Functions of first Baire class - Strange Functions in Real Analysis, Second Edition. Pure and A

Let E be a topological space and let f : E → R be a function.

We shall say that f is of Baire zero class if f is continuous at all points of E (i.e., f is continuous on E).

The family of all continuous functions acting from E into R is usually denoted by the symbol C(E, R). In accordance with the definition above, we will also use the notation Ba0(E, R) for the same family of functions.

Thus, we have

Ba0(E, R) = C(E, R).

By the standard definition due to Baire (see [5], [6]), a given function f : E → R

is of first Baire class if there exists a sequence {fn : n < ω} of functions

from Ba0(E, R) such that

limn→+∞fn(x) = f (x)

for each point x ∈ E. In other words, f : E → R belongs to the first Baire class if and only if f can be represented as a pointwise limit of a sequence of functions belonging to Ba0(E, R).

It is well known that functions of first Baire class play a significant role in various topics of real analysis. The following simple but important example emphasizes this circumstance.

Example 1. Let E = R and let f : E → R be a derivative. Then there exists a continuous function g : E → R such that

g0(x) = f (x) (x ∈ E).

Define a sequence {gn : 1 ≤ n < ω} of real-valued functions on R by the

formula

Obviously, we have

limn→+∞gn(x) = f (x) (x ∈ R).

Because each gn is a continuous function on R, we conclude that f belongs

to the first Baire class.

For any topological space E, the family of all functions f : E → R belonging to the first Baire class will be denoted by Ba1(E, R).

Note that the other Baire classes Baξ(E, R) of real-valued functions on

E can be naturally introduced by iterating the limit process and using the method of transfinite recursion on ξ < ω1(in this connection, see [6], [120],

[154] andChapter 7of the present book). Here we are mainly interested in properties of functions belonging to Ba1(E, R).

The following simple properties follow directly from the definition of the class Ba1(E, R):

(1) Ba1(E, R) is a linear algebra over the field R; in other words, if

f ∈ Ba1(E, R), g ∈ Ba1(E, R), a ∈ R and b ∈ R, then

af + bg ∈ Ba1(E, R),

f · g ∈ Ba1(E, R);

(2) if f ∈ Ba1(E, R), g ∈ Ba1(E, R) and g(x) 6= 0 for all x ∈ E, then

f /g ∈ Ba1(E, R);

(3) if f ∈ Ba1(E, R) and φ : ]a, b[ → R is a continuous function such

that ran(f ) ⊂ ]a, b[, then

φ ◦ f ∈ Ba1(E, R).

Exercise 1. Verify the validity of the assertions (1), (2), and (3) above.

Exercise 2. Suppose that E is a separable topological space (i.e., E contains a countable everywhere dense subset). Show that

card(Ba1(E, R)) ≤ c.

In fact, show that if E is nonempty and separable, then card(Ba1(E, R)) = c.

functions of first baire class 65

Lemma 1. Let E be a topological space, {an : 1 ≤ n < ω} be a sequence

of strictly positive real numbers, such that X

1≤n<ω

an< +∞,

and let {fn: 1 ≤ n < ω} ⊂ Ba1(E, R) be a sequence of functions such that

|fn(x)| < an (x ∈ E, 1 ≤ n < ω).

Define a function

f : E → R by the formula

f (x) = f1(x) + f2(x) + ... + fn(x) + ... (x ∈ E).

Then f also belongs to Ba1(E, R).

Proof. First, note that the function f is well defined since the series f1(x) + f2(x) + ... + fn(x) + ... (x ∈ E)

converges uniformly with respect to x ∈ E (in view of the relations |fn(x)| <

an for all x ∈ E and n ≥ 1). Further, because fn ∈ Ba1(E, R) for any

natural number n ≥ 1, we can write

fn(x) = limk→+∞fn,k(x) (x ∈ E),

where fn,k (k = 1, 2, ...) are some real-valued continuous functions on E.

Without loss of generality, we may assume that

|fn,k(x)| ≤ an (x ∈ E, n = 1, 2, ..., k = 1, 2, ...).

Now, let us put

hk(x) = f1,k(x) + f2,k(x) + ... + fk,k(x) (x ∈ E, k = 1, 2, ...).

Clearly, all functions hk are continuous on E and it suffices to show that

f (x) = limk→+∞hk(x) (x ∈ E).

For this purpose, fix a real ε > 0. There exists a natural number m such that

Consequently, we have

|fm+1(x)| + |fm+2(x)| + ... + |fi(x)| + ... < ε/3 (x ∈ E),

For any x ∈ E and k > m, we can write the inequalities

|f (x) − hk(x)| ≤ |f1(x) − f1,k(x)| + |f2(x) − f2,k(x)| + ... + |fm(x) − fm,k(x)|+

|fm+1(x)| + |fm+2(x)| + ... + |fi(x)| + ... +

≤ |f1(x) − f1,k(x)| + |f2(x) − f2,k(x)| + ... + |fm(x) − fm,k(x)| + 2ε/3.

If x ∈ E is fixed, then we can find k0 < ω so large that for all natural

numbers k > k0the relation

|f1(x) − f1,k(x)| + |f2(x) − f2,k(x)| + ... + |fm(x) − fm,k(x)| < ε/3

will be satisfied. But this relation immediately yields the inequality |f (x) − hk(x)| < ε

for all integers k > k0. Therefore, we get

limk→+∞hk(x) = f (x) (x ∈ E),

which completes the proof ofLemma 1.

We need this lemma in order to prove the following result due to Baire.

Theorem 1. Let E be a topological space. Suppose that a sequence {fn: n < ω} ⊂ Ba1(E, R)

is given uniformly convergent to a function f : E → R. Then we have f ∈ Ba1(E, R).

Proof. According to our assumption, for any natural number k, there exists a natural number nk such that

|f (x) − fnk(x)| <

functions of first baire class 67

Evidently, we may assume that

n0< n1< ... < nk < ... .

Let us consider the series of functions

(fn1− fn0) + (fn2− fn1) + ... + (fnk+1− fnk) + ... .

Because the inequalities

|fnk+1(x)−fnk(x)| ≤ |fnk+1(x)−f (x)|+|fnk(x)−f (x)| < 1 2k+2+ 1 2k+1 < 1 2k

hold for all x ∈ E, we can applyLemma 1 to this series. In this way, we obtain that the function

g = (fn1− fn0) + (fn2− fn1) + ... + (fnk+1− fnk) + ...

belongs to Ba1(E, R). But it is easy to see that

g = limk→+∞fnk+1− fn0 = f − fn0.

According to property (1), we finally get

f = g + fn0∈ Ba1(E, R).

The theorem has thus been proved.

Remark 1. Theorem 1implies that, for E 6= ∅, the family of all bounded functions from Ba1(E, R) becomes a Banach space with respect to the norm

of uniform convergence or, equivalently, with respect to the standard sup- norm

||f || = supx∈E|f (x)|.

Remark 2. Theorem 1 can be directly generalized to the case of the Baire class Baξ(E, R), where ξ is an arbitrary ordinal number strictly less

than ω1. The proof essentially remains the same as above (cf. [120]).

Lemma 2. Let E be a topological space and let g ∈ Ba1(E, R). Then,

for every t ∈ R, the sets

g−1(] − ∞, t[) = {x ∈ E : g(x) < t}, g−1(]t, +∞[) = {x ∈ E : g(x) > t} are Fσ-subsets of E.

Proof. Take any t ∈ R. Because g ∈ Ba1(E, R), there exists a sequence

{gn: n < ω} ⊂ Ba0(E, R) such that

limn→+∞gn(x) = g(x) (x ∈ E).

It is not difficult to verify the following relations:

g(x) < t ⇔ (∃k < ω)(∃n < ω)(∀m ∈ [n, ω[)(gm(x) ≤ t − 1/k),

g(x) > t ⇔ (∃k < ω)(∃n < ω)(∀m ∈ [n, ω[)(gm(x) ≥ t + 1/k).

These relations yield at once that the above-mentioned sets {x ∈ E : g(x) < t}, {x ∈ E : g(x) > t} are Fσ-subsets of E, and the lemma is proved.

Lemma 3. Let E be a normal topological space, g : E → R be a function and suppose that

ran(g) = {t1, t2, ..., tk}.

If, for any integer i ∈ [1, k], the set

Xi= {x ∈ E : g(x) = ti} = g−1(ti)

is an Fσ-subset of E, then g ∈ Ba1(E, R).

Proof. Obviously, we can write

E = X1∪ X2∪ ... ∪ Xk,

Xi= Fi,0∪ Fi,1∪ ... ∪ Fi,n∪ ... (i = 1, 2, ..., k),

where all Fi,n (1 ≤ i ≤ k, n < ω) are closed subsets of E and

Fi,0⊂ Fi,1⊂ ... ⊂ Fi,n⊂ ... .

Let us put

Fn = F1,n∪ F2,n∪ ... ∪ Fk,n (n = 0, 1, 2, ...)

and define a function

gn : Fn → R (n = 0, 1, 2, ...)

by the formula

functions of first baire class 69

Because the finite family of closed sets {F1,n, F2,n, ..., Fk,n} is disjoint, the

function gn is continuous on the closed set Fn. By the Tietze-Urysohn

theorem (see, e.g., [85] or [120]), gn admits a continuous extension

g_n∗ : E → R. Now, it is easy to check that

limn→+∞g∗n(x) = g(x)

for all points x ∈ E. This finishes the proof of the lemma.

Lemma 4. Let E be a topological space in which every open set is an Fσ-subset (or, equivalently, in which every closed set is a Gδ-subset). Let a

set X ⊂ E be representable in the form

X = A1∪ A2∪ ... ∪ Ak,

where all Aj (j = 1, 2, ..., k) are Fσ-subsets of E. Then X is representable

in the form

X = B1∪ B2∪ ... ∪ Bk,

where

Bj ⊂ Aj (j = 1, 2, ..., k),

all Bj are also Fσ-subsets of E and, in addition, they are pairwise disjoint.

Proof. Obviously, we have the equality X = F1∪ F2∪ ... ∪ Fi∪ ...,

where all sets Fi (1 ≤ i < ω) are closed in E and each Fi is contained in

some set Aj(i). Let us put

C1= F1, C2= F2\ F1, ..., Ci= Fi\ (F1∪ ... ∪ Fi−1), ... .

The family of sets {Ci : 1 ≤ i < ω} is disjoint and, in view of our assumption

on E, all Ci are Fσ-subsets of E. Moreover, we have

X = C1∪ C2∪ ... ∪ Ci∪ ... .

Now, for any natural number j ∈ [1, k], define the set Bj by the formula

Clearly, the family {B1, B2, ..., Bk} is disjoint, all sets Bj(j = 1, 2, ..., k) are

Fσ-subsets of E and

Bj ⊂ Aj (j = 1, 2, ..., k),

X = B1∪ B2∪ ... ∪ Bk.

Lemma 4has thus been proved.

Recall that a topological space is perfectly normal if E is normal and each open set in E is an Fσ-subset of E. For such spaces the following

important statement due to Lebesgue is true.

Theorem 2. Let E be a perfectly normal space and let f : E → R

be a function. These three assertions are equivalent: (1) f ∈ Ba1(E, R);

(2) for any t ∈ R, both sets {x ∈ E : f (x) < t} and {x ∈ E : f (x) > t} are Fσ-subsets of E;

(3) for any open set U ⊂ R, the preimage f−1(U ) is an Fσ-subset of E.

Proof. The equivalence (2) ⇔ (3) is trivial and the implication (1) ⇒ (2) was established by Lemma 2 (even for an arbitrary topological space E). Consequently, it remains to prove the implication (2) ⇒ (1). Suppose that (2) is valid and suppose first that ran(f ) ⊂ ]0, 1[, i.e.,

0 < f (x) < 1 (x ∈ E).

For any integer n ≥ 1, consider the sequence {t0, t1, ..., tn} of points of R

determined by conditions:

t0= 0, tj+1− tj= 1/n (j = 0, ..., n − 1).

In particular, we have tn = 1. Further, introduce the sets:

A0= {x ∈ E : f (x) < t1},

An= {x ∈ E : f (x) > tn−1},

Aj = {x ∈ E : tj−1< f (x) < tj+1} (j = 1, ..., n − 1).

Obviously, we have the equality

functions of first baire class 71

and all Aj (j = 0, 1, ..., n) are Fσ-subsets of E. ApplyingLemma 4, we get

another representation

E = B0∪ B1∪ ... ∪ Bn,

where all Bj (j = 0, 1, ..., n) are also Fσ-subsets of E, are pairwise disjoint

and

Bj ⊂ Aj (j = 0, 1, ..., n).

Now, define a function

fn: E → R

by putting

fn(x) = tj if f x ∈ Bj.

According toLemma 3, the function fn belongs to Ba1(E, R).

Take an arbitrary point x ∈ E. Then x ∈ Bj for some integer j ∈ [0, n].

These relations show that

limn→+∞fn(x) = f (x)

uniformly with respect to x ∈ E. By virtue ofTheorem 1, we obtain that f ∈ Ba1(E, R).

Suppose now that f : E → R is an arbitrary function satisfying (2). Fix any increasing homeomorphism

φ : R → ]0, 1[

and consider the function φ ◦ f . This function also satisfies (2) and ran(φ ◦ f ) ⊂ ]0, 1[.

As demonstrated above, φ ◦ f ∈ Ba1(E, R). Consequently,

which completes the proof ofTheorem 2.

Example 2. Let E be a perfectly normal space and let X be a subset of E. Denote by fX the characteristic function (i.e., indicator) of X. It is

easy to verify that if X is closed in E, then relation (2) of Theorem 2 is satisfied for f = fX. Therefore, according to this theorem, we have

fX∈ Ba1(E, R).

Now, if Y is an open subset of E, then, taking into account the equality fY = 1 − fE\Y,

we see that fY ∈ Ba1(E, R), too.

The above-mentioned facts follow also fromLemma 3. Example 3. Let E be a subinterval of R and let

f : E → R

be a monotone function. It is easy to check that, for any t ∈ R, both sets {x ∈ E : f (x) < t}, {x ∈ E : f (x) > t}

are some subintervals of E. Because each interval in E is an Fσ-subset of

E, we infer (in view of Theorem 2) that f ∈ Ba1(E, R).

The same conclusion is true for those f : E → R which are of finite variation on E. Indeed, such functions are representable in the form of the difference of two increasing functions on E (see, e.g., [154], [180]) and it suffices to refer to property (1) of the class Ba1(E, R).

Example 4. Let E be again a perfectly normal space and let f : E → R

be an upper semicontinuous function. According to the definition of upper semicontinuous functions, for any t ∈ R, the set {x ∈ E : f (x) < t} is open in E and, hence, is an Fσ-subset of E. At the same time, the set

{x ∈ E : f (x) > t} = ∪n<ω{x ∈ E : f (x) ≥ t + 1/(n + 1)}

is the union of countably many closed sets, i.e., is also an Fσ-subset of E.

Applying again Theorem 2, we deduce that f ∈ Ba1(E, R).

functions of first baire class 73

From this fact it immediately follows that g ∈ Ba1(E, R) for any lower

semicontinuous function g : E → R.

Actually, the characteristic function fX of a closed set X ⊂ E (see

Example 2) is upper semicontinuous.

Exercise 3. Show that the set Q ⊂ R of all rational numbers is not a Gδ-subset of R. Deduce from this circumstance that the characteristic

function fQ, the so-called Dirichlet function, does not belong to Ba1(R, R)

(actually, fQ is of second Baire class).

More generally, prove that if E is an uncountable Polish space without isolated points and X is a countable everywhere dense subset of E, then the characteristic function fX does not belong to Ba1(E, R).

Exercise 4. Let E be a topological space and let f : E → R

be a function. For any x ∈ E, define

Ωf(x) = infV ∈V(x)diam(f (V )),

where V(x) denotes the filter of all neighborhoods of x and diam(f (V )) stands for the diameter of the set f (V ). Denote also by D(f ) the set of all discontinuity points of f .

Verify that:

(a) Ωf(x) = 0 if and only if f is continuous at x (equivalently, Ωf(x) > 0

if and only if x is a discontinuity point of f ); (b) for any t ∈ R, the set

{x ∈ E : Ωf(x) ≥ t}

is closed in E;

where

En = {x ∈ E : Ωf(x) ≥ 1/n}

for each integer n ≥ 1.

Conclude from (b) and (c) that D(f ) is an Fσ-subset of E (therefore,

the set C(f ) of all continuity points of f is a Gδ-subset of E).

Let B be a base of open sets in R and let F = {Y ⊂ R : R \ Y ∈ B}.

Prove the equality

D(f ) = ∪{cl(f−1(Y )) \ f−1(Y ) : Y ∈ F }.

Generalize these results to an arbitrary function f : E → E0, where E0 is a metric space. Some applications of these facts will be presented below (see, for instance,Chapter 7).

Exercise 5. Let E be a topological space. We say that E is resolvable if E admits a representation in the form E = A ∪ B, where A and B are some disjoint everywhere dense subsets of E (this notion was first introduced by Hewitt).

As a rule, topological spaces used in various questions of mathematical analysis turn out to be resolvable. In particular, prove that:

(i) every locally compact topological space without isolated points is resolvable;

(ii) every Hausdorff topological vector space over R (whose dimension is not equal to zero) is resolvable.

Fix a resolvable space E and let X be an Fσ-subset of E. Show that

there exists a function

f : E → R

such that X coincides with the set D(f ) of all discontinuity points of f . To do this, first represent X in the form

X = F1∪ F2∪ ... ∪ Fn∪ ...,

where all sets Fn are closed in E and Fn ⊂ Fn+1 for each natural number

n ≥ 1. Further, put F0= ∅ and, for any integer n ≥ 1, define a function

fn: E → {0, 1}

satisfying the relations:

(a) fn is equal to zero at all points of the set E \ Fn;

(b) Ωfn(x) = 1 if x ∈ Fn.

Now, take a sequence {an: n ≥ 1} of strictly positive real numbers, such

that

an+1+ an+2+ ... + ak+ ... < an (n = 1, 2, ...).

For example, it suffices to put

an =

functions of first baire class 75

Finally, consider the function

f = a1f1+ a2f2+ ... + anfn+ ... .

This function is well defined since the series on the right-hand side of the above equality converges uniformly on E.

Verify that:

(d) for any integer n ≥ 1 and for all points x ∈ Fn\ Fn−1, we have

Ωf(x) ≥ an−

k>n

ak > 0.

Conclude from (c) and (d) that D(f ) = X.

Exercise 6. Let E be a perfectly normal topological space and let f : E → R

be a function whose graph is closed in the product space E×R. Applying the Kuratowski lemma on closed projections (i.e.,Lemma 1 of the Introduction), show that f is of first Baire class.

Give an example of a function f : R → R whose graph is closed in R×R and whose discontinuity points constitute a nonempty perfect set in R.

Recall that a topological space E is Baire if no nonempty open subset of E is of first category in E. For such an E, an important result (due to Baire) is well known, which yields an essential information about the structure of the set D(f ) of discontinuity points of an arbitrary function f : E → R belonging to the class Ba1(E, R).

Theorem 3. Let E be a Baire space and let f ∈ Ba1(E, R). Then

the set D(f ) of all discontinuity points of f is of first category in E. In particular, we have

C(f ) ∩ U 6= ∅ for any nonempty open set U ⊂ E.

Proof. As we know (seeExercise 4 above), the relation C(f ) = E \ D(f ) = \

1≤n<ω

{x ∈ E : Ωf(x) < 1/n}

is valid, where all sets {x ∈ E : Ωf(x) < 1/n} are open in E. So it suffices to

it suffices to show that, for any ε > 0 and for any nonempty open set U ⊂ E, there exists a nonempty open set W ⊂ U such that

(∀x ∈ W )(∀y ∈ W )(|f (x) − f (y)| < ε).

Taking into account the fact that f belongs to Ba1(E, R), choose a sequence

{fk : k < ω} ⊂ Ba0(E, R) satisfying the relation

f (x) = limk→+∞fk(x) (x ∈ E).

Further, for any natural number k, define the set

Xk = {x ∈ E : (∀i ≥ k)(∀j ≥ k)(|fi(x) − fj(x)| ≤ ε/3)}.

All sets Xk (k < ω) are closed in E and

(∀k < ω)(Xk⊂ Xk+1), E = ∪{Xk: k < ω}.

Consequently, we have

U = (U ∩ X0) ∪ (U ∩ X1) ∪ ... ∪ (U ∩ Xk) ∪ ... .

Because E is a Baire space, there exists a natural number n such that int(U ∩ Xn) 6= ∅.

Let V ⊂ U ∩ Xn be a nonempty open subset of E. If x is an arbitrary point

of V , then

(∀i ≥ n)(∀j ≥ n)(|fi(x) − fj(x)| ≤ ε/3).

Putting j = n and tending i to +∞, we get

(∀x ∈ V )(|f (x) − fn(x)| ≤ ε/3).

Therefore, we can write

|f (y) − f (x)| ≤ |f (y) − fn(y)| + |fn(y) − fn(x)| + |fn(x) − f (x)| ≤

2ε/3 + |fn(y) − fn(x)|

for any two points x and y from V . Finally, because fn is a continuous

function, there exists a nonempty open set W ⊂ V such that (∀x ∈ W )(∀y ∈ W )(|fn(y) − fn(x)| < ε/3).

functions of first baire class 77

This gives at once the relation

(∀x ∈ W )(∀y ∈ W )(|f (y) − f (x)| < ε), which ends the proof of the Baire theorem.

Remark 3. The proof presented above is based on the classical argument due to Baire (cf. also [154], [162]).

Remark 4. More general versions of the Baire theorem (with further information about it) can be found in [120].

The next exercise shows that the main part of the Baire theorem remains true for an arbitrary topological space E.

Exercise 7. Let E be a topological space and let f ∈ Ba1(E, R).

Demonstrate that the set D(f ) is of first category in E (for this purpose, use againExercise 4 andLemma 2).

Another way to show this fact is based on the Banach statement (see

Exercise 30 from the Introduction) which leads to a representation of E in the form

E = E0∪ E00_,

where E0 is an open Baire subspace of E and E00 is a first category closed subset of E. ApplyingTheorem 3to the set D(f |E0), we get the required result.

Deduce from this result that if f ∈ Ba1(E, R) and X is an arbitrary

subspace of E, then the set D(f |X) is of first category in X. In particular, claim that if E is a complete metric space, f ∈ Ba1(E, R) and X is a

nonempty closed subspace of E, then there exist points in X at which f |X is continuous.

Exercise 8. According to the definition of Luzin, a topological space X is always of first category (or X is perfectly meager) if each nonempty dense in itself subset of X is of first category in X.

Let E be a topological space and let X be a subspace of E such that, for every perfect set P ⊂ E, the set X ∩ P is of first category in P . Show that the space X is always of first category (apply Exercise 11 from the Introduction).

Luzin proved that there exists an uncountable subspace X of R which is always of first category (see [136] or [120]). Other constructions of uncountable universally small sets can be found in [167] and [235] (cf. also

Exercise 9. Let E be a hereditarily Lindel¨of topological space always of first category and let f : E → R be a function. Demonstrate that, for each subspace X of E, the set D(f |X) is of first category in X. For this purpose, begin with establishing the fact that X admits a representation in the form

X = Y ∪ Z,

where Y is dense in itself, Z is at most countable and Y ∩ Z = ∅. Then verify the inclusion

D(f |X) ⊂ Y ∪ (X0∩ Z),

where X0 denotes the set of all accumulation points of X (in E). Finally, observe that both sets Y and X0∩ Z are of first category in X.

Exercise 10. Let E be a subspace of R satisfying the following relations:

(a) card(E) = c;

(b) E is always of first category.

The existence of such a subspace of R will be demonstrated inChapter 10under Martin’s Axiom (seeTheorem 7 of that chapter).

Prove that there exists a function f : E → R having the following properties:

(d) for any subspace Z of E, the set D(f |Z) is of first category in Z. This classical result is due to Luzin. It shows that property (d) of f does not imply the relation f ∈ Ba1(E, R).

It is useful to compare the above-mentioned result withExercise 13given below.

Exercise 11. Let E be a hereditarily Lindel¨of topological space and let F0⊃ F1⊃ ... ⊃ Fξ ⊃ ... (ξ < ω1)

be a decreasing (with respect to the inclusion relation) ω1-sequence of closed

subsets of E. Prove that there exists an ordinal α < ω1 such that

(∀ξ ∈ [α, ω1[)(Fξ = Fα).

This result is known as the Baire stationarity principle.

In particular, take an arbitrary closed subset X of E and define by transfinite recursion an ω1-sequence {Xξ: ξ < ω1} in the following manner:

functions of first baire class 79

X0= X;

Xξ+1= (Xξ)0 for any ξ < ω1, where (Xξ)0 denotes the set of all accu-

mulation points of Xξ;

Xξ= ∩{Xζ : ζ < ξ} for any limit ordinal ξ < ω1.

Applying the Baire stationarity principle to {Xξ : ξ < ω1}, show that

X admits a representation in the form

X = Y ∪ Z, Y ∩ Z = ∅,

where Y is a perfect subset of E and Z is at most countable (the Cantor- Bendixson theorem).

Give another proof of the same result that does not use the method of transfinite induction. For this purpose, consider the set of all condensation points of X and take it as Y . Then define the set Z by the equality

Z = X \ Y.

Note that a certain generalization of the Baire stationarity principle was obtained by Luzin for decreasing ω1-sequences of Fσ-subsets of R (in this

connection, see [136]).

Exercise 12. Let E be a topological space. We say that E is scattered if E does not contain a nonempty dense in itself subset.

Demonstrate that the following two assertions are equivalent: (a) E is scattered;

(b) E can be represented in the form of an injective α-sequence E = {eξ: ξ < α},

where α is some ordinal and, for any ξ < α, the element eξ is an isolated

point of the set {eζ : ξ ≤ ζ < α}.

Note that the implication (b) ⇒ (a) is trivial. Supposing now that (a) is valid, use the method of transfinite recursion for obtaining the required representation of E.

Finally, demonstrate that every topological space X can be represented in the form

X = Y ∪ Z,

where Y is a perfect subset of X, Z is a scattered subset of X and Y ∩ Z = ∅ (this classical result is due to Cantor).

Exercise 13. Let E be a separable metric space and let g : E → R be a function such that, for every nonempty closed set F ⊂ E, there exists a

point x ∈ F at which the restricted function g|F is continuous. Prove that g is of first Baire class (this result is due to Baire).

The following argument enables to establish the above-mentioned result. Take any a ∈ R and b ∈ R such that a < b, and denote

A = {x ∈ E : g(x) > a}, B = {x ∈ E : g(x) < b}. Clearly, we have the equality

E = A ∪ B.

Further, construct by transfinite recursion an ω1-sequence

F0⊃ F1⊃ ... ⊃ Fξ ⊃ ... (ξ < ω1)

of closed subsets of E. Put F0= E. Suppose that, for a given ξ < ω1, the

partial family {Fζ : ζ < ξ} has already been defined.

If ξ is a limit ordinal, then we put

Fξ = ∩{Fζ : ζ < ξ}.

If ξ = η + 1, consider the set Fη. Only two cases are possible.

1. Fη= ∅. In this case, we define

Fξ = Fη = ∅.

2. Fη6= ∅. In this case, there exists a point x ∈ Fηat which the function

g|Fη is continuous. Consequently, there exists an open neighborhood V (x)

of x such that

Fη∩ V (x) ⊂ A ∨ Fη∩ V (x) ⊂ B.

Then we define

Fξ = Fη\ V (x).

Proceeding in this way, we will be able to construct the sets Fξ (ξ < ω1).

Observe now that, for each ordinal ξ < ω1, we have

Fξ\ Fξ+1⊂ A ∨ Fξ\ Fξ+1⊂ B

and, according to the Baire stationarity principle, for some α < ω1, the

equalities

∅ = Fα= Fα+1= ... = Fξ= ... (α ≤ ξ < ω1)

are valid. Deduce from these facts that there exist two sets A0 _{and B}0 _such

that

functions of first baire class 81

and both A0 and B0 are Fσ-subsets of E.

Now, let {bn: n < ω} be a strictly decreasing sequence of real numbers

satisfying the relation limn→+∞bn = a. Put again

A = {x ∈ E : g(x) > a} and, for each n < ω, define

Bn= {x ∈ E : g(x) < bn}.

As above, show the existence of Fσ-sets A0n and Bn0 such that

A0_n⊂ A, Bn0 ⊂ Bn, An0 ∪ Bn0 = E, A0n∩ Bn0 = ∅.

Finally, denote

X = ∪{A0_n: n < ω}. Verify that

X = {x ∈ E : g(x) > a} = A and hence A is an Fσ-subset of E.

By using a similar argument, demonstrate that B = {x ∈ E : g(x) < b} is an Fσ-subset of E, too.

Conclude, in view of the Lebesgue theorem (i.e., Theorem 2 of this

In document Strange Functions in Real Analysis, Second Edition. Pure and Applied Mathematics. A. B. Kharazishvili.pdf (Page 74-100)