proceedings of theamerican mathematical societyVolume 108, Number 1, January 1990
MEASURES OF GRAPHS ON THE REALS
SETH M. MALITZ
(Communicated by R. Daniel Mauldin)
Abstract. This paper studies measure properties of graphs with infinitely many
vertices. Let [0, 1] denote the real unit interval, and y be the collection of
bijections taking [0, 1] onto itself. Given a graph G = ([0, \\,E) and / € & ,
define the f-representation of G to be the set Ef = {{f(x),f(y)):x,y e [0,1]
and (x,y) e E} . Let p be 2-dimensional Lebesgue measure. Define the
measure spectrum of G to be the set M(G) = {m 6 [0,1]:3/ e & with Ef
measurable and pEf = m) . Our main result characterizes those subsets of
reals that are the measure spectra of graphs.
1. Introduction
Consider an infinite graph G = ([0,l],E) where [0,1] denotes the real
unit interval. To each bijection of [0,1] onto itself (i.e. each relabeling of the
vertices of G ) there corresponds a subset of the unit square which is essentially
the adjacency matrix of G under the specified labeling. This paper establishes
some interesting relationships between the structure of a graph G and measure
properties of its corresponding family of adjacency matrices.
Our notation is as follows. We consider undirected graphs G = ([0, l],E)
without loops or multiple edges. Let G denote the complement graph of G.
Let c denote the power of the continuum, N the natural numbers, and N+ the
positive natural numbers. For 5 G N+ U {co} , define K (c) to be any complete
5-partite graph with color classes of cardinality c, and Kc to be any complete
graph on c vertices. Let & be the set of bijections of [0,1] onto itself. Given
a graph G = ([0, l],E) and an /e/, define the f-representation of G to
be the set Ef = {(f(x),f(y)):x,y G [0,1] and x,y G E}. Take pt to be
/-dimensional Lebesgue measure. Define the measure spectrum of a graph G to
be the set M(G) = {m6[0,l]:3/e^" with E. measurable and ß2^f ~ m} •
Observe that M(G) = 1 - M(G) = {1 - m: m G M(G)}.
This paper provides solutions to the following three problems:
(1) Describe the subsets of [0,1] that are the measure spectra of graphs
and relate M(G) to the structure of G.
Received by the editors September 23, 1986 and, in revised forms, December 3, 1987 and
February 3, 1988.1980 Mathematics Subject Classification (1985 Revision). Primary 28A05, 28A20, 05C99.
©1990 American Mathematical Society
0002-9939/90 $1.00+ $.25 per page
77
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78 SETH m. malitz
(2) Give necessary and sufficient conditions on a graph G that insure E,
is measurable for some f G3r.
(3) Give necessary and sufficient conditions on a graph G that insure Ef
is measurable for every f € SF.
Most of the paper is devoted to question (1) because (2) is essentially an-
swered in Blass [Bl], and (3) is relatively simple. As (1) is our main concern,
let us get a feeling for what some typical measure spectra look like.
Example. If G is a K2(c), then M(G) = [0, \]. To see this, let A c [0,1] be
the set of vertices in color class 1, and Ac the set of vertices in color class 2.
M(G) 2 [0,j]: For every m G [0,±], there is a tm G [0,^] such that if
fm G !? maps A onto [0,tm] and Ac onto (tm,l], then ß2E, = m. To
obtain a measure 0 representation, let f^G^F map A into the Cantor set.
M(G) ç [0,j]: Suppose f G £? and £\- is measurable. Let B = f[A].
By Fubini's Theorem, B is a measurable subset of [0,1]. Thus fi2Ef = 1 -
(ßxB)2 - (I - pxB)2 < \ .
In a similar way, one can prove that M(Ks(c)) = [0,1- l/s] for every
5 G N+ , and that M(Kw(c)) = [0, 1). It follows that if G is s-colorable, then
G has no representation of measure > 1 - 1 /5, and if G is w-colorable, then
G has no representation of measure 1. The disjoint union of a Kc and a Kc
is a graph with measure spectrum [0,1].
2. Graphs with peculiar measure spectra
Proposition 2.1. There is a graph with empty measure spectrum.
Proof. The following construction is due to Sierpinski [Si]. Pick any bijection
g:[0,1] —y c. Let E consist of those pairs (x,y) G [0,1] such that g pre-
serves the order of x and y. The reader can verify that neither G nor G
contains a Kc. Thus, looking ahead to Theorem 5.1, G has no measurable
representation. D
For « g N+ and / ,j G {0, ...,«- 1}, define S" to be the open square
(//«,(/+ 1)/«) x (j/n,(j +l)/n).
Proposition 2.2. For every n G N+ , there is a graph G" with measure spectrum
[0,1-1/«]U{1}.
Proof. Take the edge set E defined in Proposition 2.1 and delete those pairs
in the upper right and lower left quarters of [0,1 ] . Call this new set E . For
each « € N+ define
e" = e°ös2xxu y s2;0<i.j<n-l
mand G" = ([0, l],En). Fix / G 9~ and set A = f[[0, {-]] . Denote the repre-
sentation (E")f by Enf. We claim that if Enf is measurable, then ßxA = 0 or
1. For contradiction, suppose not. There are two cases to consider.
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measures of graphs on the reals 79
(i) A is measurable and 0 < ßxA < 1 . Observe that for any x g [0, j],
we have \{y G [j,l]:(x,y) & E°}\ < c, and for any y G [j,l], we have
\{xG[0,2-]:(x,y)GE°}\ <c. Consequently, neither 2?°n([0,f]x [\, 1]) nor
([0,j]x[j,l])-E contains the Cartesian product of two sets P, Q c [0,1]
where \P\ — \Q\ = c. Hence neither Efn(Ax Ac) nor (A x Ac) - E f contains
the product of such sets.
If Er is measurable, then EfC\(Ax Ac) is measurable. Since 0 < pxA < 1,
either E, n (A x Ac) or (A x Ac) - E, has measure greater than zero, say
ErC\(AxAc) without loss of generality. By Theorem 3.3, Efr\(AxAc) contains
the Cartesian product of two nonempty perfect sets. But this is a contradiction
as any nonempty perfect set has cardinality c. Thus Enf is not measurable.
(ii) A is nonmeasurable. Let Hx D A be a Gs set satisfying nxHx =
ß°nieTA, and H2 d Ac be a Ga set satisfying pxH2 = /u°xuteTAc. Let H3 =
HxnH2. Clearly, ßxHi > 0 since A is nonmeasurable. For 0 < k <
« - 1 , define Bk — f[(k/2n , (k + l)/2«)]. Again, since A is nonmeasurable,
¡u2uleT(Bk n//3) > 0 for some k G {0,1, ... ,n - 1} , say k = 0 without loss of
generality. Let H4 D B0 n H3 be a G& set satisfying p2H4 = p2nteT(BQ n H3).
Notice p°2l"[Enf n (H4 x H4)] = (ßxHA)2 because nxateT(H4 n ^c) = /¿,//4 and
^f x Ac c £" . However, n2ner[E"f n (7/4 x //4)] = 0 because p°xuler(H4 n fi0) =
//,//4 and B0x BQ c [E"]c. Thus £" is nonmeasurable. D
3. Laying the groundwork
Section 4 characterizes the subsets of [0,1] that are the measure spectra
of graphs. This characterization is obtained via the following ideas. Suppose
G has a representation 3? of measure m G (I - l/r, 1 - l/(r + I)) where
r g N+. We want to argue that G has a representation of measure m for
every m' e [0,1 - l/(r + 1)]. Let us first see that G has a representation
of measure m for every m G [m, 1 - l/(r + 1)). Fix an arbitrarily small
ô G (0, m - ( 1 -1 //•)). By Theorem 3.1, it is possible to chop the unit square into
a matrix of little squares so small that if we shade only those squares in which 'S
has density greater than 1 - ô , the set of shaded squares has measure within ô
of m . By Turán's Theorem (see [Bo]) from extremal graph theory, there exists
an (r + 1) x (r + 1) submatrix of little squares such that (1) the submatrix is
symmetric about the line y = x , and (2) all squares of the submatrix not lying
on the line y = x are shaded. At this point, it is easy to show that G has a
representation of measure m for every m G [m,(I -â)(l - l/(r+ 1))]. Since
ô can be chosen arbitrarily small, G has a representation of measure m for
every m G [m , 1 - l/(r + 1)). If we assume now that G has no representation
with measure greater than 1 - l/(r + 1), then by Theorem 3.3, which asserts
the existence of c-sized Cartesian squares in measurable sets satisfying certain
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80 SETH M. MALITZ
properties, it can be shown that G has a representation of measure m for all
m G[0,l- l/(r+l)].
We now expand on these ideas formally. Given « e N+ , i,j G {0, ... ,n -
1}, and a measurable subset F Ç [0,1] , define F"j — F n S"¡ where 5". is
the open square (/'/«,(j + 1)/«) x (j/n, (j + 1)/«). For e G [0,1], say F is
(.-dense in S", if /i2F"/Ju25'" > e . Let F"'£ be the collection of squares S"¡ in
which F is e-dense.
Theorem 3.1. Let F be a measurable subset of [0,1] , and £€(0,1). Then
lim p-,Fn'e = PjF.n—>oo z z
Proof. Let O be an arbitrary open set containing F . Since O is the countable
union of open squares, it is easy to see
Now for all «
which implies
Thus
lim p70"' = P-.0.n—y<x> ¿ l
rJi,\ . „/!,£ . „n, 1 , 1 V-l(0 - O ' ]p20 < p20 < ß20 + — • —-
lim ß-,0"'£ = ß-,0.
lim sup/ijF"^ < lim sup^20"'E = p20./-►oo „^/ /-»oo „^/n>l i^°° n>l
But since O is arbitrary,
lim sup/¿2.F ' < n2F.l-yoo „>/
In the other direction
lim infp2Fn'e = 1 - lim sup//2(Ff)"' £ > 1 - n2Fc = p2F. al—yoo n>l I—»oo „>/
Lemma 3.2. F/x « G N+ . r«ere /5 an e e (0,1) such that for every measurable
subset D ç [0,1]2, if D is en-dense in all the off-diagonal squares 5" , there
exist points v¡ g (i/n, (i + 1 )/«), /' = 0, ... , « - 1, satisfying (v¡, d .) e £> yôr
all 0< i,j <n- I with i ¿ j.
Proof. Take en = 1 - 1/2« and choose the points i>( in the appropriate inter-
vals (i/n,(i + l)/n) randomly, independently, with uniform distribution. For
each fixed i ^ j, the probability that (v¡, v.) fails to be in D is at most 1/2«2.
So the probability that such a failure occurs for at least one of the «(« -1 ) pairs
is less than 1/2 . Thus some choice of v¡ 's satisfies the conclusion. D
Given a measurable subset F ç [0,1]', let <¡>(F) denote those points of
[0,1]' at which F has density 1 with respect to Ju/.
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MEASURES OF GRAPHS ON THE REALS 81
For « g N, let {0,1}" denote the set of binary strings of length « . If a
and p are two binary strings, let op denote the concatination of o and p .
Mycielski [My] showed that if E isa subset of the unit square of measure 1,
then E contains P x P for some nonempty perfect set P c [0,1 ]. Theorem
3.3, stated below, is an improvement of this result. Although it is stated only
for dimension 2, the proof can be easily adapted to give an analogous result in
all higher dimensions.
Theorem 3.3. Let E be a measurable subset of [0,1 ]2. If px {x G [0,1 ]: (x, x) G
4>(E)} > 0, then there is a nonempty perfect set P c [0,1] such that P x P -
{{x,x):x G P} c E.
Proof. First, we show that there is a closed subset F ç E such that
px{x G [0,1]: (x,x) G <f>{F)} > 0. For each /' g N+ , define //, to be the
set of points in the plane whose distance from the line y = x is greater than
1//. Let Ti = [0, l]2r\Ecn(Hj+x -H.). Let Oi be an open set in the plane that
lies entirely within Hj+2 - HjX , contains Ti, and satisfies p-,Oj < pt2Ti + 2"'.
Now define F to be [0,1] - lj°^, Oi. Obviously F is a closed subset of E.
Furthermore, any density point (x ,x) of E is also a density point of F . To
see this, consider an ax a square Sfa rotated 45° and centered at the point
(x,x) G <t>(E). For a g (0,1), take i to be the largest integer i such that
1//' > a/2. Assuming Sf is contained in [0,1] , we have
p2(fc n fj < p.2(ec n5fa) + ¿2 2 J = ^ n^) + 2
Since
J='„
obviously
iimA2(F/n^()=Q and lim2^=0)
a-0 a2
Hence (x ,x) is a density point of F. Thus F is a closed subset of E and
Px{xg[0,1]:{x,x)g4>(F)}>0.Our task now is to prove the existence of a nonempty perfect set P c [0, 1 ]
suchthat PxP-{(x,x):xgP} c F. Let Kx = {x G [0,l]:(x,x) G(f>(F)} and
K2 = {x G[0,l]:x G <f>(Kx)}. For j G N+ , let £ be defined as in Lemma 3.2.
We want to construct a sequence of nested closed sets J0d Jx d J2d ■ ■■ whose
intersection is the desired perfect set P. The sequence is defined inductively
as follows.
For the base step, pick a point z g K2. Let
/ = [z - a0, z + a0] ;
7o = (z_ao'z);
l'x = (z,z + a0),
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82 SETH M. MALITZ
and
where a0 is chosen so small that
• fi^l'u n K2)/a0 is sufficiently close to 1 for each o G {0,1} that the
next inductive step can be performed;
• /i2[(l'a x l'p) n F]/a20 > £2, forall o,pG{0,l}x wither^/?.
Now let us proceed to the second step of the induction. By Lemma 3.2, there
is a point za G l'ar\K2 for each o G {0, l}1 such that (z0,zx), (z, ,zQ) G <¡>(F).
Forall <7 €{0,1}', let
7áo = (Za-a.'Zrr);
7ál =(Zo>Zo+a0>
and
Jr= U 7.'<re{0,I}'
where a, is chosen so small that
• Igcl'g forall rje{0,l}';
• ßyil'gD K2)/ax is sufficiently close to 1 for each a G {0, l}2 that the
next inductive step can be performed;
• n2[(l'a x l'p) n F]/a2x > e2, for all a, p G {0,1 }2 with er ̂ />.
In the third step of the induction, we again use Lemma 3.2 to find points
za el'af\K2 for each o G {0, l}2, so that (za,zp) G (p(F) forall a ,pG {0,1}2
with er / p , and continue.
Repeating this procedure indefinitely, we obtain a nested sequence of closed
sets JQ D Jx D J2 D ■ ■ ■ . Let P be the intersection of these J¡. Then P is
perfect and P x P - {{x,x): x G P} c F since F is closed. D
Remark. The conclusion of Theorem 3.3 is not true if we assume only that
{xe[0,l]:(x,x)e <i>(E)} has power the continuum.
Lemma 3.4. Suppose G = ([0, 1 ], E) has a representation 3? of measure m G
(1 - l/r,l - l/(r+ 1)) where r G N+. 77ze« M(G) D[m,l- l/(r+ I)).
Proof. Fix an arbitrarily small positive S < m-(l-l/r). By Theorem 3.1, there
is an integer «, such that \p1S'n'x~d - m\ < S holds for all « > «, . Notice
that 2?"' ~ corresponds to the adjacency matrix of an «-vertex undirected
graph Jn by viewing each shaded square as a 1, and each unshaded square as
a 0. As « grows, the fraction of entries that are 1 in the adjacency matrix of
Jn approaches m < 1 - (1/r). Hence, by Turán's Theorem (see [Bo]) from
extremal graph theory, there exists an N > «, such that for all « > ¿V the
graph Jn has a complete subgraph on r+1 vertices. In other words, there are
integers 0 < w0 < wx < ■■ ■ < w <: N — 1 such that f is (1 - ¿>)-dense in the
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MEASURES OF GRAPHS ON THE REALS 83
square (w¡/N ,(w¡ + l)/N)x(wj/N ,(w. + l)/N) forall i ,j G {0, ■ ■ ■ ,r) with
At this point, it is clear that for every m G [m,(I - ô)(l - l/(r+ 1))], G
has a representation of measure m . Since ô can be chosen arbitrarily small,
G has a representation of measure m for every m'e[«i,l-l/(r-i-l)). D
Let H" be the set of all off-diagonal squares S", where i ^ j. Let T be the
Cantor Set. Define an equivalence relation ~ on the power set of [0,1 ] as
follows. Say F ~ H if the symmetric set difference of F and // is contained
in (r x [0,1]) u ([0,1] x T). Let (F) be the equivalence class of F under ~.
For « g N+, D a subset of [0,1]2, and Sf a square region in [0,1]2
parallel with the coordinate axes, define °[SfC\D] to be the pure translation of
S* n D that takes the lower left corner of Sf to the origin.
Lemma 3.5. Suppose G — ([0, l],E) has a representation 2? of measure m G
(1 - 1/r, 1 - l/(r + 1)] where r G N+ . Suppose that for every v > 0 there is
an mv g [1 - l/(r+ 1), 1 - l/(r+ I) + v), such that G has no representation of
measure mv. Then G has a representation 2î' G (Hr+ ) and hence M(G) D
[0,l-l/(r+l)].
Proof. Proceed through the first paragraph in the proof of Lemma 3.4 with
0<o<m-(l-l/r) and I - ô > e x where £r+1 is defined as in Lemma
3.2.Construct a new subset 2?* c 2* as follows. For each a,b G [0,1], define
Ka b = {x g [0,1]: (x,x - a + b) g <t>(2?)}. Let 2?* = {(a,b):a € <KKaJb)} C\2?.
By Fubini's Theorem and the Lebesgue Density Theorem, 2?* is measurable
and p2(2? -2?*) = 0.
Applying Lemma 3.2 with n - r+l, choose points v. G (wJN ,(wj + l)/N)
where / = 0, ... ,r, so that (vt,v.) g 2?* for all 0 < i,j < r with /' ^ j.
Pick d sufficiently small that the intervals /, = [vj-d,vt + d] are all pairwise
disjoint. Let
a= fi °[(iVlxiVj)nsr'].0<i,j<r
i¥j
Since all pairs (v-,v.) coincide in A, clearly px{x G [0,2d]:(x,x)
G 4>(A)} > 0. Now it is not difficult to see that if (x,x) G <¡>{A), then
°[(7„ x 7„ ) n 2f\ must have density 0 at (x,x) for every /' g {0, ... ,r).
Otherwise, there is a v > 0 such that G has a representation of measure m
for every «i e [ 1 - 1 / ( /" +1 ), 1 - 1 / ( r + 1 ) + z/ ), which contradicts our hypothesis.
So (x,x)G(p(A) implies °[(/, x/(l)nf] has density 1 at (x,x) for every
i G{0, ... ,r}. Letr
B = Anf)°[(ivxiv)n2?c].1=0
Since px{x G [0,2d]:(x,x) G 4>(B)} = px{x G [0,2d]:(x,x) G <f>(A)} > 0,
Theorem 3.3 tells us there is a nonempty perfect set P c [0,2öf] such that
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84 SETH M. MALITZ
PxP-{(x ,x):x g P} c B . For each /' e {0, ... ,r} , define P¡ = {x+v-d:x g
P}. Observe that P¡ xPjc2? for all i^ j, and Pi x P¡ c 2?c for all i. Let
f gSF satisfy
/ [0,1]-^,i'=0
and
ItÍt.ttÍJ.r +l'r+1
Then F, e (//r+1), and hence Af(G)2[0,l- l/(r+l)]. □
4. The measure spectra of graphs
We are ready to prove our main theorem.
Theorem 4.1. For any graph G = ([0, l],E), M(G) or l-M(G) is one of the
following subsets of reals:
(i) </>;
(ii) [0,1);
(iii) [0,1];(iv) [0,1 -}] , J£N+;
(v) ¡0,1-I]U{1}, SGN+.
Conversely, each of the above subsets of reals is the measure spectrum of some
graph. If M(G) is of type (iv) or (v), then G has a representation 2? g (Hs) .
Proof. Suppose that M(G) is not one of the following subsets of reals: ay,
[0,1],(0,1],[0,1),{0},{1} or {0,1}. Looking ahead to Theorem 5.1, M(G)
^ (0,1). Hence there are m0, mx G (0,1) where m0 G M(G) and mx &
M(G).Let a = infl(0,1) - M(G)] and ß = inf[M(G) n (0,1)]. Notice a and ß
are well-defined. We have two cases to consider:
(i) a > 0. Since M(G) D (0,a) ¿ </>, Lemma 3.4 says a = \ - l/s for
some s>2. By Lemma 3.5, M(G) D [0,1 - l/s] D [0,£]. We claim that G
has no representation of measure m2 where ^ < a < m2 < 1. Suppose for
contradiction that it did. Applying Lemma 3.4 to G gives M(G) D[\-m2,j)
which implies M(G) D (j,m2]. Since we already know M(G) 3 [0, \], the
last statement implies M(G) D [0,«t2]. Hence a > m2, a contradiction.
(i_i) a = 0. If a = 0, then ß > 0 by Lemma 3.4. Since 1 -ß = sup[(0, l)n
M(G)], Lemma 3.4 says 1 - ß = 1 - l/s for some 5 > 2. So (1 — 1/5,1) c
(0,1)-M(G). By Lemma 3.5, M(G) 2 [0,1 -l/s].At the end of the introduction, we gave examples of graphs with measure
spectra of type (ii), (iii) and (iv). Proposition 2.1 yielded a graph with empty
measure spectrum, and Proposition 2.2 produced a graph with measure spec-
trum [0,1 - 1/5] U{1} for each 5 e N+ .
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MEASURES OF GRAPHS ON THE REALS 85
Finally, Lemma 3.5 implies that if M(G) is of type (iv) or (v), then G has
a representation 2? G (Hs). D
Remark. If M(G) is of type (iv) or (v), then clearly s must be the largest
integer « for which G contains Kn(c) as induced subgraph.
Remark. If we define &' = {/ G SF:f measurable} and M'(G) = {m G
[0,1]: 3f G SF* such that Ef is measurable and p2Ef = m), then Theorem
4.1 still holds for M'(G). In general however, M'(G) ^ M(G) even if F is a
measurable subset of [0,1 ] .
Proposition 4.2. If G = {[0,1],E) is a graph for which M(G) = [0,1 ), then Ghas a representation 2?n G (Hn) for every « g N+ .
Proof. It is not too difficult to construct an argument using ideas in the proof of
Lemma 3.5, the proof of Theorem 5.1, and the fact that G contains no Kc. G
Remark. If M(G) — [0,1), then G must contain a Ks(c) as an induced sub-
graph for every jeN+.
5. Insuring that some representation is measurable
The next theorem states conditions under which G has a measurable repre-
sentation.
Theorem 5.1. The following are equivalent:
(i) G = {[0,1], E) has a measurable representation.
(ii) G or G contains a Kc.
(iii) G has a representation of measure 0 or 1.
The proof is a consequence of the following result due to Galvin [Ga].
Theorem 5.2 [Ga]. Let F ç [0,1] be perfect and nonempty. Let B ç [0,1] be
a Borel set symmetric about the line y = x. Then there is a nonempty perfect
set P' CP such that P' x P' - {{x,x):x G P'} ç B or Bc.
Blass [Bl, p. 271 ] supplies the following verification of Theorem 5.1. The only
nontrivial direction, of course, is (i) => (ii): Suppose, without loss of generality,
E is measurable. Let F ç E be an Fa set such that p2(E-F) = 0 and H D E
be a Gs set such that p-,(H - E) = 0. Let V = Fl>Hc and observe p2 V = 1 .
By Theorem 3.3 (or the result of Mycielski [My]), there is a nonempty perfect
P c [0,1] satisfying P x P - {(x,x)\x G P} c V. By Theorem 5.2, there is
a nonempty perfect P' c P such that P1 x P' - {(x,x):x g P1} c F or Fc,
which implies P' x P' - {(x ,x):x G P'} c E or Ec. D
The proof of Theorem 5.1 has another interesting consequence. Let X be
any subset of [0,1] and define X* = {y - x\ x ,y G X and y > x}.
Proposition 5.3. For every measurable A ç [0,1 ] there is a nonempty perfect set
F ç [0,1] such that P* ç A or P* ç Ac.
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86 SETH m. malitz
Proof. Define E = {(u,v)\u,v g [0,1] and \u—v\ G A} . F is easily seen to be
a measurable subset of [0,1 ] , symmetric about the line y — x . The proof of
Theorem 5.1 says there is a perfect set Fe [0,1] suchthat PxP-{(x ,x)\x G
P} C E or P x P - {(x,x)\x G P} C Ec. In other words, P* C A orF* c Ac. G
Remark. Carlson [Ca] has constructed a nonmeasurable set A c [0,1] such
that for every A' ç [0,1] of cardinality c, X* nAyi0 and X* n Ac ¿ 0.
6. Insuring that every representation is measurable
The next theorem states conditions under which every representation of G
is measurable. First we need a lemma. The proof is essentially that of Theorem
3.1, p. 375 in[Ba].
Lemma 6.1. Let G be the graph with vertex set [0,1] whose only edges are
{x,x + ^} where x ranges over [0,j]. Then G has a representation of outer
measure 1.
Proof. Let F (y < c) enumerate all closed subsets of {{x ,y) G [0, lf\x < y}
having positive measure. By induction on ß < c, pick distinct pß,qß, rß so
that
• Pß,Qß,rß&\J7<ß {Py, Qy, ry} and
• (pß,qß)GFß.
Note that the choice of pß,qß,rß is possible by Fubini's Theorem. Let
^ = U«<f{P« >#«} and £:[0,5] —> c be 1-1 and onto. Observe that Ac has
cardinality c. Pick / G & such that f[[\ ,1]] = Ac, and (f(x),f(x + $)) =
(pg{x),q„(v,) for every x G [0,^]. Then the complement of £, contains no
closed set of positive measure and hence has inner measure 0. a
A subset lç[0,l] is a vertex cover for the graph G = {[0,l],E) if for
every edge (u,v) G E either u g X or v G X or both.
Theorem 6.2. The following are equivalent:
(i) Every representation of G = ([0,1], E) is measurable.
(ii) G or G has a vertex cover of cardinality smaller than every subset of
[0,1] of outer measure 1.
(iii) Every representation of G has measure 0 or every representation of G
has measure 1.
Proof, (i) => (ii). Suppose neither G nor G has a vertex cover of cardinality
y < c. Then G and G both contain a set of c disjoint edges. Therefore,
by Lemma 6.1, G and G both have representations with outer measure 1.
Suppose, for contradiction, that every representation of G is measurable. Then
G and G both have measurable representations of measure 1. Therefore, by
Theorem 3.3, there are disjoint subsets A,B c [0,1] of cardinality c, such
that Ax A- {(x,x):x G A} c F and B x B - {(x,x):x G B} c Ec. Let
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MEASURES OF GRAPHS ON THE REALS 87
D = [0,1] - (A u B). Pick / g 3? such that f[D] is a set of measure 0,
f[A] = T-f[D] and f[B] = ^C-/[F>], where V c [0,1] is a nonmeasurable
set of cardinality c whose complement also has cardinality c. Then Ef is
nonmeasurable.
(ii) => (iii) Suppose G — ([0,1],F) has a vertex cover C of cardinality
smaller than any subset of [0,1] of outer measure 1. Then obviously C has
measure 0. For any vertex v of G, if v £ C then {w: (v , w) G E) ç C. Thus
by Fubini's Theorem, every representation of G must have measure 0. D
7. Open questions
It would be interesting to obtain analogues of Theorem 4.1 for directed
graphs, hypergraphs, other measures besides Lebesgue measure, and graphs of
higher cardinality. With regard to hypergraphs, there is a conjecture known
as Turán's Conjecture which is the hypergraph analogue to Turán's Theorem
about graphs. Unfortunately this conjecture has eluded proof for over forty
years (see Kalai [Ka]). Thus, we suspect a hypergraph version of Theorem 4.1
will be difficult to establish. The situation for directed graphs looks a little more
promising.
Acknowledgments
I would like to thank Jerome Malitz for suggesting the theme of this pa-
per, and for his encouragement, enthusiasm and many helpful discussions. My
appreciation also goes to Tim Carlson, Rich Laver, Jan Mycielski and Monty
McGovern for all their assistance. Finally, I would like to thank an anonymous
referee for dramatically simplifying an earlier proof of Lemma 3.2, and for a
number of suggestions that improved the organization of the paper.
References
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[Bo] B. Bollabas, Graph theory—an introductory course, Springer-Verlag New York Inc., 1979,
p. 72.
[Ca] T. Carlson, personal communication.
[Ga] F. Galvin, Partition theorems for the real line, Notices Amer. Math. Soc. 15 (1968), 660;
Erratum 16 (1969), 1095.
[Ha] F. Harary, Graph theory, Addison-Wesley Publishing Co., 1972.
[Ka] G. Kalai, A new approach to Turón 's Conjecture, Graphs and Combinatorics 1 1986, 107-109.
[My] J. Mycielski, Algebraic independence and measure, Fund. Math. 61 (1967), 165-169.
[Si] W. Sierpinski, Sur un problème de la théorie des relations, Ann. Scuola Norm. Sup. Pisca 2
(1933), 285-287.
Ashdown House, Room 615A, 305 Memorial Drive, Cambridge, Massachusetts 02139
Current address: Department of Computer and Information Sciences, University of
Massachusetts, Amherst, Massachusetts 01003
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