MEASURES OF GRAPHS ON THE REALS · MEASURES OF GRAPHS ON THE REALS SETH M. MALITZ (Communicated by R. Daniel Mauldin) Abstract. This paper studies measure properties of graphs with

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

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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.


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


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/.


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),


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



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


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+ .



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.


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



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

[Ba] J. Barwise, ed., Handbook of mathematical logic, North-Holland, 1977.

[Bl] A. Blass, A partition theorem for perfect sets, Proc. Amer. Math. Soc. 82 (1981), 271-277.

[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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