enough
$D_{\alpha}(\mu)=\emptyset$. The supremum of all $\alpha$ such that
$D_{\alpha}(\mu)\neq\emptyset$ is denoted by $\alpha^{*}(\mu)$. We call the
set $D_{\alpha^{*}}(\mu)$ the _(halfspace) median set_ of $\mu$, and denote it
by $D^{*}\left(\mu\right)$. The median set is always non-empty and compact.
Elements of $D^{*}\left(\mu\right)$ are called _(halfspace) medians_ of $\mu$.
Any halfspace $H\in\mathcal{H}(x)$ that satisfies $D(x;\mu)=\mu(H)$ is called
a _minimizing halfspace_ of $\mu$ at $x\in\mathbb{R}^{d}$.

The so-called ray basis theorem provides a convenient characterization of a
halfspace depth median of a measure $\mu$ in terms of its minimizing
halfspaces. In its simplest form, the theorem asserts that for measures
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ with continuous and positive
density, a point $x\in\mathbb{R}^{d}$ is a halfspace median of $\mu$ if and
only if the collection of its minimizing halfspaces covers the whole space
$\mathbb{R}^{d}$. That result was first proved by Donoho and Gasko, [5, Claim
on p. 1818], and later extended in Rousseeuw and Ruts, [24, Propositions 8 and
12]. For the special case of uniform measures on convex bodies, such a
characterization of the deepest point relates to an early observation of
Grünbaum, [9, p. 251] from convex geometry. For details and additional
discussion about the ray basis theorem, its history and relevance in both
statistics and geometry see [20, Section 4.3.1] and [22].

Our initial goal is to revisit the ray basis theorem, and consolidate its
statement by extending it to any depth region $D_{\alpha}(\mu)$ of a general
measure $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, under minimal
assumptions.111The extension of these results from Borel probability measures
to finite Borel measures on $\mathbb{R}^{d}$ is minor. By our treatment of
_general_ measures we mean generalizations of the ray basis theorem and
related results to measures from $\mathcal{M}\left(\mathbb{R}^{d}\right)$ that
do not have to possess positive densities, or satisfy other simplifying
conditions concerning, e.g., their support. We do so in Section 2, where tools
from convex geometry are employed to devise an elementary proof of a
generalized version of the theorem. We provide conditions under which it is
possible to cover the complement to $D_{\alpha}(\mu)$ by halfspaces of
$\mu$-mass $\alpha$. As a special case, we obtain connections between the
median set, and the set of all points that allow covering $\mathbb{R}^{d}$ by
their minimizing halfspaces. An important part of our contribution are the
examples, that throughout the paper demonstrate that the conditions stated in
our main results cannot be avoided.

In Section 3 we thoroughly discuss the links of our general ray basis theorem
with the properties of the central regions $D_{\alpha}(\mu)$. It is known from
the literature [24, Proposition 6] that, writing $A^{\mathsf{c}}$ for the
complement to $A\subseteq\mathbb{R}^{d}$ in $\mathbb{R}^{d}$,

(2)
$D_{\alpha}(\mu)=\bigcap\left\\{H\in\mathcal{H}\colon\mu(H^{\mathsf{c}})<\alpha\right\\}.$

Each $D_{\alpha}(\mu)$ is thus an intersection of closed convex sets, and must
be closed and convex itself. We begin from (2), and specify relations between
the central regions $D_{\alpha}(\mu)$, the trimmed regions as considered in
[21, 17] and [29, Section 3.9.4.6], and the floating body known from convex
geometry [2, 20].

Our paper is concluded with three applications of our results in Section 4.
The set of depth medians $D^{*}\left(\mu\right)$ is not necessarily a single
point set. Especially for empirical measures $\mu$, that is measures
corresponding to datasets, the median set is frequently full-dimensional. In
Section 4.1 we propose to single out the collection of those medians that
satisfy an additional covering property. We obtain a smaller collection of
covering medians of $\mu$, which share qualitatively better properties than
the general elements of $D^{*}\left(\mu\right)$. An algorithm for finding
covering medians is given. In Section 4.2 we obtain a consequence of the
general ray basis theorem regarding the structure of the central regions
$D_{\alpha}(\mu)$ for $\mu$ an empirical measure. In that case, each facet $F$
of the convex polytope $D_{\alpha}(\mu)$ is shown to lie in a hyperplane
determined by data points. This observation promises applications in the
computation of the trimmed regions $D_{\alpha}(\mu)$ also in the case when
they are not full-dimensional, in a spirit similar to that lately employed in
[15]. Finally, in Section 4.3 we provide, as an interesting by-product of our
study, the minimal set of assumptions that guarantees the depth to be strictly
monotone. As argued by Dyckerhoff, [8, Example 4.2], strict monotonicity is
one of the most important properties a depth can have. For a depth w.r.t. a
probability measure $\mu$, it ensures the almost sure uniform convergence of
the depth upper level sets estimated from the data towards their population
counterparts. As such, sufficient conditions for strict monotonicity find
applications in the estimation of the depth-trimmed regions $D_{\alpha}(\mu)$
from data in multivariate statistics. The proofs of all theoretical results
are provided in an extensive appendix accompanying the paper.

###  1.1. Preliminaries and notation

In the proofs of our results, we use tools from measure theory, as well as
from the theory of convex sets. Our general reference to the concepts used
from convexity theory is [25]. We now set the most important notations used in
the paper, and state a preliminary observation about the halfspace depth that
will be useful later. For a set $A\subseteq\mathbb{R}^{d}$ its interior,
closure and boundary are denoted by $\operatorname{int}\left(A\right)$,
$\operatorname{cl}\left(A\right)$ and $\operatorname{bd}\left(A\right)$,
respectively. Denote by $\operatorname{aff}\left(A\right)$ and
$\operatorname{conv}\left(A\right)$ the affine hull and the convex hull of
$A$. The sets $\operatorname{relint}\left(A\right)$,
$\operatorname{relcl}\left(A\right)$ and $\operatorname{relbd}\left(A\right)$
represent the (relative) interior, closure and boundary of $A$ in the space
$\operatorname{aff}\left(A\right)$, and $\dim
A=\dim\left(\operatorname{aff}\left(A\right)\right)$ is the dimension of $A$.
The complement to $A$ is $A^{\mathsf{c}}=\mathbb{R}^{d}\setminus A$. For sets
$A$ and $B$ we write $A\subset B$ if $A\subseteq B$ and $A\neq B$. We say that
$H\in\mathcal{H}$ is a touching halfspace of a non-empty convex set $A$ if
$H\cap\operatorname{cl}\left(A\right)\neq\emptyset$ and
$\operatorname{int}\left(H\right)\cap A=\emptyset$. The collection of all
touching halfspaces to $A$ is denoted by $\mathcal{H}(A)$. We also define
$\mathcal{H}(\emptyset)=\mathcal{H}$. In this notation, $\mathcal{H}(x)$ is
the same as $\mathcal{H}(\\{x\\})$ for $x\in\mathbb{R}^{d}$.

Well-behaved measures. We say that
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ is _smooth_ if
$\mu(\operatorname{bd}\left(H\right))=0$ for each $H\in\mathcal{H}$. We call
$\mu$ _smooth at_ a convex set $A\subset\mathbb{R}^{d}$ if
$\mu(\operatorname{bd}\left(H\right))=0$ for all $H\in\mathcal{H}(A)$, and
smooth at a point $x\in\mathbb{R}^{d}$ if it is smooth at
$\left\\{x\right\\}$. Smoothness of $\mu$ at a point is a condition stronger
than $\mu$ being atom-less; yet, it is still weaker than smoothness in the
whole $\mathbb{R}^{d}$. A measure $\mu$ is said to have _contiguous support_
if the support of $\mu$ cannot be separated by a slab between two parallel
hyperplanes of non-empty interior with zero $\mu$-mass. Finally, $\mu$ is said
to have _contiguous support at_ a convex set $A\subset\mathbb{R}^{d}$ if

(3) for each $H^{\prime}\in\mathcal{H}(A)$ and $H^{\prime}\subset
H\in\mathcal{H}$,
$\operatorname{bd}\left(H\right)\cap\operatorname{int}\left(A\right)\neq\emptyset$
implies $\mu(H^{\prime})<\mu(H)$.

Note that the condition (3) is void if
$\operatorname{int}\left(A\right)=\emptyset$, and therefore it is enough to
consider $A$ full-dimensional. In that case, (3) means that any shift
$H\supset H^{\prime}$ of a touching halfspace $H^{\prime}$ of $A$ has
$\mu$-mass larger than $H^{\prime}$. It is satisfied if, for instance, the set
$A$ is a subset of the support of $\mu$. An absolutely continuous measure is
smooth (at any $A$ convex). A measure with connected support has contiguous
support (at any convex subset $A$ of the support of $\mu$).

Minimizing halfspaces. For $x\in\mathbb{R}^{d}$ recall that
$H\in\mathcal{H}(x)$ is a minimizing halfspace of
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ at $x$ if $D(x;\mu)=\mu(H)$. In
general, the set of minimizing halfspaces of $\mu$ at a point may be empty. It
is guaranteed to be non-empty if, for instance, the measure $\mu$ is smooth at
$x$. Our first observation that will prove to be useful in the sequel is that
it is always possible to find $H\in\mathcal{H}(x)$ with the property
$\mu(\operatorname{int}\left(H\right))\leq D\left(x;\mu\right)$. We call such
a halfspace $H$ a _generalized minimizing halfspace_ of $\mu$ at $x$.

######  Lemma 1.

For any $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ there exists a
generalized minimizing halfspace of $\mu$ at any point $x\in\mathbb{R}^{d}$.
If $\mu$ is smooth at $x$, then there exists a minimizing halfspace of $\mu$
at $x$.

Additional notations. In addition to the upper level set of the halfspace
depth $D_{\alpha}(\mu)=\left\\{x\in\mathbb{R}^{d}\colon
D\left(x;\mu\right)\geq\alpha\right\\}$, we also consider the set

$U_{\alpha}(\mu)=\left\\{x\in\mathbb{R}^{d}\colon
D(x;\mu)>\alpha\right\\}=\bigcup_{\beta>\alpha}D_{\beta}(\mu).$

It is convex, but not always closed. As will be seen later in the paper, the
set $U_{\alpha}(\mu)$ is also not open in general. Both $D_{\alpha}(\mu)$ and
$U_{\alpha}(\mu)$ are bounded for $\alpha>0$.

A face of a convex set $A$ is a convex subset $F\subseteq A$ such that $x,y\in
A$ and $(x+y)/2\in F$ implies $x,y\in F$. Denote by
$\operatorname{\mathcal{F}}\left(A\right)$ the set of all faces of
$\operatorname{cl}\left(A\right)$. A facet of $A$ is a face of $A$ of
dimension $\dim(A)-1$. For $F\in\operatorname{\mathcal{F}}\left(A\right)$ and
$A$ convex, denote by $\mathcal{H}(A,F)=\\{H\in\mathcal{H}(A)\colon
F\subseteq\operatorname{bd}\left(H\right)\\}$ the collection of all halfspaces
touching $A$ at its face $F$. We say that a sequence of halfspaces
$\left\\{H_{x_{n},u_{n}}\right\\}_{n=1}^{\infty}\subset\mathcal{H}$ converges
to $H_{x,u}\in\mathcal{H}$ if $x_{n}\to x$ in $\mathbb{R}^{d}$ and $u_{n}\to
u$ in $\mathbb{S}^{d-1}$.

We write $L(x,y)$ and $L[x,y]$ for the relatively open (that is, not
containing $x$ and $y$) and the relatively closed (containing $x$ and $y$)
line segment between distinct point $x,y\in\mathbb{R}^{d}$, respectively, and
$l(x,y)$ for the infinite line determined by $x$ and $y$.

##  2\. The general ray basis theorem

The standard ray basis theorem, as dubbed by Rousseeuw and Ruts, [24], asserts
that under certain conditions on a measure
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, the halfspace median $x$ of
$\mu$ is characterized by the fact that $\mathbb{R}^{d}$ can be covered by
minimizing halfspaces of $x$. The formal statement of the theorem is given
below.

######  Theorem 2 ([24, Propositions 8 and 12]).

For $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, the following holds true:

  1. (i)

If for some $x\in\mathbb{R}^{d}$

(4)
$\mathbb{R}^{d}=\bigcup\left\\{H\in\mathcal{H}(x)\colon\mu(H)=D(x;\mu)\right\\},$

then $x$ is a median of $\mu$, i.e. $D(x;\mu)=\alpha^{*}(\mu)$.

  2. (ii)

If $x$ is a median of $\mu$ and, in addition, $\mu$ is absolutely continuous
with a density that is continuous and positive in an open convex set, then (4)
is true for $x$.

The proof of the direct part (i) of Theorem 2 is simple: since
$\mathbb{R}^{d}$ is covered by halfspaces of $\mu$-mass exactly $D(x;\mu)$,
each $y\in\mathbb{R}^{d}$ must be contained in such a halfspace $H_{y}$, and
consequently $D(y;\mu)\leq\mu(H_{y})=D(x;\mu)$, as follows directly from the
definition of the halfspace depth (1). The proof of the so-called _inverse ray
basis theorem_ , stated in part (ii), is slightly more involved, and
technical. Our aim is to obtain a result similar to Theorem 2 for any depth
region $D_{\alpha}(\mu)$, under minimal conditions. We prove that for well-
behaved measures it is possible to cover all points outside
$\operatorname{int}\left(D_{\alpha}(\mu)\right)$ with closed halfspaces $H$
whose $\mu$-mass is bounded from above by $\alpha$. For measures that may not
be smooth the appropriate formulation of the latter condition turns out to be
$\mu(\operatorname{int}\left(H\right))\leq\alpha$. In the following lemma we
introduce a condition that will play an important role in the formulation of
the extended ray basis theorem. We treat not only the depth-trimmed regions
$D_{\alpha}(\mu)$, but also $U_{\alpha}(\mu)$, as well as their interiors.

######  Lemma 3.

Let $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, $\alpha\in\mathbb{R}$ and
$S\in\\{D_{\alpha}(\mu),U_{\alpha}(\mu),\operatorname{int}\left(D_{\alpha}(\mu)\right),\operatorname{int}\left(U_{\alpha}(\mu)\right)\\}$.
Then

(5)
$S^{\mathsf{c}}\subseteq\left(\operatorname{int}\left(S\right)\right)^{\mathsf{c}}\subseteq\bigcup\\{H\in\mathcal{H}\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\mbox{
and }\operatorname{int}\left(H\right)\cap S=\emptyset\\}.$

Application of Lemma 3 to the median set
$S=\operatorname{int}\left(D^{*}\left(\mu\right)\right)$ gives a slightly
weaker version of the standard inverse ray basis theorem. Suppose that the
measure $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ satisfies the
conditions of part (ii) of Theorem 2. First, it is not difficult to observe
that for such $\mu$ our auxiliary Lemma 1 about the existence of a generalized
minimizing halfspace ensures that the median set $D^{*}\left(\mu\right)$
cannot be full-dimensional.222This claim is proved under weaker conditions as
Corollary 7 in Section A.6 in the Appendix. Because the median set is always
convex, it is not full-dimensional if and only if its interior is empty.
Therefore, for any $\mu$ with a positive continuous density, Lemma 3 gives
that the whole space can be covered by halfspaces  (i) of $\mu$-mass at most
$\alpha^{*}(\mu)$; and  (ii) without a particular connection to any fixed
point $x\in D^{*}\left(\mu\right)$.

Our intention is now to extend the results of Lemma 3 by restricting the
halfspaces on the right hand side of (5) to only those that touch $S$, i.e. to
the collection $\mathcal{H}(S)$. For that purpose, we introduce additional
notation.

######  Definition.

For a convex set $A\subset\mathbb{R}^{d}$ and a point
$x\notin\operatorname{int}\left(A\right)$ define

$\mathfrak{F}(x,A)=\begin{cases}\left\\{F\in\operatorname{\mathcal{F}}\left(A\right)\colon\operatorname{relint}\left(\operatorname{conv}\left(F\cup\\{x\\}\right)\right)\cap\operatorname{cl}\left(A\right)=\emptyset\right\\}&\mbox{if
}x\notin\operatorname{cl}\left(A\right),\\\
\left\\{F\in\operatorname{\mathcal{F}}\left(A\right)\colon
x\in\operatorname{relint}\left(F\right)\right\\}&\mbox{if
}x\in\operatorname{bd}\left(A\right).\end{cases}$

The collection $\mathfrak{F}(x,A)$ consists of those faces of the closed
convex set $\operatorname{cl}\left(A\right)$ that are completely visible from
a point $x\notin\operatorname{int}\left(A\right)$. Note that
$F\in\mathfrak{F}(x,A)$ implies $\dim(F)<d$, and that
$\mathfrak{F}(x,\emptyset)=\\{\emptyset\\}$. The concept of visible faces
relates to the theory of illumination of convex bodies by external light
sources [1]; for its application to the statistics of the depth see [19]. Note
that any point $x\notin\operatorname{int}\left(A\right)$ illuminates at least
one non-empty face of a non-empty convex set $A$; for details see Lemma 17
presented in the Appendix.

Now we are able to extend Lemma 3 to the touching halfspaces. Starting from
(5), our intention is to find $H\in\mathcal{H}(S)$ that covers $x\notin S$
given, with the properties as on the right hand side of (5). The main idea is
to approach any point $y\in F\in\mathfrak{F}\left(x,S\right)$ by a sequence
$\left\\{y_{n}\right\\}_{n=1}^{\infty}\subset\left(\operatorname{cl}\left(S\right)\right)^{\mathsf{c}}$,
that converges to $y$ from the outside of $S$. Lemma 3 gives that for each
$y_{n}$ there is a halfspace $H_{n}\ni y_{n}$ whose interior has empty
intersection with $S$ and
$\mu(\operatorname{int}\left(H_{n}\right))\leq\alpha$, see the left hand panel
of Figure 1. Since $y_{n}\to y$, there exists a convergent subsequence of
$\\{H_{n}\\}_{n=1}^{\infty}$ whose limiting halfspace $H$ can be shown to be
touching $S$, containing $x$, and satisfying
$\mu(\operatorname{int}\left(H\right))\leq\alpha$, as we wanted to show. The
formal proof of Lemma 4 is postponed to the Appendix.

Figure 1. Left panel: Proof of Lemma 4. As $y_{n}\to y$, a subsequence of
halfspaces $\left\\{H_{n}\right\\}_{n=1}^{\infty}$ converges to
$H\in\mathcal{H}(y)$ that touches $S$, contains $x$ and the face $F$, and
$\mu(\operatorname{int}\left(H\right))\leq\alpha$. Right panel: Local
smoothness is a necessary condition for the inverse ray basis theorem. The
origin $o\in\mathbb{R}^{2}$ is the unique median of $\mu$ from Example 1, but
it fails to satisfy the covering condition (4).

######  Lemma 4.

Consider $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ and a bounded convex
set $S\subset\mathbb{R}^{d}$ satisfying (5) with $\alpha\in\mathbb{R}$. For
any $x\notin S$ and $F\in\mathfrak{F}(x,S)$ there exists
$H(x,F)\in\mathcal{H}(S,F)$ such that $x\in H(x,F)$ and
$\mu(\operatorname{int}\left(H(x,F)\right))\leq\alpha$. In particular,

$S^{\mathsf{c}}\subseteq(\operatorname{int}\left(S\right))^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}(S)\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\right\\}.$

Lemma 4 is technical, but presents an important intermediate result. It
provides multiple consequences that we explore in the sequel. Lemma 3 allows
us to apply Lemma 4 to the sets $D_{\alpha}(\mu)$, $U_{\alpha}(\mu)$ or their
interiors, for any $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$. This is the
foundation for the most general statement of the ray basis theorem that can be
devised for general measures. To obtain finer results, it is necessary to
impose additional assumptions. In the literature on the halfspace depth, two
typical assumptions are the smoothness of $\mu$, and the contiguity of its
support. We require these conditions only locally, at the considered depth
regions. Application of Lemma 4 to $S=D_{\alpha}(\mu)$ yields the following
generalization of the inverse ray basis theorem.

######  Theorem 5 (General inverse ray basis theorem).

For any $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, $\alpha\in\mathbb{R}$,
$x\notin\operatorname{int}\left(D_{\alpha}(\mu)\right)$ and
$F\in\mathfrak{F}(x,D_{\alpha}(\mu))$ there exists
$H(x,F)\in\mathcal{H}(D_{\alpha}(\mu),F)$ such that $x\in H(x,F)$ and
$\mu(\operatorname{int}\left(H(x,F)\right))\leq\alpha$. In particular,

(6)
$\left(\operatorname{int}\left(D_{\alpha}(\mu)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}(D_{\alpha}(\mu))\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\right\\}.$

Additionally, if

  1. (i)

$\mu$ is smooth at $D_{\alpha}(\mu)$, then $\mu(H(x,F))=\alpha$ and

$\left(\operatorname{int}\left(D_{\alpha}(\mu)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}(D_{\alpha}(\mu))\colon\mu(H)=\alpha\right\\}.$

  2. (ii)

$\mu$ has contiguous support at $D_{\alpha}(\mu)$, then

$\left(\operatorname{int}\left(D_{\alpha}(\mu)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\right\\}.$

  3. (iii)

$\mu$ is smooth at $D_{\alpha}(\mu)$ and has contiguous support at
$D_{\alpha}(\mu)$, then

$\left(\operatorname{int}\left(D_{\alpha}(\mu)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}\colon\mu(H)=\alpha\right\\}.$

Note that if $D_{\alpha}(\mu)$ fails to be full-dimensional, the left hand
sides in the formulas in Theorem 5 are all $\mathbb{R}^{d}$, and the theorem
therefore gives conditions under which the whole sample space can be covered
by halfspaces of limited $\mu$-mass. For $D_{\alpha}(\mu)$ contained in a
hyperplane, also the condition from part (ii) of Theorem 5 is trivially
satisfied.

Before moving to the discussion about the relevance of Theorem 5, we state an
analogous result for $U_{\alpha}(\mu)$ as another corollary of Lemma 4. Since
obviously $U_{\alpha}(\mu)\subseteq D_{\alpha}(\mu)$, its general statement is
a refinement of the first claim (6) of Theorem 5.

######  Theorem 6.

For any $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, $\alpha\in\mathbb{R}$,
$x\notin\operatorname{int}\left(U_{\alpha}(\mu)\right)$ and
$F\in\mathfrak{F}(x,U_{\alpha}(\mu))$ there exists
$H(x,F)\in\mathcal{H}(U_{\alpha}(\mu),F)$ such that $x\in H(x,F)$ and
$\mu(\operatorname{int}\left(H(x,F)\right))\leq\alpha$. In particular,

(7)
$\left(\operatorname{int}\left(U_{\alpha}(\mu)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\right\\}.$

Additionally, if $\mu$ is smooth at $U_{\alpha}(\mu)$, then
$U_{\alpha}(\mu)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$, so
$U_{\alpha}(\mu)$ is open.

Comparison of Theorem 6 and part (ii) of Theorem 5 draws connections between
the depth regions $D_{\alpha}(\mu)$ and $U_{\alpha}(\mu)$ — for $\mu$ with
contiguous support at $D_{\alpha}(\mu)$, we have
$\operatorname{int}\left(D_{\alpha}(\mu)\right)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$.
We postpone this discussion into Section 3, where connections between upper
level sets of the depth and related convex constructions are explored
thoroughly. In that section, also further applications of these observations
are found.

If $\mu$ is not smooth, the last statement of Theorem 6 certainly cannot be
claimed. This was observed already in [27, Lemma 6] where it was noted that
for $\mu$ an atomic measure on points with unit weights in general position,
$U_{\alpha}(\mu)$ equals $D_{\alpha+1}(\mu)$ for any $\alpha\in\mathbb{R}$,
and as such, $U_{\alpha}(\mu)$ is always a closed set. Our first consequence
of the general ray basis theorem is the following observation concerning the
dimensionality of the median set. It presents a refinement of [26, Proposition
3.4].

######  Corollary 7.

Let $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$.

  1. (i)

If $\mu$ has contiguous support at $D^{*}\left(\mu\right)$, then
$\dim\left(D^{*}\left(\mu\right)\right)\neq d$.

  2. (ii)

If $\mu$ is smooth at $D^{*}\left(\mu\right)$ and $d>1$, then
$\dim\left(D^{*}\left(\mu\right)\right)\neq d-1$.

Without the assumptions of smoothness and contiguous support, the median set
may be of any dimension. Consider, for instance
$\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ that gives mass $1$ to each of
the points $\left(-1,-1\right)^{\mathsf{T}}$,
$\left(-1,1\right)^{\mathsf{T}}$, $\left(1,1\right)^{\mathsf{T}}$, and mass
$2$ to $\left(1,-1\right)^{\mathsf{T}}$. It is easy to see that
$\alpha^{*}(\mu)=2$, and
$D^{*}\left(\mu\right)=L\left[\left(0,0\right)^{\mathsf{T}},\left(1,-1\right)^{\mathsf{T}}\right]$.
For additional discussion on the dimensionality of the median set for
empirical measures we refer to [14].

We are now ready to reformulate the inverse ray basis theorem from part (ii)
of Theorem 2, under minimal assumptions.

######  Corollary 8 (Inverse ray basis theorem for the median).

Suppose that $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ is smooth at
$D^{*}\left(\mu\right)$. Then

$\left(\operatorname{int}\left(D^{*}\left(\mu\right)\right)\right)^{\mathsf{c}}=\bigcup\left\\{H\in\mathcal{H}(D^{*}\left(\mu\right))\colon\mu(H)=\alpha^{*}(\mu)\right\\}.$

If, in addition, $\mu$ has contiguous support at $D^{*}\left(\mu\right)$, then
the covering condition (4) holds true for any $x\in D^{*}\left(\mu\right)$.

The assumptions of Corollary 8 are weaker than those in Theorem 2:  (i)
instead of the existence of the density $f$ of $\mu$ we require only local
smoothness of $\mu$, and  (ii) instead of the strict positivity and continuity
of $f$ in a neighbourhood of the median we need a weaker condition of locally
contiguous support at the median set.

######  Remark (Uniqueness of the median).

In Corollary 7 we show that under the assumptions of both contiguous support
and smoothness of $\mu$ at $D^{*}\left(\mu\right)$, the median set cannot be
of dimension $d$, or $d-1$. In particular, for $d=1$ and $2$ it follows that
the median must be unique. It is tempting to claim that for a smooth measure
$\mu$ with convex support (part (ii) of Theorem 2), the median set must
consist of a single point, compare with [18, Proposition 7] and the discussion
in [26, Section 3]. Surprisingly, it turns out that there exist probability
distributions with a density that is smooth and positive in a convex set in
dimension $d>2$ that fail to possess a unique halfspace median. The
appropriate conditions for the uniqueness of the halfspace median in higher
dimensions turn out to be not only smoothness and contiguous support, but,
quite surprisingly, also a certain integrability assumption. For a detailed
discussion on the problem of the uniqueness of the halfspace median we refer
to [23]. For that reason, one has to be careful when interpreting the inverse
ray basis theorem in Corollary 8. Suppose that a measure $\mu$ is smooth with
contiguous support. Then for any point $x\in D^{*}\left(\mu\right)$ it is true
that the sample space $\mathbb{R}^{d}$ is covered by minimizing halfspaces of
$\mu$ at $x$. That, however, does not mean that the median of $\mu$ must be
unique. Especially in higher dimension, the non-trivial median set may still
lie in the boundary of all the halfspaces from the covering system.

We continue by giving examples that demonstrate that the assumptions of
Corollary 8 are difficult to be weakened. In our first example we show that
without local smoothness, even in the case when $x$ is the unique median of
$\mu$, it may fail to satisfy the covering condition (4).

######  Example 1.

Consider $\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ whose support is
plotted in the right hand panel of Figure 1. It is given as a mixture of the
uniform distributions on line segments
$L_{u}=L\left[\left(0,0\right)^{\mathsf{T}},\left(0,1\right)^{\mathsf{T}}\right]$,
$L_{d}=L\left[\left(0,0\right)^{\mathsf{T}},\left(0,-1\right)^{\mathsf{T}}\right]$,
$L_{l}=L\left[\left(0,0\right)^{\mathsf{T}},\left(-1,0\right)^{\mathsf{T}}\right]$
and
$L_{r}=L\left[\left(0,0\right)^{\mathsf{T}},\left(1,0\right)^{\mathsf{T}}\right]$,
such that $\mu(L_{l})=\mu(L_{r})=\mu(L_{d})=1$ and $\mu(L_{u})=2$. The origin
$o=\left(0,0\right)^{\mathsf{T}}$ is the unique median of $\mu$ with
$D(o;\mu)=2$, since for any other point $x\in\mathbb{R}^{2}$ there is a
halfspace $H\in\mathcal{H}(x)$ that is parallel with one of the axes such that
$\mu(H)<2$. At the same time, each closed halfspace containing $L_{u}$ has
$\mu$-mass at least $3$, implying that it is impossible to cover (any point
from) $L_{u}$ by $H\in\mathcal{H}(o)$ such that $\mu(H)\leq\alpha^{*}(\mu)=2$.
On the other hand, observe that part (ii) of Theorem 5 is valid, as we easily
find open halfspaces whose boundary passes through the origin with mass at
most $\alpha^{*}(\mu)$ whose closures cover $\mathbb{R}^{2}$.

It is known that for a uniform distribution $\mu$ on a triangle $\Delta$ in
the plane, the barycentre $o\in\Delta$ of $\Delta$ is the unique median of
$\mu$ [24, Section 5.3]. We construct an example of a uniform distribution on
the set obtained by removing a narrow strip containing $o$ from $\Delta$. We
show that our measure does not satisfy the assumption of contiguous support at
$D^{*}\left(\mu\right)$, at the same time its median set is full-dimensional,
and contains points that fail to cover the plane by their minimizing
halfspaces as in (4).

######  Example 2.

Consider the equilateral triangle in $\mathbb{R}^{2}$ centred at the origin
$o=\left(0,0\right)^{\mathsf{T}}$ determined by points $a=(0,2)^{\mathsf{T}}$,
$b=(-\sqrt{3},-1)^{\mathsf{T}}$ and $c=(\sqrt{3},-1)^{\mathsf{T}}$ as
displayed in Figure 2. For $x\in(0,2)$ and $y\in(0,1)$ denote
$x_{c}=\left(0,x\right)^{\mathsf{T}}$ and
$y_{c}=\left(0,-y\right)^{\mathsf{T}}$, and let $x_{l}$ and $x_{r}$ be the
points where the horizontal line containing $x_{c}$ intersects $L(a,b)$ and
$L(a,c)$, respectively. Analogously, define points $y_{l}$ and $y_{r}$ for the
horizontal line that contains $y_{c}$. Let
$\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ be the uniform distribution on
the set
$S=\operatorname{conv}\left(\\{a,x_{l},x_{r}\\}\right)\cup\operatorname{conv}\left(\\{y_{l},b,c,y_{r}\\}\right)$
with total mass being the area of $S$. It is possible to choose $x$ and $y$
positive, but small enough so that the set $D^{*}\left(\mu\right)$ is full-
dimensional, while $y_{c}\in D^{*}\left(\mu\right)$ is the only point in
$\mathbb{R}^{2}$ that satisfies the covering property (4). For a detailed
technical proof of this claim, we refer to Section A.8 in the Appendix.

Figure 2. Example 2: For the inverse ray basis theorem, it is not enough to
assume only the smoothness of $\mu$, without the property of contiguous
support. For the uniform measure $\mu$ on the coloured region, the median is a
full-dimensional set, yet the point $y_{c}$ displayed in the left hand panel
is the only point in $\mathbb{R}^{2}$ that satisfies the covering condition
(4). On the right hand panel we see several numerically computed depth regions
$D_{\alpha}(\mu)$, with the median set $D^{*}\left(\mu\right)$ being the
smallest region, located in the closed strip removed from the triangle.

##  3\. Depth regions and floating body

Formula (2) allows us to write any depth region $D_{\alpha}(\mu)$ as an
intersection of closed halfspaces whose complements have $\mu$-mass smaller
than $\alpha$. Another important affine equivariant set is the _floating body_
of $\mu$ corresponding to $\alpha\in\mathbb{R}$

$U_{\alpha}^{FB}(\mu)=\bigcap\left\\{H\in\mathcal{H}\colon\mu(H^{\mathsf{c}})\leq\alpha\right\\},$

as defined in [2, Section 5]. According to the discussion in [20], the
floating bodies are of great interest in both geometry and probability theory.
For the special case of $\mu$ a uniform distribution on a full-dimensional
convex set $K\subset\mathbb{R}^{3}$, the floating body has a compelling
mechanical interpretation — the set $U_{\alpha}^{FB}(\mu)$ can be described as
the part of the convex solid $K$ of (volumetric mass) density
$\alpha\in(0,\alpha^{*}(\mu))$ that never submerges beneath the surface of
water of unit density, when fully rotated on the surface. The history of the
research on floating bodies goes well into the 19th century. In statistics, a
construction equivalent to the floating body of a measure is much more recent,
and sometimes referred to as the multivariate trimming [21, 17]. Being
intersections of closed sets, both $D_{\alpha}(\mu)$ and
$U_{\alpha}^{FB}(\mu)$ are closed. In what follows we use the results of
Section 2, and precise the connections between the depth regions
$D_{\alpha}(\mu)$, $U_{\alpha}(\mu)$, the floating body
$U_{\alpha}^{FB}(\mu)$, and a further set that turns out to be of interest in
our analysis

$U^{\circ}_{\alpha}(\mu)=\bigcap\left\\{\operatorname{int}\left(H\right)\colon
H\in\mathcal{H},\ \mu(H^{\mathsf{c}})\leq\alpha\right\\}.$

We demonstrate that this last region is an upper level set of a function
closely related to the halfspace depth

$D^{\circ}(x;\mu)=\inf_{x\in H}\mu(\operatorname{int}\left(H\right)),$

considered in, e.g., [18, Lemma 1]. Note also that
$(U^{\circ}_{\alpha}(\mu))^{\mathsf{c}}=\bigcup\\{H\in\mathcal{H}\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\\}$
already appeared in Theorem 6, meaning that
$U^{\circ}_{\alpha}(\mu)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$ must
be an open set. The following theorem comprehensively covers the inter-
relations between all these affine constructions, and generalizes several
results that can be found scattered in the relevant literature on multivariate
trimming concepts [26, 17, 18, 3, 20].

######  Theorem 9.

For $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ and $\alpha\in\mathbb{R}$

(8) $\displaystyle
U^{\circ}_{\alpha}(\mu)=\operatorname{int}\left(U_{\alpha}(\mu)\right)\subseteq\operatorname{cl}\left(U_{\alpha}(\mu)\right)\subseteq
U_{\alpha}^{FB}(\mu)\subseteq D_{\alpha}(\mu).$

Additionally,

  1. (i)

$U^{\circ}_{\alpha}(\mu)=\left\\{x\in\mathbb{R}^{d}\colon
D^{\circ}(x;\mu)>\alpha\right\\}$.

  2. (ii)

if $\operatorname{int}\left(U_{\alpha}(\mu)\right)\neq\emptyset$, then
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)=U_{\alpha}^{FB}(\mu)$.

  3. (iii)

if $\mu$ has contiguous support at $D_{\alpha}(\mu)$, then
$\operatorname{int}\left(D_{\alpha}(\mu)\right)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$.

  4. (iv)

if for each $H^{\prime}\in\mathcal{H}(D_{\alpha}(\mu))$ we have that
$H^{\prime}\subset H\in\mathcal{H}$ implies $\mu(H^{\prime})<\mu(H)$, then
$U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$.

  5. (v)

if $\operatorname{int}\left(U_{\alpha}(\mu)\right)\neq\emptyset$ and $\mu$ has
contiguous support at $D_{\alpha}(\mu)$, then
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)=D_{\alpha}(\mu)$.

An application of Theorem 9 to the estimation of the depth regions
$D_{\alpha}(\mu)$ from datasets is given in Section 4.3. The condition of non-
empty interior of $U_{\alpha}(\mu)$ that figures in parts (ii) and (v) of the
previous theorem is not restrictive, as shown in the next lemma.

######  Lemma 10.

If for $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ and
$\alpha\in\mathbb{R}$ there exists a point $x\in U_{\alpha}(\mu)$ such that
$\mu$ is smooth at $x$, then
$\operatorname{int}\left(U_{\alpha}(\mu)\right)\neq\emptyset$.

Combining Lemma 10 and Theorems 6 and 9 we get that for
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ with a density and
$\alpha\in(0,\alpha^{*}(\mu))$ the set $U_{\alpha}(\mu)$ is open and
$U_{\alpha}(\mu)\subset\operatorname{cl}\left(U_{\alpha}(\mu)\right)=U_{\alpha}^{FB}(\mu)$.
If the support of $\mu$ is, in addition, contiguous, we can also write
$U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$. In the simplest situation of
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ with a density that is positive
in the convex support of $\mu$, we can therefore write
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)=U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$,
and the floating bodies completely coincide with the central regions of the
depth. The last situation is common in the literature on floating bodies in
geometry, where all the above definitions are used interchangeably. In the
general setup of the depth and (probability) measures, it is however necessary
to differentiate between them.

Our statement of Theorem 9 is strict — none of the inclusions can be reversed,
in general. Each strict upper level set $U_{\alpha}(\mu)$ of a smooth measure
$\mu$ is open (Theorem 6), and thus strictly smaller than
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)$. It is easy to construct a
measure without contiguous support that violates
$U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$, see e.g. [20, Figure 7]. Even for
measures with contiguous support, equality
$U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$ from part (iv) of Theorem 9 can fail;
consider $\mu$ being the Dirac measure at the origin $o\in\mathbb{R}^{d}$, and
$\alpha=1$. In that case, $U_{\alpha}^{FB}(\mu)=\emptyset$, while
$D_{\alpha}(\mu)=\\{o\\}$. In the following example we show that in the case
when $\operatorname{int}\left(U_{\alpha}(\mu)\right)=\emptyset$, the halfspace
depth may fail to satisfy
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)=D_{\alpha}(\mu)$ (the so-called
strict monotonicity property), even under the assumption of contiguous
support. The example demonstrates that also the remaining set inclusions in
Theorem 9 cannot be reversed for general measures without local assumptions.

######  Example 3.

Consider $\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ given by a mixture of
the uniform distribution on the unit disc
$C=\left\\{x\in\mathbb{R}^{2}\colon\left\|x\right\|\leq 1\right\\}$ with mass
$1/4$, and two atoms located at $m=\left(0,1\right)^{\mathsf{T}}$ and
$z=\left(0,1/2\right)^{\mathsf{T}}$ with masses $1/2$ and $1/4$, respectively.
Point $m$ is certainly the unique halfspace median of $\mu$, with
$D\left(m;\mu\right)=\alpha^{*}(\mu)=1/2$. It is also easy to see that the
halfspace depth of $\mu$ at the point $o=\left(0,0\right)^{\mathsf{T}}$ equals
$1/8$. Because of the two large atoms of $\mu$, all points in the open line
segment $L(o,z)$ have also depth $1/8$. Similarly, all points in $L(z,m)$ have
the same depth as $z$, that is $D\left(z;\mu\right)=1/8+1/4=3/8$. Because for
any $y\notin L[o,m]$ certainly $D\left(y;\mu\right)<1/8$, we have for
$\alpha=1/8$ that $D_{\alpha}(\mu)=L[o,m]$, while
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)=L[z,m]$, meaning that the
strict monotonicity property of the halfspace depth is violated, see also the
left hand panel of Figure 3. In addition, it is easy to verify that
$U_{\alpha}^{FB}(\mu)$ is equal to
$D_{\alpha}(\mu)\supset\operatorname{cl}\left(U_{\alpha}(\mu)\right)$, while
the set $U^{\circ}_{\alpha}(\mu)$ from (8) is empty. We conclude that, in
general, one cannot write $U_{\alpha}(\mu)$ as a simple intersection of
halfspaces, as possible for $D_{\alpha}(\mu)$ in (2).

Figure 3. Left panel: Condition
$\operatorname{int}\left(U_{\alpha}(\mu)\right)\neq\emptyset$ is needed for
the strict monotonicity of the halfspace depth: Example 3 of a distribution
$\mu$ where
$D_{\alpha}(\mu)\neq\operatorname{cl}\left(U_{\alpha}(\mu)\right)$. The
halfspace depth of $\mu$ takes only values in the set
$[0,1/8]\cup\\{3/8,1/2\\}$. The upper level set $D_{\alpha}(\mu)$ with
$\alpha=1/8$ is the line segment between the origin $o$ and the black atom
$m$. Points on the thick dashed line all attain depth equal to $3/8$. Right
panel: For the atomic measure $\mu$ supported in points $a$–$f$ from Example
4, the set $C^{*}\left(\mu\right)$ (innermost coloured triangle) is a proper
subset of $D^{*}\left(\mu\right)$ (outer coloured triangle).

##  4\. Applications: Refined medians, computation, and consistency

###  4.1. Covering halfspace median

Our first application of the ray basis theorems from Section 2 is motivated by
another refinement of the main equality (6) applied to
$\alpha=\alpha^{*}(\mu)$: For any measure
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ there exists a point $x\in
D^{*}\left(\mu\right)$ that allows us to cover the whole space with halfspaces
$H\in\mathcal{H}(x)$ whose interior has $\mu$-mass at most

$\gamma^{*}(\mu)=\inf\\{\alpha>0\colon\operatorname{int}\left(D_{\alpha}(\mu)\right)=\emptyset\\}=\sup\\{\alpha>0\colon\operatorname{int}\left(D_{\alpha}(\mu)\right)\neq\emptyset\\}.$

Observe that $\gamma^{*}(\mu)\leq\alpha^{*}(\mu)$. Therefore, the following
claim is stronger than the inverse ray basis theorem not only  (i) because it
asserts the existence of such a special point in $D^{*}\left(\mu\right)$, but
also in the sense that  (ii) the masses of the halfspaces that cover
$\mathbb{R}^{d}$ are ensured to be at most $\gamma^{*}(\mu)$, and not just
$\alpha^{*}(\mu)$.

######  Theorem 11.

For any $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ there exists a point
$x\in D^{*}\left(\mu\right)$, such that

(9)
$\mathbb{R}^{d}=\bigcup\left\\{H\in\mathcal{H}(x)\colon\mu(\operatorname{int}\left(H\right))\leq\gamma^{*}(\mu)\right\\}.$

Suppose that $D^{*}\left(\mu\right)$ is not a single point set. It is
certainly of interest to single out the subset $C^{*}\left(\mu\right)$ of
$D^{*}\left(\mu\right)$ of those points that satisfy the additional covering
property (9) characteristic to centrally located points. We call such medians
the _covering halfspace medians_ (or simply _covering medians_) of the measure
$\mu$. The concept of covering medians is interesting especially in the
situation when the median set is full-dimensional, as frequently happens with
data generated from a smooth distribution. In that situation, the relatively
large median set typically reduces to a smaller subset of the most centrally
located points being the covering medians.

######  Example 4.

For an empirical measure $\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ with
atoms at points $a=\left(3,0\right)^{\mathsf{T}}$,
$b=\left(0,5\right)^{\mathsf{T}}$, $c=\left(-5,0\right)^{\mathsf{T}}$,
$d=\left(1,3\right)^{\mathsf{T}}$, $e=\left(-1,3\right)^{\mathsf{T}}$ and
$f=\left(-1,1\right)^{\mathsf{T}}$, the median set $D^{*}\left(\mu\right)$ is
equal to the triangle determined by points $d$, $e$ and $f$, while
$C^{*}\left(\mu\right)$ is a smaller triangle determined by the lines
$l(a,e)$, $l(b,f)$ and $l(c,d)$, see the right hand panel of Figure 3.

Figure 4. Example 5: The set of covering medians $C^{*}\left(\mu\right)$ may
fail to be convex. For the measure $\mu$, whose support is displayed in this
plot in colour, the points $x$ and $y$ belong to $C^{*}\left(\mu\right)$. But,
their midpoint $o$ is not contained in $C^{*}\left(\mu\right)$.

The covering medians present a genuine refinement of the halfspace medians.
They can be shown to satisfy an array of properties expected from well-behaved
location estimators, such as  (i) existence as proved in Theorem 11;  (ii)
affine equivariance; or  (iii) plausible continuity properties when considered
as a set-valued function of the measure $\mu$.  The covering medians are also
intimately connected with the robustness properties of the halfspace median
and the depth. All these results will be presented elsewhere in an appropriate
context. Here we mention only several basic observations closely linked to
Theorem 11. The first one demonstrates that unlike the standard halfspace
median set, the set of the covering medians does not have to be convex.

######  Example 5.

Let $\mu\in\mathcal{M}\left(\mathbb{R}^{2}\right)$ be the mixture of uniform
distributions on the triangles _A_ – _K_ displayed in Figure 4, such that
$\mu(A)=\mu(F)=\mu(G)=2$ and
$\mu(B)=\mu(C)=\mu(D)=\mu(E)=\mu(H)=\mu(I)=\mu(J)=\mu(K)=1$. It can be shown
that the set $C^{*}\left(\mu\right)$ contains points $x$ and $y$ displayed in
Figure 4, but it does not contain the origin $o=(x+y)/2$. Therefore, the set
of the covering halfspace medians of $\mu$ is not convex. For a detailed proof
see the Appendix, Section A.12.

The proof of Theorem 11 allows us to devise a simple algorithm for finding the
covering medians of $\mu$ listed as Algorithm 1. Although this program is
applicable to any measure including empirical measures of random samples, its
main purpose is not the computation of the sample covering medians of large
datasets. Rather, it is intended to guide a quick manual procedure for
restricting the location of possible covering medians in visual examples such
as those presented throughout this paper.

input : the full-dimensional median region $D^{*}\left(\mu\right)$ of a
measure $\mu$

input : a small positive constant $\varepsilon$ determining desired precision

output : a covering median of $\mu$

$S_{0}\leftarrow D^{*}\left(\mu\right)$ ;

$x_{0}\leftarrow$ barycentre of $S_{0}$ ;

$k\leftarrow 0$ ;

while  _$x_{k}\notin C^{*}\left(\mu\right)$ and volume of $S_{k}$ exceeds
$\varepsilon$_ do

$H_{k}\leftarrow$ generalized minimizing halfspace of $\mu$ at $x_{k}$ ;

// such a halfspace exists by Lemma 1

$S_{k+1}\leftarrow\operatorname{cl}\left(S_{k}\setminus H_{k}\right)$ ;

$k\leftarrow k+1$ ;

$x_{k}\leftarrow$ barycentre of $S_{k}$ ;

end while

return _$x_{k}$_ ;

Algorithm 1 Search for a covering halfspace median of a measure $\mu$.

Once a covering median of $\mu$ is found, it is of interest do determine
whether it is unique. Supposing that a covering median of $\mu$ and its
collection of generalized minimizing halfspaces are available as the output of
Algorithm 1, the following theorem gives a sufficient condition for the
uniqueness of this covering median.

######  Theorem 12.

Let $x\in\mathbb{R}^{d}$ be a covering median of
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$. Denote by $\mathcal{H}_{\min}$
the collection of halfspaces on the right hand side of the following display

$\mathbb{R}^{d}=\bigcup\left\\{H\in\mathcal{H}(x)\colon\mu(\operatorname{int}\left(H\right))\leq\gamma^{*}(\mu)\right\\}.$

If for each $H\in\mathcal{H}_{\min}$ and $H^{\prime}\supset H$ it follows that
$\mu(H^{\prime})>\mu(H)$, and if there is no subset
$\mathcal{H}^{\prime}\subseteq\mathcal{H}_{\min}$ such that
$\bigcup\mathcal{H}^{\prime}=\mathbb{R}^{d}$ and
$\bigcap\mathcal{H}^{\prime}\neq\\{x\\}$, then
$C^{*}\left(\mu\right)=\\{x\\}$.

Note that the point $y_{c}$ from Example 2 given in Section 3 satisfies the
conditions of Theorem 12 and $C^{*}\left(\mu\right)=\\{y_{c}\\}$, while the
median of that measure $\mu$ is full-dimensional. On the other hand, Theorem
12 does not apply to any of the two covering medians $x,y$ found in Example 5,
as the condition regarding contiguous support at minimizing halfspaces is not
satisfied.

###  4.2. Trimmed regions for atomic measures

Our second application of the generalized ray basis theorem is a necessary
condition on the depth regions $D_{\alpha}(\mu)$ of atomic measures.

######  Corollary 13.

For $\mu$ an atomic measure with finitely many atoms, $\alpha\in\mathbb{R}$,
$x\notin\operatorname{int}\left(D_{\alpha}(\mu)\right)$ and
$F\in\mathfrak{F}(x,D_{\alpha}(\mu))$ there exists
$H(x,F)\in\mathcal{H}(D_{\alpha}(\mu),F)$ such that $x\in H(x,F)$ and
$\mu(\operatorname{int}\left(H(x,F)\right))<\alpha$. Moreover, each face $F$
of the convex polytope $D_{\alpha}(\mu)$ of dimension $\dim(F)<d$ is contained
in the convex hull of at least $\min\\{\dim(D_{\alpha}(\mu))+1,d\\}$ atoms
located in a hyperplane in $\mathbb{R}^{d}$.

A special case of this result in the situation when $D_{\alpha}(\mu)$ is full-
dimensional was used in the derivation of a fast algorithm for the computation
of the depth regions of datasets in [15]. The present general version may find
applications in the computation of the halfspace depth of datasets in the
situation when $\dim\left(D_{\alpha}(\mu)\right)<d$. For example, suppose that
for given $\alpha\in\mathbb{R}$ and $\mu$ corresponding to a dataset the
algorithm from [15] fails to find an interior point of the region
$D_{\alpha}(\mu)$. The reason may be twofold: either  (i) $D_{\alpha}(\mu)$is
an empty set, or  (ii) it is less than full-dimensional. Corollary 13 asserts
that in the latter case, the region $D_{\alpha}(\mu)$ must be contained in a
hyperplane spanned by $d$ data points. Thus, the search for $D_{\alpha}(\mu)$
may continue in the intersection of data-determined hyperplanes, and the last
known non-empty region $D_{\beta}(\mu)$ for $\beta<\alpha$.

###  4.3. Estimation of central regions

Part (v) of Theorem 9 yields an important equality. According to Dyckerhoff,
[8, Definition 3.1] the halfspace depth is _strictly monotone_ at
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ if for all
$\alpha\in(0,\alpha^{*}(\mu))$ we can write
$D_{\alpha}(\mu)=\operatorname{cl}\left(U_{\alpha}(\mu)\right)$. Strict
monotonicity is a crucial assumption that ensures the sample version
consistency of the halfspace depth-trimmed regions [8, Example 4.2]. As a
consequence of Lemma 10 we obtain a condition ensuring the strict monotonicity
of the halfspace depth — for a measure
$\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ to have a strictly monotone
depth it is enough to assume contiguous support and smoothness at a single
point of the median set. In particular, under these very mild conditions, it
is possible to guarantee the almost sure uniform convergence of the sample
depth trimmed regions to their population counterparts. To precise this, we
need to consider a topology on the space of compact subsets of
$\mathbb{R}^{d}$. A natural choice is that given by the Hausdorff distance
[25, Section 1.8]. For $K,L\subset\mathbb{R}^{d}$ compact the Hausdorff
distance of $K$ and $L$ is given by

$\delta_{H}(K,L)=\max\left\\{\sup_{x\in K}\inf_{y\in
L}\left|x-y\right|,\sup_{x\in L}\inf_{y\in K}\left|x-y\right|\right\\}.$

######  Corollary 14.

Let $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ be a probability measure
with contiguous support that is smooth at some $x\in D^{*}\left(\mu\right)$.
Let $X_{1},\dots,X_{n}$ be a random sample from $\mu$ defined on the
probability space $\left(\Omega,\mathcal{A},\mathsf{P}\right)$, and denote by
$\widetilde{\mu_{n}}\equiv\widetilde{\mu_{n}}(\omega)\in\mathcal{M}\left(\mathbb{R}^{d}\right)$
the empirical measure of $X_{1},\dots,X_{n}$. Then for any closed interval
$A\subset\left(0,\alpha^{*}(\mu)\right)$

$\mathsf{P}\left(\left\\{\omega\in\Omega\colon\lim_{n\to\infty}\sup_{\alpha\in
A}\delta_{H}\left(D_{\alpha}(\widetilde{\mu_{n}}),D_{\alpha}(\mu)\right)=0\right\\}\right)=1.$

Further, let $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ be any measure
that is smooth with contiguous support. Then for any closed interval
$A\subset\left(0,\alpha^{*}(\mu)\right)$ and any $\mu_{n}$ converging weakly
to $\mu$ we have

$\lim_{n\to\infty}\sup_{\alpha\in
A}\delta_{H}\left(D_{\alpha}(\mu_{n}),D_{\alpha}(\mu)\right)=0.$

This result follows directly from [8, Theorem 4.5] and our previous
discussion. The _General assumption_ in [8, p. 9] is satisfied thanks to our
Lemma 10; the assumption of compact convergence of the depth from [8] follows,
e.g., from [20, Section 3.2.7]. Corollary 14 should be compared with earlier
contributions regarding the consistency of the trimmed regions and derived
quantities [21, 10, 34, 11, 16, 31, 30]. In those references analogous
consistency results are proved under more restrictive conditions.

##  Appendix A Proofs of the theoretical results

We begin with a lemma collecting several properties of the convergence of
halfspaces.

######  Lemma 15.

Consider $\mu\in\mathcal{M}\left(\mathbb{R}^{d}\right)$, a sequence of closed
halfspaces $\\{H_{n}\\}_{n=1}^{\infty}\subset\mathcal{H}$, and
$\alpha\in\mathbb{R}$. The following claims hold true:

  1. (i)

If there is a sequence $\\{x_{n}\\}_{n=1}^{\infty}$ such that
$x_{n}\in\operatorname{bd}\left(H_{n}\right)$ and $x_{n}\rightarrow x$, then
there exists a subsequence $\\{H_{n_{k}}\\}_{k=1}^{\infty}$ converging to a
closed halfspace $H\in\mathcal{H}(x)$.

  2. (ii)

If $H_{n}\rightarrow H$ and
$\mu(\operatorname{int}\left(H_{n}\right))\leq\alpha$ for each $n$, then
$\mu(\operatorname{int}\left(H\right))\leq\alpha$.

  3. (iii)

If $H_{n}\rightarrow H$, $\mu(\operatorname{int}\left(H_{n}\right))\leq\alpha$
for each $n$, $H\cap D_{\alpha}(\mu)\neq\emptyset$ and
$\mu(\operatorname{bd}\left(H\right))=0$, then $\mu(H)=\alpha$.

  4. (iv)

If $H_{n}\rightarrow H$ and $x\notin\operatorname{int}\left(H_{n}\right)$ for
each $n$, then $x\notin\operatorname{int}\left(H\right)$.

###### Proof.

For part (i), denote $H_{n}=H_{x_{n},v_{n}}$. The set of unit vectors
$\mathbb{S}^{d-1}$ is bounded, meaning that
$\left\\{v_{n}\right\\}_{n=1}^{\infty}\subset\mathbb{S}^{d-1}$ contains a
convergent subsequence $\\{v_{n_{k}}\\}_{k=1}^{\infty}$, whose limit point we
denote by $v\in\mathbb{S}^{d-1}$. We obtain that $H_{n_{k}}\rightarrow
H_{x,v}=H\in\mathcal{H}(x)$ as $k\to\infty$ as desired. Part (ii) is a
consequence of the Fatou lemma [6, Lemma 4.3.3] and the fact that
$\operatorname{int}\left(H\right)\subseteq{\lim\inf}_{k\to\infty}\operatorname{int}\left(H_{n_{k}}\right)$.
Under the assumptions of part (iii) there exists $x\in D_{\alpha}(\mu)\cap H$.
For this point we can directly write by part (ii) of this lemma $\alpha\leq
D\left(x;\mu\right)\leq\mu(H)=\mu(\operatorname{int}\left(H\right))\leq\alpha$.
To prove (iv), assume that $x\in\operatorname{int}\left(H\right)$. Then
$x\in\operatorname{int}\left(H_{n}\right)$ starting from some index, a
contradiction. ∎

######  Lemma 16.

If $H\in\mathcal{H}$ and $S\subset\mathbb{R}^{d}$ is a convex set such that
$\operatorname{int}\left(H\right)\cap S=\emptyset$ and
$\operatorname{relint}\left(S\right)\cap H\neq\emptyset$, then
$\operatorname{aff}\left(S\right)\subseteq\operatorname{bd}\left(H\right)$.

###### Proof.

Consider $x\in\operatorname{relint}\left(S\right)\cap H$. Since
$x\in\operatorname{relint}\left(S\right)$, there exists a ball $B_{x}$ centred
at $x$ in space $\operatorname{aff}\left(S\right)$ so that $B_{x}\subset S$.
Since $\operatorname{int}\left(H\right)\cap S=\emptyset$ we get
$x\in\operatorname{bd}\left(H\right)$ and
$B_{x}\subset\operatorname{bd}\left(H\right)$, meaning that
$\operatorname{aff}\left(S\right)=\operatorname{aff}\left(B_{x}\right)\subseteq\operatorname{bd}\left(H\right)$.
∎

######  Lemma 17.

For $A\subset\mathbb{R}^{d}$ non-empty convex and
$x\notin\operatorname{int}\left(A\right)$ the collection $\mathfrak{F}(x,A)$
contains a non-empty set.

###### Proof.

In the case that $x\in\operatorname{bd}\left(A\right)$, there is a face
$F\neq\emptyset$ of $\operatorname{bd}\left(A\right)$ containing $x$ in its
relative interior [25, Theorem 2.1.2], so $\mathfrak{F}(x,A)$ contains a non-
empty set. Otherwise, let $x\notin\operatorname{cl}\left(A\right)$. The Hahn-
Banach theorem [25, Theorem 1.3.4] guarantees that there exists a touching
halfspace $H\in\mathcal{H}(A)$ of $\operatorname{cl}\left(A\right)$ such that
$x\in\operatorname{int}\left(H\right)$. Denote
$F=\operatorname{cl}\left(A\right)\cap\operatorname{bd}\left(H\right)$. Then,
$F$ is a non-empty face of $\operatorname{cl}\left(A\right)$, and certainly
also $\operatorname{cl}\left(A\right)\setminus F\subseteq H^{\mathsf{c}}$,
$F\subseteq\operatorname{bd}\left(H\right)$, and
$x\in\operatorname{int}\left(H\right)$, meaning that $F\in\mathfrak{F}(x,A)$.
∎

###  A.1. Proof of Lemma 1

Let $\alpha=D\left(x;\mu\right)$. The definition of the halfspace depth
ensures that for each $n=1,2,\dots$ there exists $H_{n}\in\mathcal{H}(x)$ such
that $\mu(H_{n})<\alpha+1/n$. Applying part (i) of Lemma 15 we obtain a
subsequence $\\{H_{n_{k}}\\}_{k=1}^{\infty}$ converging to
$H\in\mathcal{H}(x)$. From the Fatou lemma [6, Lemma 4.3.3] we get
$\mu(\operatorname{int}\left(H\right))\leq{\lim\inf}_{k\to\infty}\mu(\operatorname{int}\left(H_{n_{k}}\right))\leq\alpha$.

Under the additional assumption of smoothness of $\mu$ at $x$ we know that
$\mu\left(\operatorname{bd}\left(H\right)\right)=\mu\left(\operatorname{bd}\left(H_{n}\right)\right)=0$
for each $n=1,2,\dots$. In that case, the Fatou lemma guarantees
$\mu(H)=\lim_{k\to\infty}\mu(H_{n,k})\leq\alpha$. Since also
$\alpha=D\left(x;\mu\right)\leq\mu(H)\leq\alpha$, we obtain the desired
result.

###  A.2. Proof of Lemma 3

For any $S\subset\mathbb{R}^{d}$ convex and $H\in\mathcal{H}$, the condition
$\operatorname{int}\left(H\right)\cap S=\emptyset$ implies
$\operatorname{int}\left(H\right)\cap\operatorname{int}\left(S\right)=\emptyset$.
Furthermore $\operatorname{int}\left(S\right)\subseteq S$. Thus, it is enough
to show the claim (5) with the smaller set $\operatorname{int}\left(S\right)$
on the left hand side, and the condition $\operatorname{int}\left(H\right)\cap
S=\emptyset$ in the expression on the right hand side.

We start with $D_{\alpha}(\mu)$. First note that by the very definition of
$D_{\alpha}(\mu)$

(10) $\mu(\operatorname{int}\left(H\right))<\alpha\mbox{ implies
}\operatorname{int}\left(H\right)\cap D_{\alpha}(\mu)=\emptyset.$

Consider any $x\notin\operatorname{int}\left(D_{\alpha}(\mu)\right)$. Suppose
first that $x\notin D_{\alpha}(\mu)$, meaning that $D(x;\mu)<\alpha$. The
auxiliary Lemma 1 implies that there exists $H\in\mathcal{H}(x)$ such that
$\mu(\operatorname{int}\left(H\right))<\alpha$, and (10) ensures that $x$ is
covered by the union on the right hand side of (5). If
$x\in\operatorname{bd}\left(D_{\alpha}(\mu)\right)$, we can approach $x$ by a
sequence
$\\{x_{n}\\}_{n=1}^{\infty}\subset\left(D_{\alpha}(\mu)\right)^{\mathsf{c}}$
such that $x_{n}\to x$. We apply Lemma 1 to each $x_{n}$ to obtain halfspaces
$H_{n}\in\mathcal{H}(x_{n})$ such that
$\mu(\operatorname{int}\left(H_{n}\right))<\alpha$, and consequently
$\operatorname{int}\left(H_{n}\right)\cap D_{\alpha}(\mu)=\emptyset$ for each
$n$ by (10). From Lemma 15, parts (i), (ii) and (iv) we conclude that there
exists $H\in\mathcal{H}(x)$ such that
$\mu(\operatorname{int}\left(H\right))\leq\alpha$ and
$\operatorname{int}\left(H\right)\cap D_{\alpha}(\mu)=\emptyset$, as desired.

The proof for $U_{\alpha}(\mu)$ is analogous. First, observe that
$\mu(\operatorname{int}\left(H\right))\leq\alpha$ implies
$\operatorname{int}\left(H\right)\cap U_{\alpha}(\mu)=\emptyset$ by the
definition of $U_{\alpha}(\mu)$. Consider any
$x\notin\operatorname{int}\left(U_{\alpha}(\mu)\right)$. If $x\notin
U_{\alpha}(\mu)$, then $D(x;\mu)\leq\alpha$, so there is $H\in\mathcal{H}(x)$
such that $\mu(\operatorname{int}\left(H\right))\leq\alpha$ by Lemma 1.
Otherwise, $x\in\operatorname{bd}\left(U_{\alpha}(\mu)\right)$ may be
approached by a sequence
$\\{x_{n}\\}_{n=1}^{\infty}\subset\left(U_{\alpha}(\mu)\right)^{\mathsf{c}}$
such that $x_{n}\to x$, and Lemma 15 again gives the desired result.

###  A.3. Proof of Lemma 4

Since each $H\in\mathcal{H}(S)$ has empty intersection with
$\operatorname{int}\left(S\right)$, it is only left to prove
$(\operatorname{int}\left(S\right))^{\mathsf{c}}\subseteq\bigcup\left\\{H\in\mathcal{H}(S)\colon\mu(\operatorname{int}\left(H\right))\leq\alpha\right\\}$.
We take any $x\notin\operatorname{int}\left(S\right)$ and consider three
cases.

Case (i): $\operatorname{int}\left(S\right)=\emptyset$. By definition,
$\mathcal{H}\left(\emptyset\right)=\mathcal{H}$. Therefore, the statement
reduces directly to condition (5).

Case (ii): $x\notin\operatorname{cl}\left(S\right)$ and
$\operatorname{int}\left(S\right)\neq\emptyset$. We consider only the
situation when $F$ is non-empty, which is possible by Lemma 17, as the other
case follows trivially from condition (5). Denote
$C=\operatorname{relint}\left(\operatorname{conv}\left(F\cup\\{x\\}\right)\right)$.
Take $y\in\operatorname{relint}\left(F\right)$, and the sequence
$y_{n}=\left(1-1/n\right)y+x/n\in C$, for $n=1,2,\dots$, that converges to
$y$. Because $x\notin S$, the choice of $F$ ensures that
$C\cap\operatorname{cl}\left(S\right)=\emptyset$. Thus, for each $n$ we have
$y_{n}\notin\operatorname{cl}\left(S\right)$ and condition (5) implies the
existence of $H_{n}\in\mathcal{H}$ such that $y_{n}\in H_{n}$,
$\operatorname{int}\left(H_{n}\right)\cap S=\emptyset$ and
$\mu\left(\operatorname{int}\left(H_{n}\right)\right)\leq\alpha$. Applying
Lemma 15, parts (i), (ii) and (iv), we conclude that there is a subsequence
$\\{H_{n_{k}}\\}_{k=1}^{\infty}$ converging to $H\in\mathcal{H}(y)$, such that
$\mu(\operatorname{int}\left(H\right))\leq\alpha$ and
$\operatorname{int}\left(H\right)\cap S=\emptyset$. Then
$y\in\operatorname{relint}\left(F\right)\cap H$. From Lemma 16 it follows that
$F\subset\operatorname{bd}\left(H\right)$, so $H\in\mathcal{H}(S,F)$.

Case (iii): $x\in\operatorname{bd}\left(S\right)$ and
$\operatorname{int}\left(S\right)\neq\emptyset$. The face $F$ was chosen so
that $x\in\operatorname{relint}\left(F\right)$. It is enough to follow the
lines of Case (ii) of this proof with $y$ replaced by $x$, and $x$ replaced by
$z=x+u\notin S$ for $u\in\mathbb{S}^{d-1}$ (any) outer normal to $S$ at $x$.
The latter choice assures that $F\in\mathfrak{F}(y,S)$, which allows us to
proceed exactly as in Case (ii).

###  A.4. Proof of Theorem 5

The general statement follows from Lemmas 3 and 4. Thus, it is enough to prove
only the results under additional assumptions. We take $x\in\mathbb{R}^{d}$
and consider the construction of the halfspace
$H=H(x,F)\in\mathcal{H}(D_{\alpha}(\mu),F)$ from the proof of Lemma 4. Only
the situation when both $D_{\alpha}(\mu)$ and $F$ are non-empty is considered;
the other case is straightforward and trivial.

Part (i): Under the assumption of smoothness at $D_{\alpha}(\mu)$ we have
$\mu(H(x,F))=\mu(\operatorname{int}\left(H(x,F)\right))$. Since $\emptyset\neq
F\subseteq D_{\alpha}(\mu)$ we also have that $H(x,F)\cap
D_{\alpha}(\mu)\neq\emptyset$, which implies $\mu(H(x,F))=\alpha$.

Part (ii): It is enough to show that for any $H\in\mathcal{H}$ the condition
$\mu(\operatorname{int}\left(H\right))\leq\alpha$ implies
$\operatorname{int}\left(D_{\alpha}(\mu)\right)\cap H=\emptyset$. Suppose
$x\in H\cap\operatorname{int}\left(D_{\alpha}(\mu)\right)$ and denote by $v$
the unit inner normal of $H$. Then $H_{x,v}\subseteq H$ and
$\operatorname{bd}\left(H_{x,v}\right)\cap\operatorname{int}\left(D_{\alpha}(\mu)\right)\neq\emptyset$.
Since $D_{\alpha}(\mu)$ is closed and convex, there exists
$y\in\operatorname{int}\left(H_{x,v}\right)\cap D_{\alpha}(\mu)$, such that
$H_{y,v}\in\mathcal{H}(D_{\alpha}(\mu))$, and $H_{y,v}\subset H_{x,v}$.
Because $\mu$ has contiguous support at $D_{\alpha}(\mu)$ it follows that
$\mu(H_{y,v})<\mu(H_{x,v})\leq\mu(H)\leq\alpha$, which is in contradiction
with $y\in D_{\alpha}(\mu)$.

Part (iii): The result follows from parts (i) and (ii).

###  A.5. Proof of Theorem 6

The inclusion “$\subseteq$” in (7) is a direct consequence of Lemmas 3 and 4.
For the other inclusion, suppose for contradiction that
$x\in\operatorname{int}\left(U_{\alpha}(\mu)\right)$ is contained in the right
hand side of (7). Then we can find $H\in\mathcal{H}$ with $x\in H$ and
$\mu(\operatorname{int}\left(H\right))\leq\alpha$, and since both
$\operatorname{int}\left(H\right)$ and
$\operatorname{int}\left(U_{\alpha}(\mu)\right)$ are open, there must exist
$y\in\operatorname{int}\left(U_{\alpha}(\mu)\right)\cap\operatorname{int}\left(H\right)$.
Denote by $v\in\mathbb{S}^{d-1}$ the inner unit normal of $H$. We obtain that
$H_{y,v}\subset\operatorname{int}\left(H\right)$, meaning that
$\alpha<D\left(y;\mu\right)\leq\mu\left(H_{y,v}\right)\leq\mu(\operatorname{int}\left(H\right))\leq\alpha$,
a contradiction. We have verified (7).

If $\mu$ is smooth at $U_{\alpha}(\mu)$, then for
$H(x,F)\in\mathcal{H}(U_{\alpha}(\mu),F)$ from the proof of Lemma 4 we have
$\mu(H(x,F))=\mu(\operatorname{int}\left(H(x,F)\right))\leq\alpha$ and
consequently $H(x,F)\cap U_{\alpha}(\mu)=\emptyset$, meaning that $F\cap
U_{\alpha}(\mu)=\emptyset$. This holds true for any
$x\notin\operatorname{int}\left(U_{\alpha}(\mu)\right)$ and any face $F$ from
$\mathfrak{F}(x,U_{\alpha}(\mu))$. In particular, it must be true for any face
$F$ of $U_{\alpha}(\mu)$ that is not full-dimensional, by considering
$x\in\operatorname{relint}\left(F\right)\subset\operatorname{bd}\left(U_{\alpha}(\mu)\right)$.
That implies
$U_{\alpha}(\mu)\cap\operatorname{bd}\left(U_{\alpha}(\mu)\right)=\emptyset$,
meaning that $U_{\alpha}(\mu)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$
and $U_{\alpha}(\mu)$ is open.

###  A.6. Proof of Corollary 7

Part (i): Suppose for contradiction that
$x\in\operatorname{int}\left(D^{*}\left(\mu\right)\right)$. Lemma 1 gives
$H\in\mathcal{H}(x)$ with
$\mu(\operatorname{int}\left(H\right))\leq\alpha^{*}(\mu)$. We can shift $H$
in the direction of its inner normal to obtain
$H^{\prime}\in\mathcal{H}\left(D^{*}\left(\mu\right)\right)$ such that
$H^{\prime}\subset\operatorname{int}\left(H\right)$. Since
$D^{*}\left(\mu\right)$ is closed, there exists $y\in
D^{*}\left(\mu\right)\cap H^{\prime}$. The assumption of contiguous support at
$D^{*}\left(\mu\right)$ then gives
$D\left(y;\mu\right)\leq\mu(H^{\prime})<\mu(\operatorname{int}\left(H\right))=\alpha^{*}(\mu)$,
which is impossible as $y\in D^{*}\left(\mu\right)$.

Part (ii): If $\dim\left(D^{*}\left(\mu\right)\right)=d-1$, for any $x$ not
lying in the hyperplane $\operatorname{aff}\left(D^{*}\left(\mu\right)\right)$
we have $D^{*}\left(\mu\right)\in\mathfrak{F}(x,D^{*}\left(\mu\right))$.
Theorem 5 gives that there exists a halfspace $H\in\mathcal{H}$ that contains
$D^{*}\left(\mu\right)$ in its boundary hyperplane, and $x$ in its interior,
with
$\mu\left(H\right)=\mu\left(\operatorname{int}\left(H\right)\right)\leq\alpha^{*}(\mu)$
due to the assumption of the smoothness of $\mu$ at $D^{*}\left(\mu\right)$.
Take any $y\notin H$. Again, by Theorem 5 we obtain $G\in\mathcal{H}$ such
that $D^{*}\left(\mu\right)\subset\operatorname{bd}\left(G\right)$,
$y\in\operatorname{int}\left(G\right)$, and
$\mu\left(G\right)\leq\alpha^{*}(\mu)$. The conditions
$\dim(D^{*}\left(\mu\right))=d-1$,
$D^{*}\left(\mu\right)\subset\operatorname{bd}\left(H\right)\cap\operatorname{bd}\left(G\right)$
and $H\neq G$ determine that $H$ and $G$ must be complementary, i.e. $H\cup
G=\mathbb{R}^{d}$. Necessarily,
$\mu\left(\mathbb{R}^{d}\right)=\mu(H)+\mu(G)\leq 2\alpha^{*}(\mu)$, meaning
that the measure $\mu$ is halfspace symmetric [33]. But, by [33, Theorem 2.1]
we know that the median of a halfspace symmetric measure is either unique, or
$\mu$ is concentrated in an infinite line $L$ in $\mathbb{R}^{d}$. The former
case contradicts $\dim\left(D^{*}\left(\mu\right)\right)=d-1$. In the latter
case, also the median set of $\mu$ is contained in $L$, which contradicts our
assumption of smoothness of $\mu$.

###  A.7. Proof of Corollary 8

The first part of the statement follows directly from part (i) of Theorem 5
with $\alpha=\alpha^{*}(\mu)$. For the second part, note that due to the
contiguous support of $\mu$ at $D^{*}\left(\mu\right)$,
$\dim\left(D^{*}\left(\mu\right)\right)<d$ by Corollary 7. Therefore,
$\operatorname{int}\left(D^{*}\left(\mu\right)\right)=\emptyset$, so
$(\operatorname{int}\left(D^{*}\left(\mu\right)\right))^{\mathsf{c}}=\mathbb{R}^{d}$.
To obtain the result, apply Theorem 5 with $x$ replaced by
$y\notin\operatorname{aff}\left(D^{*}\left(\mu\right)\right)$. That theorem
allows to choose $F$ to be
$D^{*}\left(\mu\right)\in\mathfrak{F}\left(y,D^{*}\left(\mu\right)\right)$. We
obtain that
$\mathbb{R}^{d}\setminus\operatorname{aff}\left(D^{*}\left(\mu\right)\right)$
can be covered by halfspaces $H$ from
$\mathcal{H}\left(D^{*}\left(\mu\right),D^{*}\left(\mu\right)\right)$ with
$\mu(H)=\alpha^{*}(\mu)$, but at the same time
$H\in\mathcal{H}\left(D^{*}\left(\mu\right),D^{*}\left(\mu\right)\right)$
implies
$D^{*}\left(\mu\right)\subset\operatorname{aff}\left(D^{*}\left(\mu\right)\right)\subseteq\operatorname{bd}\left(H\right)$.
In particular, we see that also
$\operatorname{aff}\left(D^{*}\left(\mu\right)\right)$ is covered by our
collection of halfspaces, and any point in the median set $x\in
D^{*}\left(\mu\right)$ is contained in the boundary of $H$ for each such $H$.
We conclude that $\mathbb{R}^{d}$ can be covered by halfspaces $H$ from
$\mathcal{H}(x)$ with $\mu(H)=D\left(x;\mu\right)$, as desired.

###  A.8. Proof of Example 2

Throughout the proof we adopt the following notation. For $m=1,2,\dots$ and
points $x_{1},\dots,x_{m}\in\mathbb{R}^{2}$, by $x_{1}\dots x_{m}$ we mean the
polygon with vertices $x_{1},\dots,x_{m}$. For $m=2$ we obtain a line segment
$x_{1}x_{2}$. Where no confusion arises, by $x_{1}x_{2}$ we also denote the
length of that line segment. For polygons with non-empty interior, or more
generally, for measurable sets $V\subseteq\mathbb{R}^{2}$ we denote by
$\operatorname{\lambda}(V)$ the two-dimensional Lebesgue measure (the area) of
$V$. Recall that for $a,b\in\mathbb{R}^{2}$ we write $L[a,b]$ for the closed
line segment between points $a$ and $b$ (that is, $ab$),
$L(a,b)=L[a,b]\setminus\\{a,b\\}$ is the corresponding open line segment, and
$l(a,b)$ the infinite line spanned by $a$ and $b$.

We suppose the distances between the origin $o=\left(0,0\right)^{\mathsf{T}}$
and the points $x_{c}$ and $y_{c}$ take values in ranges $x\in(0,1/4)$ and
$y\in(0,1/5)$, respectively. Set $d=\left(0,-1\right)^{\mathsf{T}}$. Because
$\mu$ is a uniform distribution on $S$ with total mass
$\operatorname{\lambda}(S)$, the $\mu$-mass of any measurable set $V$ is
proportional to $\operatorname{\lambda}(S\cap V)$. For $z\in(0,y]$,
$z_{c}=(0,-z)^{\mathsf{T}}$, and $\theta\in(-\pi/2,\pi/2]$ denote by
$l_{z,\theta}$ the infinite line
$l\left(z_{c},z_{c}+\left(\cos\left(\theta-\pi/2\right),\sin\left(\theta-\pi/2\right)\right)^{\mathsf{T}}\right)$
that passes through $z_{c}$ and determines the angle $\theta$ with line
$l(a,d)$. In Figure 6 the line $l_{z,\theta}$ is drawn for
$\theta=\pi/4-3/10$. Any $l_{z,\theta}$ determines two closed halfplanes. We
write $H_{z}^{\theta+}$ for the one that contains $d$, and write
$H_{z}^{\theta-}$ for the complementary closed halfplane. Denote

$\operatorname{\lambda}_{z}^{\theta+}=\operatorname{\lambda}(S\cap
H_{z}^{\theta+})\quad\mbox{ and
}\quad\operatorname{\lambda}_{z}^{\theta-}=\operatorname{\lambda}(S\cap
H_{z}^{\theta-}).$

By Lemma 1 a minimizing halfplane of $\mu$ at $z_{c}$ exists. In order to
determine one, we find $\theta$ that minimizes
$\min\left\\{\operatorname{\lambda}_{z}^{\theta+},\operatorname{\lambda}_{z}^{\theta-}\right\\}$.
The condition
$\operatorname{\lambda}_{z}^{\theta+}+\operatorname{\lambda}_{z}^{\theta-}=\operatorname{\lambda}(S)$
reduces the problem to the search for the minimum and the maximum of
$\operatorname{\lambda}_{z}^{\theta+}$. Because of the symmetry of the figure
$S$, it is enough to consider $\theta\in[0,\pi/2]$.

As depicted in Figure 5, denote by $o_{l}$ and $o_{r}$ the points of
intersection of the line $l\left(o,(0,1)^{\mathsf{T}}\right)$ with $L(a,b)$
and $L(a,c)$, respectively. Computing the lengths of the corresponding line
segments we obtain $oo_{r}=2dc/3=2/\sqrt{3}$. Using the similarity of
triangles, $ax_{c}:ao=xx_{r}:oo_{r}$, which together with $ao=2$ and
$ax_{c}=2-x$ gives

$x_{c}x_{r}=x_{c}x_{l}=\frac{2-x}{\sqrt{3}}.$

Further, we obtain $x:y=(oo_{r}-x_{c}x_{r}):(y_{c}y_{r}-oo_{r})$, and
consequently

$y_{c}y_{r}=y_{c}y_{l}=\frac{y+2}{\sqrt{3}}.$

Part (I): $y_{c}\in D^{*}\left(\mu\right)$ and $y_{c}$ satisfies the covering
property (4).

We start by considering those values of $\theta\in[0,\pi/2]$ that correspond
to the change of the behaviour in the formula for the area
$\operatorname{\lambda}_{y}^{\theta+}$. In the considered range for the angle
$\theta$, this happens twice:  (i) at $\theta_{1}$ when the line
$l_{y,\theta_{1}}$ passes through the vertex $c$, see the left hand panel of
Figure 5; and  (ii) at $\theta_{2}$ when the line $l_{y,\theta_{2}}$ passes
through the point $x_{l}$, see the right hand panel of Figure 5.  Denote
$e=l_{y,\theta}\cap l(a,b)$ and $f=l_{y,\theta}\cap l(b,c)$. Note that
$\theta_{1}$ and $\theta_{2}$ correspond to $f=c$ and $e=x_{l}$, respectively.
A simple computation yields

$\displaystyle\tan\theta_{1}$ $\displaystyle=dc/y_{c}d=\sqrt{3}/(1-y),$
$\displaystyle\tan\theta_{2}$
$\displaystyle=x_{c}x_{l}/y_{c}x_{c}=(2-x)/(\sqrt{3}(x+y)).$

We see that $\theta_{1}\rightarrow\pi/3$ and $\theta_{2}\rightarrow\pi/2$ as
$x$ and $y$ both decrease to $0$. Thus, for $x$ and $y$ small enough,
$\theta_{1}<\theta_{2}$; in particular, it is straightforward to establish
that for our choices of $x\in(0,1/4)$ and $y\in(0,1/5)$, we have
$\theta_{1}<\theta_{2}$. We distinguish three cases, according to the value of
$\theta$, and evaluate the function $\operatorname{\lambda}_{y}^{\theta+}$ in
each situation.

Figure 5. Proof of Example 2: Angles $\theta_{1}$ (left panel) and
$\theta_{2}$ (right panel) represent the two values in the interval
$[0,\pi/2]$ where the function $\operatorname{\lambda}_{y}^{\theta+}$ fails to
be differentiable.

Case (i): $0\leq\theta<\theta_{1}$.

Figure 6. Proof of Example 2: The situation that corresponds to the Case (i)
of Part (I) of the proof (left panel), and Case (ii) of Part (I) of the proof
(right panel).

For $x^{\prime}$ the point of intersection of $L(e,f)$ and $L(x_{l},x_{r})$,
we derive $\operatorname{\lambda}_{y}^{\theta+}$ by computing

(11)
$\operatorname{\lambda}_{y}^{\theta+}=\operatorname{\lambda}(ebf)-\operatorname{\lambda}(x^{\prime}x_{l}y_{l}y_{c}).$

Denote $\operatorname{\lambda}_{x}=\operatorname{\lambda}(oo_{r}x_{r}x_{c})$
and $\operatorname{\lambda}_{y}=\operatorname{\lambda}(y_{c}y_{r}o_{r}o)$.
Then $\operatorname{\lambda}_{x}=(x_{c}x_{r}+oo_{r})\cdot
x/2=x(4-x)/(2\sqrt{3})$ and
$\operatorname{\lambda}_{y}=(y_{c}y_{r}+oo_{r})\cdot y/2=y(4+y)/(2\sqrt{3})$.
Denote by $h$ the length of the normal from $e$ to line $l(y_{l},y_{c})$. We
compute $h:(h+1-y)=y_{c}y_{l}:bf$. On the other hand,
$bf=bd+df=\sqrt{3}+(y_{c}d)\tan{\theta}=\sqrt{3}+(1-y)\tan{\theta}$. From
here, we get $h=(y+2)/(1+\sqrt{3}\tan{\theta})$, and consequently

(12)
$\operatorname{\lambda}(ebf)=(h+1-y)\frac{bf}{2}=\frac{(\sqrt{3}+(1-y)\tan{\theta})^{2}}{2(\frac{1}{\sqrt{3}}+\tan{\theta})}.$

Since $x_{c}y_{c}=x+y$ and the angle between $l(y_{c},x^{\prime})$ and
$l(y_{c},x_{c})$ equals $\theta$, it holds true that
$\operatorname{\lambda}(x_{c}x^{\prime}y_{c})=(x+y)^{2}\tan{\theta}/2$ and

(13)
$\operatorname{\lambda}(x^{\prime}x_{l}y_{l}y_{c})=\operatorname{\lambda}_{x}+\operatorname{\lambda}_{y}-\operatorname{\lambda}(x_{c}x^{\prime}y_{c})=\frac{(4-x)x}{2\sqrt{3}}+\frac{(4+y)y}{2\sqrt{3}}-\frac{\tan{\theta}}{2}(x+y)^{2}.$

Finally, substituting (12) and (13) into (11) gives

$\operatorname{\lambda}_{y}^{\theta+}=\frac{(\sqrt{3}+(1-y)\tan{\theta})^{2}}{2(\frac{1}{\sqrt{3}}+\tan{\theta})}-\frac{2}{\sqrt{3}}(x+y)-\frac{1}{2\sqrt{3}}(y^{2}-x^{2})+\frac{\tan{\theta}}{2}(x+y)^{2}.$

Now we take $t={1}/{\sqrt{3}}+\tan{\theta}$, and substitute $\tan{\theta}$ by
$t-1/{\sqrt{3}}$ in the last equation. Consolidating the terms with $t$, we
get

(14)
$\operatorname{\lambda}_{y}^{\theta+}=\frac{1}{\sqrt{3}}(2+y)(1-x-2y)+F_{1}(t)$

where

$F_{1}(t)=\frac{\left(2+y\right)^{2}}{6}\frac{1}{t}+\left((1-y)^{2}+(x+y)^{2}\right)\frac{t}{2}.$

The derivative of $F(t)$ is given by

$F_{1}^{\prime}(t)=\frac{(1-y)^{2}+(x+y)^{2}}{2}-\frac{\left(2+y\right)^{2}}{6}\frac{1}{t^{2}}.$

Condition $F_{1}^{\prime}(t)=0$ gives
$t^{2}={\left(2+y\right)^{2}}/{\left(3\left((1-y)^{2}+(x+y)^{2}\right)\right)}$.
Denote

$t_{m}=\frac{2+y}{\sqrt{3\left((1-y)^{2}+(x+y)^{2}\right)}}.$

The derivative $F_{1}^{\prime}(t)$ is negative for $t<t_{m}$ and positive for
$t>t_{m}$. Therefore, the function $F_{1}(t)$ decreases with $t$ until
$t=t_{m}$, and then increases again, so it is only left to check whether
$t_{m}<\tan{\theta_{1}}+1/\sqrt{3}$. In order to do so, we should verify

$\frac{2+y}{\sqrt{3\left((1-y)^{2}+(x+y)^{2}\right)}}<\frac{\sqrt{3}}{1-y}+\frac{1}{\sqrt{3}}=\frac{4-y}{\sqrt{3}(1-y)},$

which can be rewritten as

(15) $\frac{2+y}{\sqrt{1+\left(\frac{x+y}{1-y}\right)^{2}}}<4-y.$

Since the left hand side of the last equation is smaller than $2+y$, it is
enough to show $2+y<4-y$, which is satisfied because we started with $y<1/5$.
We conclude that $F_{1}(t)$ attains its minimum value at $t=t_{m}$, or
equivalently, at $\tan{\theta_{\min}}=t_{m}-1/\sqrt{3}$. The appropriate
minimum area is given by

(16)
$\operatorname{\lambda}_{y}^{\theta_{\min}+}=\frac{1}{\sqrt{3}}(2+y)(1-x-2y)+\frac{1}{\sqrt{3}}(2+y)\sqrt{(1-y)^{2}+(x+y)^{2}}.$

Case (ii): $\theta_{1}\leq\theta<\theta_{2}$.

Now $e\in L[a,x_{l}]$ and $f\notin L(d,c)$, see the right hand panel of Figure
6. Denote $g=L[y_{c},f]\cap L[a,c]$. In order to get the formula for
$\operatorname{\lambda}_{y}^{\theta+}$ in this case, one simply subtracts
$\operatorname{\lambda}(gcf)$ from formula (14). Denote by $h$ the orthogonal
projection of $g$ onto $l(d,f)$. Then
$\tan\theta=fd/y_{c}d=(fc+\sqrt{3})/(1-y)$, so $fc=(1-y)\tan\theta-\sqrt{3}$,
and $hc=hg/\sqrt{3}$. At the same time,
$\tan\theta=hf/hg=(hc+cf)/hg=cf/hg+1/\sqrt{3}$, implying
$hg=cf/(\tan\theta-1/\sqrt{3})$. Now we may calculate
$\operatorname{\lambda}(gcf)$ as $cf\cdot hg/2$, which, after the substitution
$t=1/\sqrt{3}+\tan\theta$, gives

$\operatorname{\lambda}(gcf)=\frac{(1-y)^{2}}{2}t+\frac{(2+y)^{2}}{6}\frac{1}{t-\frac{2}{\sqrt{3}}}-\sqrt{3}(1-y).$

For $\theta=\theta_{1}$ one gets $\operatorname{\lambda}(gcf)=0$, as expected.
Finally, we derive

(17)
$\operatorname{\lambda}_{y}^{\theta+}=\frac{1}{\sqrt{3}}(5-6y-2y^{2}-2x-xy)+F_{2}(t),$

where

$F_{2}(t)=(x+y)^{2}\frac{t}{2}+\frac{\left(2+y\right)^{2}}{6}\left(\frac{1}{t}-\frac{1}{t-\frac{2}{\sqrt{3}}}\right).$

The above formula holds true for $\tan\theta_{1}+1/\sqrt{3}\leq
t<\tan\theta_{2}+1/\sqrt{3}$. By introducing
$s=t-\tan\theta_{1}-1/\sqrt{3}\geq 0$, we may write

(18) $F_{2}(t)=\frac{(x+y)^{2}}{2\sqrt{3}}\frac{4-y}{1-y}+G(s),$

where

(19)
$G(s)=\frac{(x+y)^{2}}{2}s-\frac{\left(2+y\right)^{2}}{3\sqrt{3}}\frac{1}{(s+\frac{\sqrt{3}}{1-y})^{2}-\frac{1}{3}}.$

One may investigate $G(s)$ instead of $F_{2}(t)$, because $t\mapsto s$ is a
simple shift by a constant. The derivative of $G(s)$,

(20)
$G^{\prime}(s)=\frac{(x+y)^{2}}{2}+\frac{2(2+y)^{2}}{3\sqrt{3}}\frac{1}{((s+\frac{\sqrt{3}}{1-y})^{2}-\frac{1}{3})^{2}}\left(s+\frac{\sqrt{3}}{1-y}\right),$

is always positive for $s\geq 0$, so $\operatorname{\lambda}_{y}^{\theta+}$
grows with $\theta$ for $\theta_{1}\leq\theta<\theta_{1}$.

Case (iii): $\theta_{1}\leq\theta\leq\pi/2$.

As $e\in L\left[x_{l},y_{l}\right]$, $\operatorname{\lambda}_{y}^{\theta+}$
obviously grows with increasing $\theta$.

Summary of Part (I).

Now that we established the behaviour of the mapping
$\theta\mapsto\operatorname{\lambda}_{y}^{\theta+}$, we are able to find its
extreme values. The value of $\operatorname{\lambda}_{y}^{\theta+}$ decreases
for $0\leq\theta<\theta_{\min}$ and increases for
$\theta_{\min}<\theta\leq\pi/2$, so it attains minimum value at
$\theta=\theta_{\min}$, and maximum value at $\theta=0$ or $\theta=\pi/2$.
Therefore, the depth of the point $y_{c}$ with respect to $\mu$ is equal to
$\min\left\\{\operatorname{\lambda}_{y}^{\theta_{\min}+},\operatorname{\lambda}_{y}^{0-},\operatorname{\lambda}_{y}^{\pi/2-}\right\\}$.
Note that $\operatorname{\lambda}_{y}^{0-}=\lambda(S)/2$ and
$\operatorname{\lambda}_{y}^{\pi/2-}=\operatorname{\lambda}(ax_{l}x_{r})$. As
$x,y\rightarrow 0$ from the right, $\operatorname{\lambda}(ax_{l}x_{r})$ goes
to $4/9\cdot\operatorname{\lambda}(abc)$, and $\operatorname{\lambda}(S)$ goes
to $\operatorname{\lambda}(abc)$. In particular, it is easy to show that for
$x\in(0,1/4)$ and $y\in(0,1/5)$ we have
$\operatorname{\lambda}_{y}^{\pi/2-}<\operatorname{\lambda}_{y}^{0-}$. In what
follows we show that it is possible to choose $x$ and $y$ such that
$\operatorname{\lambda}_{y}^{\pi/2-}=\operatorname{\lambda}_{y}^{\theta_{\min}+}$.

Note that
$\operatorname{\lambda}_{y}^{\pi/2-}=\operatorname{\lambda}(ax_{l}x_{r})=x_{c}x_{r}\cdot
ax_{c}=(2-x)^{2}/\sqrt{3}$. Therefore, the condition
$\operatorname{\lambda}_{y}^{\pi/2-}=\operatorname{\lambda}_{y}^{\theta_{\min}+}$
is, using (16), equivalent with

$f(x,y)=-2-3y-2y^{2}+2x-x^{2}-xy+(2+y)\sqrt{(1-y)^{2}+(x+y)^{2}}=0.$

Consider $x\in(0,1/4)$ fixed. As $y\rightarrow 0$, $f(x,y)$ converges to
$f_{0}(x)=(\sqrt{2x}+\sqrt{1+x^{2}}-1)(\sqrt{2x}+1-\sqrt{1+x^{2}})$. To show
that $f_{0}(x)>0$ for any choice of $x\in(0,1/4)$, we should prove
$\sqrt{2x}+1>\sqrt{1+x^{2}}$, which is equivalent with
$2x+2\sqrt{2x}-x^{2}>0$. The latter inequality follows from
$2x+2\sqrt{2x}-x^{2}=2\sqrt{2x}+x(2-x)>0$. In the case $y\rightarrow 1/5$,
$f(x,y)$ converges to
$f_{1/5}(x)=-67/25+9x/5-x^{2}+11\sqrt{15/25+\left(x+1/5\right)^{2}}/5\leq-67/25+9x/5+11\sqrt{15/25+\left(x+1/5\right)^{2}}/5<0$
for each $x\in(0,1/4)$. Because the mapping $y\mapsto f(x,y)$ is continuous in
$y$ for each fixed $x$, the intermediate value theorem [6, Problem 2.2.14(d)]
allows us to conclude the existence of $y\in(0,1/5)$ such that $f(x,y)=0$ for
each $x\in(0,1/4)$.

Finally, from all the considerations above, we conclude that for each $x<1/4$,
there exists $y$ such that $D(y_{c};\mu)=\operatorname{\lambda}(ax_{l}x_{r})$
and there are three minimizing halfspaces of $\mu$ at $y_{c}$, being
$H_{y}^{\theta_{\min}+}$, $H_{y}^{-\theta_{\min}+}$ and $H_{y}^{\pi/2-}$.
Obviously, those halfspaces cover the whole $\mathbb{R}^{2}$, so $y_{c}$
satisfies the covering property (4), and by part (i) of Theorem 2, $y_{c}$
must be a median of $\mu$. The course of the function
$\operatorname{\lambda}_{y}^{\theta+}$ can be observed in Figure 7.

Figure 7. Example 2: Functions $\operatorname{\lambda}_{y}^{\theta+}$ (blue
curve) and $\operatorname{\lambda}_{0}^{\theta+}$ (thick orange curve) on
their full domain $[0,2\pi]$ (left panel) and the same functions zoomed into
the interval $[0,\pi/2]$ (right panel). Both functions attain the same minimum
value. On the right hand panel, the point $\theta_{\min}$ where
$\operatorname{\lambda}_{y}^{\theta+}$ attains its minimum, along with the
points $\theta_{1}$ and $\theta_{2}$ of non-differentiability of function
$\operatorname{\lambda}_{y}^{\theta+}$ are marked by dashed vertical lines.

Part (II): $\dim\left(D^{*}\left(\mu\right)\right)=2$.

We first show that there exists a point
$z_{c}=\left(0,-z\right)^{\mathsf{T}}\neq y_{c}$ in $\mathbb{R}^{2}$ that is
also a halfspace median of $\mu$. For that reason, we use the results from
Part (I) to calculate $\operatorname{\lambda}_{z}^{\theta+}$. We consider only
$x$ and $y$ from Part (I) that satisfy
$\operatorname{\lambda}_{y}^{\pi/2-}=\operatorname{\lambda}_{y}^{\theta_{\min}+}=D(y_{c};\mu)=\alpha^{*}(\mu)$.

Denote $q=x_{r}y_{l}\cap x_{l}y_{r}$ and consider point
$z_{c}=(0,-z)^{\mathsf{T}}$ for $z\in(0,y)$ on the line segment $L(q,y_{c})$,
see the left panel of Figure 8. Because $z_{c}\in L(q,y_{c})$, the line
$l(x_{l},z_{c})$ intersects $L[y_{c},y_{r}]$. Denote $\delta=y-z>0$ and
$y^{\prime}=l(x_{l},z_{c})\cap L[y_{c},y_{r}]$.

Figure 8. Proof of Example 2: A scheme of the configuration that corresponds
to Part (II) of the proof, with point $z_{c}\in D^{*}\left(\mu\right)$
displayed (left panel), and the region where the additional median $w$ is
searched for (right panel).

Denote $x_{z}=l_{z,\theta}\cap l(x_{l},x_{r})$, $y_{z}=l_{z,\theta}\cap
l(y_{l},y_{r})$, $e_{z}=l_{z,\theta}\cap l(a,b)$ and $f_{z}=l_{z,\theta}\cap
l(b,c)$. Consider angles $\widetilde{\theta_{1}}$, $\widetilde{\theta_{2}}$
and $\widetilde{\theta_{3}}$ corresponding to cases $f_{z}=c$, $e_{z}=x_{l}$
and $y_{z}=y_{r}$, respectively. It is not difficult to calculate

$\displaystyle\tan\widetilde{\theta_{1}}$
$\displaystyle=dc/z_{c}d=\sqrt{3}/(1-z),$
$\displaystyle\tan\widetilde{\theta_{2}}$
$\displaystyle=x_{c}x_{l}/z_{c}x_{c}=(2-x)/(\sqrt{3}(x+z)),$
$\displaystyle\tan\widetilde{\theta_{3}}$
$\displaystyle=y_{c}y_{r}/y_{c}z_{c}=(y+2)(\sqrt{3}\delta).$

If $\delta$ is small enough, then
$\widetilde{\theta_{1}}<\widetilde{\theta_{2}}<\widetilde{\theta_{3}}$. We
consider four different cases and calculate
$\operatorname{\lambda}_{z}^{\theta+}$.

For $0\leq\theta<\widetilde{\theta_{1}}$, one obtains
$\operatorname{\lambda}_{z}^{\theta+}$ by substituting $y$ by $z$ in (14), and
subsequently subtracting $\operatorname{\lambda}(y_{l}y_{z}z_{c}z_{l})$, where
$z_{l}=l_{z,\pi/2}\cap l(a,b)$. Note that
$\operatorname{\lambda}(z_{c}y_{c}y_{z})=\delta^{2}\tan\theta/2$ and
$\operatorname{\lambda}(y_{l}y_{c}z_{c}z_{l})=(4+y+z)\delta/(2\sqrt{3})$, so

(21)
$\operatorname{\lambda}(y_{l}y_{z}z_{c}z_{l})=\frac{\delta^{2}}{2}t+\frac{4+y+z}{2\sqrt{3}}\delta-\frac{1}{2\sqrt{3}}\delta^{2},$

where $t=1/\sqrt{3}+\tan\theta$.

Case (i): $0\leq\theta<\widetilde{\theta_{1}}$.

The previous calculation leads to

$\operatorname{\lambda}_{z}^{\theta+}=\frac{1}{\sqrt{3}}\left(2-3z-2z^{2}-2x-xz-2\delta-\frac{y+z}{2}\delta+\frac{\delta^{2}}{2}\right)+\widetilde{F}_{1}(t),$

where

$\widetilde{F}_{1}(t)=\frac{\left(2+z\right)^{2}}{6}\frac{1}{t}+\left((1-z)^{2}+(x+z)^{2}-\delta^{2}\right)\frac{t}{2}.$

For $\delta$ small, $\widetilde{F}_{1}(t)$ behaves similarly to $F_{1}(t)$. In
that case $\operatorname{\lambda}_{z}^{\theta+}$ decreases with $\theta$ for
small positive values of $\theta$, reaches its minimum value at

$\tan\widetilde{\theta}_{\min}=\frac{2+z}{\sqrt{3\left((1-z)^{2}+(x+z)^{2}-\delta^{2}\right)}}-\frac{1}{\sqrt{3}},$

and then increases again as $\theta$ grows. We also check
$\tan\widetilde{\theta}_{\min}<\tan\widetilde{\theta_{1}}$ in the same way as
we did in Part (I) of this proof — analogously to (15), we should verify

$\frac{2+z}{\sqrt{1+\left(\frac{x+z}{1-z}\right)^{2}-\frac{\delta^{2}}{(1-z)^{2}}}}<4-z.$

For $\delta$ positive small enough the left hand side of the previous display
is smaller than $2+z$, and we may conclude
$\widetilde{\theta}_{\min}<\widetilde{\theta_{1}}$.

Case (ii): $\widetilde{\theta_{1}}\leq\theta<\widetilde{\theta_{2}}$.

In this situation, analogously to Case (i) we substitute $y$ by $z$ in (17)
and subtract (21) to obtain

$\operatorname{\lambda}_{z}^{\theta+}=\frac{1}{\sqrt{3}}(5-6z-2z^{2}-2x-xz-2\delta-\frac{z+y}{2}\delta+\frac{\delta^{2}}{2})+\widetilde{F}_{2}(t),$

where

$\widetilde{F}_{2}(t)=\left((x+z)^{2}-\delta^{2}\right)\frac{t}{2}+\frac{\left(2+z\right)^{2}}{6}\left(\frac{1}{t}-\frac{1}{t-\frac{2}{\sqrt{3}}}\right).$

We introduce again $s=t-\tan\widetilde{\theta_{1}}-1/\sqrt{3}\geq 0$ and
investigate $\widetilde{G}(s)$, analogously as in (18) and (19), instead of
$\widetilde{F}_{2}(t)$. As in (20), we obtain the derivative of
$\widetilde{G}(s)$ of the form

$\widetilde{G}^{\prime}(s)=\frac{(x+z)^{2}-\delta^{2}}{2}+\frac{2(2+z)^{2}}{3\sqrt{3}}\left(\left(s+\frac{\sqrt{3}}{1-z}\right)^{2}-\frac{1}{3}\right)^{-2}\left(s+\frac{\sqrt{3}}{1-z}\right),$

which is positive for small $\delta$, meaning that
$\operatorname{\lambda}_{z}^{\theta+}$ increases as a function of
$\theta\in\left[\widetilde{\theta_{1}},\widetilde{\theta_{2}}\right)$.

Case (iii): $\widetilde{\theta_{2}}\leq\theta<\widetilde{\theta_{3}}$.

For these values of the parameter, it is easy to see that
$\operatorname{\lambda}_{z}^{\theta+}$ grows with increasing $\theta$, for the
same reason as in Part (I) of this proof.

Case (iv): $\widetilde{\theta_{3}}\leq\theta\leq\frac{\pi}{2}$.

In this range for $\theta$, the function
$\operatorname{\lambda}_{z}^{\theta+}$ is clearly constant, and equal to
$\operatorname{\lambda}(y_{r}y_{l}bc)$.

Summary of Part (II).

We conclude our analysis by observing that
$\operatorname{\lambda}_{z}^{\theta+}$ attains its minimum at
$\theta=\widetilde{\theta}_{\min}$, and its maximum at $\theta=\pi/2$. Because
$\widetilde{\theta}_{\min}\in[0,\widetilde{\theta_{1}})$, $y_{c}$ lies in the
interior of $H_{z}^{\widetilde{\theta}_{\min}+}$ and therefore
$\operatorname{\lambda}_{z}^{\widetilde{\theta}_{\min}+}>\operatorname{\lambda}_{y}^{\theta_{\min}+}=\operatorname{\lambda}_{y}^{\pi/2-}=\alpha^{*}(\mu)$.
On the other hand,
$\operatorname{\lambda}_{z}^{\pi/2-}=\operatorname{\lambda}_{y}^{\pi/2-}=\alpha^{*}(\mu)$.
Because by Part (I) we know that $y_{c}$ is a median of $\mu$, this means that
$z_{c}$ and $y_{c}$ have the same depth, and the set $D^{*}\left(\mu\right)$
contains at least two distinct points $z_{c}$ and $y_{c}$. The minimizing
halfspaces of $\mu$ at $z_{c}$ are those determined by the angle
$\theta\in[-\pi/2,-\widetilde{\theta_{3}}]\cup[\widetilde{\theta_{3}},\pi/2]$.
Therefore, it is impossible to cover $\mathbb{R}^{2}$ with minimizing
halfspaces of $\mu$ at point $z_{c}$, meaning that $z_{c}\in
D^{*}\left(\mu\right)$ is a point that fails to satisfy the covering property
(4).

Due to the convexity of the median set $D^{*}\left(\mu\right)$ we already know
that the line segment $L[y_{c},z_{c}]$ is contained in
$D^{*}\left(\mu\right)$. To see that the set $D^{*}\left(\mu\right)$ is in
fact two-dimensional, now we find an additional median of $\mu$ of the form
$w=(-\varepsilon,\tau)^{\mathsf{T}}\in\left(\left(-\infty,0\right)\times(-y,-z)\right)\setminus
H_{y}^{\theta_{\min}+}$, see the right hand panel of Figure 8. The smoothness
of $\mu$ implies that for $\tau$ fixed, $\inf\left\\{\mu(H)\colon
H\in\mathcal{H}(w)\mbox{ and }\\{y_{c},z_{c}\\}\subset
H^{\mathsf{c}}\right\\}\to 1/2$ as $\varepsilon\to 0$. Since
$D\left(w;\mu\right)\leq\alpha^{*}(\mu)<1/2$ we obtain that for any $\tau$
there exists $\varepsilon$ small enough so that each minimizing halfspace of
$\mu$ at $w$ has to contain at least one of the points $y_{c}$ and $z_{c}$.
Each halfspace $H\in\mathcal{H}(w)$ whose boundary passes through $w$ that
contains either $y_{c}$ or $z_{c}$, however, must have $\mu$-mass at least
$\alpha^{*}(\mu)$, since both $y_{c}$ and $z_{c}$ belong to
$D^{*}\left(\mu\right)$. Therefore, for any $\tau\in(-y,-z)$ and all
$\varepsilon$ small enough, we conclude that the point
$w=\left(-\varepsilon,\tau\right)^{\mathsf{T}}$ is contained in the median set
$D^{*}\left(\mu\right)$, and the latter set must therefore be two-dimensional.

Part (III): The single point that satisfies the covering property (4) is
$y_{c}$.

In Part (I) of this proof we demonstrated that $y_{c}$ satisfies (4). The fact
that no other point in $\mathbb{R}^{2}$ shares that property is a consequence
of Theorem 12 proved in Section 4.1, whose assumptions are satisfied by
$y_{c}$.

###  A.9. Proof of Theorem 9

First note that all the considered sets are convex, meaning that each of them
is of dimension $d$ if and only if its interior is non-empty [25, Theorem
1.1.13]. The equality
$U^{\circ}_{\alpha}(\mu)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$
follows from Theorem 6; the interior of any set is a subset of its closure,
which implies
$\operatorname{int}\left(U_{\alpha}(\mu)\right)\subseteq\operatorname{cl}\left(U_{\alpha}(\mu)\right)$;
the floating body $U_{\alpha}^{FB}(\mu)$ is an intersection of a larger
collection of halfspaces than the depth region $D_{\alpha}(\mu)$, meaning that
$U_{\alpha}^{FB}(\mu)\subseteq D_{\alpha}(\mu)$. We prove the remaining non-
trivial statements of the proposition in several steps.

Inclusion $\operatorname{cl}\left(U_{\alpha}(\mu)\right)\subseteq
U_{\alpha}^{FB}(\mu)$. Suppose that there exists $x\in
U_{\alpha}(\mu)\setminus U_{\alpha}^{FB}(\mu)$. Since $x\notin
U_{\alpha}^{FB}(\mu)$, there must exist $H_{y,v}\in\mathcal{H}$ such that
$x\notin H_{y,v}$ and $\mu(H_{y,v}^{\mathsf{c}})\leq\alpha$. Its complement
$H_{y,v}^{\mathsf{c}}$ is an open halfspace whose boundary passes through $y$
and has inner normal $-v$, i.e.
$H_{y,v}^{\mathsf{c}}=\operatorname{int}\left(H_{y,-v}\right)$. We know that
$x\in\operatorname{int}\left(H_{y,-v}\right)$ and
$\mu(\operatorname{int}\left(H_{y,-v}\right))=\mu(H_{y,v}^{\mathsf{c}})\leq\alpha$.
Therefore, $H_{x,-v}\subset\operatorname{int}\left(H_{y,-v}\right)$ and
consequently
$\mu(H_{x,-v})\leq\mu(\operatorname{int}\left(H_{y,-v}\right))\leq\alpha$,
which is in contradiction with $D(x;\mu)>\alpha$, as well as $x\in
U_{\alpha}(\mu)$. We conclude that $U_{\alpha}(\mu)\subseteq
U_{\alpha}^{FB}(\mu)$ and consequently
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)\subseteq U_{\alpha}^{FB}(\mu)$,
because $U_{\alpha}^{FB}(\mu)$ is closed.

Part (i): Inclusion
$U^{\circ}_{\alpha}(\mu)\subseteq\left\\{x\in\mathbb{R}^{d}\colon
D^{\circ}(x;\mu)>\alpha\right\\}$. Pick $x\in U^{\circ}_{\alpha}(\mu)$, and
suppose that $D^{\circ}\left(x;\mu\right)=\beta\leq\alpha$. Analogously to
Lemma 1 it is possible to show that if $D^{\circ}\left(x;\mu\right)=\beta$,
there must exist $G\in\mathcal{H}(x)$ such that
$\mu\left(\operatorname{int}\left(G\right)\right)=\beta$. Define
$H=\left(\operatorname{int}\left(G\right)\right)^{\mathsf{c}}\in\mathcal{H}$
and note that $x\in\operatorname{bd}\left(H\right)$ and
$\mu\left(H^{\mathsf{c}}\right)=\mu(\operatorname{int}\left(G\right))=\beta\leq\alpha$.
We reached a contradiction, since for such a halfspace $H$ the point $x$ must
be contained in $\operatorname{int}\left(H\right)$.

Part (i): Inclusion $\left\\{x\in\mathbb{R}^{d}\colon
D^{\circ}(x;\mu)>\alpha\right\\}\subseteq U^{\circ}_{\alpha}(\mu)$. If
$D^{\circ}(x;\mu)>\alpha$, then for each $H\in\mathcal{H}$ that contains $x$
it has to be $\mu(\operatorname{int}\left(H\right))>\alpha$ and therefore
$x\notin\left(U^{\circ}_{\alpha}(\mu)\right)^{\mathsf{c}}$.

Part (ii): Inclusion
$U_{\alpha}^{FB}(\mu)\subseteq\operatorname{cl}\left(U_{\alpha}(\mu)\right)$
if $\operatorname{int}\left(U_{\alpha}(\mu)\right)\neq\emptyset$. Note that
$\operatorname{int}\left(U_{\alpha}(\mu)\right)\subseteq\operatorname{cl}\left(U_{\alpha}(\mu)\right)\subseteq
U_{\alpha}^{FB}(\mu)$, meaning that
$\operatorname{int}\left(U_{\alpha}^{FB}(\mu)\right)\neq\emptyset$. Therefore,
if
$U_{\alpha}^{FB}(\mu)\setminus\operatorname{cl}\left(U_{\alpha}(\mu)\right)$
is non-empty, there must exist
$x\in\operatorname{int}\left(U_{\alpha}^{FB}(\mu)\right)\setminus\operatorname{cl}\left(U_{\alpha}(\mu)\right)$.
To see this, denote by $y$ any point in
$U_{\alpha}^{FB}(\mu)\setminus\operatorname{cl}\left(U_{\alpha}(\mu)\right)$.
Since $U_{\alpha}^{FB}(\mu)$ is full-dimensional and convex, there exists an
open ball $B_{1}\subset U_{\alpha}^{FB}(\mu)$ [25, Theorem 1.1.13], and
$\operatorname{conv}\left(B_{1}\cup\\{y\\}\right)\subset
U_{\alpha}^{FB}(\mu)$. Since $\operatorname{cl}\left(U_{\alpha}(\mu)\right)$
is a closed set that does not contain $y$, there exists an open ball $B_{2}$
containing $y$ that does not intersect
$\operatorname{cl}\left(U_{\alpha}(\mu)\right)$. The set
$\operatorname{conv}\left(B_{1}\cup\\{y\\}\right)\cap B_{2}$ is then a convex,
full-dimensional subset of
$U_{\alpha}^{FB}(\mu)\setminus\operatorname{cl}\left(U_{\alpha}(\mu)\right)$,
and as such has to contain a point in its interior
$x\in\operatorname{int}\left(U_{\alpha}^{FB}(\mu)\right)\setminus\operatorname{cl}\left(U_{\alpha}(\mu)\right)$,
as we needed to show. Now, because $x\notin U_{\alpha}(\mu)$, Lemma 1 gives
that there exists $H\in\mathcal{H}(x)$ such that
$\mu(\operatorname{int}\left(H\right))\leq D(x;\mu)\leq\alpha$. Denote by
$H_{x}=(\operatorname{int}\left(H\right))^{\mathsf{c}}$ the closed halfspace
that satisfies
$\mu(H_{x}^{\mathsf{c}})=\mu(\operatorname{int}\left(H\right))\leq\alpha$. We
obtain $U_{\alpha}^{FB}(\mu)\subseteq H_{x}$, which is in contradiction with
$x\in\operatorname{int}\left(U_{\alpha}^{FB}(\mu)\right)$.

Part (iii):
$\operatorname{int}\left(D_{\alpha}(\mu)\right)=\operatorname{int}\left(U_{\alpha}(\mu)\right)$.
Under the considered condition of $\mu$ having contiguous support at
$D_{\alpha}(\mu)$ the result follows by Theorem 6 and part (ii) of Theorem 5.

Part (iv): $U_{\alpha}^{FB}(\mu)=D_{\alpha}(\mu)$. In dimension $d=1$, the
equality of the two expressions is straightforward to verify by rewriting the
conditions defining both $U_{\alpha}^{FB}(\mu)$ and $D_{\alpha}(\mu)$ in terms
of the function $F(t)=\mu\left([t,\infty)\right)$, for $t\in\mathbb{R}$. The
general statement follows by projecting $\mu$ via
$\pi_{u}\colon\mathbb{R}^{d}\to\mathbb{R}\colon y\mapsto\left\langle
y,u\right\rangle$, for $u\in\mathbb{S}^{d-1}$, into its pushforward measure
$\mu_{u}\in\mathcal{M}\left(\mathbb{R}\right)$. The image of a halfspace
$H_{x,u}$ by $\pi_{u}$ is the interval $[\left\langle
x,u\right\rangle,\infty)$. Knowing the equality of the two concepts of central
regions after projecting into $\mathbb{R}$, it is enough to realise that
$D_{\alpha}(\mu)=\bigcap_{u\in\mathbb{S}^{d-1}}\pi_{u}^{-1}\left(D_{\alpha}(\mu_{u})\right)$
for $\pi_{u}^{-1}$ the inverse map to $\pi_{u}$, and analogously for the
floating body $U_{\alpha}^{FB}(\mu)$. For details see [7, Theorem 2].

Part (v): $\operatorname{cl}\left(U_{\alpha}(\mu)\right)=D_{\alpha}(\mu)$.
Using part (iii) and the assumption of non-empty interior of both involved
sets, [25, Theorem 1.1.15] ensures that
$D_{\alpha}(\mu)=\operatorname{cl}\left(\operatorname{int}\left(D_{\alpha}(\mu)\right)\right)=\operatorname{cl}\left(\operatorname{int}\left(U_{\alpha}(\mu)\right)\right)=\operatorname{cl}\left(U_{\alpha}(\mu)\right)$
as needed to verify.

###  A.10. Proof of Lemma 10

Suppose that $\operatorname{int}\left(U_{\alpha}(\mu)\right)=\emptyset$. We
may apply Theorem 6 with $F=U_{\alpha}(\mu)$ and conclude that there exists
$H\in\mathcal{H}$ such that
$U_{\alpha}(\mu)\subset\operatorname{bd}\left(H\right)$ and
$\mu(\operatorname{int}\left(H\right))\leq\alpha$. At the same time, existence
of $x\in U_{\alpha}(\mu)$ implies $\mu(H)\geq D(x;\mu)>\alpha$ and
consequently $\mu(\operatorname{bd}\left(H\right))>0$. Therefore, for any
$x\in U_{\alpha}(\mu)$, $\mu$ is not smooth at $x$.

###  A.11. Proof of Theorem 11

We distinguish two cases according to whether the median set
$D^{*}\left(\mu\right)$ is full-dimensional, or not.

Case (i): $\operatorname{int}\left(D^{*}\left(\mu\right)\right)=\emptyset$. We
write $a=\gamma^{*}(\mu)$ and start with
$U_{a}(\mu)=\bigcup_{\beta>a}D_{\beta}(\mu)$. Since
$\operatorname{int}\left(D_{\beta}(\mu)\right)=\emptyset$ for each $\beta>a$
and sets $D_{\beta}(\mu)$ are nested, we conclude that also
$\operatorname{int}\left(U_{a}(\mu)\right)=\emptyset$. Choose any
$x\in\operatorname{aff}\left(U_{a}(\mu)\right)$. From Theorem 6 applied with
$F=U_{a}(\mu)$ it follows that $\mathbb{R}^{d}$ can be covered by halfspaces
$H\in\mathcal{H}(U_{a}(\mu))\subset\mathcal{H}(x)$ such that
$\mu(\operatorname{int}\left(H\right))\leq a=\gamma^{*}(\mu)$ for each such
$H$. Since
$\operatorname{aff}\left(D^{*}\left(\mu\right)\right)\subseteq\operatorname{aff}\left(U_{a}(\mu)\right)$,
we obtain that also all points of $D^{*}\left(\mu\right)$ are possible to be
used for such a covering, as we needed to show.

Case (ii):
$\operatorname{int}\left(D^{*}\left(\mu\right)\right)\neq\emptyset$. In this
case, $\gamma^{*}(\mu)=\alpha^{*}(\mu)$. Denote $S_{0}=D^{*}\left(\mu\right)$.
This is a full-dimensional convex set, and for its barycentre $x_{0}\in S_{0}$
and $\nu_{0}\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ the uniform probability
distribution on $S_{0}$ we know that $D(x_{0};\nu_{0})>\mathrm{e}^{-1}$, due
to a result of Grünbaum, see [20, Theorem 3]. Let $H_{0}\in\mathcal{H}(x_{0})$
be a generalized minimizing halfspace of $\mu$ at $x_{0}$ in the sense of
Lemma 1, i.e.
$\mu\left(\operatorname{int}\left(H_{0}\right)\right)\leq\alpha^{*}(\mu)$.
Writing $\operatorname{vol}(A)$ for the $d$-dimensional volume of a measurable
set $A$, this implies $\nu_{0}(H_{0})=\operatorname{vol}(S_{0}\cap
H_{0})/\operatorname{vol}(S_{0})\geq D(x_{0};\nu_{0})>\mathrm{e}^{-1}$. Denote
$S_{1}=\operatorname{cl}\left(S_{0}\setminus H_{0}\right)$ with
$\operatorname{vol}(S_{1})=\operatorname{vol}(S_{0})(1-\nu_{0}(H_{0}))<\operatorname{vol}(S_{0})(1-\mathrm{e}^{-1})$.
We iterate the previous procedure and for $k\geq 1$ define $x_{k}\in S_{k}$ to
be the barycentre of $S_{k}$. In each step, we can find a generalized
minimizing halfspace $H_{k}\in\mathcal{H}(x_{k})$ of $\mu$ at $x_{k}$, put
$S_{k+1}=\operatorname{cl}\left(S_{k}\setminus H_{k}\right)$, and again
conclude

$\operatorname{vol}(S_{k})<\operatorname{vol}(S_{0})\left(1-\mathrm{e}^{-1}\right)^{k}.$

Necessarily, $\operatorname{vol}(S_{k})\rightarrow 0$ as $k\rightarrow\infty$.
Denote $S=\bigcap_{k=0}^{\infty}S_{k}$. Then $S\subset D^{*}\left(\mu\right)$
is a convex set, $\operatorname{vol}(S)=0$, and consequently
$\operatorname{int}\left(S\right)=\emptyset$.

Note that $D^{*}\left(\mu\right)=S_{0}$ satisfies the assumption (5) of our
main Lemma 4. Recursively, if $S_{k}$ satisfies (5) for some $k=0,1,2,\dots$,
then the same holds true for $S_{k+1}$, because

$S_{k+1}^{\mathsf{c}}=S_{k}^{\mathsf{c}}\cup\operatorname{int}\left(H_{k}\right).$

By the induction step, $S_{k}^{\mathsf{c}}$ is covered by halfspaces
$H\in\mathcal{H}$ with
$\mu\left(\operatorname{int}\left(H\right)\right)\leq\alpha^{*}(\mu)$ and
$\operatorname{int}\left(H\right)\cap
S_{k+1}\subseteq\operatorname{int}\left(H\right)\cap S_{k}=\emptyset$. Adding
to that collection the halfspace $H_{k}$ that also satisfies
$\mu\left(\operatorname{int}\left(H_{k}\right)\right)\leq\alpha^{*}(\mu)$ and
$\operatorname{int}\left(H_{k}\right)\cap S_{k+1}=\emptyset$, we obtain a
covering of $S_{k+1}^{\mathsf{c}}$ with the desired property, i.e. also
$S_{k+1}$ satisfies (5). We finally show that also $S$ satisfies (5). Take any
$x\notin S$ and the first index $k=0,1,2,\dots$ such that $x\notin S_{k}$.
There exists $H\in\mathcal{H}$ with $x\in H$,
$\mu(\operatorname{int}\left(H\right))\leq\alpha$ and
$\operatorname{int}\left(H\right)\cap S_{k}=\emptyset$. Since $S\subseteq
S_{k}$, it follows that $\operatorname{int}\left(H\right)\cap S=\emptyset$.
Lemma 4 can now be applied to the set $S$ to give the desired result, as
follows by $\operatorname{int}\left(S\right)=\emptyset$.

###  A.12. Proof of Example 5

Each point in $\mathbb{R}^{2}$ is contained in one of the four halfspaces of
$\mu$-mass $6$ determined by the dashed lines presented in Figure 9. Therefore
$\alpha^{*}(\mu)\leq 6$. To prove that both points $x$ and $y$ are medians, it
is enough to show $D(x;\mu)=D(y;\mu)=6$. Because of the symmetry of $\mu$, the
problem reduces to showing $D(y;\mu)=6$.

Figure 9. Example 5: Each point in $\mathbb{R}^{2}$ is contained in one of the
four halfspaces of $\mu$-mass $6$ determined by the dashed lines (left hand
panel). Thus, $\alpha^{*}(\mu)\leq 6$. On the right hand panel, several
numerically computed contours $D_{\alpha}(\mu)$ are displayed as sets with
thick coloured boundaries. The median set $D^{*}\left(\mu\right)$ is the
innermost convex set in this collection.

Consider an infinite line $l$ passing through $y$ and denote by $H^{+}$ and
$H^{-}$ the two closed halfspaces determined by $l$. We need to show that
neither $\mu(H^{+})$ nor $\mu(H^{-})$ decreases below $6$ as we rotate $l$
around $y$. Equivalently, as $\mu(\mathbb{R}^{2})=14$, it is enough to show
that $6\leq\mu(H^{+})\leq 8$ when rotating $l$ by an angle of $\pi$. Several
specific positions of $l$ are shown in the left hand panel of Figure 10; the
corresponding halfspace $H^{+}$ for line $l_{1}$ is the one that does not
contain the origin $o$. We start from $l=l_{1}$ when $\mu(H^{+})=6$. As we
rotate $l$ around $y$ counter-clockwise, $\mu(H^{+})$ remains the same for $l$
being between $l_{1}$ and $l_{2}$, since the triangles $A$ and $G$ in Figure
10 are symmetric with respect to $y$. Continuing the rotation, $\mu(H^{+})$
remains constant and then increases from $l=l_{3}$ to $l=l_{4}$, when
$\mu(H^{+})=7$. For $l$ between $l_{4}$ and $l_{5}$, $H^{+}$ contains the
triangles $A$, $B$, $I$, $J$, $K$, so $\mu(H^{+})\geq 6$ and $H^{-}$ contains
$D$, $E$, $F$, $G$, meaning that $\mu(H^{-})\geq 6$. For $l=l_{5}$,
$\mu(H^{+})$ equals $7$ and then decreases until $l=l_{6}$, when
$\mu(H^{+})=6$. Afterwards $\mu(H^{+})$ increases again and reaches value $8$
for $l=l_{7}$. Finally, past $l=l_{7}$ the $\mu$-mass decreases and becomes
equal to $6$ when $l$ is horizontal and then increases again and becomes $8$
for $l=-l_{1}$, where “$-$” in front of $l_{1}$ means that the orientation of
the halfspace $H^{+}$ is the opposite from that in the beginning of the
rotation. We have shown that $x,y\in D^{*}\left(\mu\right)$ and
$\alpha^{*}(\mu)=6$. The convexity of $D^{*}\left(\mu\right)$ implies that
$o\in D^{*}\left(\mu\right)$, as well.

Figure 10. Example 5: In the left hand panel we display specific positions of
halfspaces when rotating the line $l$ that contains the point $y$. Boundaries
of these halfspaces are shown as dashed lines. In the right hand panel we see
several halfspaces whose boundary passes through $o$, as described in the
proof. Each halfspace from $\mathcal{H}(o)$ that contains the point $a$ has
$\mu$-mass at least $6$.

Consider now the covering medians of $\mu$. Certainly $\gamma^{*}(\mu)=6$. We
are able to cover the whole space by halfspaces whose boundary contains $y$.
To see this, consider the halfspaces determined by the horizontal and the
vertical line containing $y$, respectively, and the one determined by line $q$
in Figure 9. For $x$, the halfspaces are determined by the line $p$, and the
horizontal and vertical line. Therefore, $x,y\in C^{*}\left(\mu\right)$. In
the right hand panel of Figure 10 we see that each halfspace whose boundary
passes through the origin $o$ and contains point $a$ has $\mu$-mass greater
than $6$. Therefore, it is not possible to cover $a$ by minimizing halfspaces
of $o$, and consequently $o\notin C^{*}\left(\mu\right)$.

###  A.13. Proof of Theorem 12

For any $y\in\mathbb{R}^{d}$ denote by $\mathcal{H}_{\min}(y)$ the collection
of those halfspaces $H\in\mathcal{H}(y)$ that satisfy
$\mu\left(\operatorname{int}\left(H\right)\right)\leq\gamma^{*}(\mu)$. In
particular, $\mathcal{H}_{\min}=\mathcal{H}_{\min}(x)$. Suppose there is $y\in
C^{*}\left(\mu\right)$ such that $y\neq x$. Denote $z=(x+y)/2$. Since
$\mathbb{R}^{d}=\bigcup\left\\{H\in\mathcal{H}_{\min}(y)\right\\}$, there
exists $v\in\mathbb{S}^{d-1}$ and $H_{y,v}\in\mathcal{H}_{\min}(y)$ such that
$x\in H_{y,v}$. Then $H_{x,v}\subseteq H_{y,v}$ and
$\mu(\operatorname{int}\left(H_{x,v}\right))\leq\mu(\operatorname{int}\left(H_{y,v}\right))\leq\gamma^{*}(\mu)$,
meaning that $H_{x,v}\in\mathcal{H}_{\min}(x)$. If
$x\in\operatorname{int}\left(H_{y,v}\right)$, then $H_{x,v}\subset
H_{z,v}\subset\operatorname{int}\left(H_{y,v}\right)$. Our contiguity
assumption then implies
$\mu(H_{x,v})<\mu(H_{z,v})\leq\mu(\operatorname{int}\left(H_{y,v}\right))\leq\gamma^{*}(\mu)\leq\alpha^{*}(\mu)$,
which contradicts $x\in D^{*}\left(\mu\right)$. Thus,
$x\in\operatorname{bd}\left(H_{y,v}\right)$ and
$H_{y,v}\in\mathcal{H}_{\min}(x)$.

Consider any $w\in\mathbb{R}^{d}$. If $w\in H_{y,v}$, choose $H_{w}=H_{y,v}$.
Otherwise, the sequence of points $z_{n}=(1-1/n)z+w/n$, $n=1,2,\dots$
converges to $z$ and $z_{n}\notin H_{y,v}$. For each $z_{n}$ there is
$H_{n}\in\mathcal{H}_{\min}(y)$ such that $z_{n}\in H_{n}$ and
$\mu(\operatorname{int}\left(H_{n}\right))\leq\gamma^{*}(\mu)$. Using the same
argument as for the halfspace $H_{y,v}$ in the first part of the proof, we
conclude that $x\notin\operatorname{int}\left(H_{n}\right)$, because of
$H_{n}\in\mathcal{H}_{\min}(y)$ and because of our contiguity assumption.
Lemma 15 implies the existence of a convergent subsequence
$H_{n_{k}}\rightarrow H_{w}\in\mathcal{H}(z)$ such that
$\mu(\operatorname{int}\left(H_{w}\right))\leq\gamma^{*}(\mu)$. Note that also
$y\notin\operatorname{int}\left(H_{n}\right)$ because
$y\in\operatorname{bd}\left(H_{n}\right)$. Using Lemma 15, part (iv), we
conclude that $\\{x,y\\}\cap\operatorname{int}\left(H_{w}\right)=\emptyset$.
Because $z\in L(x,y)$ and $z\in\operatorname{bd}\left(H_{w}\right)$, it has to
be $\\{x,y\\}\subset\operatorname{bd}\left(H_{w}\right)$, meaning that
$H_{w}\in\mathcal{H}_{\min}(x)\cap\mathcal{H}_{\min}(y)$. Finally, we conclude
that for each $w\in\mathbb{R}^{d}$, there is
$H_{w}\in\mathcal{H}_{\min}(x)\cap\mathcal{H}_{\min}(y)$ such that $w\in
H_{w}$. Then $\mathcal{H}^{\prime}=\\{H_{w}\colon
w\in\mathbb{R}^{d}\\}\subseteq\mathcal{H}_{\min}(x)$ covers $\mathbb{R}^{d}$
by halfspaces whose interior has mass at most $\gamma^{*}(\mu)$, that at the
same time all contain $y\neq x$, which violates the assumption of our theorem.

###  A.14. Proof of Corollary 13

The first part of the corollary is a direct consequence of Theorem 5. Because
$\mu$ is atomic with finitely many atoms, there are only finitely many unique
$\mu$-masses of halfspaces $\mathcal{H}$. Therefore, when applying the Fatou
lemma of Lemma 15 in the proof of Lemma 4, one obtains a strict inequality
$\mu\left(\operatorname{int}\left(H(x,F)\right)\right)<\alpha$.

As for the second claim, first note that $D_{\alpha}(\mu)$ is a convex
polytope by [12, Lemma 1]. If $D_{\alpha}(\mu)$ is full-dimensional, each face
$F$ of $D_{\alpha}(\mu)$ is a subset of a $(d-1)$-dimensional face
$\widetilde{F}$ of $D_{\alpha}(\mu)$; if $\dim\left(D_{\alpha}(\mu)\right)<d$,
$F\subseteq\widetilde{F}=D_{\alpha}(\mu)$. If we prove our claim for
$\widetilde{F}$, it is necessarily true also for $F$. Without loss of
generality, we may therefore suppose that $F$ is of dimension
$\dim(\widetilde{F})=\min\left\\{\dim(D_{\alpha}(\mu)),d-1\right\\}$. Let
$H(x,F)$ be the halfspace from the first part of the proof. Denote by
$\mu_{1}\in\mathcal{M}\left(\mathbb{R}^{d}\right)$ the restriction of $\mu$ to
$\operatorname{bd}\left(H(x,F)\right)$ for
$H(x,F)\in\mathcal{H}\left(D_{\alpha}(\mu),F\right)$ from the first part of
the proof, and consider any $x\in F$. If $D(x;\mu_{1})=0$, then there exists a
closed $(d-1)$-dimensional halfspace $H_{1}$ in the hyperplane
$\operatorname{bd}\left(H(x,F)\right)$ such that $x\in H_{1}$ and
$\mu_{1}(H_{1})=0$. The existence of that minimizing halfspace follows because
$\mu$ contains only finitely many atoms, thus there are only finitely many
possible values of $\mu_{1}(H)$ for $H\in\mathcal{H}$. Consider now a slight
perturbation of the halfspace $H(x,F)$, in the sense that the unit normal
$v\in\mathbb{S}^{d-1}$ of $H(x,F)$ is perturbed, but  (i) the
$(d-2)$-dimensional affine space $\operatorname{relbd}\left(H_{1}\right)$
remains in the boundary of the perturbed halfspace
$H^{\prime}\in\mathcal{H}(x)$, and  (ii) $H_{1}\subset H^{\prime}$.  Because
there are only finitely many atoms of $\mu$, it is certainly possible to
obtain $H^{\prime}$ such that
$\mu(H^{\prime})=\mu(\operatorname{int}\left(H(x,F)\right))+\mu(H_{1})=\mu(\operatorname{int}\left(H(x,F)\right))<\alpha$,
which contradicts $x\in F\subseteq D_{\alpha}(\mu)$. We obtain that
$F\subseteq U_{\beta}(\mu_{1})$ for $\beta=0$. Since $U_{0}(\mu_{1})$ is,
again by the assumption of only finitely many atoms of $\mu$, a polytope whose
vertices are atoms of $\mu_{1}$, $F$ has to be contained in a convex hull of
at least $\dim(F)+1$ atoms of $\mu$, all lying in the hyperplane
$\operatorname{bd}\left(H(x,F)\right)$.

### Acknowledgement

This research was supported by the grant 19-16097Y of the Czech Science
Foundation, and by project PRIMUS/17/SCI/3 of Charles University. P. Laketa
was supported by the OP RDE project “International mobility of research,
technical and administrative staff at the Charles University”
CZ.02.2.69/0.0/0.0/18_053/0016976.

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