ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE S I X (1976) ROCZNIKI POLSKIEGO TOWAEZYSTWA MATEMATYCZNEGO
Séria I: PRACE MATEMATYCZNE X IX (1976)
C zeslaw B ylka (Poznan)
S. Saks in [ 1 ] lias constructed strictly monotonie function with dérivative equal zero almost everywhere. A. C. Zaanen along with A. J.
Luxemburg [ 2 ] gave an example of such a function (however, let us remark this example is wrong, the function is not monotonie and, in ad
dition to this, at points x = 2~n (n = 1, 2, ...) is not continuous). The construction is based on the Cantor function.
In this note we give an example of function strictly increasing, differen
tiable almost everywhere with derivative equal zero almost everywhere.
L emma . Let A denote a set of such real numbers x that there exists a se-, quence (hn), hn > 0, lim hn = 0 and there exists sequence of integers (pn), (qn) such that
for n — 1,2, ... Then (л{А) = 0, where ju denotes the measure defined on the real line.
P ro o f. Write
where c is any integer number, Q set of all rational numbers. To prove th e lemma it is sufficient to show th a t ja(Ac) = 0 for each c.
O n some class oî monotonie functions
й п < К * 1 \ Х~Рп1йп\ < К
A c = {An[ c, c + l ] ) \ $
a [c, c + 1], b > 3e 1 we set So for e > 0 and for w =
b
€Iw(e) = {»•* \<o — w \ < b 4).
Then
МЫ* ) ) = 26“
and
00
e < « > < o + l}) < У 2 У , 2 гг5 < ! г У
РЛ- Л
С<№<С+ 1
n 2 < e.
24 C. B y l k a
Because for any x e A c there exists I w{s) such th a t we I w(e), hence Ac <= 1(e) = U {Iw(s) : c < w < c + 1 }. Consequently
e > 0}) = 0 .
This implies /u(Ac) — 0 for each integer c.
T heorem . Function
/ ( « ) Z - i У — 2p+e
— Q
( a > 0 ) j
where p, q Ф 0 are positive integers, {p, q) = 1 , has the following properties:
(i) f(x) is bounded and strictly increasing,
(ii) f(æ) is differentiable a.e. and its derivative is egual to zero a.e.
P ro o f. Property (i) is evident. For proving (ii) let x<{A, where A is a set defined as in th e above lemma. Then æ is non-rational and
о /(«+>)-/(») =jr, y
x i 2 p+a
x < —*Cx+h
a
Because \æ—p / q \ < h so for sufficiently small h we have q ^ h Hence
к- 1 Z Уj Л -< V i 2
p+
q aL
i 2®
x<-<x+h »
Q
q > h ~ iLet h ->0 and q ^ h~l. Then £ a~^ 0. Therefore f ( x ) = 0 for x$A .
q>h~i
This ends th e proof.
R e m a rk . When a function is of the form
/0 » ) = P^ x Q
ap + q l
00
where sequence (nan) tends monotonically to zero and series nan is
n= 1
