PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 51, Number 1, August 1975
ON THE SUMOF DARBOUX FUNCTIONS
A. M. BRUCKNERl AND j. CEDER
ABSTRACT.
a continuous
The
function
main
purpose
with
is to indicate
a Darboux
function
how
badly
fail
to be a Darboux
can
the
sum of
function.
1. Introduction.
questions
In recent
concerning
with the Darboux
example,
further
property
Sierpinski
conditions
functions,
this
Fast
functions.
can be found
property).
For
that in the absence
function
conditions
sums,
for each
of results
of
is the sum of two such
additional
on the possible
A survey
with
value
function
in §7 of [lj.
dealt
functions
certain
restrictions
have
two real
[4]) showed
every
If one imposes
imposes
of articles
of adding
the intermediate
also
that f + d is a Darboux
ate the constant
functions
a number
outcomes
on the functions,
functions.
/ such
(i.e.
[8] (see
(Darboux)
tions
years
the possible
but the only func-
Darboux
involving
Some more recent
on the
function
sums
d,
of Darboux
work on the subject
can be found in [2], [5], and 16].
Since
the sum of a nonconstant
tion might
fail to be a Darboux
it can fail.
Not every
the uniform
limit
severe
functions
obtained
/ and sets
is unbounded
purpose
extends
except
which
two theorems
in every
and almost
that
yet the range
is to obtain
two theorems
by placing
necessary
no restriction
functions,
show
on the continuous
just
such
that
that
function.
Svarc
19]
continuous
a dense
that / + d
M. The
Theorem
no restrictions
M have
be
imposes
d such
of f + d misses
almost
how badly
hand,
to certain
functions
func-
a sum must
and this
On the other
in §3 a result,
condition
and a Darboux
to ask
In fact,
Darboux
paper
the obviously
it is natural
in [3].
correspond
interval
function
a sum.
of Darboux
are analyzed
M there
of the present
Svarc's
function,
is such
of a sequence
restrictions
has recently
function
continuous
main
1, which
on the set M
complement
In §4 we obtain
Received by the editors April 27, 1974.
AMS (MOS) subject classifications
(1970). Primary 26A15; Secondary
a
26A21,
26A24.
Key words
and phrases.
Darboux
1 This author was supported
functions.
in part by NSF grant GP 18968.
Copyright © 1975. American Mathematical Society
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97
98
A. M. BRUCKNER AND J. CEDER
general
theorem
rem also
related
contains
to the types
the two theorems
2. Preliminaries.
functions
defined
Darboux
functions,
real
value
take
on every
real
interval,
real
value
numbers
projection
we shall
tinguish
denote
interval,
that
its cardinality
by CJ0. Finally,
a function
from its graph.
3. The main result.
have
property
the Darboux
Theorem
we obtain
rather
for each
let M be a set
countable
dense
its x-
denote
we shall
to obtain
the
measure
denote
We shall
convenient,
which
If A is a set of
its Lebesgue
we shall
a bit more,
than
1. Let f be a Darboux
of R and
set,
of
on every
c denotes
C Ju C D.
by rng B.
where
take
of functions
where
.V
by |A|,
We are now ready
Actually,
theo-
with real
the class
which
the class
If ß is a planar
numbers
the Introduction.
Then,
in every
It is clear
by XA.
This
entirely
let -ß denote
of functions
denote
by dorn B and its y-projection
of natural
interval
the class
c times
consideration.
be concerned
We shall
and JJ
of the continuum.
(if it is measurable)
we shall
line R.
~D denote
under
of Svarc.
In the sequel
on the real
in every
cardinality
of sums
the result
the set
not dis-
mentioned
as we assume
in
only that /
that / be continuous.
function
which
of real
numbers
subset
D of R\M
is constant
whose
on no sub-
complement
there
exists
is dense.
a function
d £ 9) such that the range of f + d is D.
Proof.
Let D be
a countable
dense
subset
\(x, r - f(x)): r £ D\. We shall define a function
This will guarantee
Let K consist
of all horizontal
lines
H(z) or V(z) will denote
It is clear
for each
For
for which
and U consist
V £ U, V n ß is
/ is not constant
is dense
in R, so there
exists
an r £ D such
r - f(x.)
and r — ¡(xA).
Since
t - f(x) = X. Therefore,
V(z) n B n H A0\.
B =
/ is Darboux
H n ß is dense
set A, let
of all vertical
exist
x
on any subinterval.
that X = rng H lies
there
lines.
z.
in V.
of R there
since
planar
let
of K or ü containing
NQ-dense
4 f(x.),
For each
and
d £ 3) such that d C B.
that member
H £ H and / any open subinterval
f(x.)
R\M
that f(x) + d(x) £ D for all x.
If z e R , then
that
of
exists
and x
in /
Now D
between
an x £ J for which
in H.
M(A) = ¡ß Pi H: there
exists
z £ A such
that
Let GQ(A) = M(A) and define
oo
Gn +1,(A) = M(Gn (A)) and
Let
Jl = \G(H):
H £ K}.
In order
to define
G(A) = U Gn (A).
n =0
the desired
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function
d, we need
THE SUM OF DARBOUX FUNCTIONS
the following
99
facts:
(1) |rng N\ = N0 for all Neîl;
(2) UK=ß;
(3) N is dense in/VnJV for all H £ K and N £ Jl;
(A) if Nv N2 £ Jl and dom Nj n dom N2 ¿0,
Parts
(2), (3) and (4) can be verified
of (1) it suffices
Nn.
To show
such
that
to show that
this
let
r £ D
(x, r - f(x))
to have cardinality
NQ for each
without
|rng G ,.(H)\
then Nj = Nr
difficulty.
For the proof
= NQ whenever
|rng G (H)\ =
and K £ K and put A(K, r) = iy: there
£ K and (x, y) £ B\.
Nn. Putting
The set
A(K)= {J\A(K,
exists
A(K, r) is easily
x
seen
r): r £ D\ we have |Á(K)| =
K £ J(. Then for H £ K and |rng Gn(H)\ = KQ we have
rng Gn + 1(H)= \J\A(K):
rng K Ç rng G„(H)i so that rng G„ + 1(H) is a count-
able union
sets
of countable
and hence
|rng G
AH)\ = N. finishing
the
proof of (1).
We can now define
function
dN C N such
To do this
intervals
the function
that
^ut
^N ~ 'u;':
in each
By induction
because
where
pick
in any nonempty
N £ Ji a
H n N foi H £ SX.
of all horizontal
rational
n N and z^n + j £ 0n
Then
family
open
*■
C\N-
rfw is easily
seen
to be a func-
of uncountable
sets
of the form
J is a rational
a sequence
e Tî: there
the range
for each
H H N.
the countable
i(x, y): x £ J, y = r - /(x)|,
e. 6 C¿\(JÍ/V
z e ^nl'
nonempty
Next enumerate
sible
is dense
we define
Í0 n }n —u„ be an enumeration
which intersect
N. Pick wQ £ O
tion dense
and
First
let
U ■_n^(M'P"
|C¿S¿_0.
«^
d.
exists
open
of points
interval
íe¿i°l0
/ < i with e . e N\.
of C . is an interval
whereas
and r £ D, as
such
This
that eQ £ CQ
choice
the range
is pos-
of the other
set is countable.
Define
the function
e by c = \e¿: z e ûj0I.
Then
|e n /V| < 1 for any
N £ Jl so that |e n ¿N| < 1 for any N £ Ti.
Now define
e(x)
d(x) = / «\,(x)
a function
¿ on ü as follows:
if x £ dom e,
if x e dom ii .Adorn e,
any point in V((x, 0)) n ß
It is clear
N and 0 n dN\e
that d C B.
is dense
Since
each
if x /Uidom
horizontal
in 0, it follows
¿N: ;V £ îi! u dom e.
open
interval
that ^ takes
finitely
oftenrestrictions
overmayeach
subinterval.
Therefore,
d e J.) .
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on each
0 hits
value
some
in-
100
A. M. BRUCKNER AND J. CEDER
For any open interval
set
This finishes
Remark
R\M
1. It is easy
be dense
M meet certain
additional
Svarc
Theorem
It follows
constant
from Theorem
a function
can miss
in JJ
in every
respects,
Baire.
for example,
be chosen
there
suppose
2 below
absolutely
shows
continuous,
Theorem
2.
I, and if d £ S)
Proof.
of / + d were
then d cannot
even
// / is defined
first
that since
this
of the sets
set.
and enumerated
is
the
well
it takes
behaved
but this
does
is countable
be Lebesgue
condition
loss
to be true.
and / is
measurable.
on an interval
an uncountable
Suppose
of generality
• Let
of
then d can
not seem
continuous,
(T.).
as \miXi_
class
continuous
have
on every
in other
behaved,
if R\M
/
Now
range.
it satisfies
that
the range
we may suppose
E^ - \x: d(x) + f(x) =
and E, n E, = 0 if k /= I. Since d is measurable,
E is measurable.
set is measurable
that
lines.
in the second
f + d must
Without
/ which
such
because
/ is absolutely
(N) and Banach's
function
can be well
on R and absolutely
then
of / are
of horizontal
functions
that
sets
d £ I)
pathological
such
falsifying
that / and/or
of the sum of a continuous
number
well behaved,
m, S. Then / = U,_.£,,
each
continuous
functions
that
and M is nowhere
a function
in particular
a countable
set is infinite
the
are unnecessary.
if / is sufficiently
is measurable,
We note
condition
exist
that
the level
constant
the graph
somewhat
But such
to be reasonably
Theorem
Lusin's
Thus,
is already
One might
to each
corresponds
all but a countable
interval.
that
assumptions
1 that
there
without
in his two theorems
namely,
these
of / + d is countable.
value
d hits
drop the requirements
from our hypotheses
assumed
or / is nowhere
1 now shows
on no interval
and a d £ D
e and hence
rng(/ + d)~\ I = D.
that we cannot
conditions
and M is countable,
dense.
range
to verify
or that / be constant
in general.
countable
Hence,
the proof.
the conclusion
this
/ and r £ D the function
\(x, y): x £ I, y = r - f(x)\.
because
Now d(Ek)
/ satisfies
Xd(Ek) = Xf(Ek)
= (mk - j)(Ek)
Lusin's
condition
tot each
k, and
(N). It follows
that
for each k.
Now let
A = Sx: / has
no finite
or infinite
derivative
at x\,
B = \x: f'(x) is finite!,
and
C = \x: f'(x) is infinite!.
Then (1) Xf(A)=
0 because
/ satisfies
Banach's
condition
(Tj)
(see
Saks
[7, p. 278]),(2) Xf(B)< ¡B\f'\dX (see [7, p. 272]),and (3) Xf(C) = 0 because
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THE SUM OF DARBOUX FUNCTIONS
XC = 0 and / satisfies
Lusin's
oo
condition
(N).
oo
101
Thus
oo
~ = Xd(l)< £ Xd(Ek)= Z Xf(Ekn B) < £ fE nB l/'l dX= f l/'l dX< -,
¿1= 1
fe=l
a contradiction.
Remark
valid
Thus
4. A general
functions
the range
2. We have
if we weaken
been
result.
/ which,
Svarc's
R.
Then
there
to determine
continuity
for a given
observe
3.
set
gives
follow
readily
Let M be any proper
a d £ T)
such
that
Proof.
(«=) Suppose
number
X.
Well-order
the set
ate disjoint.
d £ JÔ
range
induction.
the range
ix o> y a) e
We have
of f + d misses
open
ii xd
Note
R\M.
for all a < ß.
M if
I.
interval
/
in R.
that
R x â and
d through
will be used
the
to guarantee
to guarantee
Suppose
on
interval
intervals
function
M) will be used
equals
of
and we
that
that
the
a point
We wish to define
(x ß, y ß).
two cases:
Case I. z ߣ&x
Since
(f x (R\
interval
nas Deen chosen
the range
X £ R and every
the required
The product
and the product
of / + d over each
define
those
theorem.
(R x Q) U (â x (R - M)) as \z¿a<c-
use of transfinite
that
|ix £ I: X + f(x) ¿ M\\ = c for each
We shall
2 remains
of R and f be a real function
Let (l be the set of all nonvoid
Cl x R\M
that
that
such
a characterization,
from this
subset
Theorem
of /.
of characterizing
a d £ S)
such
and only if \\x £ I: X + f(x) ¿ Mi| = c ¡or every
and real
whether
of / to continuity
M, admit
3 below
results
exists
be countable.
We now turn to the question
M. Theorem
Theorem
k
of d + / cannot
unable
absolute
f + d misses
that
k=l
\\x
: a < ß\\
(R\M) and z „= (I, X) where / £ <2 and X £ r\m.
< c we may pick
x „ £ l\\x
: a < ß\.
Then
put
yß-
A - f(xß).
Case IL z „ £ R x d and z „= (A, /) where I £ d and A £ P.. Since
|{x e /: A + f(x) i Mi| = c, we may pick
xß £ /\l*a:
a < /3i so that A +
f(xB) i M. Then put yß~ X.
Our construction
namely
d(x
thus
not in E, choose
all,
a function
this
function
d on the set E = \x : a < c\,
to all
a\x) in R so that d(x) + f(x) Í M. This
We now show that
for any nonvoid
intervals
far defines
) = y . We now extend
the function
subinterval
of J is c so that
\\ß:
d has the desired
of R.
is possible
if x is
because
properties.
/ and X £ R the cardinality
z „= (A, /) where
Thus,
First
of all open
/ £ U and / C /!| = c.
for
each
/3 restrictions
in thismayset
„= A sosee that
|i* £ J: d(x) = A!| = c.
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Hence,
M £ R.
of
subBut
/ e 3)
.
102
A. M. BRUCKNER AND J. CEDER
Secondly,
when zß=
for any A £ R\
(I, X) by Case I.
M and any interval
hand,
we have
have
d(x A) + f(x „) ¿ M whenever
f(x „) + d(x A) = X £ R\M
d(x) + f(x) i M. Hence,
rng(/ + d) j /= R\M
(=») Suppose
stated
in any case
in the statement
Suppose
whenever
z „ £ R x (Í.
d(x A) + f(x A) = A
exists
On the other
z „ £ U x R\M
Moreover,
rng(/ + d)
fot all subintervals
there
/ we have
Hence, rng(/ + d) | / D R\M.
| / C P\
and we
for x ff dom E,
M. Therefore
/.
a Darboux
function
d with the properties
of the theorem.
there
exists
X £ R and an open interval
X + f(x) i Mi| < c.
Since
||x £ I: d'x) = X\\ = c there
d(x) = X and A + f(x) £ M. Therefore,
/ for which
exists
|{* £ I:
x £ I such
d(x) + f(x) £ M, a contradiction.
\{x £ I: X + f(x) ¿ MÍ| = c for all A £ R and intervals
/.
This
that
Hence,
completes
the
proof of Theorem 3.
Theorem
suppose
3 does
not hold if we replace
S)
M = Í0! and f(x) = 0 if x is irrational
/.
However,
Moreover
Then rng(/ + d) | / = R - \0\
|ix £ I: d(x) = X\\ < c and in fact
\\x £ l: 0+/(x)¿Í0}}|
We have
been unable
the existence
of a d £ 3)
each interval
I.
For example,
and f(x) = 1 when x is rational.
Let d £ 3) with d(x) = 0 when x is rational.
fot each
with S) .
|!x £ /: d(x) = 0!| < c.
- Kn-
to find a necessary
and sufficient
having
that
the property
condition
rng(/ + d)\
for
I D R\M
fot
REFERENCES
1. A. M. Bruckner
and J. G. Ceder,
Verein. 67 (1964/65), Abt. 1, 93-117.
tions
Darboux
continuity,
Jber.
Deutsch.
2. A. M. Bruckner,
J. G. Ceder and R. Keston,
Representations
by Darboux functions
in the first class of Baire, Rev. Roumaine
Appl. 13 (1968), 1246-1254.
3. A. M. Bruckner,
4. H. Fast,
and M. Weiss,
Une remarque
sur la propriété
D-continuous
Zu zwei
Appl. 12 (1967), 849-860.
7. S. Saks,
de Weierstrasse,
Sierpinski
of Darboux
func-
Colloq.
Math. 7
Sätzen
components,
Akad.
Nauk
von W. Sierpinski,
Rev.
Roumaine
Math.
Pures
MR 36 #6557.
Theory of the integral,
8. W. Sierpinski,
Dokl.
MR 34 #4424
2nd rev. ed., Monografie
PWN, Warsaw, 1937; reprint, Dover, New York, 1964.
Sur une propriété
matiche (Catania) 8 (1953), no. 2, 43-48.
9. R. Svarc,
limits
MR 22 #11087.
5. A. B. Gurevic,
L. Misik,
Uniform
MR 33 #5794.
BSSR 10 (1966), 539-541. (Russian)
6.
and approximaMath. Pures
MR 39 #4333.
J. G. Ceder
tions, Colloq. Math. 15 (1966), 65-77.
(1959), 75-77.
Math.-
MR 32 #4217.
On the range
of values
de fonctions
Math., vol. VII,
MR 29 #4850.
reelles
quelconques,
Mate-
MR 16, 229.
of the sum of a continuous
function, Casopis Pest. Mat. 98 (1973), 178-180, 213.
and a Darboux
MR 48 #470.
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA, SANTA BARBARA,
CALIFORNIA 93106
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