SUPERPOSITION ON MONOTONIC FUNCTIONS

BY ESTHER McCORMICK TORRANCE

The object of this paper is the study of the superposition of a function on ageneral monotonic function y g(x) or on an inverse of such a function,x g-l(y), and the associated problem of what properties the set A can havewhen A’ has a stipulated property and g(At) A or g-l(A’) A. The firstpart of the paper deals with the functions h(x) f[g(x)] and k(x) f[g-l(x)]when restrictions are put on f(y), and g(x) is any monotonic function. Theresults obtained are contrasted with the results for superposition on continuousfunctions, and are used to determine relations between classes of sets. Thesecond part of the paper deals with the function h(x) f[g(x)] when restrictionsare put on g(x). The results obtained include a generalization of a well-knowntheorem concerning such superposition and give a good example of the Baire-measure duality. The paper closes with an application of the results to theStielties integral.The functions studied in this paper have as their domain of definition and

range of values the interval [0,1]. The terminology is that of Kuratowski [3].

1. A monotonic function considered as a transformation consists of a homeo-morphism between two G’s, H and H’, the relation g(K) K’, and the relationg-l(K’) K, where K’ is a denumerable set of points and K is the sum of adenumerable set of points and a denumerable set of closed intervals (see [11]).To every point of K’ corresponds one point or one interval of K, but the intervalcan be open, semi-closed, or closed.A property P of sets is said to be a restricted intrinsic invariant property if

any set homeomorphic to a set having property P also has property P, and ifwhenever a set X has property P, GX + F also has property P.

THEOREM 1. A general monotonic transformation and its inverse carry setshaving a restricted intrinsic invariant property P into sets having the sameproperty P.

Let X have a restricted intrinsic invariant property P. Then

g(X) g(XH - XK) g(XH) @ g(XK) X’H’ + g(XK).

Since X has a restricted intrinsic invariant property P and H is a G, XH has arestricted intrinsic invariant property P. Since the monotonic function g(x)is a homeomorphism between H and H’, it transforms XH into a set X’H’

Received June 13, 1939; presented in part to the American Mathematical Society, Sep-tember 7, 1938.

Numbers in brackets refer to the bibliography at the end of the paper.

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3O8 ESTHER McCORMICK TORRANCE

homeomorphic to XH, and thus into a set with property P. Since g(XK) mustbe a subset of K’, it is at most a denumerable set of points, and so it is an F.Hence X’H’ - g(XH) is the sum of a set having property P and an F thusit is a set having property P, so that g(x) has property P.

Let X’ have a restricted intrinsic invariant property P. Then

g-l(Z’) g-l(X’H’ T X’K’) g-I(X’H’) @ g-l(X’K’).

The set X’H’ is the product of a set having the property P and a G, and henceis a set with property P. Since g-(X’H’) is a homeomorph of X’H’, it musthave property P. The set g-I(X’K’) is a subset of K, so it is at most the sumof a denumerable set of points and a denumerable set of intervals, and henceis an F. Then g-(X’) is the sum of a set having property P and an F henceit is a set having property P.A linear set E is perfectly measurable if it is non-denumerable and if every

set homeomorphic to E is measurable [4].Let El, E, be a denumerable sequence of sets on the x-axis. Let r,

r, be the rational points on the y-axis ordered in some arbitrary but fixedmanner. In the XOY-plane erect perpendiculars to the y-axis through everypoint r and mark off on this line the points whose abscissas belong to Ecall this set en The crible C is the set e. Let P be the line perpendicularto the x-axis at x, and R the set of points in Px( e). Consider the set c ofpoints v such that R has a subset of points pl, p, with pk+ =< pk forevery k. Call this set the positively cribled set by means of crible C. Call itscomplement the negatively cribled set.An operation which permits us to pass from sets E, E, to the positively

and negatively cribled sets will be called the crible operation.The crible sets C are the sets formed by the crible operation operating on

intervals [5].A family Fa or Ga of Borel sets is called a multiplicative family of class a if the

product of a denumerable number of members of this family is still a memberof this family.A funct.ion y f(x) is a Baire function of class , if every closed set F in the

space 0 of the range of values of the function is transformed by f-(y) xinto a multiplicative Borel set of class a; i.e., for every closed set F,f-I(F) A F,

where A is a multiplicative Borel set of class a [3].THEOREM 2. The following classes of sets have a restricted intrinsic invariant

property P as their defining property: (1) additive Borel sets of class a, a > 1;(2) multiplicative Borel sets of class a, a > 1; (3) sets having the Baire propertyin the restricted sense; (4) perfectly measurable sets; (5) projective sets of class n.

The property distinguishing these classes is a property invariant underhomeomorphisms (see [3], pp. 217 and 243, and [4]). Furthermore, sets Gand F have each of these properties (1)-(5), and the class of sets having eachof these properties is closed with respect to finite addition and multiplication,

SUPERPOSITION ON MONOTONIC I)’UNCTIONS 309

so if X has any one of these properties, GX + F has it also. Hence each ofthese classes of sets has a restricted intrinsic invariant property P as its definingproperty.

In like manner it can be shown that crible sets C (see [5]) have a restrictedintrinsic invariant property.From Theorems 1 and 2 we can deduce the fact that all classes of sets listed

in Theorem 2 are such that every monotonic transformation transforms a setof this class into a set of the same class. From this result we can sharpen formonotonic functions the well-known theorem which says that if f(y) is a Baircfunction of class a and g(x) is a Baire function of class 1, then h(x) fig(x)]is a Baire function of class a -t- 1.

THEOREM 3. If f(y) is a Baire function of class a, a > 1, then f[g(x)] is aBaire function of class .

Let h(x) f[g(x)]. Then h-l(F) g-[f-(F)] g-(Z), where Z is amultiplicative Borel set of class a, a > 1, and F is an arbitrary closed set.Since multiplicative Borel sets of class a, a > 1, are transformed by the inversemonotonic function into sets of the same class, g-(X) is a set of this class, soh(x) must be a Baire function of class a.

A non-denumerable set of points E is said to be concentrated in the neighborhoodof a denumerable set H if any open set containing the set H contains also theset E with the exception of at most a denumerable set of points [1].

If we confine ourselves to spaces in the interval [0, 1], we can give an equiva-lent definition of concentrated sets: a non-denumerable set C is a concentratedset if there exists a denumerable set {A (j) of increasing or decreasing sequencesof points A(j) {a(j)} such that if for every j an open set G contains aninterval I. which contains an infinite number of points of the sequence A (j),then the open set G contains all but a denumerable set of points of C [12]. Ifwe use this definition, it is easy to show that the property of being a concentratedset is hereditary and invariant under monotonic homeomorphisms, and wehave the following

LEMMA 1. A monotonic transformation from space to space transformsconcentrated sets into concentrated sets or denumerable sets. A monotonic trans-formation from space ) to space transforms concentrated sets into concentratedsets, or denumerable sets, plus open sets.

LEMM/ 2. Every set having property L is a concentrated set.

Let E be a set which has property L. Let x, be a denumerable subset of Ewhich is dense in E. Let G be any open set containing x. Consider theset of points A (1 G)E. Set A is non-dense in E because in every setG open in E with G1E 0 there exists a point of the set x which is containedin the set G, so GG is a set open in E and lying in the set G and containing nopoint of A. Then set A is non-dense in the entire space, and so also is the closureof A. The closure of A is either a denumerable set of points B or the sum

ESTHER MCCORMICK TORRANCE

of a perfect non-dense set P and a denumemble set, of points B, P -? B.From the definition of A, A is the product, of a closed set, (1 G) and theset E. Then A E. If is the sum (f perfect, non-dense se P and adenumerable set of points B, then A (P + B)E. Since E has property L,it can have at most a denumerable number of points in common with theperfect non-dense set P, so that A is at most a denumerable set of points. ThenE is a concentrated set.A totally imperfect set is a set containing no perfect subset.

LEMMA 3. The property of being a totally imperfect set is an invariant of themonotonic transformation from the space to the space , and lhe monotdnictransformation from the space to the space carries a totally imperfect set intothe sum of a totally imperfect set and an open set.

Consider a totally imperfect set X. It is transformed into H’X’ by thehomeomorphism associated with the monotonic function. If H’X’ containsany perfect subset, HX must contain a non-denumerable Borcl set and hencea perfect subset. By definition, X does not contain any perfect subset. The set(1 H’)X’ consists of some of the points of K’, p, ,so X’ H’X’ + p:is totally imperfect since it contains no perfect subset.

In like manner consider a totally imperfect set X’. It is transformed intoHX + AK, where HX is totally imperfect, and AK is at most the sum of adenumerable set of points and an open set.From Lemma 2 we deduce the fact that monotonic functions transform sets

having property L into concentrated sets or denumerable sets. It is easy togive an example of a monotonic function which does not transform a set havingproperty L into a set having property L.

Sierpinski [7] has defined and proved the existence of a class of sets of power Nwhose homeomorphs are of measure zero and always of the first category. Letus call the class of all non-denumerable subsets of sets having this propertythe class of sets having property S. It can be shown very easily that amonotonic function transforms sets having property S into sets having thisproperty, and an inverse monotonic function transforms any set having propertyS into the sum of a set of this type and an open set.A set always of the first category is a set which is totally imperfect and which

has the restricted Baire property ([3], p. 269).LEMMA 4. The monotonic transformation from the space to the space takes

sets always of the first category into sets always of the first category, and the mono-tonic transformation from the space to the space taes any set always of thefirst category into the sum of a set always of the first category and an open set.

A totally imperfect set’ is transformed into a totally imperfect set by themonotonic transformation from the space to the space , and a totally im-perfect set is transformed into the sum of a totally imperfect set and an openset by the monotonic transformation from the space to the space by Lemma3. A set having the restricted Baire property is transformed into a set with

SUPERPOSITION ON MONOTONIC FUNCTIONS 311

the restricted Baire property by Theorems 1 and 2. It follows that a sethaving both these properties, that is, a set always of the first category, is trans-formed into a set always of the first category by the monotonic transformationfrom the space to the space ), and a set always of the first category in thespace ) is transformed into the sum of a set always of the first category andan open set.

Collecting together these results, and noting that Theorems 1 and 2 implythat all the classes of sets mentioned in Theorem 2 are invariant under mono-tonic functions and their inverses, we obtain

THEOREM 4. The following classes of sets have as their defining property aproperty invariant under every monotonic transformation from the space to thespace and from the space to the space : (1) projective sets of class n; (2) multi-plicative Borel sets of class a, a 1; (3) additive Borel sets of class a, o 1;(4) sets having the restricted Baire property; (5) crible sets C; (6) perfectly measur-able sets.

The following classes of sets have as their defining property a property invariantunder every monotonic transformation from the space to the space , and amonotonic transformation from the space to the space transforms a set of one

of these classes into the sum of a set of the same class and an open set: (1) con-centrated sets and denumerable sets, (2) totally imperfect sets, (3) sets with prop-erty S and denumerable sets, (4) sets always of the first category.

For a large number of classes of sets we can find examples of members A and Bof these classes and monotonic functions gi(x) and g2(x) for which gl(A) A’does not give a member of the same class and g(B) B’ does not give amember of the same class or the sum of a member of such a class and an openset [12]. The classes for which such examples exist are listed in Theorem 5.

THEOREM 5. The following classes of sets are classes which are not invariantunder every monotonic transformation, whether the monotonic transformation isconsidered as a transformation from the space to the space or as a transforma-tion from the space to the space : (1) open sets, (2) closed sets, (3) G, (4) F,(5) dense sets, (6) frontier sets, (7) non-dense sets, (8) sets dense in themselves,(9) sets of the first category, (10) Baire sets, (11) sets with property L and denu-merable sets.

Let us compare these results with the results for continuous functions.There exists a continuous function c(x) and a projective set A of class n,

where n is even, such that c(A) is a projective set of class n 1 ([3], p. 239).By Theorem 4 every monotonic function g(x) transforms A into a projectiveset of class n.

Sierpinski [8] has shown that there exists a totally imperfect set A and thereexists a continuous function c(x) such that c(A) is not totally imperfect. ByTheorem 4, every monotonic function is such that g(A) is totally imperfectwhen A is totally imperfect.

Sierpinski [9] has shown that there exists a continuous function c(x) and a func-

312 ESTHER MCCORMICK TORRANCE

tionf(y) having the restricted Baire property which is such that h(x) f[c(x)]does not have the restricted Baire property. Suppose f(y) has the restrictedBaire property and consider h(x) fig(x)], where g(x) is a monotonic function.For every closed set F, h-l(F) g-i[f-(F)] g-(A), where A has the restrictedBaire property; so by Theorem 4, g-i(A) has the restricted Baire property andthe function h(x) has the restricted Baire property.

2. Let us consider the function h(x) f[g(x)] when restrictions are put on g(x).

LEMMA 5. If a homeomorphism between H and H is not absolutely continuous

from H to H’, it takes a point set of measure zero in H into a point set of positivemeasure in H.

Suppose a homeomorphism is not absolutely continuous from H to H’. Thengiven an arbitrary small positive quantity e there is no i such that (X) < ti

implies t[f(X)] < e; that is, for every ti there exists a set X such that t(X) < iand [f(X)] > e. Call the class of all the sets X such that (X) < ti andt[f(X)] > e the class X. Every set in this class has measure less than /t.

Then for any e consider the ensemble /X} of these classes X. Consider somesequence /in} of i’s which is such that tim is finite. Take one set X fromeach of the classes X.. The measure of Xn is less than i. Then the set

Gm X has a measure which is less than ti, and the measure of the

set lim G,,. is zero since lim t is zero and the G form a decreasing sequence

G1 G. :::) Corresponding to the set Gm= X is a set G’ which con-

rains X:. Recall that X was a set with t(X) < ti and t[f(X)] > e, so X:has a measure greater than e. Then G G is a decreasing sequenceof sets each having measure greater than e, so lim G’ has measure greater than

or equal to e; hence it is a set of positive measure. Then a non-absolutelycontinuous transformation from H to H’ takes a set of measure zero into a setof positive measure in H’.

THEORE 6. A necessary and sucient condition that h(x) fig(x)] be measur-able for every measurable function f(y) is that the inverse monotonic functionx g-(y) be absolutely continuous on H’, the G associated with the monotonicfunction y g(x).

The function y g(x) sets up a homeomorphism between H’ in the spaceand H in the space . If g-(y) is absolutely continuous as far as the homeo-morphism set up by y g(x) is concerned on H, then given an arbitrary smallpositive quantity e, there exists a /t such that if t(Y) ti and Y H’, thent[g-(Y)] < e, where g-(Y) is a set in H since Y is in H’. Suppose we have apoint set Z of measure zero in H’. Consider its transform f(Z) in H. Sincet(Z) < i, where i is any positive number, then for every e, [f(Z)] < e, so theset f(Z) is a point set of measure zero. Any measurable set A in the space

SUPERPOSITION ON MONOTONIC FUNCTIONS 313

is composed of a Borel set B and a set of measure zero Z. Then A B - Zand AH’ (B -t- Z)H BH ZH’. The product of two Borel sets being aBorel set, BH’ is a Borel set. The property of being a set of measure zero beinghereditary, ZH’ is of measure zero. Since the homeomorphism between Hand H takes every Borel set into a Borel set, BH’ has as correspondent a Borelset B1 in H. Since a set which is a Borel set with respect to a Borel set in ametric separable compact space is a Borel set in this space, B1 is a Borel set inthe space in question. The homeomorphism between H’ and H carries pointsets of measure zero in H’ into point sets of measure zero, so ZH is carried into apoint set of measure zero, Z. Then the part of A lying in H’ is transformed intothe sum of a Borel set and a set of measure zero in the space E which is a meas-urable set in the space .Any point of A not lying in H’ will either have no corresponding point in the

space or will correspond to a point in the space or will correspond to an inter-val in the space . There will be at most a denumerable number of such pointsand intervals and since the sum of a denumerable number of points and intervalsis measurable, the part of A lying outside of H’ is transformed into a measurableset in the space .

Since A AH’ A (1 H’) is the sum of two sets both of which are trans-formed into measurable sets in the space , A is transformed into a measurableset in the space . Then if a set is measurable, it is transformed into a measur-able set by any inverse monotonic function x g-’(y) which is absolutely con-tinuous on the G associated with the monotonic function y g(x). Considerany measurable function f(y). The condition that h(x) fig(x)] be measurableis that h-’ (F) be measurable for every closed set F. Then h-l(E) g-[f-(E)]g-l(A), where A is a measurable set. For functions y g(x) satisfying ourconditions we have just proved that g-i(A) is measurable, so our conditions aresufficient that fig(x)] be measurable.Now to prove that the conditions are necessary. If the homeomorphism

between H and H associated with y g(x) is not absolutely continuous fromH to H, then it takes a point set of measure zero, X, in H into a point set ofpositive measure, X, in H, by Lemma 5. Consider some non-measurable subsetof X, call it N. This corresponds to some subset N’X’ of X which must be ofmeasure zero since X’ is. Then the transformation y g(x) takes a measurableset (here NX) into a non-measurable set. Consider the characteristic functionf(y) of N’X’. This is a measurable function since NX is measurable. TakeF as the closed set from 1/2 to 1. Then

h-’(F) g-l[f-’(F)] g-’(N’X’) g-’(H’N’X’) + g-’(N’X’K’) N -t- DK.

We thus have the sum of two disjoint sets, one a non-measurable set and theother a measurable set, so that the sum is non-measurable. Hence if the homeo-morphism between H and H’ associated with y g(x) is not absolutely continu-ous from H’ to H, the compounded function fig(x)] is not measurable for everymeasurable function f(y). This completes the proof.

This theorem is a generalization of the well-known theorem" in order that a

314 ESTHER MCCORMICK TORRANCE

strictly increasing continuous function y g(x) be such that h(x) f[g(x)] ismeasurable for every measurable function f(y) it is necessary and sufficient thatg-l(y) be absolutely continuous.

THEOREM 7. A necessary and sufficient condition that h(x) f[g(x)] be a meas-urable function for every function f(y), where g(x) is a monotonic function, is thatthe Gn over which the function y g(x) defines a homeomorphism be of measure zero.

First, we prove that if the G set H associated with g(x) is of measure zero,every set is carried into a measurable set by the monotonic transformation fromthe space to the space associated with g(x), and any function f(y) is suchthat h(x) f[g(x)] is measurable. Let X’ be any set and let X be its corre-spondent under the monotonic transformation. Then g-l(x’) g-I(H’X’ +X’K’) g-I(H’X’) + g-(X’K’) HX + AK X. Then X is a measurableset, being the sum of a set HX of measure zero and a Borel set.

Consider any function f(y). Then h-l(F) g-[f-’(F)] g-(X’), X’ beingany set. We have just shown that g-(X’) X is a measurable set, so h(x)f[g(x)] is measurable.Now we prove that if every function f(y) is such that h(x) f[g(x)] is measur-

able, where g(x) is a monotonic function, then the G set H over which y g(x)defines a homeomorphism is of measure zero. Suppose the Gn set H is not ofmeasure zero. Then it contains a non-measurable set M. Let M’ be the trans-form of M by the monotonic homeomorphism defined by y g(x). Let f(y) bethe characteristic function of M’, taking on the value 1/2 at the points y e M’, andzero elsewhere. Then h-(1/2) g-’[f-(1/2)] g-(M’) M is not a measurableset, so h(x) is not a measurable function. Then if H is not of measure zero, thereexist functions h(x) f[g(x)] which are not measurable, and the hypothesis iscontradicted. This completes the proof.

r[-IEORE{ 8. A necessary and sucient condition that the function h(x)f[g(x)] be non-measurable for every non-measurable function f(y) is that the homeo-morphism between the sets H and H’ be absolutely continuous on H.

Suppose the homeomorphism between H and H’ is absolutely continuous onH. Then if h(x) is a measurable function, for every closed set F, h-(F)g-l[f-l(F)] g-(X’) X is a measurable set, so X B - Z, where B is a Borelset and Z is a set of measure zero. Since the homeomorphism is absolutelycontinuous on H, g(Z) is a set of measure zero. By Theorem 4, g(B) is a Borelset. Thenf-(F) X’ g(X) is a measurable set, so f(y) must be a measurablefunction. Then if the homeomorphism between H and H’ is absolutely continu-ous on H, h(x) must be a non-measurable function if f(y) is a non-measurablefunction.The condition is also necessary, because if the homeomorphism between

H and H’ is not absolutely continuous, it transforms a set of measure zero intoa set of positive measure by Lemma 5, and hence it transforms a set of measurezero into a non-measurable set. Suppose f(y) is the characteristic function ofthis non-measurable set. Then f(y) is non-measurable and h(x) is measurable.

SUPERPOSITION ON MONOTONIC FUNCTIONS 315

It is obvious that theorems similar to Theorems 6-8 could be proved for thefunction l(x) f[g-l(x)].THEOREM 9. A necessary and sucient condition that f[g(x)] h(x) have the

Baire property whenever f(y) has the Baire property is that the inverse functiong-l(y) take sets in H’ which are of the first category in 0 into sets of the first categoryin .The condition is necessary. For, suppose it is not fulfilled. Then there is a

set A’ of the first category such that g-I(A’H) A is not of the first category.Then there exists a subset of A, call it B, which does not have the Baire property.Now g(B) B’ is a subset of A’. Let f(y) be the characteristic function of B’.Then f(y) has the Baire property since B’ is a set of the first category. Nowh-(F) g-l[f-(F)] g-l(B’) B is a set not having the Baire property, soh(x) does not have the Baire property.The condition is sufficient. For h-(F) g-[f-l(F)] g-(A’) where, since

A has the Baire property, A’ B’ C’, where B’ is a Borel set and C’ is a setof the first category. Since g(x) is a Baire function of class one, g-(B’) is a Borelset B. Then g-l(B’ + C’) g-(B’) + g-l(c’) B + g-I[C’H’ + C’K’]B - C - DK, where C is of the first category by hypothesis, and where DKis some denumerable set of points and denumerable set of intervals in K. ThenB - DK C is the sum of a Borel set and a set of the first category, and istherefore a set with the Baire property.

THEOREM 10. A necessary and suicient condition that h(x) fig(x)] havethe Baire property for every function f(y) is that H be a set of the first category.

Suppose H is not a set of the first category. Then let X and 1 X be adivision of the interval into two totally imperfect sets; XH and (1 X)H arethen totally imperfect sets. Suppose both of these sets are sets having the Baireproperty. Then both are the sum of a Borel set and a set of the first category.Since both are totally imperfect, the Borel set in both cases must be denumerable,and both XH and (1 X)H must be sets of the first category, so their sum Hmust be a set of the first category, and this is contrary to the hypothesis. Thenthere exists a subset of H which does not have the Baire property. Call thissubset the set A, and call its transform A’. Let f(y) be the characteristic func-tion of A’. Then fig(x)] h(x) is such that the closed set F" [1/2, 1] has h-(F)g-l[f-(F)] g-(A’) g-I(A’H) A, so h(x) is not a Baire function. Then Hmust be a set of the first category in order that for every function f(y) the func-tion fig(x)] shall be a Baire function. The sufficiency is obvious.

TEORnM 11. A necessary and suigicient condition that fig(x)] h(x) fail tohave the Baire property whenever f(y) does not have the Baire property is that thefunction g(x) take sets in H which are of the first category in the space into setsof the first category in the space .Suppose f(y) does not have the Baire property. Then for some closed set F,

f-(F) X’ does not have the Baire property and h-(F) g-[f-(F)] g-l(X’)

316 ESTHER McCORMICK TORRANCE

g (X’H’) - (X’K’) XH -- AK X, where AK is some denumerableset of points and intervals of K. If X has the Baire property and g(x) takessets in H of the first category in into sets of the first category in ), then XHhas the Baire property so that XH is the sum of a Borel set B and a set of thefirst category D, and g(x) g(B -- D) g(B) -- g(DH) A’K’ B’ -- D’,and B’ is a Borel set. Since g(X) X’, X’ B’ -- D’ is a set having the Baireproperty. But X’ is a set which does not have the Baire property, so the condi-tion of the theorem is sufficient that X shall not have the Baire property, andhence that h(x) shall be a function which fails to have the Baire property.Suppose g(x) takes a set X which is of the first category in and whic5 is in H

into a set which is not of the first category in ). Then it takes some subset ofthis set into a set E which does not have the Baire property. Let f(y) be thecharacteristic function of E. Then f(y) does not have the Baire property andh(x) f[g(x)] does have the Baire property, so the condition of the theorem isnecessary.We note that Sierpinski considers measure and restricted Baire property as

dual properties ([6], Chapter III), and studies applications of this duality. AsKond5 and Szpilrajn [2, 10] point out, there is to some extent a duality betweenthe restricted Baire property and perfect measure, and it is that duality and theduality between measure and the Baire property, between sets with perfectmeasure zero and sets always of the first category, that we have found useful,Theorems 9-11 being the duals of Theorems 6-8. It would be interesting tosee how many of the theorems which Sierpinski did not dualize could be dualizedby using this latter duality. For example, consider Sierpinski’s theorem" Everymeasurable function of a real variable transforms sets with property into setsalways of the first category ([6], p. 85).A set has property $ if it has at most a denumerable set of points in common

with every set of measure zero. The dual of a set with property $ is obviouslya set with property L. Then the dual of Sierpinski’s theorem is Every functionhaving the Baire property transforms sets with property L into sets of perfect measurezero. This theorem I cannot prove, but it suggests a weaker theorem which iseasy to prove.

THEOE 12. Every function having the Baire property transforms sets withproperty L into sets of measure zero.

Let E, the domain of definition, be a complete metric separable space. Iff(x) has the Baire property, there exists a set Z such that Z is of the firstcategory and f(Z) has measure zero. Let E be a set with property L. Thenf(E) f[E(, Z) + EZ] f[E( Z)] + f(EZ). The set E( Z) is atmost denumerable since E has property L, so f(E)

_p f(EZ) which

must have measure zero since f(Z) has measure zero.

3. Consider a Lebesgue-Stieltjes integral f f(x) dg(x), where g(x) is a mono-tonic function. When does this integral exist?

SUPERPOSITION ON MONOTONIC FUNCTIONS 317

Consider the substitution

?! g(x), x g-l(y), f f(x) dg(x) f f[g-l(y)] dy.

The integral exists when/(y) f[g-l(y)] is measurable.Szpilrajn has shown that a necessary and sufficient condition that h(x)

f[g(x)] be measurable for every monotonic function g(x) is that f(x) be perfectlymeasurable, that is, f-(F) A be a perfectly measurable set for every closedset F. The same condition is necessary and sufficient for k(y) f[g-l(y)] tobe measurable. The sufficiency follows from Theorem 4. Associated withevery monotonic function is a homeomorphism between H and H’, and asso-ciated with this homeomorphism is an inverse monotonic function setting upthe same homeomorphism, and vice versa. If a set is such that every inversemonotonic function transforms it into a measurable set, then every monotonicfunction must transform it into a measurable set, and hence the set must beperfectly measurable by Szpilrajn’s theorem, so the condition is necessary. Itis easy to show that if f(x) is perfectly measurable, k(y) is also, and vice versa.We have noted that theorems similar to Theorems 6-8 hold for k(y)

f[g-(y)]. From these results we can say that(1) A necessary and sufficient condition that the integral ff(x) dg(x) exists

for every monotonic function g(x) is that f(x) be perfectly measurable (Szpilrajn).(2) A necessary and sufficient condition that f f(x) dg(x) exist for every meas-

urable function f(x) and g(x) a monotonic function is that g(x) be absolutelycontinuous on H.

(3) A necessary and sufficient condition that f f(x) dg(x) exist for every func-tion f(x) and g(x) a monotonic function is that H have measure zero.

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