The "ham sandwich" theorem has been proven only for measures that are absolutely continuous with respect to Lesbeque measure. We prove a generalized version of the ham sandwich theorem which is applicable to arbitrary finite measures, and we give some sufficient conditions for uniqueness of the hyperplane identified by the theorem

Cox, Gary W.

McKelvey, Richard D.

Caltech Authors - Main

DIVISION OF THE HUMANITIES AND SOCIAL SCIENCES 
CALIFORNIA INSTITUTE OF TECHNOLOGY 
PASADENA, CALIFORNIA 91125 
A HAM SANDWICH THEOREM FOR GENERAL 11EASURES 
Gary W. Cox and R.ichard D. ,McKel vey 
SOCIAL SCIENCE WORKING PAPER 337 
July 1980 
ABSTRACT 
The "ham sandwich" theorem has been proven only for 
measures that are absolutely continuous with respect to Lesbeque 
measure. We prove a generalized version of the ham sandwich 
theorem which is applicable to arbitrary finite measures, and we 
give some sufficient conditions for uniqueness of the hyperplane 
identified by the theorem. 
A HAM SANDWICH THEOREM FOR GENERAL MEASURES 
Gary W. Cox and Richard D. McKelvey 
I. INTRODUCTION 
The Ham Sandwich problem was first posed by Ulam [1930], 
and has since been examined by Borsuk [1933], Steinhaus [1945], 
Stone and Tukey [1942], Tucker [1945], and Dubins and Spanier [1961]. 
The problem derives its name from Steinhaus' picturesque formulation 
of the problem as that of dividing a ham, butter, and bread sandwich bv a 
plane into two parts each containing exactly one half of the ham, 
one half of the butter, and one half of the bread. 
The theorem has an n-dimensional generalization which 
uses the following definitions: A hyperpl.ane is any set of the form 
H { x £ nfl x • v > c} 
where v £ Sn-l, and c £ JR • n-1 Here S is the n - 1 sphere of unit length 
vectors in lRn . We use the notation 
and 
2 
to denote the positive and negative open half spaces defined by H. 
If µ is a finite measure on the Borel sets of ]Rn, we say H bisects 
µ iff µ (H
+
) = µ (H-), We have the following statement of the 
theorem: 
Theorem 1 Given any n finite measures, µ1 , .. . ,µn defined on the Borel 
subsets of n-dimensional Euclidian Space, JR
n
, if each µi 
is absolutely 
continuous with respect to Lebesgue measure, there exists a hyperplane 
which simultaneously bisects each measure. 
The usual proof goes as follows: 
( 1 )  Consider the measure µn. W e  know from measure-theoretic 
considerations that, for each unit vector v, there exists 
a real number cv such that the hyperplane Hv = 
{x E JRn I x • v 
bisects µn. 
(2) Now define a mapping f from the unit n - 1 sphere Sn- l
t o  JRn-l as: 
(3) 
f. (v) J l,. . .,n - 1 .  
Note that f i s  continuous and that f (v) = - f  (- v). 
Use the Borsuk-Ulam theorem
1 
to infer that there exists 
a v E Sn- l which f maps into the origin in JR
n- l
, implying 
µj(H:) = µj(H�) for j = 1, . . .  ,n and proving the theorem. 
c } v 
3 
The Ham sandwich theorem and its proof depend on the absolute 
continuity of the measures µ
i
, Otherwise, the function f defined 
above need not be continuous. In fact, the theorem as stated is 
not true for general measures, as is illustrated by the example of 
Figure 1 ,  where µ1 is the atomic measure defined by setting 
µ2(y) = 0 otherwise. 
Here, any bisecting line for µ
1 
must pass thru one and only one xi' 
with the remaining x.'s lying on either side of the line. A bisecting J 
line for µ2 must have similar properties. But no line passing thru 
one xi and one yi splits the remaining points in the desired fashion. 
I\ 
;;_ ' 
I 2 \ /µl 
I \ 
Figure 1 
-+--L 
II. A GENERALIZED VERSION OF THE HAM SANDWICH THEOREM 
4 
Although there is no bisecting hyperplane for the example 
of the previous section, note that the line L is a "median" hyperplane 
for both measures. A median hyperplane for µi is defined as a 
µ. (JRn) 
hyperplane H for which µi(H
+) < 1 2 We 
prove that with this modification of the notion of bisection, the ham 
sandwich theorem is true whether or not the measures are absolutely 
continuous. Specifically, we prove the following theorem: 
Theorem 2 Given any n finite measures µ
1, . .. , µn 
defined on the B orel 
sets of JR
n, there exists a hyperplane H {x E JRnl x • v = c} with 
n-1 v E S , c E JR, such that for all 1 .:':, i .:':, n, 
Proof: For each o > O, and 1 .:':, i .:':, n, we define the derived measure 
dx 
where, 
{x E JR
n
l llx ll < o} 
5 
and m(B 0) is the Lebesgue measure of B 0. 
0 
Now µi is absolutely continuous with respect to Lebesgue 
measure, m, because if m(A) = O, then µ,(A+ x) = 0 for almost l 
0 
every x - µi(A)
3 H0 = {x E1R
n lx 
for i 1, ... , n. 
O. Hence by Theorem 1 for each o, 
} 
n-1 v0 = c0 with v0 E s , c0 E JR such that 
For each o, define g(o) = c0 • v0 and pick r E JR so that 
µi(B r) > 1/2 µi( JR
n
) for i = 1,2, ... , n. Then for 0 < o < r, g 
2 
associates with o a vector g(o) in the compact set Br. Now if {ok} 
is a monotone decreasing infinite sequence converging to zero, then 
g(ok) is an infinite sequence in the compact set Br. Hence, there 
is a subsequence n(k) such that g(o
n(k) ) 
converges to a point in 
B r. For notational convenience, we will assume that the original
sequence is such that g(ok) converges, and we write
where llv* ll 1, and c* < r. We will show that the hyperplane 
H {x E JR
n Ix • v* c*} 
v1<, 
satisfies the conditions of the theorem. Suppose, to the contrary that 
H* does not satisfy the conditions of the theorem. Then for some i, 
either 
µ. (JR
n) 
> -1 __ 
2 
We assume, without loss of generality that 
n Now, we define a sequence of sets, � :__ 1R as follows: 
It is easily verified that 
+ 
H c li1Il inf A_ 
k -,c 
Thus, from Fatou's Lemma, (see e.g., Kingman and Taylor (1966] 
p. 20), we have the following relation 
lim inf 
k 
lim
k 
inf f x� (x)dµi (x) 
� f lim
k 
inf x� (
x) dµi (x) 
So we can pick k* such that for k > k*, 
6 
(*) 
Further, by construction, for all x EB A t\' k 
so for x E B0 , 
k 
7 
(**) 
But now using (*) and (**) together with the definition of µ�, we have 
ok + µ. (H,. ) l U k 
+ 
µ. (H0 + x) 
l k -�-'-'--- dx > m(B0 ) 
k 
But this implies that H0 k 
does not bisect µ?k, a contradiction. 
l 
So the hyperplane H must satisfy the conditions of the theorem, 
Q,E.D. 
III. UNIQUENESS 
In this section we give some sufficient conditions for 
the uniqueness of the bisecting (or median) hyperplane identified by 
the ham sandwich theorem. That the bisecting hyperplane need not in 
general be unique be easily illustrated in Figure 2, using the two 
atomic measures µ1 and µ2 in JR.
2 
defined by setting µ1 (xi
) ; µ2(yi) 1/
3 
8 
for i = l,Z,3. Here we have three distinct bisecting lines, Similar 
examples can clearly be constructed for absolutely continuous measures. 
-
Yz 
Figure Z 
For this section, we let Ai be the support set for the 
measure µi' i.e. , x s Ai 
*l> µi(Bs(x)) f 
0 for every opens ball 
B (x) around x, and we let co(A.) denote the convex hull of Ai, The s 1 
9 
measure µi has a unique median in the direction v s Sn
-l iff there 
is a unique c s JR for which H = {xix • v = c} is a median hyperplane 
for µ
i
. We deal here only with measures which have a unique median 
in every direction. This would include any absolutely continuous 
measure. It would also include any measure µi whose support set 
cannot be partitioned into two seperable sets (see below for 
definition of separability) each containing half the measure. For 
example, a completely atomic measure with f of the measure at each of
k points has a unique median in every direction, as long as k 
is odd. 
Lemma 1 Let µi have a unique median in every direction, and let 
H1 and Hz be two distinct median hyperplanes for µi then
Proof: �· Then, since H1 n Hz is 
convex, and co(Ai) is convex, there is a separating hyperplane, say 
Write H
1 = {x s JR
n I x • v1 = c1}, and Hz = {x s JR
nl x • vz = cz}, where 
n-1 v1, vz s S , and c1, cz s JR. Also pick v1 and vz so that v1, vz 
and v0 positively span O. This is possible since v1, v2 and v0 are 
10 
linearly dependent, any pair is linearly independent, and v
1 
and vz 
can either one or both be replaced by -v1, or -vz (with an appropriate 
change to c1 or cz), So we can write 
with ao, al
, az all positive. 
b H n H c H+ =l> x ecause x E 1 z _ 0 
+ Now Hl n Hz.::_ HO =l> a0c0 + a1c1 + azcz < 
0, 
' VO > CO =l> X ' a0
v0 > a0c0 
Now, we pick E > O, and define 
For all E, we have H+ c H+ so E - l' 
and for small enough E, we have 
To see this, set E < l__ then x E a ' 1 
H: n HO � x . 
x . VO < co· Since x E H�, we need only show x E 
vl 
> cl + E 
+ Hz. But 
(*) 
(**) 
and 
> 
x • ( - v -0 
(aoco + al cl) 
az 
azcz + 
0 0 
az a2 
11 
al E 
az 
cz 
+ so x E H2 as required. Since Ai 
:=_ H0, it follows from (*) and (**) that 
and 
Thus 
and 
H+ n A c H+ n A E i 1 i 
µi(H;) � µi(H1) µi (JR
n
) 
< ---2 
µi(H�) 
+ _:::_ µi(H2) < 
\1. (JRn) 1 
2 
so HE is a median hyperplane. But then µi 
does not have a unique 
median in the direction v
1, a contradiction. 
Two sets, B1, 
hyperplane H in ]Rn with 
Q.E.D. 
Bz :=_ ]R
n are separable iff there is a 
+ B1 :=_ H and B2 :=_ H . An n-2 dimensional 
hyperplane is any set L of the form 
12 
c
1 
and x • v2 
n-1 where v
1
, v2 E S , �1,c2 E JR, and v1, v2 are linearly independent. 
We now have the following definition 
Definition A collection of sets {Bi }�=l with Bi 5:_ 1Rn is
(S) Separable iff V i, B
i 
and B - Bi 
are separable, where 
n 
B = U Bi. 
i=l 
(N) NonDegenerate iff there is no n-2 dimensional hyperplane, L, 
with L n co(Bi) f ¢ for all i. 
The following relation holds for any sets B
1, ... ,Bn 
5:_ 1Rn 
Lemma 2 For all n > 2, N � S, for n < 3, N '** S. 
Proof: We prove the first assertion first. Define B 
n 
U Bi i=l 
and Di= co(B - Bi) n co(B1). Now Di =¢ for all i, because if not, 
then for any x E Di' x can be written as a convex combination of 
n - 1 points x. E co (B. ), j f i. But these n - 1 points determine a J J 
n - 2 dimensional hyperplane which contains all x. as well as J 
x E co(Bi)' and this contradicts nondegenercy. Hence nondegenercy 
implies Di = ¢ for all i. But then co(B - Bi) and co(Bi
) are disjoint 
convex sets, and invoking the separating hyperplane theorem, it follows 
that separability is satisfied. Hence N � S, as we wished to show. 
To prove the second part, we need only show for n 2, 3, 
that S � N. This is proven by Steinhaus (1945) whose proof is 
13 
sketched here. For n = 2, the proof is trivial. For n = 3, suppose 
the line L passes through the points x
1
, x2, x3, belonging to 
co(Bi)' co(B2), co(B3) respectively. Since B1
, B
2, B3 are separable, 
their convex hulls have no common points, so x
1, x2
, and x3 are distinct. 
We can suppose that x2 is between x1 and x3. But the plane H which 
separates B
2 
from B
1 
U B3 puts x2 on one side, and x1 and x3 on the 
other, contradicting the order of the points on the line L. 
Q.E.D. 
Theorem 3 Let µ1
, . .. ,µn be finite measures on the Borel sets of 
1R
n
, with each µi having unique medians in all directions. 
Then if the 
support sets, A
1
, • . .  ,A
n 
for the measures ]Jl, .. . , µn satisfy nondegeneracy, 
the Ham Sandwich hyperplane is unique. 
Proof: Let H
1 
and H
2 
be two distinct hyperplanes satisfying the 
conditions of Theorem 2. Then H
1 
n H2 is a n - 2 dimensional hyperplane, 
and by Lemma 1, we have, for all i, H1 n H2 n co(Ai) f ¢. But then 
the sets A
1
, . . .  ,An are degenerate, a contradiction. 
Corollary 
Q. E. D. 
If n � 3, and µ1, . . •  ,µn are as in Theorem 3, then if the 
support sets A
1
, . .. ,An 
for the measures µ1
, . .. ,µ
n 
are separable, the 
Ham Sandwich hyperplane is unique. 
14 
Proof: This follows directly from Theorem 3, along with the observation 
(proven in Lellllila 2) that for n < 3, separability and nondegeneracy 
are equivalent, 
Q.E.D. 
To show that separability is not enough to guarantee uniqueness 
of the ham sandwich hyperplane for n > 4, the following example is 
given. Here, note S =f N. We set 
al 
(0, o, 1, 0) 
a2 (0, O, -1, 0)
a3 (0, 0, O, 1) 
a4 (0, 0, 0, -1) 
and for 1 < i ..'.: 4, we define µi to be the measure for which 
Clearly Ai 
otherwise. 
1, ... ,4. Now define 
1/2} for j 1,2,3,4 
The hyperplane Hj separates Aj from A - Aj for all j, where A 
But clearly A is contained in the plane (a n - 2 hyperplane of 
L = {x s JR4jsl x = 0, €2 x = o}, where €1 and r2 are the 
first two basis vectors. So N is violated, Also, there are 
15 
4 
U A
i. i=l 
JR4 )
4 infinitely many three spaces (hyperplanes in JR ) which contain 
L, each of which is a bisecting (median) hyperplane, so uniqueness 
of the ham sandwich plane is violated. This example can be 
extended straightforwardly for n > 4. 
16 
FOOTNOTES 
1. The Borsuk-Ulam Theorem is as follows: If f is a continuous 
mapping of the n- sphere S
n 
into 1Rn such that diametrically 
opposed points of S
n 
map into points symmetric about the origin 
in lln (i.e,, f(x) f (-x) ) ,  then there exists a point of S
n 
which maps into the origin of lRn, 
17 
REFERENCES 
Borsuk, K., "Drei Satze liben Die n-Dimensionale Euklidische Sphare," 
Fund. Math, 20 (1933) 1 7 7-190. 
Dubins, L. E., and E. H. Spanier, "How to Cut a Cake Fairly," 
American Mathematical Monthly, 68 (1961) !157-459. 
Kingman, J, F. C., and S. J. Taylor, Introduction to Probability and 
Measure, (Cambridge: Cambridge University Press, 1966). 
Steinhaus, H., "Sur la Division der Ensembles de l'Espace par les 
Plans et des Ensembles Plans par les Cercles," Fund. 
Math., 33 (1945), 245-263. 
Stone, A. H., and J. W. Tukey, "Generalized Sandwich Theorems," 
Duke }!athematics Journal, 9 (1942) 356-359. 
Tucker, A. W. , "Some Topological Properties of Disk and Sphere," 
Proceedings of the Canadian Mathematical Congress (1945) 
285-309. 
Ulam, S., "Zur Masstheorie un der allgemeinein Megenlehere," Fund. 
Math., 16 (1930) 140-150. 


A Ham Sandwich Theorem for General Measures

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A Ham Sandwich Theorem for General Measures

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