On the Beer Index of Convexity and Its Variants
On the Beer Index of Convexity and Its Variants
Martin Balko 0 1 2
Vít Jelínek 0 1 2
Pavel Valtr 0 1 2
Bartosz Walczak 0 1 2
Bartosz Walczak 0 1 2
Editor in Charge: János Pach
0 Computer Science Institute, Faculty of Mathematics and Physics, Charles University , Malostranské nám. 25, 118 00 Prague 1 , Czech Republic
1 Department of Applied Mathematics, Faculty of Mathematics and Physics, Charles University , Malostranské nám. 25, 118 00 Prague 1 , Czech Republic
2 Theoretical Computer Science Department, Faculty of Mathematics and Computer Science, Jagiellonian University , Kraków , Poland
Let S be a subset of Rd with finite positive Lebesgue measure. The Beer index of convexity b(S) of S is the probability that two points of S chosen uniformly independently at random see each other in S. The convexity ratio c(S) of S is the Lebesgue measure of the largest convex subset of S divided by the Lebesgue measure of S. We investigate the relationship between these two natural measures of convexity. We show that every set S ⊆ R2 with simply connected components satisfies b(S) α c(S) for an absolute constant α, provided b(S) is defined. This implies an affirmative answer to the conjecture of Cabello et al. that this estimate holds for simple

polygons. We also consider higherorder generalizations of b(S). For 1 k d, the
kindex of convexity bk (S) of a set S ⊆ Rd is the probability that the convex hull of a
(k + 1)tuple of points chosen uniformly independently at random from S is contained
in S. We show that for every d 2 there is a constant β(d) > 0 such that every
set S ⊆ Rd satisfies bd (S) β c(S), provided bd (S) exists. We provide an almost
matching lower bound by showing that there is a constant γ (d) > 0 such that for
every ε ∈ (
0, 1
) there is a set S ⊆ Rd of Lebesgue measure 1 satisfying c(S) ε and
bd (S) γ log2ε1/ε γ log2c1(S/)c(S) .
Mathematics Subject Classification 52A27
1 Introduction
For positive integers k and d and a Lebesgue measurable set S ⊆ Rd , we use λk (S)
to denote the kdimensional Lebesgue measure of S. We omit the subscript k when it
is clear from the context. We also write “measure” instead of “Lebesgue measure”, as
we do not use any other measure in the paper.
For a set S ⊆ Rd , let smc(S) denote the supremum of the measures of convex
subsets of S. Since all convex subsets of Rd are measurable [11], the value of smc(S)
is well defined. Moreover, Goodman’s result [8] implies that the supremum is achieved
on compact sets S, hence it can be replaced by maximum in this case. When S has
finite positive measure, let c(S) be defined as smc(S)/λd (S). We call the parameter
c(S) the convexity ratio of S.
For two points A, B ∈ Rd , let A B denote the closed line segment with endpoints
A and B. Let S be a subset of Rd . We say that points A, B ∈ S are visible one from
the other or see each other in S if the line segment A B is contained in S. For a point
A ∈ S, we use Vis( A, S) to denote the set of points that are visible from A in S. More
generally, for a subset T of S, we use Vis(T , S) to denote the set of points that are
visible in S from T . That is, Vis(T , S) is the set of points A ∈ S for which there is a
point B ∈ T such that A B ⊆ S.
Let Seg(S) denote the set {( A, B) ∈ S × S : A B ⊆ S} ⊆ (Rd )2, which we call the
segment set of S. For a set S ⊆ Rd with finite positive measure and with measurable
Seg(S), we define the parameter b(S) ∈ [0, 1] by
b(S) :=
λ2d (Seg(S))
λd (S)2
.
If S is not measurable, or if its measure is not positive and finite, or if Seg(S) is not
measurable, we leave b(S) undefined. Note that if b(S) is defined for a set S, then c(S)
is defined as well.
We call b(S) the Beer index of convexity (or just Beer index) of S. It can be interpreted
as the probability that two points A and B of S chosen uniformly independently at
random see each other in S.
1.1 Previous Results
The Beer index was introduced in the 1970s by Beer [1–3], who called it “the index of
convexity”. Beer was motivated by studying the continuity properties of λ(Vis( A, S))
as a function of A. For polygonal regions, an equivalent parameter was later
independently defined by Stern [19], who called it “the degree of convexity”. Stern was
motivated by the problem of finding a computationally tractable way to quantify how
close a given set is to being convex. He showed that the Beer index of a polygon P
can be approximated by a Monte Carlo estimation. Later, Rote [16] showed that for a
polygonal region P with n edges the Beer index can be evaluated in polynomial time
as a sum of O(n9) closedform expressions.
Cabello et al. [6] have studied the relationship between the Beer index and the
convexity ratio, and applied their results in the analysis of their nearlineartime
approximation algorithm for finding the largest convex subset of a polygon. We
describe some of their results in more detail in Sect. 1.3.
1.2 Terminology and Notation
We assume familiarity with basic topological notions such as pathconnectedness,
simple connectedness, Jordan curve, etc. The reader can find these definitions, for
example, in Prasolov’s book [15].
Let ∂ S, S◦, and S denote the boundary, the interior, and the closure of a set S,
respectively. For a point A ∈ R2 and ε > 0, let Nε( A) denote the open disc centered
at A with radius ε. For a set X ⊆ R2 and ε > 0, let Nε(X ) := A∈X Nε( A). A
neighborhood of a point A ∈ R2 or a set X ⊆ R2 is a set of the form Nε( A) or
Nε(X ), respectively, for some ε > 0.
A closed interval with endpoints a and b is denoted by [a, b]. Intervals [a, b] with
a > b are considered empty. For a point A ∈ R2, we use x ( A) and y( A) to denote the
x coordinate and the ycoordinate of A, respectively.
A polygonal curve Γ in Rd is a curve specified by a sequence ( A1, . . . , An ) of
points of Rd such that Γ consists of the line segments connecting the points Ai and
Ai+1 for i = 1, . . . , n − 1. If A1 = An, then the polygonal curve Γ is closed. A
polygonal curve that is not closed is called a polygonal line.
A set X ⊆ R2 is polygonally connected, or pconnected for short, if any two
points of X can be connected by a polygonal line in X , or equivalently, by a
selfavoiding polygonal line in X . For a set X , the relation “ A and B can be connected
by a polygonal line in X ” is an equivalence relation on X , and its equivalence classes
are the pcomponents of X . A set S is pcomponentwise simply connected if every
pcomponent of S is simply connected.
A line segment in Rd is a bounded convex subset of a line. A closed line segment
includes both endpoints, while an open line segment excludes both endpoints. For two
points A and B in Rd , we use A B to denote the open line segment with endpoints A
and B. A closed line segment with endpoints A and B is denoted by A B.
We say that a set S ⊆ Rd is starshaped if there is a point C ∈ S such that
Vis(C, S) = S. That is, a starshaped set S contains a point which sees the entire S.
(
0, 1
)
(
1, 1
)
Similarly, we say that a set S is weakly starshaped if S contains a line segment such
that Vis( , S) = S.
1.3 Results
We start with a few simple observations. Let S be a subset of R2 such that Seg(S)
is measurable. For every ε > 0, S contains a convex subset K of measure at least
(c(S) − ε)λ2(S). Two points of S chosen uniformly independently at random both
belong to K with probability at least (c(S) − ε)2, hence b(S) (c(S) − ε)2. This
yields b(S) c(S)2. This simple lower bound on b(S) is tight, as shown by a set S
which is a disjoint union of a single large convex component and a large number of
small components of negligible size.
It is more challenging to find an upper bound on b(S) in terms of c(S), possibly
under additional assumptions on the set S. This is the general problem addressed in
this paper.
As a motivating example, observe that a set S consisting of n disjoint convex
components of the same size satisfies b(S) = c(S) = n1 . It is easy to modify this
example to obtain, for any ε > 0, a simple starshaped polygon P with b( P) n1 − ε
and c( P) n1 , see Fig. 1. This shows that b(S) cannot be bounded from above by a
sublinear function of c(S), even for simple polygons S.
For weakly starshaped polygons, Cabello et al. [6] showed that the above example
is essentially optimal, providing the following linear upper bound on b(S).
Theorem 1.1 [6, Thm. 5] For every weakly starshaped simple polygon P, we have
b( P) 18 c( P).
For polygons that are not weakly starshaped, Cabello et al. [6] gave a superlinear
bound.
Theorem 1.2 [6, Thm. 6] Every simple polygon P satisfies
b( P)
Moreover, Cabello et al. [6] conjectured that even for a general simple
polygon P, b( P) can be bounded from above by a linear function of c( P).
(The question whether b( P) = O(c( P)) for simple polygons P was originally asked
by S. Cabello and M. Saumell, personal communication to G. Rote.) The next theorem,
which is the first main result of this paper, verifies this conjecture. Recall that b(S) is
defined for a set S if and only if S has finite positive measure and Seg(S) is
measurable. Recall also that a set is pcomponentwise simply connected if its pcomponents
are simply connected. In particular, every simply connected set is pcomponentwise
simply connected.
Theorem 1.3 Every pcomponentwise simply connected set S ⊆ R2 whose b(S) is
defined satisfies b(S) 180 c(S).
Clearly, every simple polygon satisfies the assumptions of Theorem 1.3. Hence we
directly obtain the following, which verifies the conjecture of Cabello et al. [6].
Corollary 1.4 Every simple polygon P ⊆ R2 satisfies b( P)
180 c( P).
The main restriction in Theorem 1.3 is the assumption that S is pcomponentwise
simply connected. This assumption cannot be omitted, as shown by the set
S := [0, 1]2 Q2, where it is easy to verify that c(S) = 0 and b(S) = 1, see
Proposition 3.7.
A related construction shows that Theorem 1.3 fails in higher dimensions. To see
this, consider again the set S := [0, 1]2 Q2, and define a set S ⊆ R3 by
S := {(t x , t y, t ) : t ∈ [0, 1] and (x , y) ∈ S}.
Again, it is easy to verify that c(S ) = 0 and b(S ) = 1, although S is simply
connected, even starshaped.
Despite these examples, we will show that meaningful analogues of Theorem 1.3
for higher dimensions and for sets that are not pcomponentwise simply connected are
possible. The key is to use higherorder generalizations of the Beer index, which we
introduce now.
For k ∈ {1, . . . , d} and a set S ⊆ Rd , we define the set Simpk (S) ⊆ (Rd )k+1 by
Simpk (S) := {( A0, . . . , Ak ) ∈ Sk+1 : Conv({ A0, . . . , Ak }) ⊆ S},
where the operator Conv denotes the convex hull of a set of points. We call Simpk (S)
the ksimplex set of S. Note that Simp1(S) = Seg(S).
For k ∈ {1, . . . , d} and a set S ⊆ Rd with finite positive measure and with
measurable Simpk (S), we define bk (S) by
bk (S) :=
λ(k+1)d (Simpk (S))
λd (S)k+1
.
Note that b1(S) = b(S). We call bk (S) the kindex of convexity of S. We again leave
bk (S) undefined if S or Simpk (S) is nonmeasurable, or if the measure of S is not
finite and positive.
We can view bk (S) as the probability that the convex hull of k + 1 points chosen
from S uniformly independently at random is contained in S. For any S ⊆ Rd , we
have b1(S) b2(S) · · · bd (S), provided all the bk (S) are defined.
We remark that the set S := [0, 1]d Qd satisfies c(S) = 0 and b1(S) = b2(S) =
· · · = bd−1(S) = 1, see Proposition 3.7. Thus, for a general set S ⊆ Rd , only the
dindex of convexity can conceivably admit a nontrivial upper bound in terms of c(S).
Our next result shows that such an upper bound on bd (S) exists and is linear in c(S).
Theorem 1.5 For every d 2, there is a constant β = β(d) > 0 such that every set
S ⊆ Rd with bd (S) defined satisfies bd (S) β c(S).
We do not know if the linear upper bound in Theorem 1.5 is best possible. We can,
however, construct examples showing that the bound is optimal up to a logarithmic
factor. This is our last main result.
Theorem 1.6 For every d 2, there is a constant γ = γ (d) > 0 such that for every
ε ∈ (
0, 1
), there is a set S ⊆ Rd satisfying c(S) ε and bd (S) γ log2ε1/ε , and in
particular, we have bd (S) γ log2c1(S/)c(S) .
The proof of Theorem 1.3 is given in Sect. 2. In Sect. 3, we will prove Theorems
1.5 and 1.6. We conclude, in Sect. 4, with some further remarks and a collection of
open problems.
2 Bounding the Mutual Visibility in the Plane
The goal of this section is to prove Theorem 1.3. Since the proof is rather long and
complicated, we first present a highlevel overview of its main ideas.
We first show that it is sufficient to prove the estimate from Theorem 1.3 for bounded
open simply connected sets. This is formalized by the next lemma, whose proof can
be found in Sect. 2.2.
Lemma 2.1 Let α > 0 be a constant such that every bounded open simply connected
set T ⊆ R2 satisfies b(T ) α c(T ). It follows that every pcomponentwise simply
connected set S ⊆ R2 with b(S) defined satisfies b(S) α c(S).
In the proof of Lemma 2.1, we first show that the set S can be reduced to a bounded
open set S whose Beer index b(S ) can be arbitrarily close to b(S) from below. This
is done by considering a part S of S that is contained in a sufficiently large disc and
by showing that all segments in S are in fact contained in the interior of S , except for
a set of measure zero. The proof is then finished by choosing S as the interior of S
and by applying the assumption of the lemma to every pcomponent of S .
Suppose now that S is a bounded open simply connected set. We seek a bound of
the form b(S) = O(c(S)). This is equivalent to a bound of the form λ4(Seg(S)) =
O(smc(S)λ2(S)). We therefore need a suitable upper bound on λ4(Seg(S)).
We first choose in S a diagonal (i.e., an inclusionmaximal line segment in S),
and show that the set S is a union of two open simply connected sets S1 and S2
(Lemma 2.4). It is not hard to show that the segments in S that cross the diagonal
contribute to λ4(Seg(S)) by at most O(smc(S)λ2(S)) (Lemma 2.8). Our main task is
to bound the measure of Seg(Si ∪ ) for i = 1, 2. The two sets Si ∪ are what we call
rooted sets. Informally, a rooted set is a union of a simply connected open set S and
an open segment r ⊆ ∂ S , called the root.
To bound λ4(Seg(R)) for a rooted set R with root r , we partition R into levels
L1, L2, . . ., where Lk contains the points of R that can be connected to r by a
polygonal line with k segments, but not by a polygonal line with k − 1 segments. Each
segment in R is contained in a union Li ∪ Li+1 for some i 1. Thus, a bound of the
form λ4(Seg(Li ∪ Li+1)) = O(smc(R)λ2(Li ∪ Li+1)) implies the required bound
for λ4(Seg(R)).
We will show that each pcomponent of Li ∪ Li+1 is a rooted set, with the extra
property that all its points are reachable from its root by a polygonal line with at most
two segments (Lemma 2.5). To handle such sets, we will generalize the techniques
that Cabello et al. [6] have used to handle weakly starshaped sets in their proof of
Theorem 1.1. We will assign to every point A ∈ R a set T( A) of measure O(smc(R)),
such that for every ( A, B) ∈ Seg(R), we have either B ∈ T( A) or A ∈ T(B)
(Lemma 2.7). From this, Theorem 1.3 will follow easily.
2.1 Proof of Theorem 1.3 for Bounded Open Simply Connected Sets
First, we need a few auxiliary lemmas.
Lemma 2.2 For every positive integer d, if S is an open subset of Rd , then the set
Seg(S) is open and the set Vis( A, S) is open for every point A ∈ S.
Proof Choose a pair of points ( A, B) ∈ Seg(S). Since S is open and A B is compact,
there is ε > 0 such that Nε( A B) ⊆ S. Consequently, for any A ∈ Nε( A) and
B ∈ Nε(B), we have A B ⊆ S, that is, ( A , B ) ∈ Seg(S). This shows that the set
Seg(S) is open. If we fix A = A, then it follows that the set Vis( A, S) is open.
Lemma 2.3 Let S be a simply connected subset of R2 and let and be line segments
in S. It follows that the set Vis( , S) ∩ is a (possibly empty) subsegment of .
Proof The statement is trivially true if and intersect or have the same supporting
line, or if Vis( , S) ∩ is empty. Suppose that these situations do not occur. Let
A, B ∈ and A , B ∈ be such that A A , B B ⊆ S. The points A, A , B , B form
a (possibly selfintersecting) tetragon Q whose boundary is contained in S. Since S
is simply connected, the interior of Q is contained in S. If Q is not selfintersecting,
then clearly A B ⊆ Vis( , S). Otherwise, A A and B B have a point D in common,
and every point C ∈ A B is visible in R from the point C ∈ A B such that D ∈ CC .
This shows that Vis( , S) ∩ is a convex subset and hence a subsegment of .
Now, we define rooted sets and their treestructured decomposition, and we explain
how they arise in the proof of Theorem 1.3.
A set S ⊆ R2 is halfopen if every point A ∈ S has a neighborhood Nε( A) that
satisfies one of the following two conditions:
1. Nε( A) ⊆ S,
2. Nε( A) ∩ ∂ S is a diameter of Nε( A) splitting it into two subsets, one of which
(including the diameter) is Nε( A) ∩ S and the other (excluding the diameter) is
Nε( A) S.
The condition 1 holds for points A ∈ S◦, while the condition 2 holds for points A ∈ ∂ S.
A set R ⊆ R2 is a rooted set if the following conditions are satisfied:
1. R is bounded,
2. R is pconnected and simply connected,
3. R is halfopen,
4. R ∩ ∂ R is an open line segment.
The open line segment R ∩ ∂ R is called the root of R. Every rooted set, as the union
of a nonempty open set and an open line segment, is measurable and has positive
measure.
A diagonal of a set S ⊆ R2 is a line segment contained in S that is not a proper
subset of any other line segment contained in S. Clearly, if S is open, then every
diagonal of S is an open line segment. It is easy to see that the root of a rooted set R
is a diagonal of R.
The following lemma allows us to use a diagonal to split a bounded open simply
connected subset of R2 into two rooted sets. It is intuitively clear, and its formal proof
is postponed to Sect. 2.3.
Lemma 2.4 Let S be a bounded open simply connected subset of R2, and let be a
diagonal of S. It follows that the set S has two pcomponents S1 and S2. Moreover,
S1 ∪ and S2 ∪ are rooted sets, and is their common root.
Let R be a rooted set. For a positive integer k, the kth level Lk of R is the set of
points of R that can be connected to the root of R by a polygonal line in R consisting
of k segments but cannot be connected to the root of R by a polygonal line in R
consisting of fewer than k segments. We consider a degenerate onevertex polygonal
line as consisting of one degenerate segment, so the root of R is part of L1. Thus
L1 = Vis(r, R), where r denotes the root of R. A kbody of R is a pcomponent of Lk .
A body of R is a kbody of R for some k. See Fig. 2 for an example of a rooted set
and its partitioning into levels and bodies.
We say that a rooted set P is attached to a set Q ⊆ R2 P if the root of P is
subset of the interior of P ∪ Q. The following lemma explains the structure of levels
and bodies. Although it is intuitively clear, its formal proof requires quite a lot of work
and can be found in Sect. 2.4.
Lemma 2.5 Let R be a rooted set and (Lk )k 1 be its partition into levels. It follows
that
1. R = k 1 Lk ; consequently, R is the union of all its bodies;
2. every body P of R is a rooted set such that P = Vis(r, P), where r denotes the
root of P;
d(A⎨ )⎧
A
A⎧
d(A, r ) ⎨
⎩
r
B = B
⎫
⎪⎪⎪⎬⎪
⎪⎪⎪⎪⎭
R
d(B ) = d(B, r )
3. L1 is the unique 1body of R, and the root of L1 is the root of R;
4. every j body P of R with j 2 is attached to a unique ( j − 1)body of R.
Lemma 2.5 yields a tree structure on the bodies of R. The root of this tree is the
unique 1body L1 of R, called the root body of R. For a kbody P of R with k 2,
the parent of P in the tree is the unique (k − 1)body of R that P is attached to, called
the parent body of P.
Lemma 2.6 Let R be a rooted set, (Lk )k 1 be the partition of R into levels, be a
closed line segment in R, and k 1 be minimum such that ∩ Lk = ∅. It follows
that ⊆ Lk ∪ Lk+1, ∩ Lk is a subsegment of contained in a single kbody P of
R, and ∩ Lk+1 consists of at most two subsegments of each contained in a single
(k + 1)body whose parent body is P.
Proof The definition of the levels directly yields ⊆ Lk ∪ Lk+1. The segment
splits into subsegments each contained in a single kbody or (k + 1)body of R. By
Lemma 2.5, the bodies of any two consecutive of these subsegments are in the
parentchild relation of the body tree. This implies that ∩ Lk lies within a single kbody P.
By Lemma 2.3, ∩ Lk is a subsegment of . Consequently, ∩ Lk+1 consists of at
most two subsegments.
In the setting of Lemma 2.6, we call the subsegment ∩ Lk of the base segment
of , and we call the body P that contains ∩ Lk the base body of . See Fig. 2 for an
example.
The following lemma is the crucial part of the proof of Theorem 1.3.
Lemma 2.7 If R is a rooted set, then every point A ∈ R can be assigned a measurable
set T( A) ⊆ R2 so that the following is satisfied:
1. λ2(T( A)) < 87 smc(R);
2. for every line segment BC in R, we have either B ∈ T(C ) or C ∈ T(B);
3. the set {( A, B) : A ∈ R and B ∈ T( A)} is measurable.
Proof Let P be a body of R with the root r . First, we show that P is entirely contained
in one closed halfplane defined by the supporting line of r . Let h− and h+ be the
two open halfplanes defined by the supporting line of r . According to the definition
of a rooted set, the sets {D ∈ r : ∃ε > 0 : Nε(D) ∩ h− = Nε(D) ∩ ( P r )} and
{D ∈ r : ∃ε > 0 : Nε(D) ∩ h+ = Nε(D) ∩ ( P r )} are open and partition the entire
r , hence one of them must be empty. This implies that the segments connecting r to
P r lie all in h− or all in h+. Since P = Vis(r, P), we conclude that P ⊆ h− or
P ⊆ h+.
According to the above, we can rotate and translate the set R so that r lies on the
x axis and P lies in the halfplane {B ∈ R2 : y(B) 0}. For a point A ∈ R, we
use d( A, r ) to denote the ycoordinate of A after such a rotation and translation of R.
We use d( A) to denote d( A, r ) where r is the root of the body of A. It follows that
d( A) 0 for every A ∈ R.
Let γ ∈ (
0, 1
) be a fixed constant whose value will be specified at the end of the
proof. For a point A ∈ R, we define sets
V1( A) := {B ∈ Vis( A, R) :  A B  γ  A B, A ∈ Vis(r , R), d( A , r ) d(B , r )},
V2( A) := {B ∈ Vis( A, R) :  A B  γ  A B, A ∈/ Vis(r , R), d( A , r ) d(B , r )},
V3( A) := {B ∈ Vis( A, R) :  A B  < γ  A B,  A A 
B B },
where r denotes the root of the base body of A B and A and B denote the endpoints of
the base segment of A B such that  A A  <  A B . For every A ∈ R, the sets V1( A),
V2( A), and V3( A) are pairwise disjoint. Moreover, we have A ∈ 3i3=1 Vi (B) or
B ∈ i3=1 Vi ( A) for every line segment A B in R. If for some B ∈ i=1 Vi ( A) the
point A lies on r , then we have B ∈ V1( A) and V1( A) ⊆ r .
For the rest of the proof, we fix a point A ∈ R. We show that the union i3=1 Vi ( A)
is contained in a measurable set T( A) ⊆ R2 with λ2(T( A)) < 87 smc(R) that is a
union of three trapezoids. We let P be the body of A and r be the root of P. If P is a
kbody with k 2, then we use r to denote the root of the parent body of P.
Claim 1 V1( A) is contained in a trapezoid T1( A) with area 6γ −2 smc(R).
height d( A) with bases of length 8 smc(R) ⊆andR.4 sLmect(RT) sbuechththeartApairsaltlheel ctreanpteezrooifdthoef
Let H be a point of r such that A H
d(A) d(A)
larger base and H is the center of the smaller base. The homothety with center A and
ratio γ −1 transforms T into the trapezoid T := A + γ −1(T − A). Since the area of
T is 6 smc(R), the area of T is 6γ −2 smc(R). We show that V1( A) ⊆ T . See Fig. 3
for an illustration.
Let B be a point in V1( A). Using a similar approach to the one used by Cabello et
al. [6] in the proof of Theorem 1.1, we show that B ∈ T . Let A B be the base segment
of A B such that  A A  <  A B . Since B ∈ V1( A), we have  A B  γ  A B,
A ∈ Vis(r , R), and d(B, r ) d( A, r ), where r denotes the root of the base level
of A B. Since A is visible from r in R, the base body of A B is the body of A and
thus A = A and r = r . As we have observed, every point C ∈ { A} ∪ A B satisfies
d(C, r ) = d(C ) 0.
Let ε > 0. There is a point E ∈ A B such that B E  < ε. Since E lies on the
base segment of A B, there is F ∈ r such that E F ⊆ R. It is possible to choose F
T
E B
B
A = A
D
so that A H and E F have a point C in common where C = F, H . Let D be a point
of A H with d(D) = d(E ). The point D exists, as d(H ) = 0 d(E ) d( A). The
points A, E , F, H form a selfintersecting tetragon Q whose boundary is contained
in R. Since R is simply connected, the interior of Q is contained in R and the triangles
AC E and C F H have area at most smc(R).
The triangle AC E is partitioned into triangles A D E and C D E with areas
21 (d( A) − d(D))D E  and 21 (d(D) − d(C ))D E , respectively. Therefore, we have
21 (d( A) − d(C ))D E  = λ2( AC E ) smc(R). This implies
.
For the triangle C F H , we have 21 d(C )F H  = λ2(C F H ) smc(R). By the
similarity of the triangles C F H and C D E , we have F H  = D E d(C )/(d(E ) − d(C ))
and therefore
D E 
2 smc(R)
d(C )2 (d(E ) − d(C )).
Since the first upper bound on D E  is increasing in d(C ) and the second is decreasing
in d(C ), the minimum of the two is maximized when they are equal, that is, when
d(C ) = d( A)d(E )/(d( A) + d(E )). Then we obtain D E  2 dsm(Ac()2R) (d( A) + d(E )).
This and 0 d(E ) d( A) imply E ∈ T . Since ε can be made arbitrarily small and
T is compact, we have B ∈ T . Since  A B  γ  A B, we conclude that B ∈ T .
This completes the proof of Claim 1.
Claim 2 V2( A) is contained in a trapezoid T2( A) with area 3(1 − γ )−2γ −2 smc(R).
We assume the point A is not contained in the first level of R, as otherwise V2( A) is
empty. Let p be the r parallel line that contains the point A and let q be the supporting
line of r . Let p+ and q+ denote the closed halfplanes defined by p and q, respectively,
such that r ⊆ p+ and A ∈/ q+. Let O be the intersection point of p and q.
Let T ⊆ p+ ∩ q+ be the trapezoid of height d( A, r ) with one base of length
(1−4γs)m2cd((RA),r ) on p, the other base of length (1−2γs)m2cd((RA),r ) on the supporting line of r ,
D
C
F
H
H 0
E B
B
T
T
and one lateral side on q. The homothety with center O and ratio γ −1 transforms T
into the trapezoid T := O + γ −1(T − O). Since the area of T is 3(1 − γ )−2 smc(R),
the area of T is 3(1 − γ )−2γ −2 smc(R). We show that V2( A) ⊆ T . See Fig. 4 for an
illustration.
Let B be a point of V2( A). We use A B to denote the base segment of A B such that
 A A  <  A B . By the definition of V2( A), we have  A B  γ  A B, A ∈/ Vis(r , R),
and d(B, r ) d( A, r ), where r denotes the root of the base body of A B. By
Lemma 2.6 and the fact that A ∈/ Vis(r , R), we have r = r . The bound d( A, r )
d(B, r ) thus implies A ∈ r ∩ p+ and B ∈ q+. We have d(C, r ) = d(C ) 0 for
every C ∈ A B .
Observe that (1 − γ )d( A, r ) d( A , r ) d( A, r ). The upper bound is trivial,
as d(B, r ) d( A, r ) and A lies on A B. For the lower bound, we use the expression
A = t A + (1 − t )B for some t ∈ [0, 1]. This gives us d( A , r ) = t d( A, r ) +
(1 − t )d(B , r ). By the estimate  A B  γ  A B, we have
 A A  + B B 
(1 − γ ) A B = (1 − γ )( A B  + B B ).
This can be rewritten as  A A  (1 − γ ) A B  − γ B B . Consequently, B B  0
and γ > 0 imply  A A  (1 − γ ) A B . This implies t 1 − γ . Applying the bound
d(B , r ) 0, we conclude that d( A , r ) (1 − γ )d( A, r ).
Let (Gn)n∈N be a sequence of points from A B that converges to A . For every
n ∈ N, there is a point Hn ∈ r such that Gn Hn ⊆ R. Since r is compact, there is
a subsequence of (Hn)n∈N that converges to a point H0 ∈ r . We claim that H0 ∈ q.
Suppose otherwise, and let q = q be the supporting line of A H0. Let ε > 0 be
small enough so that Nε( A ) ⊆ R. For n large enough, Gn Hn is contained in an
arbitrarily small neighborhood of q . Consequently, for n large enough, the supporting
line of Gn Hn intersects q at a point Kn such that Gn Kn ⊆ Nε( A ), which implies
Kn ∈ r ∩ Vis(r , R), a contradiction.
Again, let ε > 0. There is a point E ∈ A B such that B E  < ε. Let D be
a point of q with d(D , r ) = d(E ), and let δ > 0. There are points G ∈ A B
and H ∈ r such that G ∈ Nδ( A ) and G H ⊆ R ∩ Nδ(q). If δ is small enough,
then d(E ) d( A , r ) − δ d(G) d( A , r ). Let D be the point of G H with
d(D) = d(E ). The point E lies on A B and thus it is visible from a point F ∈ r .
Again, we can choose F so that the line segments E F and G H have a point C in
common where C = F, H . The points E , F, H, G form a selfintersecting tetragon
Q whose boundary is in R. The interior of Q is contained in R, as R is simply
connected. Therefore, the area of the triangles C E G and C F H is at most smc(R).
The argument used in the proof of Claim 1 yields
D E 
2 smc(R)
d(G)2 (d(G) + d(E ))
2 smc(R)
(d( A , r ) − δ)2 (d( A , r ) + d(E )).
This and the fact that δ (and consequently D D) can be made arbitrarily small yield
D E  d2(sAmc,r(R)2) (d( A , r ) + d(E )). This together with d( A , r ) (1 − γ )d( A, r )
yield D E  (1−2γs)m2dc((AR,)r )2 (d( A, r ) + d(E )). Finally, from 0 d(E ) d( A, r ) it
follows that E ∈ T .
Since ε can be made arbitrarily small and T is compact, we have B ∈ T . Since
 A B  γ  A B γ  A B, we conclude that B ∈ T . This completes the proof of
Claim 2.
Claim 3 V3( A) is contained in a trapezoid T3( A) with area (4(1−γ )−2 −1) smc(R).
By Lemma 2.3, the points of r that are visible from A in R form a subsegment
C D of r . The homothety with center A and ratio 2(1 − γ )−1 transforms the triangle
T := AC D into the triangle T := A + 2(1 − γ )−1(T − A). See Fig. 5 for an
illustration. We claim that V3( A) is a subset of the trapezoid T := T T .
Let B be an arbitrary point of V3( A). Consider the segment A B with the base
segment A B such that  A A  <  A B . Since B ∈ V3( A), we have  A B  < γ  A B
and  A A  B B . This implies  A A  1−2γ  A B > 0 and hence A = A and
B ∈/ P. From the definition of C and D, we have A ∈ C D. Since  A A  1−2γ  A B
and B ∈/ P, we have B ∈ T .
The area of T is (4(1 − γ )−2 − 1)λ2(T ). The interior of T is contained in R, as
all points of the open segment C D are visible from A in R. The area of T is at most
smc(R), as its interior is a convex subset of R. Consequently, the area of T is at most
(4(1 − γ )−2 − 1) smc(R). This completes the proof of Claim 3.
To put everything together, we set T( A) := i3=1 Ti ( A). Then, it follows that
3
i=1 Vi ( A) ⊆ T( A) for every A ∈ R. Clearly, the set T( A) is measurable. Summing
the three estimates on areas of the trapezoids, we obtain
λ2(T( A))
6γ −2 + 3(1 − γ )−2γ −2 + 4(1 − γ )−2 − 1 smc(R)
for every point A ∈ R. We choose γ ∈ (
0, 1
) so that the value of the coefficient is
minimized. For x ∈ (
0, 1
), the function x → 6x −2 + 3(1 − x )−2x −2 + 4(1 − x )−2 − 1
attains its minimum 86.7027 < 87 at x ≈ 0.5186. Altogether, we have λ2(T( A)) <
87 smc(R) for every A ∈ R.
It remains to show that the set {( A, B) : A ∈ R and B ∈ T( A)} is measurable. For
every body P of R and for i ∈ {1, 2, 3}, the definition of the trapezoid Ti ( A) in Claim i
implies that the set {( A, B) : A ∈ P and B ∈ Ti ( A)} is the intersection of P × R2 with
a semialgebraic (hence measurable) subset of (R2)2 and hence is measurable. There
are countably many bodies of R, as each of them has positive measure. Therefore,
{( A, B) : A ∈ R and B ∈ T( A)} is a countable union of measurable sets and hence is
measurable.
Let S be a bounded open subset of the plane, and let be a diagonal of S that lies
on the x axis. For a point A ∈ S, we define the set
S( A, ) := {B ∈ Vis( A, S) : A B ∩
= ∅ and y( A)
y(B)}.
The following lemma is a slightly more general version of a result of Cabello et al. [6].
Lemma 2.8 Let S be a bounded open simply connected subset of R2, and let be its
diagonal. It follows thatλ2(S( A, )) 3 smc(S) for every A ∈ S.
Proof We can assume without loss of generality that lies on the x axis. Using
an argument similar to the proof of Lemma 2.2, we can show that the set
{B ∈ Vis( A, S) : A B ∩ = ∅} is open. Therefore, S( A, ) is the intersection of an
open set and the closed halfplane {(x , y) ∈ R2 : y −y( A)} or {(x , y) ∈ R2 : y
−y( A)}, whichever contains A. Consequently, the set S( A, ) is measurable for every
A ∈ S.
We clearly have λ2(S( A, )) = 0 for points A ∈ S Vis( , S). By Lemma
2.3, the set Vis( A, S) ∩ is an open subsegment C D of . The interior T ◦ of the
triangle T := AC D is contained in S. Since T ◦ is a convex subset of S, we have
λ2(T ◦) = 21 C D·y( A) smc(S). Therefore, every point B ∈ S( A, ) is contained
in a trapezoid of height y( A) with bases of length C D and 2C D. The area of this
trapezoid is 23 C D · y( A) 3 smc(S). Hence we have λ2(S( A, )) 3 smc(S) for
every point A ∈ S.
Proof of Theorem 1.3 In view of Lemma 2.1, we can assume without loss of generality
that S is a bounded open simply connected set. Let be a diagonal of S. We can assume
without loss of generality that lies on the x axis.
According to Lemma 2.4, the set S has exactly two pcomponents S1 and S2,
the sets S1 ∪ and S2 ∪ are rooted sets, and is their common root. By Lemma 2.7,
for i ∈ {1, 2}, every point A ∈ Si ∪ can be assigned a measurable set Ti ( A) so
that λ2(Ti ( A)) < 87 smc(Si ∪ ) 87 smc(S), every line segment BC in Si ∪
satisfies B ∈ Ti (C ) or C ∈ Ti (B), and the set {( A, B) : A ∈ Si ∪ and B ∈ Ti ( A)}
is measurable.
We set S( A) := Ti ( A) ∪ S( A, ) for every point A ∈ Si with i ∈ {1, 2}. We set
S( A) := T1( A) ∪ T2( A) for every point A ∈ = S (S1 ∪ S2). Let
S := {( A, B) : A ∈ S and B ∈ S( A)} ∪ {(B, A) : A ∈ S and B ∈ S( A)} ⊆ (R2)2.
It follows that the set S is measurable.
Let A B be a line segment in S, and suppose y( A) y(B). Then either A and B
are in distinct pcomponents of S or they both lie in the same component Si with
i ∈ {1, 2}. In the first case, we have B ∈ S( A), since A B intersects and S( A, ) ⊆
S( A). In the second case, we have B ∈ Ti ( A) ⊆ S( A) or A ∈ Ti (B) ⊆ S(B).
Therefore, we have Seg(S) ⊆ S. Since both Seg(S) and S are measurable, we have
λ4(Seg(S))
λ4(S)
2
λ2(S( A)),
A∈S
where the second inequality is implied by Fubini’s Theorem. The bound λ2(S( A))
90 smc(S) implies
S
λ4(Seg(S))
2
90 smc(S) = 180 smc(S)λ2(S).
Finally, this bound can be rewritten as b(S) = λ4(Seg(S))λ2(S)−2
180 c(S).
2.2 Proof of Lemma 2.1
In this section, we prove Lemma 2.1, which reduces the general setting of Theorem 1.3
to the case that S is a bounded open simply connected subset of R2.
Lemma 2.9 Let S ⊆ R2 be a set whose b(S) is defined. For every ε > 0, there is a
bounded set S ⊆ S such that λ(S ) (1 − ε)λ(S) and b(S ) b(S) − ε. Moreover,
if S is pcomponentwise simply connected, then so is S .
Proof Let B be an open ball in R2 centered at the origin. Consider the sets S = S ∩ B
and S0 = S B partitioning the set S. Fix the radius of B large enough, so that S0 has
measure at most ελ(S)/2. We claim that S has the properties stated in the lemma.
Clearly λ(S ) (1 − ε/2)λ(S) > (1 − ε)λ(S). Moreover, Seg(S ) = Seg(S)
(S0 × S) ∪ (S × S0) , and hence Seg(S ) is measurable and we have λ4(Seg(S ))
λ4(Seg(S)) − ελ(S)2. Therefore,
λ4(Seg(S ))
b(S ) = λ4(S × S )
λ4(Seg(S ))
λ4(S × S)
λ4(Seg(S)) − ελ(S)2
λ4(S × S)
= b(S) − ε,
as claimed. It is clear from the construction that if S is pcomponentwise simply
connected, then so is S .
Lemma 2.10 Let S ⊆ R2 be a bounded pcomponentwise simply connected measur
able set with measurable segment set. Then λ4(Seg(S) Seg(S◦)) = 0. In other words,
all the segments in S are in fact contained in S◦, except for a set of measure zero.
Proof Let B denote the set Seg(S) Seg(S◦), that is, B is the set of segments in
S containing at least one point of ∂ S. Note that B is measurable, since Seg(S) is
measurable by assumption and Seg(S◦) is an open set by Lemma 2.2, hence it is
measurable as well.
Let A B be a segment contained in S, and let C be a point of A B. We say that C is
an isolated boundary point of the segment A B, if C ∈ ∂ S, but there is an ε > 0 such
that no other point of A B ∩ Nε(C ) belongs to ∂ S.
We partition the set B into four parts as follows:
B := {( A, B) ∈ B : A = B or A B is a vertical segment},
B := {( A, B) ∈ B
B := {( A, B) ∈ B
B : A is an isolated boundary point of A B},
(B ∪ B ) : B is an isolated boundary point of A B},
B• := B
(B ∪ B
∪ B ).
We claim that each of these sets has measure zero. For B, this is clear, since B is
a subset of {( A, B) ∈ R2 × R2 : A = B or A B is a vertical segment}, which clearly
has λ4measure zero.
Consider now the set B . We first argue that it is measurable. For a set α ⊆ [0, 1]
and a pair of points ( A, B), define A B[α] := {t B + (1 − t ) A : t ∈ α}, and let
S(α) be the set {( A, B) ∈ R2 × R2 : A B[α] ⊆ S◦}. In particular, if α = [0, 1] then
A B[α] = A B and S(α) = Seg(S◦). If α is a closed interval, then A B[α] is a segment,
and it is not hard to see that S(α) is an open set, and, in particular, it is measurable. If α
is an open interval, say α = (s, t ) ⊆ [0, 1], then S(α) = n∈N S([s +n−1, t −n−1]),
and hence S(α) is measurable as well. We then see that
= B ∩ (∂ S × S) ∩
S((0, n−1)) ,
n∈N
showing that B is measurable. An analogous argument shows that B is measurable,
and hence B• is measurable as well.
In the rest of the proof, we will use two basic facts of integral calculus, which we
now state explicitly.
Fact 1 (see [17, Lem. 7.25 and Thm. 7.26]) Let X, Y ⊆ Rd be two open sets, and let
σ : X → Y be a bijection such that both σ and σ −1 are continuous and differentiable
on X and Y , respectively. Then, for any X0 ⊆ X , the set X0 is measurable if and only
if σ (X0) is measurable. Moreover, λ(X0) = 0 if and only if λ(σ (X0)) = 0.
Fact 2 (Fubini’s Theorem, see [17, Thm. 8.12]) Let M ⊆ Rk × R be a measurable
set. For x ∈ Rk , define Mx := {y ∈ R : (x , y) ∈ M }. Then, for almost every x ∈ Rk ,
the set Mx is λ measurable, and
λk+ (M ) =
x∈Rk
λ (Mx ).
Let us prove that λ4(B ) = 0. The basic idea is as follows: suppose that we have
fixed a nonvertical line L and a point B ∈ L. It can be easily seen that there are at
most countably many points A ∈ L such that ( A, B) ∈ B . Since a line L with a point
B ∈ L can be determined by three parameters, we will see that B has λ4measure
zero.
Let us describe this reasoning more rigorously. Let La,b denote the line {(x , y) ∈
R2 : y = ax + b}. Define a mapping σ : R4 → R2 × R2 as follows: σ (a, b, x , x ) =
( A, B), where A = (x , ax + b) and B = (x , ax + b). In other words, σ (a, b, x , x )
is the pair of points on the line La,b whose horizontal coordinates are x and
x , respectively. For every nonvertical segment A B, there is a unique
quadruple (a, b, x , x ) with x = x , such that σ (a, b, x , x ) = ( A, B). In particular, σ
is a bijection from the set {(a, b, x , x ) ∈ R4 : x = x } to the set {( A, B) ∈
R2 × R2 : A, B not on the same vertical line}.
Define B = σ −1(B ). Note that σ satisfies the assumptions of Fact 1, and
therefore B is measurable. Moreover, λ4(B ) = 0 if and only if λ4(B ) = 0.
For a fixed triple (a, b, x ) ∈ R3, let Xa,b,x denote the set {x ∈ R : (a, b, x , x ) ∈ B }.
We claim that Xa,b,x is countable. To see this, choose a point x ∈ Xa,b,x and define
( A, B) := σ (a, b, x , x ). Since ( A, B) ∈ B , we know that A is an isolated boundary
point of A B, which implies that there is a closed interval β ⊆ R of positive length
such that β ∩ Xa,b,x = {x }. This implies that Xa,b,x is countable and thus of measure
zero.
Since B is measurable, we can apply Fubini’s Theorem to get
λ4(B ) =
(a,b,x )∈R3
λ1(Xa,b,x ).
Therefore λ4(B ) = 0 as claimed. A similar argument shows that λ4(B ) = 0.
It remains to deal with the set B . We will use the following strategy: we will fix
•
two parallel nonhorizontal lines L1, L2, and study the segments orthogonal to these
two lines, with one endpoint on L1 and the other on L2. Roughly speaking, our goal is
to show that for “almost every” choice of L1 and L2, there are “almost no” segments
of this form belonging to B .
•
Let La,b denote the (nonhorizontal) line {(ay + b, y) : y ∈ R}. Let us say that a
pair of distinct points ( A, B) has type (a, b, c), if A ∈ La,b, B ∈ La,c, and the segment
A B is orthogonal to La,b (and therefore also to La,c). The value a is then called the
slope of the type t = (a, b, c).
Note that every pair of distinct points ( A, B) defining a nonvertical segment has
a unique type (a, b, c), with b = c. Define a mapping τ : R4 → R2 × R2, where
τ (a, b, c, y) is the pair of points ( A, B) of type (a, b, c) such that A = (ay + b, y).
Note that τ is a bijection from the set {(a, b, c, y) ∈ R4 : b = c} to the set
{( A, B) ∈ R2 × R2 : A, B not on the same vertical line}. We can easily verify that
τ satisfies the assumptions of Fact 1.
Define B• = τ −1(B•). From Fact 1, it follows that B• is measurable, and
λ4(B•) = 0 if and only if λ4(B•) = 0. For a type t = (a, b, c) ∈ R3, define
Yt = {y ∈ R : (a, b, c, y) ∈ B•}. Furthermore, for a set α ⊆ [0, 1], define B•(α) =
B• ∩ S(α), B•(α) = τ −1(B•(α)), and Yt (α) = {y ∈ R : (a, b, c, y) ∈ B•(α)}. In
our applications, α will always be an interval (in fact, an open interval with rational
endpoints), and in such case we already know that B•(α) is measurable, hence B•(α)
is measurable.
By Fubini’s Theorem, we have
λ4(B•) =
a∈R (b,c)∈R2
λ1(Y(a,b,c)),
(*)
and Yt is measurable for all t ∈ R3 up to a set of λ3measure zero. An analogous
formula holds for B•(α) and Yt (α) for any open interval α ⊆ [0, 1] with rational
endpoints. Since there are only countably many such intervals, and a countable union
of sets of measure zero has measure zero, we know that there is a set T0 ⊆ R3 of
measure zero, such that for all t ∈ R3 T0 the set Yt is measurable, and moreover for
any rational interval α the set Yt (α) is measurable as well.
Our goal is to show that there are at most countably many slopes a ∈ R for
which there is a (b, c) ∈ R2 such that λ1(Y(a,b,c)) > 0. From (*) it will then follow
that λ4(B ) = 0. To achieve this goal, we will show that to any type t for which
•
λ1(Yt ) > 0, we can assign a set Rt ⊆ ∂ S of positive λ2measure (the region of t ),
so that if t and t have different slopes and if Yt and Yt both have positive measure,
then Rt and Rt are disjoint. Since there cannot be uncountably many disjoint sets of
positive measure, this will imply the result.
Let us fix a type t = (a, b, c) ∈ R3 T0 such that λ1(Yt ) > 0. Let us say that an
element y ∈ Yt is halfisolated if there is an ε > 0 such that [y, y + ε] ∩ Yt = {y} or
[y − ε, y] ∩ Yt = {y}. Clearly, Yt has at most countably many halfisolated elements.
Define Yt∗ := {y ∈ Yt : y is not halfisolated}. Of course, λ1(Yt∗) = λ1(Yt ). See Fig. 6
for an illustration.
L a,c
B 2
A 1
S
B 1
y1 ∈ Yt
y2 ∈ Yt∗
Fig. 6 An illustration for the proof of Lemma 2.10. The element y1 of Yt is halfisolated while y2 is not
Choose y ∈ Yt∗, and define ( Ay , By ) := τ (a, b, c, y). We claim that Ay By ∩ S◦
is either empty or a single interval. Let us choose any two points C, D ∈ Ay By ∩
S◦. We will show that the segment C D is inside S◦. For ε > 0 small enough, the
neighborhoods Nε(C ) and Nε(D) are subsets of S. Since y is not halfisolated in
Yt , we can find two segments P, Q ∈ B• of type t that intersect both Nε(C ) and
Nε(D), with Ay By being between P and Q. We can then find a closed polygonal
curve Γ ⊆ P ∪ Q ∪ Nε(C ) ∪ Nε(D) whose interior region contains C D. Since S is
pcomponentwise simply connected, we see that C D ⊆ S◦. Therefore, Ay By ∩ S◦ is
indeed an interval.
Since ∂ S is a closed set, we know that for every y ∈ Yt∗, the set Ay By ∩ ∂ S is
closed as well. Moreover, neither Ay nor By are isolated boundary points of Ay By ,
because then ( Ay , By ) would belong to B or B . We conclude that Ay By ∩ ∂ S is
either equal to a single closed segment of positive length containing Ay or By , or it
is equal to a disjoint union of two closed segments of positive length, one of which
contains Ay and the other contains By .
For an integer n ∈ N, define two sets Yt (n) and Yt (n) by
Yt (n) := {y ∈ Yt∗ : Ay By [(0, n−1)] ⊆ ∂ S} and
Yt (n) := {y ∈ Yt∗ : Ay By [(1 − n−1, 1)] ⊆ ∂ S}.
Note that these sets are measurable: for instance, Yt (n) is equal to Yt∗ α Yt (α) ,
where we take the union over all rational intervals α intersecting (0, n−1). Moreover,
we have Yt∗ = n∈N(Yt (n) ∪ Yt (n)). It follows that there is an n such that Yt (n)
or Yt (n) has positive measure. Fix such an n and assume, without loss of generality,
that λ1(Yt (n)) is positive. Define the region of t , denoted by Rt , by
Rt :=
Ay By (0, n−1) .
y∈Yt (n)
The set Rt is a bijective affine image of Yt (n) × (0, n−1), and in particular it is
λ2measurable with positive measure. Note that Rt is a subset of ∂ S.
Consider now two types t, t ∈ R3 T0 with distinct slopes, such that both Yt and
Yt have positive measure. We will show that the regions Rt and Rt are disjoint.
For contradiction, suppose there is a point C ∈ Rt ∩ Rt . Let A B and A B be
the segments containing C and having types t and t , respectively. Fix ε > 0 small
enough, so that none of the four endpoints A, B, A , B lies in Nε(C ). Since Yt∗ has
no halfisolated points of Yt , we know that B• has segments of type t arbitrarily close
to A B on both sides of A B, and similarly for segments of type t close to A B . We can
therefore find four segments P, Q, P , Q ∈ B• { A B, A B } with these properties:
– P and Q have type t , and P and Q have type t .
– A B is between P and Q (i.e., A B ⊆ Conv( P ∪ Q)) and A B is between P
and Q .
– Both P and Q intersect both P and Q inside Nε(C ).
We see that the four points where P ∪ Q intersects P ∪ Q form the vertex set
of a parallelogram W whose interior contains the point C . Moreover, the boundary
of W is a closed polygonal curve contained in S. Since S is pcomponentwise simply
connected, W is a subset of S and C belongs to S◦. This is a contradiction, since all
points of Rt (and Rt ) belong to ∂ S.
We conclude that Rt and Rt are indeed disjoint. Since there cannot be uncountably
many disjoint sets of positive measure in R2, there are at most countably many values
a ∈ R for which there is a type t = (a, b, c) with λ1(Yt ) positive. Consequently, the
righthand side of (*) is zero, and so λ4(B•) = 0, as claimed.
Proof of Lemma 2.1 Observe that the inequalities b(S) α c(S) and λ4(Seg(S))
α smc(S)λ2(S) are equivalent. Call a set S bad if Seg(S) is measurable and b(S) >
α c(S) or equivalently λ4(Seg(S)) > α smc(S)λ2(S). To prove the lemma, we suppose
for the sake of contradiction that there exists a bad pcomponentwise simply connected
set S ⊆ R2 of finite positive measure.
By Lemma 2.9, for each ε > 0, there is a bounded pcomponentwise simply
connected set S ⊆ S such that λ2(S ) (1 − ε)λ2(S) and b(S ) b(S) − ε. In
particular, such a set S satisfies c(S ) c(S)/(1 − ε). Hence, for ε small enough, the
set S is bad.
Let S be the interior of S . By Lemma 2.10, λ4(Seg(S )) = λ4(Seg(S )). Clearly,
λ2(S ) λ2(S ) and smc(S ) smc(S ), and therefore S is bad as well.
Note that S is pcomponentwise simply connected. Since S is an open set, all
its pcomponents are open as well. In particular, S has at most countably many
pcomponents. Let C be the set of pcomponents of S . Each T ∈ C is a bounded
open simply connected set, and therefore cannot be bad. Therefore,
λ4(Seg(S )) =
λ4(Seg(T ))
α smc(T )λ2(T )
α smc(S )λ2(S ),
T ∈C
T ∈C
Here we prove Lemma 2.4, which says that every bounded open simply connected
subset of R2 can can be split by a diagonal into two rooted sets.
Lemma 2.11 Let S be a bounded open simply connected subset of R2, and let be a
diagonal of S. Let h− and h+ be the open halfplanes defined by the supporting line
of . It follows that the set S has exactly two pcomponents S1 and S2. Moreover,
for every point A ∈ and every neighborhood Nε( A) ⊆ S, we have Nε( A)∩h− ⊆ S1
and Nε( A) ∩ h+ ⊆ S2.
Proof Notice first that any pcomponent of an open set is also open. This implies that
any pathconnected open set is also pconnected, and therefore every open simply
connected set is pconnected as well.
Let A ∈ , and let Nε( A) be a neighborhood of A contained in S. We choose
arbitrary points B ∈ Nε( A) ∩ h− and C ∈ Nε( A) ∩ h+. Suppose for a contradiction
that S has a single pcomponent. Then there exists a polygonal curve Γ in S
with endpoints B and C . Let ⊆ S be the closed polygonal curve Γ ∪ BC . We can
assume that the curve is simple using a local redrawing argument. See Fig. 7.
The curve separates R2 into two regions. The closure of the diagonal is a
closed line segment that intersects in exactly one point. It follows that one endpoint
of is in the interior region of . Since the endpoints of do not belong to S, this
contradicts the assumption that S is simply connected.
Now, we show that the set S has at most two pcomponents. For a point
D ∈ , let Nε(D) be a neighborhood of D in S. The set Nε(D) ∩ h− is contained
in a unique pcomponent S1 of S , and Nε(D) ∩ h+ is contained in a different
pcomponent S2. Choose another point E ∈ with a neighborhood Nε (E ) ⊆ S. We
claim that Nε (E ) ∩ h− also belongs to S1. To see this, note that since D E is a compact
subset of the open set S, it has a neighborhood Nδ(D E ) which is contained in S.
Clearly, Nδ(D E ) ∩ h− is pconnected and therefore belongs to S1, hence Nε (E ) ∩ h−
belongs to S1 as well. An analogous argument can be made for the halfplane h+ and
the pcomponent S2.
Since for every pcomponent S of S , there is a point A ∈ and a neighborhood
Nε( A) ⊆ S such that Nε( A) ∩ S = ∅, we see that S1 and S2 are the only two
pcomponents of S .
Proof of Lemma 2.4 By Lemma 2.11, the set S has of exactly two pcomponents
S1 and S2. It remains to show that S1 ∪ and S2 ∪ are rooted sets.
Since S1 and S2 are pconnected, S1 ∪ and S2 ∪ are pconnected as well. To
show that S1 ∪ and S2 ∪ are simply connected, choose a Jordan curve Γ in, say,
S1 ∪ , and let Z be the interior region of Γ . Suppose for a contradiction that Z is
not a subset of S1 ∪ . Since S is simply connected, we have Z ⊆ S. Hence there is a
point A ∈ Z ∩ S2. Since both S2 and Z are open, we can assume that A does not lie
on the supporting line of . Let A B be the minimal closed segment parallel to such
that B ∈ Γ . Then B belongs to S1, A belongs to S2, and yet A and B are in the same
pcomponent of S . This contradiction shows that S1 ∪ and S2 ∪ are simply
connected.
As subsets of the bounded set S, the sets S1 ∪ and S2 ∪ are bounded. Lemma 2.11
and the fact that Si is open imply that the set Si ∪ is halfopen and Si ∩ ∂ Si = for
i ∈ {1, 2}. Therefore, the sets S1 ∪ and S2 ∪ are rooted, and is their root.
2.4 Proof of Lemma 2.5
Here we prove Lemma 2.5, which explains the tree structure of rooted sets. For this
entire section, let R be a rooted set and (Lk )k 1 be the partition of R into levels. We
will need several auxiliary results in order to prove Lemma 2.5.
For disjoint sets S, T ⊆ R2, we say that the set S is T halfopen if every point
A ∈ S has a neighborhood Nε( A) that satisfies one of the following two conditions:
1. Nε( A) ⊆ S,
2. Nε( A) ∩ ∂ S is a diameter of Nε( A) splitting it into two subsets, one of which
(including the diameter) is Nε( A) ∩ S and the other (excluding the diameter) is
Nε( A) ∩ T .
The only difference with the definition of S being halfopen is that we additionally
specify the “other side” of the neighborhoods Nε( A) for points A ∈ S ∩ ∂ S in the
condition 2. A rooted set R is T halfopen if and only if it is attached to T according
to the definition of attachment from Sect. 2.
Lemma 2.12 The set L1 is (R2
R)halfopen and L1 ∩ ∂ L1 = R ∩ ∂ R.
Proof We consider two cases for a point A ∈ L1. First, suppose A ∈ L1 ∩ ∂ R. It
follows that A has a neighborhood Nε( A) that satisfies the condition 2 of the definition
of a halfopen set. By the definition of L1, the same neighborhood Nε( A) satisfies
the condition 2 for L1 being an (R2 R)halfopen set. In particular, A ∈ ∂ L1. Since
R ∩ ∂ R ⊆ L1 by the definition of L1, we have R ∩ ∂ R ⊆ L1 ∩ ∂ L1.
Now, suppose A ∈ L1 ∩ R◦. Let B be a point of the root of R such that A B ⊆
R. We have A B {B} ⊆ R◦, as otherwise the point t A + (1 − t )B for t :=
sup{t ∈ [0, 1] : At + (1 − t )B ∈ A B ∩ ∂ R} would contradict the fact that R is
halfopen. There is a family of neighborhoods {NεC (C )}C∈AB such that all NεD (D) with
D ∈ A B {B} satisfy the condition 1 and NεB (B) satisfies the condition 2 for R
being halfopen. Since A B is compact, there is a finite set X ⊆ A B such that A B ⊆
C∈X NεC /2(C ). Hence Nε( A B) ⊆ C∈X NεC (C ), where ε := minC∈X εC /2. It
follows that Nε( A B) ∩ ∂ R is an open segment Q containing B but not A and splitting
Nε( A B) into two subsets, one of which (including Q) is Nε( A B) ∩ R and the other
(excluding Q) is Nε( A B) R. Let ε be the minimum of ε and the distance of A
to the line containing Q. It follows that Nε ( A) ⊆ Nε( A B) ∩ R. Therefore, for
every A ∈ Nε ( A), we have A B ⊆ R, hence Nε ( A) ⊆ L1. It also follows that
L1 ∩ ∂ L1 ⊆ R ∩ ∂ R.
We say that a set P ⊆ R is Rconvex when the following holds for any two points
A, B ∈ P: if A B ⊆ R, then A B ⊆ P.
Lemma 2.13 The set L1 is Rconvex.
Proof This follows directly from Lemma 2.3.
A branch of R is a pcomponent of
k 2 Lk .
Lemma 2.14 Every branch of R is Rconvex.
Proof Let P be a branch of R, and let A, B ∈ P be such that A B ⊆ R. Since R is
halfopen, it follows that A B ⊆ R◦. Suppose A B P. It follows that A B ∩ L1 = ∅. Since
L1 is (R2 R)halfopen (Lemma 2.12) and Rconvex (Lemma 2.13), we see that
A B ∩ L1 is an open segment A B for some A , B ∈ A B. It follows that A , B ∈ P.
There is a simple polygonal line in P connecting A with B , which together with
A B forms a Jordan curve Γ in R. Now, let C ∈ A B . Since C ∈ L1, there is a
point D on the root of R such that C D ⊆ R. Since A , B ∈/ L1, D does not lie on
the supporting line of A B . Extend the segment DC beyond C until hitting ∂ R at a
point C . Here we use the fact that R is bounded. Since R is simply connected, the
entire interior region of Γ is contained in R, so the points D and C both lie in the
exterior region of Γ . However, since Γ ∩ L1 = A B , the line segment DC crosses
Γ at exactly one point, which is C . This is a contradiction.
Lemma 2.15 The set L1 and every branch of R are pconnected and simply
connected.
Proof Let P be the set L1 or a branch of R. It follows directly from the definitions
of L1 and a branch of R that P is pconnected. To see that P is simply connected,
let Γ be a Jordan curve in P, A be a point in the interior region of Γ , and BC be an
inclusionmaximal open line segment in the interior region of Γ such that A ∈ BC . It
follows that B, C ∈ Γ and BC ⊆ R, as R is simply connected. Since B, C ∈ P and
P is Rconvex (Lemmas 2.13 and 2.14), we have A ∈ P.
Lemma 2.16 Every branch of R is L1halfopen.
Proof Let P be a branch of R. It is enough to check the condition 1 or 2 for P being
L1halfopen for points in ∂ P ∩ P. Let A ∈ ∂ P ∩ P. Since R is halfopen, A has a
neighborhood Nε( A) that satisfies the condition 1 or 2 for S being halfopen. It cannot
be 2, as then A would lie on the root of R and thus in L1. Hence Nε( A) ⊆ R.
Since L1 is (R2 R)halfopen (Lemma 2.12) and Rconvex (Lemma 2.13) and
A ∈/ L1, the set Nε( A) ∩ L1 lies entirely in some open halfplane h whose boundary
line passes through A. The set Nε( A) h is pconnected and contains A, so it lies
entirely within P. The set Nε( A) ∩ h is disjoint from P. Indeed, if there was a point
B ∈ Nε( A) ∩ h ∩ P, then by the Rconvexity of P (Lemma 2.14), the convex hull
of Nε( A) h and B would lie entirely within P and would contain A in its interior,
which would contradict the assumption that A ∈ ∂ P. It follows that Nε( A) ∩ ∂ P is
an open segment that partitions Nε( A) into two halfdiscs, one of which (including
Nε( A) ∩ ∂ P) is Nε( A) ∩ P.
We show that Nε( A) ∩ ∂ P ⊆ ∂ L1. Suppose to the contrary that there is a point
A ∈ Nε( A)∩∂ P L1. It follows that A has a neighborhood Nε ( A ) ⊆ Nε( A) L1.
Since Nε ( A ) is pconnected and contains a point of P, it lies entirely within P. This
contradicts the assumption that A ∈ ∂ P.
Since Nε( A) ∩ ∂ P ⊆ ∂ L1, there is a point B ∈ Nε( A) ∩ L1. Let A ∈ Nε( A) ∩ ∂ P.
Since L1 is (R2 R)halfopen and Rconvex and A ∈/ L1, there is a point C ∈ A B
such that C B {C } ⊆ L1 while A C is disjoint from L1. The latter implies that
A C ⊆ P, as A ∈ P. Hence C = A . This shows the whole triangle T spanned by
Nε( A) ∩ ∂ P and B excluding the open segment Nε( A) ∩ ∂ P is contained in L1.
Since A lies in the interior of Nε( A) ∩ ( P ∪ T ), it has a neighborhood Nε ( A) that
lies entirely within Nε( A) ∩ ( P ∪ T ). This neighborhood witnesses the condition 2
for P being L1halfopen.
Lemma 2.17 Let P be a branch of R. If A0, A1 ∈ P ∩ ∂ P, then A0 A1 ⊆ R.
Proof Let A0, A1 ∈ P ∩ ∂ P. By Lemma 2.16, P is L1halfopen, hence there are
B0, B1 ∈ L1 such that A0 B0 { A0} ⊆ L1 and A1 B1 { A1} ⊆ L1. There is a polygonal
line Γ1 in P connecting A0 with A1, and a polygonal line Γ2 in L1 connecting B0
with B1. These polygonal lines together with the line segments A0 B0 and A1 B1 form
a closed polygonal curve Γ in R. We can assume without loss of generality that Γ is
simple (see Fig. 7) and that the x coordinates of A0 and A1 are equal to 0. We also
assume that no two vertices of Γ except A0 and A1 have the same x coordinates.
We color the points of Γ ∩ L1 red and the points of Γ ∩ P blue. For convenience,
we assume that A0 and A1 have both colors. Let Z denote the interior region delimited
by Γ including Γ itself. Since R is simply connected, we have Z ⊆ R.
Let x1 < · · · < xn be the x coordinates of all vertices of Γ . We use [n] to denote
the set of indices {1, . . . , n}. Since the x coordinates of A0 and A1 are zero, there is
j ∈ [n] such that x j = 0. For i ∈ [n], we let i be the vertical line {xi } × R. Since the
x coordinates of the vertices of Γ { A0, A1} are distinct, there is at most one vertex
of Γ on i for every i ∈ [n] { j }. For i ∈ [n], the intersection of Z with i is a family
of closed line segments with endpoints from Γ ∩ i . Some of the segments can be
trivial, that is, consisting of a single point, and some segments can contain a point of
Γ in their interior.
For i ∈ [n] and a point A ∈ Γ ∩ i , we say that a point B is a left neighbor of A if B
lies on Γ ∩ i−1 and A B ⊆ Γ . Similarly, B is a right neighbor of A if B ∈ Γ ∩ i+1
and A B ⊆ Γ . Note that every point A ∈ Γ ∩ i has exactly two neighbors and if
A ∈/ { A0, A1}, then the neighbors of A have the same color as A. We distinguish two
types of points of Γ ∩ i . We say that a point A ∈ Γ ∩ i is onesided if it either has
two right or two left neighbors. Otherwise, we say that A is twosided. That is, A is
twosided if it has one left and one right neighbor. See Fig. 8.
l1
Z
Γ
x n
x 3
· · ·
x j = 0
· · ·
Fig. 8 Situation in the proof of Lemma 2.17. Here A is a left neighbor of C and B is a left neighbor of D.
The points B, C, and D are twosided, the points A and E are onesided
Note that every onesided point is a vertex of Γ and that onesided points from
Γ ∩ i are exactly the points of Γ ∩ i that either form a trivial line segment or that
are contained in the interior of some line segment of Z ∩ i . Consequently, every line
segment in Z ∩ i contains at most one point of Γ in its interior.
For 2 i n and C, D ∈ i ∩ Γ , let C D be a line segment in Z ∩ i whose interior
does not contain a point of Γ with a left neighbor. Let A and B be left neighbors of C
and D, respectively, such that there is no left neighbor of C and D between A and B
on i−1. Since no point between A and B on i−1 can have a right neighbor, we have
A B ⊆ Z ∩ i−1 and A, B, C, D are vertices of a trapezoid whose interior is contained
in Z . An analogous statement holds for right neighbors of C and D provided that the
interior of C D does not contain a point of Γ with a right neighbor.
Claim Let i ∈ [n] { j }, and let C and D be points of Γ ∩ i satisfying C D ⊆ Z ∩ i .
Then C and D have the same color.
First, we will prove the claim by induction on i for all i < j . The claim clearly
holds for i = 1, as Z ∩ 1 contains only a single vertex of Γ . Fix i with 1 < i < j
and suppose that the claim holds for i − 1. Let C, D ∈ Γ ∩ i be points satisfying
C D ⊆ Z ∩ i . We show that C and D have the same color. Obviously, we can assume
that the line segment C D is nontrivial. Assume first that the points C and D are
twosided.
Suppose the interior of C D does not contain a point of Γ with a left neighbor. Let
A, B ∈ Γ ∩ i−1 be the left neighbors of C and D, respectively. Then A B ⊆ Z ∩ i−1.
Thus A and B have the same color by the induction hypothesis. Since C, D ∈/ { A0, A1},
the points A and C have the same color as well as the points B and D. This implies
that C and D have the same color too. If there is a point E of Γ in the interior of C D,
then it follows from Rconvexity of P (Lemma 2.14) and L1 (Lemma 2.13) that E
has the same color as C and D.
Now, suppose the interior of C D contains a point E of Γ with a left neighbor.
We have already observed that there is exactly one such point on C D. We also know
that E has two left neighbors. The points C and E with their left neighbors A and B,
respectively, where there is no left neighbor of E between A and B on i−1, form a
trapezoid in Z such that A B ⊆ Z ∩ i−1. From induction hypothesis A and B have
the same color which implies that C and E have the same color as well. Similarly, D
and E have the same color which implies that C and D have the same color as well.
The case where either C or D is onesided is covered by the previous cases. The
same inductive argument but in the reverse direction shows the claim for all i with
j < i n. This completes the proof of the claim.
Now, consider the inclusionmaximal line segment C D of Z ∩ j that contains A0.
We can assume that either C and D are twosided. Suppose for a contradiction that A1
is not contained in C D. If C D is trivial, that is, C = D = A0, then A0 is onesided
and its neighbors A and B have different colors, as Γ changes color in A0. This is
impossible according to the claim, since we have A B ⊆ Z ∩ i−1 or A B ⊆ Z ∩ i+1.
Therefore C D is nontrivial.
First, we assume that A0 is an endpoint of C D, say C = A0. Then A0 is twosided.
By symmetry, we can assume that the left neighbor A of A0 and the left neighbor B
of D have different colors. If there is no point of Γ with a left neighbor in the interior
of A0 D, then A B ⊆ Z ∩ i−1. This is impossible according to the claim. If there is
a point E ∈ Γ with a left neighbor in the interior of A0 D, then we can use a similar
argument either for the line segment A0 E or for E D, as the neighbors of E have the
same color. The last case is when A0 is an interior point of C D. Since Γ does not
change color in C nor in D, we apply the claim to one of the line segments A0C ,
A0 D, and C D and show, again, that none of the cases is possible. Altogether, we have
derived a contradiction.
Therefore, A0 and A1 are contained in the same line segment of Z ∩ i . This
completes the proof, as Z ⊆ R.
Lemma 2.18 For every branch P of R, the set P ∩ ∂ P is an open segment.
Proof Let P be a branch of R. First, we show that the set P ∩ ∂ P is convex. Let
A0, A1 ∈ P ∩ ∂ P. By Lemma 2.17, we have A0 A1 ⊆ R. It follows that A0 A1 is
disjoint from the root of R and thus is contained in R◦. By compactness, A0 A1 has a
neighborhood Nε( A0 A1) contained in R◦. Since P is L1halfopen by Lemma 2.16,
there are B0, B1 ∈ Nε( A0 A1) ∩ L1 such that A0 B0 { A0} ⊆ L1 and A1 B1 { A1} ⊆
L1. For t ∈ [0, 1], let At = (1 − t ) A0 + t A1 and Bt = (1 − t )B0 + t B1. We have
At ∈ A0 A1 and Bt ∈ B0 B1, hence At , Bt ∈ Nε( A0 A1), for all t ∈ [0, 1]. Now, it
follows from the Rconvexity of P (Lemma 2.14) and L1 (Lemma 2.13) that At ∈ P
and At Bt { At } ⊆ L1, hence At ∈ P ∩ ∂ P, for all t ∈ [0, 1]. This shows that P ∩ ∂ P
is convex.
If P ∩ ∂ P had three noncollinear points, then they would span a triangle with
nonempty interior contained in P ∩ ∂ P, which would be a contradiction. Since R is
bounded, the set P ∩ ∂ P is a line segment. That it is an open line segment follows
directly from Lemma 2.16.
Lemma 2.19 For every j 2, every pcomponent P of k j Lk is a rooted set
attached to L j−1. Moreover, for k 1, the kth level of P is equal to L j−1+k ∩ P.
Proof The proof proceeds by induction on j . For the base case, let P be a pcomponent
of i 2 Li , that is, a branch of R. It follows from Lemmas 2.15, 2.16 and 2.18 that P
is a rooted set attached to L1. Let be the root of P, and let (Lk )k 1 be the partition of
P into levels. We prove that Lk ⊆ ik=+11 Li and Lk+1 ∩ P ⊆ ik=1 Li for every k 1.
Let A ∈ Lk . It follows that there is a polygonal line Γ with k line segments
connecting A to a point B ∈ . Moreover, since there is no shorter polygonal line
connecting A to , the last line segment of Γ is not parallel to . Since P is
L1halfopen (Lemma 2.16), there is a neighborhood Nε(B) that is split by into two parts, one
of which is a subset of L1. Let C be a point in Nε(B)∩ L1 such that BC is an extension
of the last line segment of Γ . Since C ∈ L1, there is a point D on the root of R such
that C D ⊆ L1. The polygonal line Γ extended by BC and C D forms a polygonal line
with k +1 line segments connecting A to the root of R. This shows that Lk ⊆ ik=+11 Li .
Now, let A ∈ Lk+1 ∩ P. It follows that there is a polygonal line Γ with k + 1 line
segments connecting A to the root of R. Since L1 is an open subset of R (Lemma 2.12)
and P is a pcomponent of R L1, there is a point B ∈ Γ such that the part of Γ
between A and B (inclusive) is contained in P and is maximal with this property. It
follows that B ∈ . Since B ∈/ L1, the part of Γ between A and B consists of at most
k segments. This shows that Lk+1 ∩ P ⊆ ik=1 Li .
We have thus proved that Lk ⊆ ik=+11 Li and Lk+1 ∩ P ⊆ ik=1 Li for every k 1.
To conclude the proof of the base case, we note that a straightforward induction shows
that Lk = Lk+1 ∩ P for every k 1.
For the induction step, let j 3, and let P be a pcomponent of i j Li . Let Q
be the branch of R containing P. Let (Lk )k 1 be the partition of Q into levels. As we
have proved for the base case, we have Lk = Lk+1 ∩ Q for every k 1. Hence P is
a pcomponent of ( i j Li ) ∩ Q = i j−1 Li . By the induction hypothesis, P is a
rooted set attached to L j−2 ⊆ L j−1. Moreover, for k 1, the kth level of P is equal
to L j−2+k ∩ P = L j−1+k . This completes the induction step and proves the lemma.
Proof of Lemma 2.5 The statement 1 is a direct consequence of the definition of a
rooted set, specifically, of the condition that a rooted set is pconnected.
For the proof of the statement 2, let P be a j body of R. If j = 1, then P = L1 =
Vis(r, R) = Vis(r, L1), where r is the root of R, and by Lemmas 2.12 and 2.15, L1
is rooted with the same root r . Now, suppose j 2. Let Q be the pcomponent of
k j Lk containing P. By Lemma 2.19, Q is a rooted set and L j ∩ Q is the first
level of Q. Since P ⊆ L j ∩ Q, the definition of the first level yields P = L j ∩ Q =
Vis(r, Q) = Vis(r, P), where r is the root of Q. By Lemma 2.12, P is a rooted set
with the same root r .
Lemma 2.12 and the fact that L1 is pconnected directly yield the statement 3.
Finally, for the proof of the statement 4, let P be a j body of R with j 2. Let Q be
the pcomponent of k j Lk containing P. As we have proved above, Q is a rooted
set and P is the first level of Q and shares the root with Q. Moreover, by Lemma 2.19,
Q (and hence P) is attached to L j−1. The definition of attachment implies that P is
attached to a single pcomponent of L j−1, that is, a single ( j − 1)body of R.
3 General Dimension
This section is devoted to the proofs of Theorems 1.5 and 1.6. In both proofs, we use
the operator Aff to denote the affine hull of a set of points.
3.1 Proof of Theorem 1.5
Let T = (B0, B1, . . . , Bd ) be a (d + 1)tuple of distinct affinely independent points
in Rd . We say that a permutation B0, B1, . . . , Bd of T is a regular permutation of T
if the following two conditions hold:
1. the segment B0 B1 is the diameter of T ,
2. for i = 2, . . . , d−1, the point Bi has the maximum distance to Aff({B0, . . . , Bi−1})
among the points Bi , Bi+1, . . . , Bd .
Obviously, T has at least two regular permutations due to the interchangeability of B0
and B1. The regular permutation Bi0 , Bi1 , . . . , Bid with the lexicographically minimal
vector (i0, i1, . . . , id ) is called the canonical permutation of T .
Let T be a (d + 1)tuple of distinct affinely independent points in Rd , and let
B0, B1, . . . , Bd be the canonical permutation of T . For i = 1, . . . , d − 1, we define
Boxi (T ) inductively as follows:
1. Box1(T ) := B0 B1,
2. for i = 2, . . . , d − 1, Boxi (T ) is the box containing all the points
P ∈ Aff({B0, B1, . . . , Bi }) with the following two properties:
– the orthogonal projection of P to Aff({B0, B1, . . . , Bi−1}) lies in Boxi−1(T ),
– the distance of P to Aff({B0, B1, . . . , Bi−1}) does not exceed the distance of
Bi to Aff({B0, B1, . . . , Bi−1}),
3. Boxd (T ) is the box containing all the points P ∈ Rd such that the orthogonal
projection of P to Aff({B0, B1, . . . , Bd−1}) lies in Boxd−1(T ) and
λd (Conv({B0, B1, . . . , Bd−1, P}))
λd (S) c(S).
The definition of Boxd (T ) is independent of Bd , so we can define Boxd (T
by
{Bd })
Boxd (T
{Bd }) := Boxd (T ).
It is not hard to see that this gives us a proper definition of Boxd (T −) for every dtuple
T − of d distinct affinely independent points in Rd .
Lemma 3.1 1. For i = 1, . . . , d − 1, the box Boxi (T ) contains the orthogonal
projection of any point of T to Aff({B0, B1, . . . , Bi−1}).
2. If Conv(T ) ⊆ S then Boxd (T ) contains the point Bd .
Proof We prove the statement 1 by induction on i . First, let i = 1. Then the segment
Box1(T ) must contain every point A j ∈ T since otherwise one of the segments B0 A j
and B1 A j would be longer than the segment B0 B1. Further, if a point A j ∈ T satisfies
the statement 1 for a parameter i ∈ {1, . . . , d − 2} then it also satisfies the statement 1
for the parameter i + 1 since otherwise A j = Bi+1 and A j should have been chosen
for Bi+1.
The statement 2 follows from the fact that Conv(T ) ⊆ S implies λd (Conv(T ))
smc(S) = λd (S) c(S).
For i = 1, . . . , d − 1, let di be the distance of Bi to Aff({B0, B1, . . . , Bi−1}).
In particular, d1 is the diameter of T . The following observation follows from the
definition of the canonical permutation of T and from the construction of the boxes
Boxi (T ).
Proof of Theorem 1.5 To estimate bd (S), we partition Simpd (S) into the following
d + 2 subsets:
X := {T ∈ Simpd (S) : T is affinely dependent},
Yi := {T = ( A0, . . . , Ad ) ∈ Simpd (S) : T is affinely independent, and Ai is the last
element of the canonical permutation of T },
for i = 0, . . . , d.
We point out that T is considered to be affinely dependent in the above definitions of
X and Yi also in the degenerate case when some point of S appears more than once
in T . We have λd(d+1)(X ) = 0. Let i ∈ {0, . . . , d}. The set Yi is a subset of the set
Yi := {T = ( A0, . . . , Ad ) ∈ Sd+1 : T
{ Ai } is affinely independent and we have
Ai ∈ Boxd (T
{ Ai })}.
By Observation 3.2.2, λd (Boxd (T { Ai })) is equal to z := 2d−2d!λd (S) c(S) for
every set T { Ai } appearing in the definition of Yi . Therefore, by Fubini’s Theorem,
the set Yi is λd(d+1)measurable and, moreover,
λd(d+1)(Yi ) = (λd (S))d z = (λd (S))d+12d−2d! c(S).
Thus,
bd (S) =
λd(d+1)(Simpd (S))
λd (S)d+1
λd(d+1)(X ) + id=0 λd(d+1)(Yi ) = 2d−2(d + 1)! c(S).
λd (S)d+1
This completes the proof of Theorem 1.5.
3.2 Proof of Theorem 1.6
In the following, we make no serious effort to optimize the constants. As the first step
towards the proof of Theorem 1.6, we show that if we remove an arbitrary ntuple of
points from the open ddimensional box (
0, 1
)d , then the dindex of convexity of the
resulting set is of order Ω( n1 ).
Lemma 3.3 For every positive integer n and every ntuple N of points from (
0, 1
)d ,
the set S := (
0, 1
)d N satisfies bd (S) 1/2n.
Proof Let S and N = {B1, . . . , Bn} be the sets from the statement and let 0 be the
origin. We use Sd−1 to denote the set of (d − 1)tuples ( A1, . . . , Ad−1) ∈ Sd−1
∗
that satisfy the following: for every B ∈ N the points A1, . . . , Ad−1, B are affinely
0 ) = {B}. Note that the set Sd−1
independent and Aff({ A1, . . . , Ad−1, B}) ∩ (N ∪ { } ∗
is measurable and λd(d−1)(Sd−1) = 1. If h is a hyperplane in Rd that does not contain
∗
the origin, we use h− and h+ to denote the open halfspaces defined by h such that
0 ∈ h−.
Let ( A1, . . . , Ad−1) ∈ Sd−1. For a point B ∈ N , we let h A1,...,Ad−1,B be the
∗
hyperplane determined by the dtuple ( A1, . . . , Ad−1, B). Since ( A1, . . . , Ad−1) ∈
cSodn−t1a,inwtehseeoertihgaint .hTAh1,e.r..e,Afod−re1,tBhesahtiaslfifesspahcAe1s,..h.,−Ad−1,B ∩ N = {B} and that it does not
∗
A1,...,Ad−1,B and h+A1,...,Ad−1,B are well
and P2( A1, . . . , Ad−1) := S ∩ h+A1,...,Ad−1,B1 . Suppose we have
sSp∩lit ht−hAe1,..s.,eAtd−S1,iBn1to sets Pi ( A1, . . . , Ad−1) for 1 i 2(n − 1) and n 2.
Consider the hyperplane h A1,...,Ad−1,Bn . Since for every k ∈ {1, . . . , n − 1} the
intersection h A1,...,Ad−1,Bk ∩ h A1,...,Ad−1,Bn is the affine hull of A1, . . . , Ad−1, we see that
h A1,...,Ad−1,Bn Aff({ A1, . . . , Ad−1}) is contained in two sets Pi ( A1, . . . , Ad−1)
and Pj ( A1, . . . , Ad−1) for some 1 i < j 2(n − 1). We restrict these sets to
their intersection with the halfspace h−A1,...,Ad−1,Bn and set P2n−1( A1, . . . , Ad−1) and
P2n( A1, . . . , Ad−1) as the intersection of h+A1,...,Ad−1,Bn with Pi ( A1, . . . , Ad−1) and
Pj ( A1, . . . , Ad−1), respectively. See Fig. 9 for an illustration.
Since none of the sets Pi ( A1, . . . , Ad−1) contains a point from N , it can be
regarded as an intersection of (
0, 1
)d with open halfspaces. Therefore every set
Pi ( A1, . . . , Ad−1) is an open convex subset of S. Let P( A1, . . . , Ad−1) be the
set S B∈N h A1,...,Ad−1,B . Clearly, λd ( P( A1, . . . , Ad−1)) = 1. Since the
sets Pi ( A1, . . . , Ad−1) form a partitioning of P( A1, . . . , Ad−1), we also have
2n
i=1 λd ( Pi ( A1, . . . , Ad−1)) = 1.
For i = 1, . . . , 2n, we let Ri be the subset of S∗d−1 × S2 defined as
Ri := {( A1, . . . , Ad+1) ∈ S∗d−1 × S2 : Ad , Ad+1 ∈ Pi ( A1, . . . , Ad−1)},
and we let R := i2=n1 Ri . The sets Ri are pairwise disjoint and it is not difficult to
argue that these sets are measurable. If a (d + 1)tuple ( A1, . . . , Ad+1) is contained
in Ri for some i ∈ {1, . . . , 2n}, then ( A1, . . . , Ad+1) is contained in Simpd (S), as
Pi ( A1, . . . , Ad−1) ∪ (Aff({ A1, . . . , Ad−1}) ∩ S) is a convex subset of S. Therefore, to
find a lower bound for bd (S) = λd(d+1)(Simpd (S)), it suffices to find a lower bound
for λd(d+1)(R), because R is a subset of Simpd (S). By Fubini’s Theorem, we obtain
B 1
P2
P2
P4
P2
P4
P1
P1
P3
P1 P5
P3
Fig. 9 The inductive splitting of S = (
0, 1
)2 N with respect to the point A1. The points B1, B2, B3 of
N are denoted as empty circles and we use a shorthand Pi for Pi (A1)
(Ad ,Ad+1)∈S2
= (A1,...,Ad−1)∈S∗d−1 λd ( Pi ( A1, . . . , Ad−1))2,
Ad , Ad+1 ∈ Pi ( A1, . . . , Ad−1)
where [φ] is the characteristic function of a logical expression φ, that is, [φ] equals 1
if the condition φ holds and 0 otherwise. For the measure of R we then derive
2n
λd(d+1)(Ri ) =
(A1,...,Ad−1)∈S∗d−1 i=1
2n
2
λd ( Pi ( A1, . . . , Ad−1) .
Since the function x → x 2 is convex, we can apply Jensen’s inequality and bound the
last term from below by
(A1,...,Ad−1)∈S∗d−1
2n
i2=n1 λd ( Pi ( A1, . . . , Ad−1)) 2
2n
The next step in the proof of Theorem 1.6 is to find a convenient ntuple N of
points from (
0, 1
)d whose removal produces a set with sufficiently small convexity
ratio. We are going to find N using a continuous version of the wellknown Epsilon
Net Theorem [9]. Before stating this result, we need some definitions.
Let X be a subset of Rd and let U be a set system on X . We say that a set
T ⊆ X is shattered by U if every subset of T can be obtained as the intersection of
some U ∈ U with T . The VapnikChervonenkis dimension (or VCdimension) of U ,
denoted by dim(U ), is the maximum n (or ∞ if no such maximum exists) for which
some subset of X of cardinality n is shattered by U .
Let U be a system of measurable subsets of a set X ⊆ Rd with λd (X ) = 1, and let
ε ∈ (
0, 1
) be a real number. A set N ⊆ X is called an εnet for (X, U ) if N ∩ U = ∅
for every U ∈ U with λd (U ) ε.
Theorem 3.4 [13, Thm. 10.2.4] Let X be a subset of Rd with λd (X ) = 1. Then for
every system U of measurable subsets of X with dim(U ) v, v 2, there is a
r1 net for (X, U ) of size at most 2vr log2 r for r sufficiently large with respect to v.
To apply Theorem 3.4, the VCdimension of the set system U has to be finite.
However, it is known that the VCdimension of all convex sets in Rd is infinite (see e.g.
[13, p. 238]). Therefore, instead of considering convex sets directly, we approximate
them by ellipsoids.
A ddimensional ellipsoid in Rd is an image of the closed ddimensional unit ball
under a nonsingular affine map. A convex body in Rd is a compact convex subset of Rd
with nonempty interior. The following result, known as John’s Lemma [10], shows
that every convex body can be approximated by an inscribed ellipsoid.
Lemma 3.5 [13, Thm. 13.4.1] For every ddimensional convex body K ⊆ Rd , there
is a ddimensional ellipsoid E with the center C that satisfies
E ⊆ K ⊆ C + d(E − C ).
In particular, we have λd (K )/dd
λd (E ).
As the last step before the proof of Theorem 1.6, we mention the following fact,
which implies that the VCdimension of the system E of ddimensional ellipsoids in
Rd is at most d +d2 .
Lemma 3.6 [13, Prop. 10.3.2] Let R[x1, . . . , xd ] t denote the set of real polynomials
in d variables of degree at most t , and let
Pd,t = {x ∈ Rd : p(x )
0} : p ∈ R[x1, . . . , xd ] t .
Then dim(Pd,t )
dd+t .
Proof of Theorem 1.6 Suppose we are given ε > 0 which is sufficiently small with
respect to d. We show how to construct a set S ⊆ Rd with λd (S) = 1 satisfying
c(S) ε and
bd (S)
1 ε
8 d+2 dd · log2 1/ε
d
.
Without loss of generality we assume that ε = dd /r for some integer r 2d2d .
Consider the open ddimensional box (
0, 1
)d and the system E (
0, 1
)d of
ddimensional ellipsoids in (
0, 1
)d . Since the restriction of E to (
0, 1
)d does not
increase the VCdimension, Lemma 3.6 implies dim(E (
0, 1
)d ) d +d2 .
If we set n := 2 d+2 r log2 r , then, by Theorem 3.4, there is a r1 net N for the
d
system ((
0, 1
)d , E (
0, 1
)d ) of size n, having r sufficiently large with respect to d.
Let S be the set (
0, 1
)d N . Clearly, we have λd (S) = 1.
Suppose K is a convex subset of (
0, 1
)d with λd (K ) > ε. Since the measure of K is
positive, we can assume that K is a convex body of measure at least ε. By Lemma 3.5,
the convex body K contains a ddimensional ellipsoid E with λd (E ) ε/dd = r1 .
Therefore E ∩ N = ∅. Since we have E ⊆ K and N ∩ S = ∅, we see that K is not a
subset of S. In other words, we have c(S) ε.
By Lemma 3.3, we have bd (S) 21n . According to the choice of n and r , the term
21n is bounded from below by
4 d +d2 r log2 2r = 4 d +d2 dd log2 (2dd /ε)
1
ε
,
where the last inequality follows from the estimate 2dd
proof of Theorem 1.6.
1/ε. This completes the
It is a natural question whether the bound for bd (S) in Theorem 1.6 can be improved
to bd (S) = Ω(c(S)). In the plane, this is related to the famous problem of Danzer and
Rogers (see [5,14] and Problem E14 in [7]) which asks whether for given ε > 0 there
is a set N ⊆ (
0, 1
)2 of size O( 1ε ) with the property that every convex set of area ε
within the unit square contains at least one point from N .
If this problem was to be answered affirmatively, then we could use such a set N
to stab (
0, 1
)2 in our proof of Theorem 1.6 which would yield the desired bound for
b2(S). However it is generally believed that the answer is likely to be nonlinear in 1ε .
3.3 A Set with Large kIndex of Convexity and Small Convexity Ratio
Proposition 3.7 For every integer d 2, the set S := [0, 1]d
and bk (S) = 1 for every positive integer k < d.
Qd satisfies c(S) = 0
Proof Since Qd is countable and λd ([0, 1]d ) = 1, we have λd (S) = 1. Every
convex subset K of [0, 1]d with positive ddimensional measure contains an open
ddimensional ball B with positive diameter, as there is a (d + 1)tuple of affinely
independent points of K . Since Qd is a dense subset of Rd , we see that B ∩ Qd = ∅
and thus c(S) = 0.
It remains to estimate bk (S). By Fubini’s Theorem, we have
bk (S) =
(B1,...,Bk )∈Sk
λd ({ A ∈ S : Conv({B1, . . . , Bk , A}) ⊆ S}) .
λd (S)k+1
If A is a point of S such that Conv({B1, . . . , Bk , A}) is not contained in S, then A
is a point of the affine hull Aff({B1, . . . , Bk , Q}) of B1, . . . , Bk and some Q ∈ Qd .
Therefore, bk (S) is at least
(B1,...,Bk )∈Sk
− λd
λd (S)k+1
Q∈Qd Aff({B1, . . . , Bk , Q})
.
A countable union of affine subspaces of dimension less than d has ddimensional
measure zero and we already know that λd (S) = 1 = λd ([0, 1]d ), hence bk (S) = 1.
4 Other Variants and Open Problems
We have seen in Theorem 1.3 that a pcomponentwise simply connected set S ⊆ R2
whose b(S) is defined satisfies b(S) α c(S), for an absolute constant α 180.
Equivalently, such a set S satisfies smc(S) b(S)λ2(S)/180.
By a result of Blaschke [4] (see also Sas [18]), every convex set K ⊆ R2 contains
√
a triangle of measure at least 34π3 λ2(K ). In view of this, Theorem 1.3 yields the
following consequence.
Corollary 4.1 There is a constant α > 0 such that every pcomponentwise simply
connected set S ⊆ R2 whose b(S) is defined contains a triangle T ⊆ S of measure at
least α b(S)λ2(S).
A similar argument works in higher dimensions as well. For every d 2, there is a
constant β = β(d) such that every convex set K ⊆ Rd contains a simplex of measure
at least βλd (K ) (see e.g. Lassak [12]). Therefore, Theorem 1.5 can be rephrased in
the following equivalent form.
Corollary 4.2 For every d 2, there is a constant α = α(d) > 0 such that every set
S ⊆ Rd whose bd (S) is defined contains a simplex T of measure at least α bd (S)λd (S).
What can we say about sets S ⊆ R2 that are not pcomponentwise simply con
nected? First of all, we can consider a weaker form of simple connectivity: we call a
set S pcomponentwise simply connected if for every triangle T such that ∂ T ⊆ S
we have T ⊆ S. We conjecture that Theorem 1.3 can be extended to pcomponentwise
simply connected sets.
Conjecture 4.3 There is an absolute constant α > 0 such that every pcomponentwise
simply connected set S ⊆ R2 whose b(S) is defined satisfies b(S) α c(S).
What does the value of b(S) say about a planar set S that does not satisfy even a weak
form of simple connectivity? As Proposition 3.7 shows, such a set may not contain
any convex subset of positive measure, even when b(S) is equal to 1. However, we
conjecture that a large b(S) implies the existence of a large convex set whose boundary
belongs to S.
Conjecture 4.4 For every ε > 0, there is a δ > 0 such that if S ⊆ R2 is a set
with b(S) ε, then there is a bounded convex set C ⊆ R2 with λ(C ) δλ(S) and
∂C ⊆ S.
Theorem 1.3 shows that Conjecture 4.4 holds for pcomponentwise simply
connected sets, with δ being a constant multiple of ε. It is possible that even in the general
setting of Conjecture 4.4, δ can be taken as a constant multiple of ε.
Motivated by Corollary 4.1, we propose a stronger version of Conjecture 4.4, where
the convex set C is required to be a triangle.
Conjecture 4.5 For every ε > 0, there is a δ > 0 such that if S ⊆ R2 is a set with b(S) ε, then there is a triangle T ⊆ R2 with λ(T ) δλ(S) and ∂ T ⊆ S.
Note that Conjecture 4.5 holds when restricted to pcomponentwise simply
connected sets, as implied by Corollary 4.1.
We can generalize Conjecture 4.5 to higher dimensions and to higherorder indices
of convexity. To state the general conjecture, we introduce the following notation: for
a set X ⊆ Rd , let Xk be the set of kelement subsets of X , and let the set Skelk ( X )
be defined by
Skelk ( X ) :=
Conv(Y ).
Y ∈(k +X1)
If X is the vertex set of a ddimensional simplex T = Conv( X ), then Skelk ( X ) is
often called the kdimensional skeleton of T . Our general conjecture states, roughly
speaking, that sets with large kindex of convexity should contain the kdimensional
skeleton of a large simplex. Here is the precise statement.
Conjecture 4.6 For every k, d ∈ N such that 1 k d and every ε > 0, there is
a δ > 0 such that if S ⊆ Rd is a set with bk (S) ε, then there is a simplex T with
vertex set X such that λd (T ) δλd (S) and Skelk ( X ) ⊆ S.
Corollary 4.2 asserts that this conjecture holds in the special case of k = d 2,
since Skeld ( X ) = Conv( X ) = T . Corollary 4.1 shows that the conjecture holds for
k = 1 and d = 2 if S is further assumed to be pcomponentwise simply connected. In
all these cases, δ can be taken as a constant multiple of ε, with the constant depending
on k and d.
Finally, we can ask whether there is a way to generalize Theorem 1.3 to higher
dimensions, by replacing simple connectivity with another topological property. Here
is an example of one such possible generalization.
Conjecture 4.7 For every d 2, there is a constant α = α(d) > 0 such that if
S ⊆ Rd is a set with bd−1(S) defined whose every pcomponent is contractible, then
bd−1(S) α c(S).
A modification of the proof of Theorem 1.5 implies that Conjecture 4.7 is true for
starshaped sets S.
Acknowledgements The first three authors were supported by the Grant GACˇ R 1414179S. The first
author acknowledges the support of the Grant Agency of the Charles University, GAUK 690214 and the
Grant SVV2015260223. The last author was supported by the Ministry of Science and Higher Education of
Poland Mobility Plus Grant 911/MOB/2012/0. The authors would like to thank Marek Eliáš for interesting
discussions about the problem and participation in our meetings during the early stages of the research.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution,
and reproduction in any medium, provided you give appropriate credit to the original author(s) and the
source, provide a link to the Creative Commons license, and indicate if changes were made.
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