# Properties of equi-Baire 1 and equi-Lebesgue families of functions

Marek Balcerzak  Institute of Mathematics, Lodz University of Technology, al.
Politechniki 8, 93-590 Łódź, Poland<EMAIL_ADDRESS>,  Ľubica Holá
Academiy of Sciences, Institute of Mathematics, Štefánikova 49, 814 73
Bratislava, Slovakia<EMAIL_ADDRESS>and  Dušan Holý  Department of
Mathematics and Computer Science, Faculty of Education, Trnava University,
Priemyselná, 918 43 Trnava, Slovakia<EMAIL_ADDRESS>
###### Abstract.

We study several properties of equi-Baire 1 families of functions between
metric spaces. We consider the related equi-Lebesgue property for such
families. We examine the behaviour of equi-Baire 1 and equi-Lebesgue families
with respect to pointwise and uniform convergence. In particular, we obtain a
criterion for a choice of a uniformly convergent subsequence from a sequence
of functions that form an equi-Baire 1 family, which solves a problem posed in
[3]. Finally, we discuss the notion of equi-cliquishness and relations between
equi-Baire 1 families and sets of equi-continuity points.

###### Key words and phrases:

equi-Baire 1 families, equi-Lebesgue families, pointwise convergence, uniform
convergence, equi-cliquishness, equi-continuity points

###### 2020 Mathematics Subject Classification:

54H05, 26A21

##  1\. Introduction

The notion of equi-Baire 1 families of functions introduced by Lecomte [15]
and rediscovered by Alikhani-Koopaei [1] was motivated by the epsilon-delta
characterization of Baire 1 functions due to Lee, Tang and Zhao [17]. Another
old characterization of Baire 1 functions, due to Lebesgue [14], turns out
useful in this context since, in some general cases, equi-Baire 1 families
coincide with equi-Lebesgue ones. The notion of equi-Lebesgue family is
introduced in our paper, however it appeared without this name in [15,
Proposition 32] and [1, Theorem 3.6]).

Equi-Baire 1 families of functions have been recently studied in [3]. In our
paper we continue these studies. We weaken some assumptions in several results
of [3]. We consider properties concerning pointwise and uniform convergence.
In particular, we present a complete solution of the following Question 4.3
posed in [3]:

> Given a uniformly bounded sequence $(f_{n})$ of real-valued Baire 1
> functions defined on a Polish space $X$, assume that the family
> $\\{f_{n}:n\in\mathbb{N}\\}$ is equi-Baire 1. What more should be assumed to
> guarantee that $(f_{n})$ contains a subsequence which is uniformly
> convergent on $X$?

Then we investigate other related properties. We prove the following result:
Let $(X,d_{X})$ be a separable Baire metric space and $(Y,d_{Y})$ be a metric
space. If $\mathcal{F}$ is an equi-Baire 1 family of functions from $X$ to
$Y$, then its set $EC(\mathcal{F})$ of equi-continuity points is dense
$G_{\delta}$. Also we consider the reverse situation where functions from a
$G_{\delta}$ set $H\subset X$ to $Y$, that are equi-continuous on $H$, can be
extended to functions from $X$ to $Y$ that form an equi-Baire 1 family.

In what follows let $\mathbb{N}$ be the set of positive integers, $\mathbb{R}$
be the space of real numbers with the usual Euclidean metric and
$\mathbb{R}^{+}$ be the set of positive reals.

If $(Y,d)$ is a metric space, the open $d$-ball with center $y_{0}\in Y$ and
radius $\varepsilon>0$ will be denoted by $S(y_{0},\varepsilon)$ and the
closed $d$-ball with center $y_{0}\in Y$ and radius $\varepsilon>0$ will be
denoted by $B(y_{0},\varepsilon)$. For a nonempty set $A\subset Y$, the symbol
$S(A,\varepsilon)$ denotes the generalized open ball
$\\{y\colon\operatorname{dist}(y,A)<\varepsilon\\}$. The closure of $A\subset
Y$ will be denoted by $\overline{A}$. Given $\varepsilon>0$, we say that a set
$A\subset Y$ is $\varepsilon$-discrete, if $d(x,y)\geq\varepsilon$ for any
distinct $x,y\in A$. A topological space is called Polish if it is separable
and completely metrizable.

The paper is organized as follows. In Section 1 we recall two
characterizations of Baire 1 functions between metric spacess $X$ and $Y$, and
establish minimal assumptions on $X$ and $Y$ under which these
characterizations work. In Section 2 we define equi-Baire 1 and equi-Lebesgue
families of functions and discuss relations between them. Sections 4 and 5 are
devoted to results concerning the behaviour of equi-Baire 1 and equi-Lebesgue
families with respect to pointwise and uniform convergence, respectively. The
final Section 6 contains other related results. We discuss the notion of equi-
cliquishness and relations between equi-Baire 1 families and sets of equi-
continuity points.

##  2\. Two characterizations of Baire 1 functions

A function $f$ between metric spaces $X$ and $Y$ is called Baire 1 if is
$F_{\sigma}$-measurable, that is, whenever the preimage of any open set is
$F_{\sigma}$ (cf. [11, Def. 24.1]). Several characterizations of Baire 1
functions are known; some of them are due to Baire, see [11, Thm 24.15].
Usually some additional properties of $X$ and $Y$ are assumed. We will focus
on two characterizations: one classical characterization is due to Lebesgue
[14], and the other one was proposed quite recently in [17] and [6].

The Lebesgue characterization was originally formulated for real-valued Borel
measurable functions of arbitrary class $\alpha>0$. A more general version can
be found in [13]. Below, we present it for $\alpha=1$, in a bit modified form,
and we give its proof for a reader’s convenience.

######  Theorem 2.1.

(cf. [13, p. 375]) Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces. For
$f\colon X\to Y$ consider the following conditions:

  * (1)

A function $f$ is Baire 1;

  * (2)

For each $\varepsilon>0$ there is a cover $(X_{i})_{i\in\mathbb{N}}$ of $X$
consisting of closed sets such that
$\operatorname{diam}f(X_{i})\leq\varepsilon$, for all $i\in\mathbb{N}$.

Then (2) $\Rightarrow$ (1) holds. If $(Y,d_{Y})$ is separable, then (1)
$\Rightarrow$ (2) is true.

###### Proof.

(1) $\Rightarrow$ (2): Let $f$ be Baire 1. Since $(Y,d_{Y})$ is separable, for
a fixed $\varepsilon>0$ we can find open balls $B_{1},B_{2},\dots$ in $Y$ such
that $Y=\bigcup_{n\in\mathbb{N}}B_{n}$ and
$\operatorname{diam}B_{n}\leq\varepsilon$ for all $n\in\mathbb{N}$. For
$n\in\mathbb{N}$, the set $f^{-1}(B_{n})$ is of type $F_{\sigma}$, so it can
be expressed as the union $\bigcup_{j\in\mathbb{N}}F_{n,j}$ of closed sets
$F_{n,j}$, $j\in\mathbb{N}$. Enumerate $\\{F_{n,j}:n,j\in\mathbb{N}\\}$ as
$\\{X_{i}:i\in\mathbb{N}\\}$ and observe that the respective condition in (2)
holds.

(2) $\Rightarrow$ (1): By (2), for each $k\in\mathbb{N}$ we find closed sets
$X_{n,k}$, $n\in\mathbb{N}$ such that $X=\bigcup_{n\in\mathbb{N}}X_{k,n}$ and
$\operatorname{diam}f(X_{k,n})<1/k$ for all $n\in\mathbb{N}$. Let $U$ be an
open set in $Y$. Define $\mathcal{F}=\\{X_{k,n}:f(X_{k,n})\subset U\\}$. It is
enough to show that $f^{-1}(U)=\bigcup\mathcal{F}$ since then $f^{-1}(U)$ is
of type $F_{\sigma}$.

Firstly, if $x\in\bigcup\mathcal{F}$, then $x\in X_{k,n}$ for some
$k,n\in\mathbb{N}$ with $f(X_{k,n})\subset U$. Hence $f(x)\in U$ and so, $x\in
f^{-1}(U)$. Secondly, if $f(x)\in U$, we can find $k\in\mathbb{N}$ such that
every $y\in Y$ with $d_{Y}(f(x),y)<1/k$ is in $U$. Find $n\in\mathbb{N}$ such
that $x\in X_{k,n}$. Since $\operatorname{diam}f(X_{k,n})<1/k$, we have
$f(X_{k,n})\subset U$. Hence $x\in\mathcal{F}$ as desired. ∎

If a function $f$ between metric spaces $X$ and $Y$ satisfies condition (2)
from Theorem 2.1, we will say that $f$ has the Lebesgue property.

Thus, by implication (2) $\Rightarrow$ (1) in Theorem 2.1, if $f\colon X\to Y$
has the Lebesgue property, then it is Baire 1. We observe that the
separability of $Y$ is necessary for implication (1) $\Rightarrow$ (2) in
Theorem 2.1.

######  Proposition 2.1.

Let $(Y,d_{Y})$ be a metric space. The following are equivalent:

  * (i)

$(Y,d_{Y})$ is separable;

  * (ii)

For each metric space $(X,d_{X})$, every Baire 1 function $f\colon X\to Y$ has
the Lebesgue property.

###### Proof.

We only need to show (ii) $\Rightarrow$ (i). Suppose that $(Y,d_{Y})$ is not
separable. Then there is $\varepsilon>0$ and an uncountable
$\varepsilon$-discrete set $L\subset Y$. Put $L=\\{y_{i}:i\in I\\}$. Then
$\\{B(y_{i},\varepsilon/4):i\in I\\}$ is an uncountable family of pairwise
disjoint closed balls.

Let $X=Y$, $d_{X}=d_{Y}$ and $f\colon X\to Y$ be the identity function. Then
$f$ is Baire 1. By (ii) there is a cover $(X_{n})_{n\in\mathbb{N}}$ of $X$
consisting of closed sets such that
$\operatorname{diam}f(X_{n})\leq\varepsilon/4$, for all $n\in\mathbb{N}$.
There must exist $m\in\mathbb{N}$ such that $X_{m}\cap
B(y_{i},\varepsilon/4)\neq\emptyset$ for uncountable many balls
$B(y_{i},\varepsilon/4)$, a contradiction. ∎

Another characterization of Baire 1 functions, formulated in epsilon-delta
style, is due to Lee, Thang and Zhao [17]. Their result concerns the case of
functions between Polish spaces. The following generalization was obtained by
Fenecios and Cabral in [6].

######  Theorem 2.2.

[6] Let $(X,d_{X})$ and $(Y,d_{Y})$ be separable metric spaces. The following
are equivalent:

  * (1)

A function $f\colon X\to Y$ is Baire 1;

  * (2)

For each $\varepsilon>0$ there is a function $\delta:X\to\mathbb{R}^{+}$ such
that for any $x,y\in X$, if $d_{X}(x,y)<\min\\{\delta(x),\delta(y)\\}$, then
$d_{Y}(f(x),f(y))<\varepsilon$.

If a function $f$ between metric spaces $X$ and $Y$ satisfies condition (2)
from Theorem 2.2, we will say that $f$ has LTZ-property. What is interesting,
the proof of Theorem 5 in [6] uses the Lebesgue property and a careful
analysis of this proof shows that the following implications are valid.

######  Proposition 2.2.

Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and $f\colon X\to Y$.

  * (i)

If $X$ is separable and $f$ has LTZ-property, then $f$ has the Lebesgue
property.

  * (ii)

If $f$ has the Lebesgue property, then $f$ has LTZ-property.

######  Remark 2.1.

Using Theorem 2.1 and Proposition 2.2, we obtain Theorem 2.2 where the
separability of $Y$ can be omitted. Also, the separability of $X$ is needed
only for implication (2) $\Rightarrow$ (1) in Theorem 2.2.

We observe that the separability of $X$ is necessary for implication in
Proposition 2.2 (i).

######  Proposition 2.3.

Let $(X,d_{X})$ be a metric space. The following are equivalent:

  * (1)

$(X,d_{X})$ is separable;

  * (2)

For each metric space $(Y,d_{Y})$, every function $f\colon X\to Y$ with LTZ-
property has the Lebesgue property.

###### Proof.

We only need to show (2) $\Rightarrow$ (1). We mimic the proof of Proposition
2.1. Suppose that $(X,d_{X})$ is not separable. Then there is $\varepsilon>0$
and an uncountable $\varepsilon$-discrete set $L\subset X$. Put
$L=\\{x_{i}:i\in I\\}$. Then $\\{B(x_{i},\varepsilon/4):i\in I\\}$ is an
uncountable family of pairwise disjoint closed balls.

Put $Y=X$, $d_{Y}=d_{X}$ and let $f\colon X\to Y$ be the identity function
which surely has LTZ-property. Thus by (2) it has the Lebesgue property.
However, for $\varepsilon/4$ there is no cover $(X_{i})_{i\in\mathbb{N}}$ of
$X$ consisting of closed sets such that
$\operatorname{diam}f(X_{i})\leq\varepsilon/4$ for all $i\in\mathbb{N}$, a
contradiction. ∎

######  Remark 2.2.

Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and $(X,d_{X})$ is hereditary
Baire. By [10, Thm C], a function $f\colon X\to Y$ is Baire 1 if and only if
it has LTZ-property.

##  3\. Equi-Baire 1 and equi-Lebesgue families

The characterization of Baire 1 functions by LTZ-property motivated Lecomte in
his paper [15] to introduce the notion of equi-Baire 1 family of functions.
Then it was rediscovered by Alighani-Koopaei [1] who showed its applications
in dynamical systems; see also his recent paper [2].

######  Definition 3.1.

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces. We say that a family
$\mathcal{F}$ of functions from $X$ to $Y$ is equi-Baire 1 if, for each
$\varepsilon>0$, there exists a function $\delta\colon X\to\mathbb{R}^{+}$
such that, for all $f\in\mathcal{F}$ and $x,y\in X$ the condition
$d_{X}(x,y)<\min\\{\delta(x),\delta(y)\\}$ implies
$d_{Y}(f(x),f(y))\leq\varepsilon$.

######  Remark 3.1.

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces.

  * (i)

By Proposition 2.2(i), if $(X,d_{X})$ is separable, then every member of an
equi-Baire 1 family of functions from $X$ to $Y$ is a Baire 1 function.

  * (ii)

By Remark 2.2, if $(X,d_{X})$ is hereditarily Baire, then every member of an
equi-Baire 1 family of functions from $X$ to $Y$ is a Baire 1 function.

We suggest the following definition of an equi-Lebesgue family.

######  Definition 3.2.

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces. We say that a family
$\mathcal{F}$ of functions from $X$ to $Y$ is equi-Lebesgue if, for every
$\varepsilon>0$, there is a cover $(X_{i})_{i\in\mathbb{N}}$ of $X$ consisting
of closed sets such that $\operatorname{diam}f(X_{i})\leq\varepsilon$ for all
$i\in\mathbb{N}$ and $f\in\mathcal{F}$.

An advantage of the notion of an equi-Lebesgue family is that every member of
an equi-Lebesgue family of functions from $X$ to $Y$ is a Baire 1 function
(see Theorem 2.1).

From the proof of Theorem 3.6 in [1] we can deduce a general fact where the
assumptions on spaces $X$ and $Y$ are weakened.

######  Fact 3.1.

[1, Theorem 3.6] Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces and
$\mathcal{F}$ be a family of functions from $X$ to $Y$.

  * (i)

If $\mathcal{F}$ is equi-Lebesgue, then it is equi-Baire 1.

  * (ii)

Assume that $(X,d_{X})$ is separable. If $\mathcal{F}$ is equi-Baire 1, then
$\mathcal{F}$ is equi-Lebesgue.

The separability of $X$ in Fact 3.1 (ii) is necessary. Indeed, we use the idea
from the proof of Proposition 2.3. Assume that $(X,d_{X})$ is not separable
and take the identity function $f\colon X\to X$. We know that $f$ is a
function with the LTZ-property and without the Lebesgue property. Thus
$\mathcal{F}=\\{f\\}$ is equi-Baire 1 and it is not equi-Lebesgue.

##  4\. Properties concerning pointwise convergence

Let $X$ be a topological space, $(Y,d)$ be a metric space and $Y^{X}$ be the
space of all functions from $X$ to $Y$. We use some notions of general
topology, for instance, uniformities and nets. For more information on them,
we refer the reader to [12].

Recall that the topology $\tau_{p}$ of pointwise convergence on $Y^{X}$ is
induced by the uniformity $\mathfrak{U}_{p}$ of pointwise convergence which
has a base consisting of sets of the form

$W(A,\varepsilon)=\\{(f,g):\ \forall\ x\in A\ \ d(f(x),g(x))<\varepsilon\\},$

where $A$ is a finite subset of $X$ and $\varepsilon>0$. The general
$\tau_{p}$-basic neighborhood of $f\in Y^{X}$ will be denoted by
$W(f,A,\varepsilon)$, i.e. $W(f,A,\varepsilon)=W(A,\varepsilon)[f]$. In other
words,

$W(f,A,\varepsilon)=\\{g\in Y^{X}:\ d(f(x),g(x))<\varepsilon\ \text{for
every}\ x\in A\\}.$

The following result is classical.

######  Theorem 4.1.

[12, Chapter 7, Theorem 14] Let $X$ be a topological space and $(Y,d)$ be a
metric space. If a family $\mathcal{F}$ of functions from $X$ to $(Y,d)$ is
equicontinuous at $x\in X$, then the closure of $\mathcal{F}$ relative to the
topology of pointwise convergence is also equicontinuous at $x$.

Thus if a net $(f_{\sigma})$ of continuous functions from a topological space
$X$ to a metric space $(Y,d)$ pointwise converges to a function $f\colon X\to
Y$ and the net $(f_{\sigma})$ is equicontinuous, then $f$ is continuous. We
will find an analogy for Baire 1 functions.

The following proposition is easy to verify.

######  Proposition 4.1.

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces. Let $\mathcal{F}$ be an
equi-Lebesgue family of functions from $X$ to $Y$. The closure of
$\mathcal{F}$ relative to the topology of pointwise convergence is equi-
Lebesgue too.

We have the following corollaries.

######  Corollary 4.1.

Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and $(f_{\sigma})$ be a net of
Baire 1 functions from $X$ to $Y$ which pointwise converges to a function
$f\colon X\to Y$. If the net $(f_{\sigma})$ is equi-Lebesgue, then $f$ is
Baire 1.

###### Proof.

Use Proposition 4.1 and Theorem 2.1. ∎

######  Corollary 4.2.

Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and $X$ be separable. Let
$(f_{\sigma})$ be a net of Baire 1 functions from $X$ to $Y$ which pointwise
converges to a function $f\colon X\to Y$. If the net $(f_{\sigma})$ is equi-
Baire 1, then $f$ is Baire 1.

###### Proof.

Use Fact 3.1 (ii) and Corollary 4.1. ∎

As Example 4.1 in [3] shows, the condition of equi-Lebesgue (equi-Baire 1) is
not necessary to guarantee that the pointwise limit of a sequence of Baire 1
functions is Baire 1. This example uses $X=Y=\mathbb{R}$ with the Euclidean
metric. Let $\\{q_{n}:n\in\mathbb{N}\\}$ be an enumeration of rational
numbers. For every $n\in\mathbb{N}$ put $f_{n}=\chi_{\\{q_{n}\\}}$. The
sequence $(f_{n})$ is pointwise convergent to the zero function. However,
$\\{f_{n}:n\in\mathbb{N}\\}$ is not equi-Baire 1, and it is not equi-Lebesgue
by Fact 3.1 (i).

This example can be generalized as follows.

######  Proposition 4.2.

Let $(X,d)$ be a metric space. Consider a non-empty family $\mathcal{A}$ of
pairwise disjoint subsets of $(X,d)$. Then the family
$\\{\chi_{A}:A\in\mathcal{A}\\}$ is equi-Lebesgue if and only if $\mathcal{A}$
is countable, each element of $\mathcal{A}$ is $F_{\sigma}$ and
$\bigcup\mathcal{A}$ is $G_{\delta}$.

###### Proof.

Suppose that $\mathcal{A}$ is a family of pairwise disjoint subsets of $(X,d)$
such that $\mathcal{A}$ is countable, each element of $\mathcal{A}$ is
$F_{\sigma}$ and $\bigcup\mathcal{A}$ is $G_{\delta}$. Then it is easy to
verify that the family $\\{\chi_{A}:A\in\mathcal{A}\\}$ is equi-Lebesgue.

Suppose now that $\mathcal{A}$ is a family of pairwise disjoint subsets of
$(X,d)$ such that the family $\\{\chi_{A}:A\in\mathcal{A}\\}$ is equi-
Lebesgue. Since for every $A\in\mathcal{A}$ the function $\chi_{A}$ must be
Baire 1, every $A\in\mathcal{A}$ has to be $F_{\sigma}$. By the assumption
there is a countable cover $\mathcal{C}$ of $X$ consisting of closed sets such
that $\operatorname{diam}\chi_{A}(C)<1/2$, for all $C\in\mathcal{C}$ and
$A\in\mathcal{A}$. Suppose that $\mathcal{A}$ is uncountable. There must exist
$C\in\mathcal{C}$ such that $C\cap A\neq\emptyset$ for uncountable many
$A\in\mathcal{A}$, a contradiction.

Let us show that $\bigcup\mathcal{A}$ is $G_{\delta}$. Since $\mathcal{C}$ is
a cover of $X$, for every $A\in\mathcal{A}$ there is $C\in\mathcal{C}$ such
that $A\cap C\neq\emptyset$, and this implies that $C\subset A$ because
$\operatorname{diam}\chi_{A}(C)<1/2$ means that $\chi_{A}(C)=\\{1\\}$. Hence
for every $A\in\mathcal{A}$ we have $A=\bigcup\\{C\in\mathcal{C}:C\subset
A\\}$. Thus $X\setminus\bigcup\mathcal{A}$ has to be an $F_{\sigma}$ set. ∎

Note that Proposition 4.2 recovers Corollary 3.17 in [3] which however works
when $X$ is a separable hereditary Baire metric space since its proof is a
conequence of Theorem 3.16 in [3]. Here we deal with a more general case and
we give a direct proof.

######  Proposition 4.3.

Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and let $(f_{n})$ be a sequence
of functions $f_{n}:X\to Y$, $n\in\mathbb{N}$, that form an equi-Lebesgue
family. If $\overline{\\{f_{n}(x):n\in\mathbb{N}\\}}$ is compact in
$(Y,d_{Y})$ for every $x\in X$, then there exists a subsequence $(f_{n_{k}})$
which is pointwise convergent to a Baire 1 function.

###### Proof.

For every $x\in X$ put $K_{x}=\overline{\\{f_{n}(x):n\in\mathbb{N}\\}}$. The
set $\Pi_{x\in X}K_{x}$ is compact and $f_{n}\in\Pi_{x\in X}K_{x}$ for every
$n\in\mathbb{N}$. There is a function $f\in\Pi_{x\in X}K_{x}$ such that $f$ is
a cluster point of $(f_{n})$ in $\Pi_{x\in X}K_{x}$. By Proposition 4.1,
$\mathcal{F}=\\{f_{n}:n\in\mathbb{N}\\}\cup\\{f\\}$ is equi-Lebesgue and thus
$f$ is Baire 1.

Since $\mathcal{F}$ is equi-Lebesgue, for every $k\in\mathbb{N}$ there is a
cover $(X_{i}^{k})_{i\in\mathbb{N}}$ of $X$ consisting of closed sets such
that $\operatorname{diam}g(X_{i}^{k})\leq 1/k$ for all $i\in\mathbb{N}$ and
$g\in\mathcal{F}$. There is a countable set $H\subset X$ such that $H\cap
X_{i}^{k}\neq\emptyset$ for all $k,i\in\mathbb{N}$.

The set $\Pi_{x\in H}K_{x}$ is first countable (even metrizable) and $f|H$ is
a cluster point of $(f_{n}|H)$ in $\Pi_{x\in H}K_{x}$. Thus there is a
subsequence $(f_{n_{k}}|H)$ which converges to $f|H$ in $\Pi_{x\in H}K_{x}$.
We will show that the sequence $(f_{n_{k}})$ pointwise converges to $f$.

Let $x\in X$ and $\varepsilon>0$. Let $k\in\mathbb{N}$ be such that
$3/k<\varepsilon$. There is a cover $(X_{i}^{k})_{i\in\mathbb{N}}$ of $X$
consisting of closed sets such that $\operatorname{diam}g(X_{i}^{k})\leq 1/k$
for all $i\in\mathbb{N}$ and $g\in\mathcal{F}$. Let $i\in\mathbb{N}$ be such
that $x\in X_{i}^{k}$. Let $z\in X_{i}^{k}\cap H$. There is
$l_{0}\in\mathbb{N}$ such that $d_{Y}(f_{n_{l}}(z),f(z))<1/k$ for every $l\geq
l_{0}$. Let $l\geq l_{0}$. Then

$d_{Y}(f_{n_{l}}(x),f(x))\leq
d_{Y}(f_{n_{l}}(x),f_{n_{l}}(z))+d_{Y}(f_{n_{l}}(z)),f(z))+d_{Y}(f(z),f(x))<3/k<\varepsilon.$

∎

We have the following generalization of Proposition 4.5 in [3].

######  Corollary 4.3.

([3] for $Y=\mathbb{R}$) Let $(X,d_{X})$ be a separable metric space,
$(Y,d_{Y})$ be a metric space and let $(f_{n})$ be a sequence of functions
$f_{n}:X\to Y$, $n\in\mathbb{N}$, that form an equi-Baire 1 family. If
$\overline{\\{f_{n}(x):n\in\mathbb{N}\\}}$ is compact in $(Y,d_{Y})$ for every
$x\in X$, then there exists a subsequence $(f_{n_{k}})$ which is pointwise
convergent to a Baire 1 function.

###### Proof.

From Fact 3.1 we know that if $(X,d_{X})$ is separable then the notions of
equi-Baire 1 family and equi-Lebesgue family coincide. Thus we can use
Proposition 4.3. ∎

The following result was obtained as Theorem 3.3 in [3] where $X$ and $Y$ are
Polish spaces. By Fact 3.1 we observe that the same proof works if $X$ is only
a metric space and $Y$ is a separable metric space.

######  Proposition 4.4.

Let $(X,d_{X})$ be a metric space and $(Y,d_{Y})$ be a separable metric space.
Let $\mathcal{F}=\\{f_{n}:n\in\mathbb{N}\\}$ be a family of continuous
functions from $X$ to $Y$, such that the sequence $(f_{n})$ is pointwise
convergent on $X$. Then $\mathcal{F}$ is an equi-Lebesgue family.

##  5\. Properties concerning uniform convergence

Let us define the topology $\tau_{U}$ of uniform convergence on $Y^{X}$. This
topology is induced by the uniformity $\mathfrak{U}_{U}$ which has a base
consisting of sets of the form

$W(\varepsilon)=\\{(f,g):\ \forall\ x\in X\ \ d(f(x),g(x))<\varepsilon\\},$

where $\varepsilon>0$. The general $\tau_{U}$-basic neighborhood of $f\in
Y^{X}$ will be denoted by $W(f,\varepsilon)$, i.e.
$W(f,\varepsilon)=W(\varepsilon)[f]$. In other words,

$W(f,\varepsilon)=\\{g\in Y^{X}:\ d(f(x),g(x))<\varepsilon\ \text{for every}\
x\in X\\}.$

It is easy to verify that the countable family $\\{W(1/n):n\in\mathbb{N}\\}$
is a base of the uniformity $\mathfrak{U}_{U}$. Thus the uniformity
$\mathfrak{U}_{U}$ is metrizable [12].

Finally, let $(X,d)$ be a metric space. We will define the topology
$\tau_{UB}$ of uniform convergence on bounded sets on $Y^{X}$. This topology
is induced by the uniformity $\mathfrak{U}_{UB}$ which has a base consisting
of sets of the form

$W(B,\varepsilon)=\\{(f,g):\ \forall\ x\in B\ \ d(f(x),g(x))<\varepsilon\\},$

where $B$ is a bounded subset of $X$ and $\varepsilon>0$. A general
$\tau_{UB}$-basic neighborhood of $f\in Y^{X}$ will be denoted by
$W(f,B,\varepsilon)$, i.e. $W(f,B,\varepsilon)=W(B,\varepsilon)[f]$. In other
words,

$W(f,B,\varepsilon)=\\{g\in Y^{X}:\ d(f(x),g(x))<\varepsilon\ \text{for
every}\ x\in B\\}.$

It is easy to verify that the uniformity $\mathfrak{U}_{UB}$ is metrizable.
Let $x_{0}$ be an arbitrary point in $X$. For every $n\in\mathbb{N}$ put
$B_{n}=B(x_{0},n)$. Then the countable family
$\\{W(B_{n},1/k):n,k\in\mathbb{N}\\}$ is a base of the uniformity
$\mathfrak{U}_{UB}$.

######  Proposition 5.1.

Let $(f_{n})$ be a sequence of bounded real-valued Baire 1 functions defined
on a metric space $X$. If $(f_{n})$ is uniformly convergent to $f\colon
X\to\mathbb{R}$, then for every $\varepsilon>0$ there is a finite cover
$\mathcal{B}$ of $X$ consisting of subsets of $X$ which are $F_{\sigma}$ and
such that $\operatorname{diam}f_{n}(B)\leq\varepsilon$ for all
$B\in\mathcal{B}$ and $n\in\mathbb{N}$.

###### Proof.

Let $\varepsilon>0$. Put $\mathcal{F}=\\{f,f_{1},f_{2},\dots,f_{n},\dots\\}$.
Then $\mathcal{F}$ is a compact set in $(\mathbb{R}^{X},\tau_{U})$. There are
functions $f_{k_{1}},f_{k_{2}},\dots f_{k_{n}}\in\mathcal{F}$ such that

$\mathcal{F}\subseteq\bigcup\\{W(f_{k_{j}},\varepsilon/3):j\in\\{1,\dots,n\\}\\}.$

For simplicity put $g_{j}=f_{k_{j}}$ for every $j\in\\{1,\dots,n\\}.$ Let
$M>0$ be such that

$\bigcup\\{g_{j}(X):j\in\\{1,\dots,n\\}\\}\subset[-M,M].$

Let $\mathcal{V}=\\{V_{1},\dots,V_{m}\\}$ be a finite open cover of $[-M,M]$,
where the diameter of members of this cover is less than $\varepsilon/3$. For
each $i\in\\{1,\dots,m\\}$ and $j\in\\{1,\dots,n\\}$ let
$B_{i}^{j}=g_{j}^{-1}(V_{i})$. Since $g_{j}$ is a Baire 1 function, the set
$B_{i}^{j}$ is $F_{\sigma}$ for all $j\in\\{1,\dots,n\\}$ and
$i\in\\{1,\dots,m\\}$. Put

$\mathcal{B}^{\prime}=\\{B_{i_{1}}^{1}\cap\dots\cap
B_{i_{n}}^{n}:i_{j}\in\\{1,\dots,m\\}\ \text{for each}\
j\in\\{1,\dots,n\\}\\}.$

Every set $B\in\mathcal{B}^{\prime}$ is of type $F_{\sigma}$. Let
$\mathcal{B}$ be the family consisting of all nonempty sets from
$\mathcal{B}^{\prime}$. Clearly $X=\bigcup\mathcal{B}$. One can check that
$|f_{i}(p)-f_{i}(q)|<\varepsilon$ for all $f_{i}\in\mathcal{F}$,
$B\in\mathcal{B}$ and $p,q\in B$. ∎

######  Theorem 5.1.

Let $(f_{n})$ be a sequence of bounded real-valued Baire 1 functions defined
on a Baire metric space $X$. If $(f_{n})$ is uniformly convergent to $f\colon
X\to\mathbb{R}$, then for every $\varepsilon>0$ there is a finite cover
$\mathcal{B}$ of $X$ consisting of subsets of $X$ which are either open or
$F_{\sigma}$ nowhere dense and such that
$\operatorname{diam}f_{n}(B)\leq\varepsilon$ for all $B\in\mathcal{B}$ and
$n\in\mathbb{N}$.

###### Proof.

Let $\varepsilon>0$. Put $\mathcal{F}=\\{f,f_{1},f_{2},\dots,f_{n},\dots\\}$.
Then $\mathcal{F}$ is a compact set in $(\mathbb{R}^{X},\tau_{U})$. There are
functions $f_{k_{1}},f_{k_{2}},\dots,f_{k_{n}}\in\mathcal{F}$ such that

$\mathcal{F}\subseteq\bigcup\\{W(f_{k_{j}},\varepsilon/3):j\in\\{1,\dots,n\\}\\}.$

For simplicity put $g_{j}=f_{k_{j}}$ for every $j\in\\{1,\dots,n\\}\\}.$ Let
$M>0$ be such that

$\bigcup\\{g_{j}(X):j\in\\{1,\dots,n\\}\\}\subset[-M,M].$

Let $\mathcal{V}=\\{V_{1},\dots,V_{m}\\}$ be a finite open cover of $[-M,M]$,
where the diameter of members of this cover is less than $\varepsilon/3$. For
each $i\in\\{1,\dots,m\\}$ and $j\in\\{1,\dots,n\\}$ let

$L_{i}^{j}=\\{z\in C(g_{j}):g_{j}(z)\in V_{i}\\},$

where $C(g_{j})=\\{z\in X:\ g_{j}\ \text{is continuous at}\ z\\}$. Since $X$
is a Baire space, $C(g_{j})$ is a dense $G_{\delta}$ set in $X$ for every
$j\in\\{1,\dots,n\\}$.

For each $z\in L_{i}^{j}$, let $W_{i}^{j}(z)$ be an open neighbourhood of $z$
such that $g_{j}(W_{i}^{j}(z))\subset V_{i}$. Let

$W_{i}^{j}=\bigcup\\{W_{i}^{j}(z):\ z\in L_{i}^{j}\\}$

and

$\mathcal{U}=\\{W_{i_{1}}^{1}\cap\dots\cap
W_{i_{n}}^{n}:i_{j}\in\\{1,\dots,m\\}\ \text{for each}\
j\in\\{1,\dots,n\\}\\}.$

For each $U\in\mathcal{U}$, let

$P_{i}^{j}(U)=\\{z\in\overline{U}\setminus U:\ g_{j}(z)\in V_{i}\\}$

and

$\mathcal{P}=\\{P_{i_{1}}^{1}(U)\cap\dots\cap
P_{i_{n}}^{n}(U):U\in\mathcal{U}\text{ and }i_{j}\in\\{1,\dots,m\\}\ \text{for
each}\ j\in\\{1,\dots,n\\}\\}.$

Since $g_{j}$ is a Baire 1 function, the set $P_{i}^{j}(U)$ is of type
$F_{\sigma}$ for all $j\in\\{1,\dots,n\\}$, $i\in\\{1,\dots,m\\}$ and
$U\in\mathcal{U}$. Thus also every set $P\in\mathcal{P}$ is of type
$F_{\sigma}$ and of course nowhere dense.

We show that $X=\bigcup\mathcal{U}\cup\bigcup\mathcal{P}$. The set
$\bigcup\mathcal{U}$ is open and dense. Let $G$ be a nonempty open set in $X$.
Choose some $u\in G\cap C(g_{1})\cap\dots\cap C(g_{n})$. Then for each
$j\in\\{1,\dots,n\\}$ we have $u\in W_{i_{j}}^{j}$ for some
$i_{j}\in\\{1,\dots,m\\}$. Thus $u\in G\cap W_{i_{1}}^{1}\cap\dots\cap
W_{i_{n}}^{n}$, i.e. $G\cap\bigcup\mathcal{U}\neq\emptyset.$

Let $z\in X\setminus\bigcup\mathcal{U}$. Then
$z\in\overline{\bigcup\mathcal{U}}$. Hence there is $U\in\mathcal{U}$ such
that $z\in\overline{U}\setminus U$. It follows that for every
$j\in\\{1,\dots,n\\}$ there is $i_{j}\in\\{1,\dots,m\\}$ such that $z\in
P_{i_{j}}^{j}(U)$ and then $z\in\bigcup\mathcal{P}$. Hence
$X=\bigcup\mathcal{U}\cup\bigcup\mathcal{P}$.

Finally, let $\mathcal{B}$ be the family consisting of all nonempty sets from
$\mathcal{U}$ and $\mathcal{P}$. One can check that
$|f_{i}(p)-f_{i}(q)|<\varepsilon$ for all $f_{i}\in\mathcal{F}$,
$B\in\mathcal{B}$ and $p,q\in B$. ∎

######  Theorem 5.2.

Let $(f_{n})$ be a pointwise bounded sequence of real-valued functions defined
on a topological space $X$. Suppose that for every $\varepsilon>0$ there is a
finite cover $\mathcal{B}$ of $X$ such that
$\operatorname{diam}f_{n}(B)\leq\varepsilon$ for every $B\in\mathcal{B}$ and
for every $n\in\mathbb{N}$. Then $(f_{n})$ contains a subsequence which is
uniformly convergent on $X$.

###### Proof.

For every $x\in X$ let $M_{x}>0$ be such that

$|f_{n}(x)|\leq M_{x}\;\mbox{ for every }\;n\in\mathbb{N}.$

The set $\Pi_{x\in X}[-M_{x},M_{x}]$ is compact and for every
$n\in\mathbb{N}$, $f_{n}\in\Pi_{x\in X}[-M_{x},M_{x}]$. There is a function
$f\in\Pi_{x\in X}[-M_{x},M_{x}]$ such that $f$ is a cluster point of $(f_{n})$
in $\Pi_{x\in X}[-M_{x},M_{x}]$. There is a subnet $(g_{\sigma})$ of $(f_{n})$
which pointwise converges to $f$. First, observe the following: Let $\eta>0$
and $\mathcal{B}$ be a finite cover of $X$ such that
$\operatorname{diam}g_{\sigma}(B)\leq\eta$ for all $B\in\mathcal{B}$ and every
$\sigma$. Then also $\operatorname{diam}f(B)\leq\eta$ for every
$B\in\mathcal{B}$.

We will show that $(g_{\sigma})$ uniformly converges to $f$. Let
$\varepsilon>0$. There is a finite cover $\mathcal{D}$ of $X$ such that
$\operatorname{diam}g_{\sigma}(D)\leq\varepsilon/3$ for every
$D\in\mathcal{D}$ and for every $\sigma$ and also
$\operatorname{diam}f(D)\leq\varepsilon/3$ for every $D\in\mathcal{D}$. For
each $D\in\mathcal{D}$ choose $x_{D}\in D$. For every $D\in\mathcal{D}$ there
is $\sigma_{D}$ such that $|f(x_{D})-g_{\sigma}(x_{D})|<\varepsilon/3$ for
every $\sigma\geq\sigma_{D}$. Let $\sigma_{0}$ be such that
$\sigma_{0}\geq\sigma_{D}$ for all $D\in\mathcal{D}$. Such $\sigma_{0}$ exists
since the cover $\mathcal{D}$ is finite. We claim that

$|g_{\sigma}(x)-f(x)|<\varepsilon\;\mbox{ for every
}\;\sigma\geq\sigma_{0}\;\mbox{ and for every }\;x\in X.$

Let $x\in X$. There is $D\in\mathcal{D}$ such that $x\in D$. Let
$\sigma\geq\sigma_{0}\geq\sigma_{D}$. Then

$|g_{\sigma}(x)-f(x)|\leq|g_{\sigma}(x)-g_{\sigma}(x_{D})|+|g_{\sigma}(x_{D})-f(x_{D})|+|f(x_{D})-f(x)|$
$<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon.$

Thus $(g_{\sigma})$ uniformly converges to $f$, i.e. $(g_{\sigma})$ converges
to $f$ in the metrizable space $(\mathbb{R}^{X},\tau_{U})$. Then $f$ is a
cluster point of $(f_{n})$ in $(\mathbb{R}^{X},\tau_{U})$. Since
$(\mathbb{R}^{X},\tau_{U})$ is metrizable, there is a subsequence
$(f_{n_{k}})$ of $(f_{n})$ which converges to $f$ in
$(\mathbb{R}^{X},\tau_{U})$; i.e. $(f_{n_{k}})$ uniformly converges to $f$. ∎

Notice that Theorem 5.2 works also for functions from a topological space $X$
to a metric space $(Y,d)$ under the assumption that
$\overline{\\{f_{n}(x):n\in\mathbb{N}\\}}$ is compact for every $x\in X$.

We have the following answer to Question 4.3 from [3].

######  Theorem 5.3.

Let $(f_{n})$ be a uniformly bounded sequence of real-valued Baire 1 functions
defined on a Baire metric space $X$. The following are equivalent:

  * (1)

$(f_{n})$ contains a subsequence which is uniformly convergent on $X$;

  * (2)

There is a subsequence $(f_{n_{k}})$ of $(f_{n})$ with the following property:
for every $\varepsilon>0$ there is a finite cover $\mathcal{B}$ of $X$
consisting of subsets of $X$ which are either open or $F_{\sigma}$ nowhere
dense such that $\operatorname{diam}f_{n_{k}}(B)\leq\varepsilon$ for every
$B\in\mathcal{B}$ and for every $k\in\mathbb{N}$.

###### Proof.

$(1)\Rightarrow(2)$ follows from Theorem 5.1.

$(2)\Rightarrow(1)$: By $(2)$ consider a subsequence $(f_{n_{k}})$ of
$(f_{n})$ described in (2). By Theorem 5.2 there is a subsequence of the
sequence $(f_{n_{k}})$ which uniformly converges on $X$. However, this
subsequence of $(f_{n_{k}})$ is also a subsequence of $(f_{n})$. Thus
$(f_{n})$ contains a subsequence which is uniformly convergent on $X$. ∎

We propose the following generalization of Theorem 3.1 from [3].

######  Proposition 5.2.

Let $(X,d_{X})$ be a metric space and $(Y,d_{Y})$ be a separable metric space.
Let $(f_{n})$ be a sequence of Baire 1 functions from $X$ to $Y$. If $(f_{n})$
converges uniformly on bounded sets to $f\colon X\to Y$, then the family
$\\{f_{n}:n\in\mathbb{N}\\}$ is equi-Lebesgue. Thus, by Fact 3.1, it is equi-
Baire 1.

###### Proof.

Let $x_{0}$ be an arbitrary point in $X$. For every $n\in\mathbb{N}$ put
$B_{n}=B(x_{0},n)$. Let $l\in\mathbb{N}$ and $\varepsilon>0$. Put
$\mathcal{F}=\\{f_{1},f_{2},\dots,f_{n},\dots\\}$. Then $\mathcal{F}$ is a
totally bounded set in $(Y^{X},\tau_{UB})$. There are functions
$f_{k_{1}},f_{k_{2}},\dots f_{k_{n}}\in\mathcal{F}$ such that

$\mathcal{F}\subseteq\bigcup\\{W(f_{k_{j}},B_{l},\varepsilon/3):j\in\\{1,\dots,n\\}\\}.$

For simplicity put $g_{j}=f_{k_{j}}$ for every $j\in\\{1,\dots,n\\}.$

Let $\mathcal{V}=\\{V_{1},\dots,V_{n},\dots\\}$ be a countable open cover of
$Y$, where diameter of members of this cover is less than $\varepsilon/3$. For
every $j\in\\{1,\dots,n\\}$ and $i\in\mathbb{N}$ put
$B_{i}^{j}=g_{j}^{-1}(V_{i})$. Since $g_{j}$ is a Baire 1 function,
$B_{i}^{j}$ is an $F_{\sigma}$ set for all $j\in\\{1,\dots,n\\}$ and
$i\in\mathbb{N}$. Put

$\mathcal{B}^{\prime}=\\{B_{l}\cap B_{i_{1}}^{1}\cap\dots\cap
B_{i_{n}}^{n}:i_{j}\in\mathbb{N},\;j\in\\{1,\dots,n\\}\\}.$

Every set $B\in\mathcal{B}^{\prime}$ is of type $F_{\sigma}$. Let
$\mathcal{B}$ be the family containing all nonempty sets from
$\mathcal{B}^{\prime}$. Clearly, $B_{l}=\bigcup\mathcal{B}$ and the family
$\mathcal{B}$ is countable. One can check that for all $f_{i}\in\mathcal{F}$
and for all $B\in\mathcal{B}$ we have
$\operatorname{diam}f_{i}(B)<\varepsilon$. Since
$X=\bigcup\\{B_{l}:l\in\mathbb{N}\\}$, we are done. ∎

##  6\. Further results

The notion of a cliquish function was introduced by Thielman in [18].

######  Definition 6.1.

[18] Let $X$ be a topological space and $(Y,d)$ be a metric space. A function
$f\colon X\to Y$ is called cliquish at $x\in X$ if for every $\varepsilon>0$
and for each open neighbourhood $U$ of $x$ there is a nonempty open set
$G\subset U$ such that $d(f(x),f(y))<\varepsilon$ for all $x,y\in G$. The
function $f\colon X\to Y$ is cliquish if it is cliquish at every point $x\in
X$.

In [8] the following result was proved. (We preserve our definition of a Baire
1 function in a more general case.)

######  Proposition 6.1.

Let $X$ be a Baire space and $(Y,d)$ be a separable metric space. If $f\colon
X\to Y$ be a Baire 1 function, then $f$ is cliquish.

From the proof of Proposition 6.1 given in [8] we can derive the following
stronger result.

######  Proposition 6.2.

Let $(X,d_{X})$ be a Baire metric space and $(Y,d_{Y})$ be a metric space. If
$f\colon X\to Y$ has the Lebesgue property, then $f$ is cliquish.

It is known (see [5]) that, if $X$ is a Baire space and $(Y,d)$ is a metric
space, then $f\colon X\to Y$ is cliquish if and only if the set of continuity
points of $f$ is dense in $X$. A function with this latter property is called
pointwise discontinuous, cf. [13, p. 105].

In [4, Def. 4], the notion of equi-cliquishness of a sequence of functions
from a topological space into a metric space was introduced. We suggest the
following definition for families of functions.

######  Definition 6.2.

Let $X$ be a topological space and $(Y,d)$ be a metric space. A family
$\mathcal{F}$ of functions from $X$ to $Y$ is called equi-cliquish at $x\in X$
if for every $\varepsilon>0$ and each neighbourhood $U\subset X$ of $x$ there
is a nonempty open set $G\subset U$ such that
$\operatorname{diam}f(G)\leq\varepsilon$ for all $f\in\mathcal{F}$. A family
$\mathcal{F}$ of functions from $X$ to $Y$ is equi-cliquish if it is equi-
cliquish at every $x\in X$.

For a family $\mathcal{F}$ of functions from a topological space $X$ to a
metric space $(X,d)$ we denote by $EC(\mathcal{F})$ the set of equi-continuity
points of $\mathcal{F}$. It is known that this set is $G_{\delta}$ since
$EC(\mathcal{F})=\bigcap_{n\in\mathbb{N}}O(1/n)$ where

$O(\varepsilon)=\bigcup\\{U:U\subset X,\;U\mbox{ nonempty open,
}\operatorname{diam}f(U)<\varepsilon\mbox{ for every }f\in\mathcal{F}\\}$

is open for each $\varepsilon>0$; see [7, Lemma 3.2].

Note that in [7] the authors defined the notion of almost equicontinuity for a
family of continuous functions, which is similar to the notion of equi-
cliquishness.

######  Definition 6.3.

[7] Let $X$ and $(M,d)$ be a Hausdorff, completely regular space and a metric
space, respectively, and let $C(X,M)$ denote the set of all continuous
functions from $X$ to $M$. A subset $G\subset C(X,M)$ is said to be almost
equicontinuous if $G$ is equicontinuous on a dense subset of $X$.

######  Proposition 6.3.

Let $(X,d_{X})$ be a Baire metric space and $(Y,d_{Y})$ be a metric space. If
$\mathcal{F}$ is an equi-cliquish family of functions from $X$ to $Y$, then
the set $EC(\mathcal{F})$ is dense $G_{\delta}$, and thus it is comeager.

###### Proof.

Since $\mathcal{F}$ is equi-cliquish, the set $O(\varepsilon)$ given above is
dense in $X$. Since $X$ is a Baire space, the $G_{\delta}$ set
$EC(\mathcal{F})=\bigcap_{n\in\mathbb{N}}O(1/n)$ is dense and thus it is
comeager. ∎

######  Proposition 6.4.

Let $(X,d_{X})$ be a Baire metric space and $(Y,d_{Y})$ be a metric space. If
$\mathcal{F}$ is an equi-Lebesgue family of functions from $X$ to $Y$, then
$\mathcal{F}$ is equi-cliquish.

###### Proof.

Let $x\in X$ and $U\subset X$ be an open neighbourhood of $x$ and
$\varepsilon>0$. Since the family $\mathcal{F}$ is equi-Lebesgue, there is a
cover $(X_{i})_{i\in\mathbb{N}}$ of $X$ consisting of closed sets such that
$\operatorname{diam}f(X_{i})\leq\varepsilon$ for all $i\in\mathbb{N}$ and
$f\in\mathcal{F}$. The set $U$ is $F_{\sigma}$. Let
$\\{F_{n}:n\in\mathbb{N}\\}$ be a family of closed sets in $X$ such that
$U=\bigcup_{n\in\mathbb{N}}F_{n}$. Then

$U=\left(\bigcup_{n\in\mathbb{N}}F_{n}\right)\cap\left(\bigcup_{i\in\mathbb{N}}X_{i}\right)=\bigcup_{n\in\mathbb{N}}\bigcup_{i\in\mathbb{N}}(F_{n}\cap
X_{i}).$

Since $X$ is a Baire space, there are $n,i\in\mathbb{N}$ and nonempty open set
$V\subset U$ such that $V\subset F_{n}\cap X_{i}$. Thus
$\operatorname{diam}f(V)\leq\varepsilon$ for every $f\in\mathcal{F}$. Hence
$\mathcal{F}$ is equi-cliquish at $x\in X$. Since $x$ was arbitrary,
$\mathcal{F}$ is equi-cliquish. ∎

######  Corollary 6.1.

Let $(X,d_{X})$ be a Baire metric space and $(Y,d_{Y})$ be a metric space.
Then the set $EC(\mathcal{F})$ is dense $G_{\delta}$ for every family
$\mathcal{F}$ of equi-Lebesgue family of functions from $X$ to $Y$. If
additionally, $(X,d_{X})$ is separable, the set $EC(\mathcal{F})$ is dense
$G_{\delta}$ for every equi-Baire 1 family of functions from $X$ to $Y$.

###### Proof.

For the former assertion we use Propositions 6.3 and 6.4. The latter assertion
follows from Fact 3.1 (ii). ∎

Of course, in general, we cannot expect that in Corollary 6.1 an equi-Baire 1
family of functions will be equicontinuous on the whole space.

######  Example 6.1.

Let $X=[0,1]$ and $Y=\mathbb{R}$ be equipped with the usual Euclidean metric.
Define $f\colon X\to Y$ by $f(x)=\sin(1/x)$ if $x\neq 0$ and $f(0)=0$. The
family $\\{f\\}$ is equi-Baire 1, however it cannot be equicontinuous at $0$.
Example 3.8 in [1] presents an equi-Baire1 family of continuous functions from
$X$ to $Y$ which is not equicontinuous at $0$.

Natural questions arise whether a family $\mathcal{F}$ of functions from a
metric space $(X,d_{X})$ to a metric space $(Y,d_{Y})$, with the set
$H=EC(\mathcal{F})$ dense in $(X,d_{X})$, is equi-Baire 1 or, whether the
functions $f|H$ for $f\in\mathcal{F}$ can be extended to an equi-Baire 1
family on $X$. Firstly, we show that the answer to the first question is
negative.

######  Example 6.2.

S. Marcus in his paper [16] showed that there is a quasicontinuous function
$f$ from the interval $[0,1]$ to $\mathbb{R}$ which is not Lebesgue
measurable. Thus $f$ cannot be a Baire 1 function. It is known that every
quasicontinuous function from a Baire space into a metric space is continuous
at points of a dense $G_{\delta}$ set (see [9]).

We also have a positive answer to our second question in a full generality.

######  Theorem 6.1.

Let $(X,d_{X})$ be a separable metric space and $(Y,d_{Y})$ be a separable
complete metric space. Let $H\subset X$ be a nonempty $G_{\delta}$ set, and
$\mathcal{F}$ be an equi-continuous family of functions from $H$ to
$(Y,d_{Y})$. Then all functions in $\mathcal{F}$ can be extended to an equi-
Baire 1 family of functions from $X$ to $Y$.

###### Proof.

We use ideas from [13, pp. 434–435] with $\alpha=1$. Fix $\varepsilon>0$. Then
for each $x\in H$ there is an open set $O_{x}$ in $H$ such that $x\in O_{x}$
and $\operatorname{diam}f(O_{x})<\varepsilon/2$ for every $f\in\mathcal{F}$.
Since $X$ has a countable base, we can suppose that there are open sets
$G_{1},G_{2},\dots$ in $H$ such that $H=\bigcup_{n\in\mathbb{N}}G_{n}$ and
$\operatorname{diam}f(G_{n})<\varepsilon/2$ for all $f\in\mathcal{F}$ and
$n\in\mathbb{N}$. There is a countable discrete set $I\subset Y$ such that for
every $y\in Y$ there is $z\in I$ such that $d_{Y}(y,z)<\varepsilon/2$.

For all $f\in\mathcal{F}$ and $i\in\mathbb{N}$, pick $y_{i}^{f}\in I\cap
S(f(G_{i}),\varepsilon/2)$. Then for every $f\in\mathcal{F}$ let $g^{f}\colon
h\to Y$ be defined as follows: $g^{f}(x)=y_{1}^{f}$ if $x\in G_{1}$, and
$g^{f}(x)=y_{k}^{f}$ if $x\in G_{k}\setminus\bigcup_{i<k}G_{i}$ and $k>1$. It
is easy to check that $g^{f}\colon H\to Y$ is Baire 1 and
$d_{Y}(f(x),g^{f}(x))<\varepsilon$ for each $x\in H$.

Now, for every $f\in\mathcal{F}$ and each $n\in\mathbb{N}$, let
$g_{n}^{f}\colon H\to Y$ be the function $g^{f}$ defined for
$\varepsilon=1/2^{n+1}$. It is visible that $(g_{n}^{f})_{n\in\mathbb{N}}$
converges uniformly to $f$ on $H$. Note that for all $n,k\in\mathbb{N}$ and
every $f\in\mathcal{F}$ we have

$d_{Y}(g_{n}^{f}(x),g_{k}^{f}(x))\leq 1/2^{k}\;\;\mbox{for each }x\in H\mbox{
if }n\geq k.$

By the construction, the image $g_{n}^{f}(H)$ is a countable discrete set, for
all $f\in\mathcal{F}$ and $n\in\mathbb{N}$.

Following the proof in [13], we can extend every $g_{n}^{f}$ to a Baire 1
function $h_{n}^{f}\colon X\to g_{n}^{f}(H)$ in such a way that

$d_{Y}(h_{n}^{f}(x),h_{k}^{f}(x))\leq 1/2^{k}\;\;\mbox{for each }x\in H\mbox{
if }n\geq k.$ (1)

Moreover, given $n\in\mathbb{N}$, all the functions $h_{n}^{f}$ with
$f\in\mathcal{F}$, are defined in the same way in the sense that
$X=\bigcup_{i\in\mathbb{N}}F_{i}^{n}$ where $F_{i}^{n}$ is an $F_{\sigma}$ set
in $X$ and $|h_{n}^{f}(F_{i}^{n})|=1$ for each $i\in\mathbb{N}$. Since
$(Y,d_{Y})$ is complete, condition (1) implies that, for every
$f\in\mathcal{F}$, the (uniform) limit $\lim_{n}h_{n}^{f}(x)=f^{*}(x)$ exists
for all $x\in X$ which defines the extension $f^{*}\colon X\to Y$ of $f$.
Also, if $n\to\infty$ in (1), we obtain $d_{Y}(f^{*}(x),h_{k}^{f}(x))\leq
1/2^{k}$ for all $x\in X$.

We will prove that $\\{f^{*}\colon f\in\mathcal{F}\\}$ is equi-Lebesgue. This
yields the assertion by Fact 3.1.(i) which claims that every equi-Lebesgue
family is equi-Baire. It is enough to show that, for each $k\in\mathbb{N}$, we
have $\operatorname{diam}f^{*}(F_{i}^{k})\leq 1/2^{k-1}$ for all
$i\in\mathbb{N}$ and $f\in\mathcal{F}$. So, fix $k,i\in\mathbb{N}$ and
$f\in\mathcal{F}$. For any $x,x^{\prime}\in F_{i}^{k}$ we have

$d_{Y}(f^{*}(x),f^{*}(x^{\prime}))\leq
d_{Y}(f^{*}(x),h_{k}^{f}(x))+d_{Y}(h_{k}^{f}(x),h_{k}^{f}(x^{\prime}))+d_{Y}(f^{*}(x^{\prime}),h_{k}^{f}(x^{\prime}))\leq
1/2^{k-1}$

where $d_{Y}(h_{k}^{f}(x),h_{k}^{f}(x^{\prime}))=0$ since
$|h_{k}^{f}(F_{i}^{k})|=1$. ∎

## References

  * [1] A. Alikhani-Koopaei, _Equi-Baire one family of functions on metric spaces: A generalization of equi-continuity; and some applications_ , Topol. Appl. 277 (2020) 107170.
  * [2] A. Alikhani-Koopaei, _On dynamics of families of equi-Baire one functions on metric spaces_ , Topol. Appl. 322 (2022) 108267.
  * [3] M. Balcerzak, O. Karlova, P. Szuca, _Equi-Baire 1 families of functions_ , Topol. Appl. 305 (2022) 107900.
  * [4] J. Borsík, Ľ. Holá, D. Holý, _Baire spaces and quasicontinuous mappings_ , Filomat 25 (2011), 69–83.
  * [5] J. Ewert, _Note of limits of simply continuous and cliquish functions_ , Internat. J. Math. Math. Sci. 1 (3) (1994), 447–450.
  * [6] J. P. Fenecios, E. A. Cabral, _A simpler proof for the $\varepsilon-\delta$ characterization of Baire class one function_, Real Anal. Exch. 39 (2) (2013/14), 441–446.
  * [7] M. V. Ferrer, S. Hernández, L. Tárrega, _Equicontinuity criteria for metric-valued sets of continuous functions_ , Topol. Appl. 225 (2017), 220–236.
  * [8] Ľ. Holá, _There are $2^{c}$ quasicontinuous non Borel functions on uncountable Polish space_, Results Math. (2021) 76:126.
  * [9] Ľ. Holá, D. Holý, W. Moors, _USCO and Quasicontinuous Mappings_ , De Gruyter Studies in Mathematics, 81, De Gruyter, Berlin, 2021.
  * [10] O. Karlova, V. Mykhaylyuk, _On composition of Baire functions_ , Topol. Appl. 216 (2017), 8–24.
  * [11] A. S. Kechris, _Classical Descriptive Set Theory_ , Springer-Verlag, New York, 1987.
  * [12] J. Kelley, _General Topology_ , Van Nostrand, New York, 1955.
  * [13] K. Kuratowski, _Topology_ , Academic Press, London, 1966.
  * [14] H. Lebesgue, Sur les fonctions représentables analyquement, J. Math, Pures Appl. 1 (1905), 139–216.
  * [15] D. Lecomte, _How can we recover Baire one functions?_ , Mathematika, 50 (2003), 171–198.
  * [16] S. Marcus, _Sur les fonctions quasi-continues au sense Kempisty_ , Colloq. Math. 8 (1961), 45–53.
  * [17] P. Y. Lee, W. K. Tang and D. Zhao, _An equivalent definition of functions of the first Baire class_ , Proc. Am. Math. Soc. 129 (8) (2001), 2273–2275.
  * [18] H. P. Thielman, _Types of functions_ , Am. Math. Monthly 60 (1953), 156–161.