Null-finite sets in metric groups and their applications

TL;DR

This paper introduces null-finite sets in metric groups, explores their properties and connections to Haar-null and Haar-meager sets, and applies these concepts to prove continuity of mid-point convex functions under certain boundedness conditions, resolving a long-standing problem.

Contribution

It defines null-finite sets and demonstrates their significance in relation to Haar-null and Haar-meager sets, applying these insights to functional equations and inequalities.

Findings

Null-finite Borel sets are Haar-null and Haar-meager.

Every non-discrete metric Abelian group is the union of two null-finite sets.

A mid-point convex function is continuous if bounded on a non-null-finite Borel subset.

Abstract

In the paper we introduce a new family of "small" sets which is tightly connected with two well known $\sigma$-ideals: of Haar-null sets and of Haar-meager sets. We define a subset $A$ of a topological group $X$ to be $\mathit{null}$-$\mathit{finite}$ if there exists an infinite compact subset $K\subset X$ such that for every $x\in X$ the intersection $K\cap (x+A)$ is finite. We prove that each null-finite Borel set in a complete metric Abelian group is Haar-null and Haar-meager. The Borel restriction in the above result is essential as each non-discrete metric Abelian group is the union of two null-finite sets. Applying null-finite sets to the theory of functional equations and inequalities, we prove that a mid-point convex function $f:G\to\mathbb R$ defined on an open convex subset $G$ of a metric linear space $X$ is continuous if it is upper bounded on a subset $B$ which is not null-finite and whose closure is contained in $G$. This gives an alternative short proof of a known generalization of Bernstein-Doetsch theorem (saying that a mid-point convex function $f:G\to\mathbb R$ defined on an open covex subset $G$ of a metric linear space $X$ is continuous if it is upper bounded on a non-empty open subset $B$ of $G$). Since Borel null-finite sets are Haar-meager and Haar-null, we conclude that a mid-point convex function $f:G\to\mathbb{R}$ defined on an open convex subset $G$ of a complete linear metric space $X$ is continuous if it is upper bounded on a Borel subset $B\subset G$ which is not Haar-null or not Haar-meager in $X$. The last result resolves an old problem in the theory of functional equations and inequalities posed by Baron and Ger in 1983.

Equations

$$\Sigma :\prod _{n\in \omega }{K}_n\to X,\Sigma :(x_n{)}_{n\in \omega }\mapsto \sum _{n\in \omega }{x}_n,$$

$$\Sigma :\Pi \to X,\Sigma :(p_i{)}_{i\in \omega }\mapsto \sum _{i=0}^{\infty }{p}_i{a}_i.$$

$$\sigma {\overline{\mathsf{NF}}}_X\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\prod _{n=2}^{\infty }\frac{|C_n^{\ast }|}{|C_n|}=\prod _{n=2}^{\infty }\frac{n-1}{n}=0$$

$$x_{n}(i)=\begin{cases}\theta&\mbox{if $i\leq n$}\\ g_{i}&\mbox{if $i>n$}.\end{cases}$$

$$\prod _{n\in \omega }\frac{f(n)-1}{f(n)}>0.$$

$$\{n\in \omega :x+v_n\in S+A\}=\{n\in \omega :z+v_n\in A\}\text{mbox}and\{n\in \omega :x+u_n\in S+B\}=\{n\in \omega :z+u_n\in B\}$$

$${\mathcal{I}}_A=\{I\subset X:\exists F\in [X{]}^{<\omega }I\subset FA\}$$

$$M\ge f(x+z_n)=f(x)+f(z_n)\ge f(x)+2^n\epsilon ,$$

$$f(\frac{x}{2}+\frac{y}{2})\le \frac{1}{2}f(x)+\frac{1}{2}f(y)$$

$$2^{n-k}\epsilon$$

$$\le \frac{1}{2}f(-2^{-k+1}a)+\frac{1}{2}f(2^{-k+1}a+2^{n-k+1}{x}_n)\le \frac{1}{2}f(-2^{-k+1}a)+2^{-k}f(a+2^n{x}_n)$$

$$\sup f(B)\ge \underset{n\in \Omega }{\sup }f(a+2^n{x}_n)\ge \underset{k<n\in \Omega }{\sup }(2^n\epsilon -2^{k-1}f(-2^{-k+1}a))=\infty ,$$

Full text

Introduction

In 1920 Steinhaus [29] proved that for any measurable sets $A,B$ of positive Haar measure in a locally compact Polish group $X$ the sum $A+B:=\{a+b:a\in A,\;b\in B\}$ has non-empty interior in $X$ and the difference $B-B:=\{a-b:a,b\in B\}$ is a neighborhood of zero in $X$.

In [8] Christensen extended the “difference” part of the Steinhaus results to all Polish Abelian groups proving that for a Borel subset $B$ of a Polish Abelian group $X$ the difference $B-B$ is a neighborhood of zero if $B$ is not Haar-null. Christensen defined a Borel subset $B\subset X$ to be Haar-null if there exists a Borel $\sigma$-additive probability measure $\mu$ on $X$ such that $\mu(B+x)=0$ for all $x\in X$.

A topological version of Steinhaus theorem was obtained by Pettis [25] and Piccard [26] who proved that for any non-meager Borel sets $A,B$ in a Polish group $X$ the sum $A+B$ has non-empty interior and the difference $B-B$ is a neighborhood of zero.

In 2013 Darji [9] introduced a subideal of the $\sigma$-ideal of meager sets in a Polish group, which is a topological analog to the $\sigma$-ideal of Haar-null sets. Darji defined a Borel subset $B$ of a Polish group $X$ to be Haar-meager if there exists a continuous map $f:K\to X$ from a non-empty compact metric space $K$ such that for every $x\in X$ the preimage $f^{-1}(B+x)$ is meager in $K$. By [18], for every Borel subset $B\subset X$ which is not Haar-meager in a Polish Abelian group $X$ the difference $B-B$ is a neighborhood of zero.

It should be mentioned that in contrast to the “difference” part of the Steinhaus and Piccard–Pettis theorems, the “additive” part can not be generalized to non-locally compact groups: by [23], [18] each non-locally compact Polish Abelian group $X$ contains two closed sets $A,B$ whose sum $A+B$ has empty interior but $A,B$ are neither Haar-null nor Haar-meager in $X$. In the Polish group $X=\mathbb{R}^{\omega}$, for such sets $A,B$ we can take the positive cone $\mathbb{R}_{+}^{\omega}$.

Steinhaus-type theorems has significant applications in the theory of functional equations and inequalities. For example, the “additive” part of Steinhaus Theorem can be applied to prove that a mid-point convex function $f:\mathbb{R}^{n}\to\mathbb{R}$ is continuous if it is upper bounded on some measurable set $B\subset\mathbb{R}^{n}$ of positive Lebesgue measure (see e.g. [20, p.210]). We recall that a function $f:G\to\mathbb{R}$ defined on a convex subset of a linear space is mid-point convex if $f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}$ for any $x,y\in G$. Unfortunately, due to the example of Matous̆ková and Zelený [23] we know that the “additive” part of the Steinhaus theorem does not extend to non-locally compact Polish Abelian groups. This leads to the following natural problem whose “Haar-null” version was posed by Baron and Ger in [3, P239].

Problem (Baron, Ger, 1983). Is the continuity of an additive or mid-point convex function $f:X\to\mathbb{R}$ on a Banach space $X$ equivalent to the upper boundedness of the function on some Borel subset $B\subset X$ which is not Haar-null or not Haar-meager?

In this paper we give the affirmative answer to the Baron–Ger Problem applying a new concept of a null-finite set, which will be introduced in Section 2. In Section 3 we show that non-null-finite sets $A$ in metric groups $X$ possess a Steinhaus-like property: if a subset $A$ of a metric group $X$ is not null-finite, then $A-\bar{A}$ is a neighborhood of $\theta$ and for some finite set $F\subset X$ the set $F+(A-A)$ is a neighborhood of zero in $X$. In Sections 5 and 6 we shall prove that Borel null-finite sets in Polish Abelian groups are Haar-meager and Haar-null. On the other hand, in Example 7.1 we show that the product $\prod_{n\in\mathbb{N}}C_{n}$ of cyclic groups contains a closed Haar-null set which is not null-finite and in Example 7.2 we show that the product $\prod_{n\in\mathbb{N}}C_{2^{n}}$ contains a null-finite $G_{\delta}$-set which cannot be covered by countably many closed Haar-null sets. In Section 8 we prove that each non-discrete metric Abelian group is the union of two null-finite sets and also observe that sparse sets (in the sense of Lutsenko–Protasov [22]) are null-finite. In Section 9 we apply null-finite sets to prove that an additive functional $f:X\to\mathbb{R}$ on a Polish Abelian group is continuous if it is upper bounded on some subset $B\subset X$ which is not null-finite in $X$. In Section 10 we generalize this result to additive functions with values in Banach and locally convex spaces. Finally, in Section 11, we prove a generalization of Bernstein–Doetsch theorem and next apply it in the proof of a continuity criterion for mid-point convex functions on complete metric linear spaces, thus answering the Baron–Ger Problem. In Section 12 we pose some open problems related to null-finite sets.

1. Preliminaries

All groups considered in this paper are Abelian. The neutral element of a group will be denoted by $\theta$. For a group $G$ by $G^{*}$ we denote the set of non-zero elements in $G$. By $C_{n}=\{z\in\mathbb{C}:z^{n}=1\}$ we denote the cyclic group of order $n$.

By a (complete) metric group we understand an Abelian group $X$ endowed with a (complete) invariant metric $\|\cdot-\cdot\|$. The invariant metric $\|\cdot-\cdot\|$ determines (and can be recovered from) the prenorm $\|\cdot\|$ defined by $\|x\|:=\|x-\theta\|$. So, a metric group can be equivalently defined as a group endowed with a prenorm.

Formally, a prenorm on a group $X$ is a function $\|\cdot\|:X\to\mathbb{R}_{+}:=[0,\infty)$ satisfying three axioms:

• •

$\|x\|=0$ iff $x=\theta$;

• •

$\|-x\|=\|x\|$;

• •

$\|x+y\|\leq\|x\|+\|y\|$

for any $x,y\in X$; see [1, §3.3].

Each metric group is a topological group with respect to the topology, generated by the metric.

All linear spaces considered in this paper are over the field $\mathbb{R}$ of real numbers. By a metric linear space we understand a linear space endowed with an invariant metric.

A non-empty family $\mathcal{I}$ of subsets of a set $X$ is called ideal of sets on $X$ if $\mathcal{I}$ satisfies the following conditions:

• •

$X\notin\mathcal{I}$;

• •

for any subsets $J\subset I\subset X$ the inclusion $I\in\mathcal{I}$ implies $J\in\mathcal{I}$;

• •

for any sets $A,B\in\mathcal{I}$ we have $A\cup B\in\mathcal{I}$.

An ideal $\mathcal{I}$ on $X$ is called a $\sigma$-ideal if for any countable subfamily $\mathcal{C}\subset\mathcal{I}$ the union $\bigcup\mathcal{C}$ belongs to $X$.

An ideal $\mathcal{I}$ on a group $X$ is called invariant if for any $I\in\mathcal{I}$ and $x\in X$ the shift $x+I$ of $I$ belongs to the ideal $\mathcal{I}$.

For example, the family $[X]^{<\omega}$ of finite subsets of an infinite group $X$ is an invariant ideal on $X$.

A topological space is Polish if it is homeomorphic to a separable complete metric space. A topological space is analytic if it is a continuous image of a Polish space.

2. Introducing null-finite sets in topological groups

In this section we introduce the principal new notion of this paper.

A sequence $(x_{n})_{n\in\omega}$ in a topological group $X$ is called a null-sequence if it converges to the neutral element $\theta$ of $X$.

Definition 2.1**.**

A set $A$ of a topological group $X$ is called null-finite if there exists a null-sequence $(x_{n})_{n\in\omega}$ in $X$ such that for every $x\in X$ the set $\{n\in\omega:x+x_{n}\in A\}$ is finite.

Null-finite sets in metrizable topological groups can be defined as follows.

Proposition 2.2**.**

For a non-empty subset $A$ of a metric group $X$ the following conditions are equivalent:

1. (1)

$A$* is null-finite;* 2. (2)

there exists an infinite compact set $K\subset X$ such that for every $x\in X$ the intersection $K\cap(x+A)$ is finite; 3. (3)

there exists a continuous map $f:K\to X$ from an infinite compact space $K$ such that for every $x\in X$ the preimage $f^{-1}(A+x)$ is finite.

Proof.

The implications $(1)\Rightarrow(2)\Rightarrow(3)$ are obvious, so it is enough to prove $(3)\Rightarrow(1)$. Assume that $f:K\to X$ is a continuous map from an infinite compact space $K$ such that for every $x\in X$ the preimage $f^{-1}(A+x)$ is finite. Fix any point $a\in A$. It follows that for every $x\in X$ the fiber $f^{-1}(a+x)\subset f^{-1}(A+x)$ is finite and hence the image $f(K)$ is infinite. So, we can choose a sequence $(y_{n})_{n\in\omega}$ of pairwise distinct points in $f(K)$. Because of the compactness and metrizability of $f(K)$, we can additionally assume that the sequence $(y_{n})_{n\in\omega}$ converges to some point $y_{\infty}\in f(K)$. Then the sequence $(x_{n})_{n\in\omega}$ of points $x_{n}:=y_{n}-y_{\infty}$, $n\in\omega$, is a null-sequence witnessing that the set $A$ is null-finite. ∎

Null-finite sets are “small” in the following sense.

Proposition 2.3**.**

For any null-finite set $A$ in a metric group $X$ and any finite subset $F\subset X$ the set $F+A$ has empty interior in $X$.

Proof.

Since $A$ is null-finite, there exists a null-sequence $(x_{n})_{n\in\omega}$ in $X$ such that for every $x\in X$ the set $\{n\in\omega:x+x_{n}\in A\}$ is finite.

To derive a contradiction, assume that for some finite set $F$ the set $F+A$ has non-empty interior $U$ in $X$. Then for any point $u\in U$ the set $\{n\in\omega:u+x_{n}\in U\}\subset\{n\in\omega:u+x_{n}\in F+A\}$ is infinite. By the Pigeonhole Principle, for some $y\in F$ the set $\{n\in\omega:u+x_{n}\in y+A\}$ is infinite and so is the set $\{n\in\omega:-y+u+x_{n}\in A\}$. But this contradicts the choice of the sequence $(x_{n})_{n\in\omega}$. ∎

Now we present some examples of closed sets which are (or are not) null-finite. We recall that for a group $G$ by $G^{*}$ we denote the set $G\setminus\{\theta\}$ of non-zero elements of $G$.

Example 2.4**.**

Let $(G_{n})_{n\in\omega}$ be a sequence of finite groups. The set $A:=\prod_{n\in\omega}G_{n}^{*}$ is null-finite in the compact metrizable topological group $G=\prod_{n\in\omega}G_{n}$ if and only if $\lim_{n\to\infty}|G_{n}|\neq\infty$.

Proof.

First, we assume that $\lim_{n\to\infty}|G_{n}|=\infty$ and show that the set $A$ is not null-finite in $G$. Given a null-sequence $(x_{n})_{n\in\omega}$ in the compact topological group $G$, we should find an element $a\in G$ such that the set $\{n\in\omega:a+x_{n}\in A\}$ is infinite.

It will be convenient to think of elements of the group $G$ as functions $x:\omega\to\bigoplus_{n\in\omega}G_{n}$ such that $x(k)\in G_{k}$ for all $k\in\omega$.

Taking into account that $\lim_{k\to\infty}|G_{k}|=\infty$ and $\lim_{n\to\infty}x_{n}=\theta\in G$, we can inductively construct an increasing number sequence $(n_{k})_{k\in\omega}$ such that $n_{0}=0$ and for every $k\in\mathbb{N}$ and $m\geq n_{k}$ the following two conditions are satisfied:

• (1)

$x_{m}(i)=\theta$ for all $i<n_{k-1}$;

• (2)

$|G_{m}|\geq k+3$.

We claim that for every $i\geq n_{1}$ the set $\{x_{n_{k}}(i):k\in\omega\}$ has cardinality $<|G_{i}|$. Indeed, given any $i\geq n_{1}$, we can find a unique $j\geq 1$ such that $n_{j}\leq i<n_{j+1}$ and conclude that $x_{n_{k}}(i)=\theta$ for all $k\geq j+1$ (by the condition (1)). Then the set $\{x_{n_{k}}(i):k\in\omega\}=\{\theta\}\cup\{x_{n_{k}}(i):k\leq j\}$ has cardinality $\leq 1+(j+1)<j+3\leq|G_{i}|$. The last inequality follows from $i\geq n_{k}$ and condition (2). Therefore, the set $\{x_{n_{k}}(i):k\in\omega\}$ has cardinality $<|G_{i}|$ and we can choose a point $a_{i}\in G_{i}\setminus\{-x_{n_{k}}(i):k\in\omega\}$ and conclude that $a_{i}+x_{n_{k}}(i)\in G_{i}^{*}$ for all $k\in\omega$. Then the element $a=(a_{i})_{i\in\omega}\in G$ has the required property as the set $\{n\in\omega:a+x_{n}\in A\}\supset\{n_{k}\}_{k\in\omega}$ is infinite.

Next, assuming that $\lim_{k\to\infty}|G_{k}|\neq\infty$, we shall prove that the set $A$ is Haar-finite in $G$. Since $\lim_{k\to\infty}|G_{k}|\neq\infty$, for some $l\in\omega$ the set $\Lambda=\{k\in\omega:|G_{k}|=l\}$ is infinite.

Since $\lim_{n\to\infty}(1-\frac{1}{l})^{n}=0$, there exists $n\in\mathbb{N}$ such that $(1-\frac{1}{l})^{n}<\frac{1}{l}$ and hence $l(l-1)^{n}<l^{n}$. This implies that for every $k\in\Lambda$ the set $G_{k}{\cdot}(G_{k}^{*})^{n}:=\{(x+a_{1},\cdots x+a_{n}):x\in G_{k},\;a_{1},\dots,a_{n}\in G_{k}^{*}\}$ has cardinality $|G_{k}{\cdot}(G_{k}^{*})^{n}|\leq l(l-1)^{n}<l^{n}=|G_{k}^{n}|$. Consequently, the (compact) set $G{\cdot}A^{n}=\{(x+a_{1},\dots,x+a_{n}):x\in G,\;a_{1},\dots,a_{n}\in A\}$ is nowhere dense in the compact topological group $G^{n}$. Now Theorem 2.5 ensures that $A$ is null-finite in $G$. ∎

Theorem 2.5**.**

A non-empty subset $A$ of a Polish group $X$ is null-finite in $X$ if for some $n\in\mathbb{N}$ the set $X{\cdot}A^{n}:=\{(x+a_{1},\dots,x+a_{n}):x\in X,\;a_{1},\dots,a_{n}\in A\}$ is meager in $X^{n}$.

Proof.

Let $\mathcal{K}(X)$ be the space of non-empty compact subsets of $X$, endowed with the Vietoris topology. For a compact subset $K\subset X$ let $K^{n}_{*}=\{(x_{1},\dots,x_{n})\in K^{n}:|\{x_{1},\dots,x_{n}\}|=n\}$ the set of $n$-tuples consisting of pairwise distinct points of $K$.

By Mycielski-Kuratowski Theorem [19, 19.1], the set $W_{*}(A):=\{K\in\mathcal{K}(X):K^{n}_{*}\cap(X{\cdot}A^{n})=\emptyset\}$ is comeager in $\mathcal{K}(X)$ and hence contains some infinite compact set $K$. We claim that for every $x\in X$ the set $K\cap(x+A)$ has cardinality $<n$. Assuming the opposite, we could find $n$ pairwise distinct points $a_{1},\dots,a_{n}\in A$ such that $x+a_{i}\in K$ for all $i\leq n$. Then $(x+a_{1},\dots,x+a_{n})\in K^{n}_{*}\cap X\cdot A^{n}$, which contradicts the inclusion $K\in W_{*}(A)$. By Proposition 2.2, the set $A$ is null-finite. ∎

3. A Steinhaus-like properties of sets which are not null-finite

In [29] Steinhaus proved that for a subset $A$ of positive Lebesgue measure in the real line, the set $A-A$ is a neighborhood of zero. In this section we establish three Steinhaus-like properties of sets which are not null-finite.

Theorem 3.1**.**

If a subset $A$ of a metric group $X$ is not null-finite, then

1. (1)

the set $A-\bar{A}$ is a neighborhood of $\theta$ in $X$; 2. (2)

each neighborhood $U\subset X$ of $\theta$ contains a finite subset $F\subset U$ such that $F+(A-A)$ is a neighborhood of $\theta$. 3. (3)

If $X$ is Polish (and $A$ is analytic), then $A-A$ is not meager in $X$ (and $(A-A)-(A-A)$ is a neighborhood of $\theta$ in $X$).

Proof.

1. Assuming that $A-\bar{A}$ is not a neighborhood of $\theta$, we could find a null-sequence $(x_{n})_{n\in\omega}$ contained in $X\setminus(A-\bar{A})$. Since $A$ is not null-finite, there exists $a\in X$ such that the set $\Omega=\{n\in\omega:a+x_{n}\in A\}$ is infinite. Then $a\in\overline{\{a+x_{n}\}}_{n\in\Omega}\subset\bar{A}$ and hence $x_{n}=(a+x_{n})-a\in A-\bar{A}$ for all $n\in\Omega$, which contradicts the choice of the sequence $(x_{n})_{n\in\omega}$.

2. Fix a decreasing neighborhood base $(U_{n})_{n\in\omega}$ at zero in $X$ such that $U_{0}\subset U$. For the proof by contradiction, suppose that for any finite set $F\subset U$ the set $F+(A-A)$ is not a neighborhood of zero. Then we can inductively construct a null-sequence $(x_{n})_{n\in\omega}$ such that $x_{n}\in U_{n}\setminus\bigcup_{0\leq i<n}(x_{i}+A-A)$ for all $n\in\omega$. Observe that for each $z\in X$ the set $\{n\in\omega:z+x_{n}\in A\}$ contains at most one point. Indeed, in the opposite case we could find two numbers $k<n$ with $z+x_{k},z+x_{n}\in A$ and conclude that $z\in-x_{k}+A$ and hence $x_{n}\in-z+A\subset x_{k}-A+A$, which contradicts the choice of the number $x_{n}$. The sequence $(x_{n})_{n\in\omega}$ witnesses that the set $A$ is null-finite in $X$.

3. Finally assume that the group $X$ is Polish. To derive a contradiction, assume that $A-A$ is meager. Since the map $\delta:X\times X\to X$, $\delta:(x,y)\mapsto x-y$, is open, the preimage $\delta^{-1}(A-A)$ is meager in $X\times X$. Observe that $\delta^{-1}(A-A)=X{\cdot}A^{2}$ where $X{\cdot}A^{2}=\{(x+a,x+b):x\in X,\;a,b\in A\}$. By Theorem 2.5, the set $A$ is null-finite, which is a desired contradiction. This contradiction shows that the set $A-A$ is not meager in $X$. If the space $A$ is analytic, then so is the subset $A-A$ of $X$. Being analytic, the set $A-A$ has the Baire Property in $X$; see [19, 29.14]. So, we can apply the Pettis-Piccard Theorem [19, 9.9] and conclude that $(A-A)+(A-A)$ is a neighborhood of $\theta$ in $X$. ∎

The last statement of Theorem 3.1 does not hold without any assumptions on $A$ and $X$.

Example 3.2**.**

Let $f:C_{2}^{\omega}\to C_{2}$ be a discontinuous homomorphism on the compact Polish group $X=C_{2}^{\omega}$ (thought as the vector space $C_{2}^{\omega}$ over the two-element field $\mathbb{Z}/2\mathbb{Z}$). Then the subgroup $A=f^{-1}(\theta)$ is not null-finite in $X$ (by Proposition 2.3), but $A-A=A=(A-A)-(A-A)$ is not a neighborhood of $\theta$ in $X$.**

4. A combinatorial characterization of null-finite sets in compact metric groups

In compact metric groups null-finite sets admit a purely combinatorial description.

Proposition 4.1**.**

A non-empty subset $A$ of a compact metric group $X$ is null-finite if and only if there exists an infinite set $I\subset X$ such that for any infinite subset $J\subset I$ the intersection $\bigcap_{x\in J}(A-x)$ is empty.

Proof.

To prove the “only if” part, assume that $A\subset X$ is null-finite. So, there exists a null-sequence $(x_{n})_{n\in\omega}$ such that for every $x\in X$ the set $\{n\in\omega:x+x_{n}\in A\}$ is finite. It follows that for every $x\in X$ the set $\{n\in\omega:x_{n}=x\}$ is finite and hence the set $I:=\{x_{n}\}_{n\in\omega}$ is infinite. We claim that this set has the required property. Indeed, assuming that for some infinite subset $J\subset I$ the intersection $\bigcap_{x\in J}(A-x)$ contains some point $a\in X$, we conclude that the set $\{n\in\omega:a+x_{n}\in A\}$ contains the set $\{n\in\omega:x_{n}\in J\}$ and hence is infinite, which contradicts the choice of the sequence $(x_{n})_{n\in\omega}$.

To prove the “if” part, assume that there exists an infinite set $I\subset X$ such that for every infinite set $J\subset I$ the intersection $\bigcap_{x\in J}(A-x)$ is empty. By the compactness of the metric group $X$, some sequence $(x_{n})_{n\in\omega}$ of pairwise distinct points of the infinite set $I$ converges to some point $x_{\infty}\in X$. Then the null-sequence $(z_{n})_{n\in\omega}$ consisting of the points $z_{n}=x_{n}-x_{\infty}$, $n\in\omega$, witnesses that $A$ is null-finite. Assuming the opposite, we would find a point $a\in X$ such that the set $\{n\in\omega:a+z_{n}\in A\}$ is infinite. Then the set $J:=\{x\in I:a-x_{\infty}+x\in A\}\subset\{x_{n}:n\in\omega,\;a-x_{\infty}+x_{n}\in A\}$ is infinite, too, and the intersection $\bigcap_{x\in J}(A-x)$ contains the point $a-x_{\infty}$ and hence is not empty, which contradicts the choice of the set $I$. ∎

Following Lutsenko and Protasov [22] we define a subset $A$ of an infinite group $X$ to be sparse if for any infinite set $I\subset X$ there exists a finite set $F\subset I$ such that $\bigcap_{x\in F}(x+A)$ is empty. By [22, Lemma 1.2] the family of sparse subsets of a group is an invariant ideal on $X$ (in contrast to the family of null-finite sets).

Proposition 4.2**.**

Each sparse subset $A$ of a non-discrete metric group $X$ is null-finite.

Proof.

Being non-discrete, the metric group $X$ contains a null-sequence $(x_{n})_{n\in\omega}$ consisting of pairwise distinct points. Assuming that the sparse set $A$ is not null-finite, we can find a point $a\in X$ such that the set $\Omega:=\{n\in\omega:a+x_{n}\in A\}$ is infinite. Since $A$ is sparse, for the infinite set $I:=\{-x_{n}:n\in\Omega\}$ there exists a finite set $F\subset I$ such that the intersection $\bigcap_{x\in F}(x+A)\supset\bigcap_{m\in\Omega}(-x_{n}+A)\ni a$ is empty, which is not possible as this intersection contains the point $a$. ∎

5. Null-finite Borel sets are Haar-meager

In this section we prove that each null-finite set with the universal Baire property in a complete metric group is Haar-meager.

A subset $A$ of a topological group $X$ is defined to have the universal Baire Property (briefly, $A$ is a uBP-set) if for any function $f:K\to X$ from a compact metrizable space $K$ the preimage $f^{-1}(A)$ has the Baire Property in $K$, which means that for some open set $U\subset K$ the symmetric difference $U\triangle f^{-1}(A)$ is meager in $K$. It is well-known that each Borel subset of a topological group has the universal Baire Property.

A uBP-set $A$ of a topological group $X$ is called Haar-meager if there exists a continuous map $f:K\to X$ from a compact metrizable space $K$ such that $f^{-1}(x+A)$ is meager in $K$ for all $x\in X$. By [9], for a complete metric group $X$ the family $\mathsf{HM}_{X}$ of subsets of Haar-meager uBP-sets in $X$ is an invariant $\sigma$-ideal on $X$. For more information on Haar-meager sets, see [9], [11], [10], [18].

Theorem 5.1**.**

Each null-finite uBP-set in a compete metric group is Haar-meager.

Proof.

To derive a contradiction, suppose that a null-finite uBP-set $A$ in a complete metric group $(X,\|\cdot\|)$ is not Haar-meager. Since $A$ is null-finite, there exists a null-sequence $(a_{n})_{n\in\omega}$ such that for every $x\in X$ the set $\{n\in\omega:x+a_{n}\in A\}$ is finite.

Replacing $(a_{n})_{n\in\omega}$ by a suitable subsequence, we can assume that $\|a_{n}\|<\frac{1}{2^{n}}$ for all $n\in\omega$. For every $n\in\omega$ consider the compact set $K_{n}:=\{\theta\}\cup\{a_{m}\}_{m\geq n}\subset X$. The metric restriction on the sequence $(a_{n})_{n\in\omega}$ implies that the function

[TABLE]

is well-defined and continuous (the proof of this fact can be found in [18]; see the proof of Theorem 2).

Since the set $A$ is not Haar-meager and has uBP, there exists a point $z\in X$ such that $B:=\Sigma^{-1}(z+A)$ is not meager and has the Baire Property in the compact metrizable space $K:=\prod_{n\in\omega}K_{n}$. Consequently, there exits a non-empty open set $U\subset K$ such that $U\cap B$ is comeager in $U$. Replacing $U$ by a smaller subset, we can assume that $U$ is of basic form $\{b\}\times\prod_{m\geq j}K_{m}$ for some $j\in\omega$ and some element $b\in\prod_{m<j}K_{m}$. It follows that for every $x\in K_{j}\setminus\{\theta\}$ the set $\{b\}\times\{x\}\times\prod_{m>j}K_{m}$ is closed and open in $K$ and hence the set $C_{x}:=\{y\in\prod_{m>j}K_{m}:\{b\}\times\{x\}\times\{y\}\subset U\cap B\}$ is comeager in $\prod_{m>j}K_{m}$. Then the intersection $\bigcap_{x\in K_{n}\setminus\{\theta\}}C_{x}$ is comeager and hence contains some point $c$. For this point we get the inclusion $\{b\}\times(K_{j}\setminus\{\theta\})\times\{c\}\subset B$. Let $v\in K$ be a unique point such that $\{v\}=\{b\}\times\{\theta\}\times\{c\}$. For any $m\geq j$ the point $a_{m}$ belongs to $K_{j}$ and the inclusion $\{b\}\times\{a_{m}\}\times\{c\}\subset B$ implies that $\Sigma(v)+a_{m}\in z+A$. Consequently, the element $u:=-z+\Sigma(v)$ has the property $u+a_{m}\in A$ for all $m\geq j$, which implies that the set $\{n\in\omega:u+a_{n}\in A\}$ is infinite. But this contradicts the choice of the sequence $(a_{n})_{n\in\omega}$. ∎

6. Null-finite Borel sets are Haar-null

In this section we prove that each universally measurable null-finite set in a complete metric group is Haar-null.

A subset $A$ of a topological group $X$ is defined to be universally measurable if $A$ is measurable with respect to any $\sigma$-additive Radon Borel probability measure on $X$. It is clear that each Radon Borel subset of a topological group is universally measurable. We recall that a measure $\mu$ on a topological space $X$ is Radon if for any $\varepsilon>0$ there exists a compact subset $K\subset X$ such that $\mu(X\setminus K)<\varepsilon$.

A universally measurable subset $A$ of a topological group $X$ is called Haar-null if there exists a $\sigma$-additive Radon Borel probability measure $\mu$ on $X$ such that $\mu(x+A)=0$ for all $x\in X$. By [8], for any complete metric group $X$ the family $\mathsf{HN}_{X}$ of subsets of universally measurable Haar-null subsets of $X$ is an invariant $\sigma$-ideal in $X$. For more information on Haar-null sets, see [8], [16], [17] and also e.g. [13], [15], [23], [28].

Theorem 6.1**.**

Each universally measurable null-finite set $A$ in a complete metric group $X$ is Haar-null.

Proof.

To derive a contradiction, suppose that the null-finite set $A$ is not Haar-null in $X$. Since $A$ is null-finite, there exists a null-sequence $(a_{n})_{n\in\omega}$ such that for each $x\in X$ the set $\{n\in\omega:x+a_{n}\in A\}$ is finite. Replacing $(a_{n})_{n\in\omega}$ by a suitable subsequence, we can assume that $\|a_{n}\|\leq\frac{1}{4^{n}}$ for all $n\in\omega$.

Consider the compact space $\Pi:=\prod_{n\in\omega}\{0,1,\dots,2^{n}\}$ endowed with the product measure $\lambda$ of uniformly distributed measures on the finite discrete spaces $\{0,1,2,3,\dots,2^{n}\}$, $n\in\omega$. Consider the map

[TABLE]

Since $\|p_{i}a_{i}\|\leq 2^{i}\|a_{i}\|\leq 2^{i}\frac{1}{4^{i}}=\frac{1}{2^{i}}$, the series $\sum_{i=0}^{\infty}p_{i}a_{i}$ is convergent and the function $\Sigma$ is well-defined and continuous.

Since the set $A$ is not Haar-null, there exists an element $z\in X$ such that the preimage $\Sigma^{-1}(z+A)$ has positive $\lambda$-measure and hence contains a compact subset $K$ of positive measure. For every $n\in\omega$ consider the subcube $\Pi_{n}:=\prod_{i=0}^{n-1}\{0,1,2,\dots,2^{i}\}\times\prod_{i=n}^{\infty}\{1,2,\dots,2^{i}\}$ of $\Pi$ and observe that $\lambda(\Pi_{n})\to 1$ as $n\to\infty$. Replacing $K$ by $K\cap\Pi_{l}$ for a sufficiently large $l$, we can assume that $K\subset\Pi_{l}$. For every $m\geq l$ let $s_{m}:\Pi_{l}\to\Pi$ be the “back-shift” defined by the formula $s_{m}((x_{i})_{i\in\omega}):=(y_{i})_{i\in\omega}$ where $y_{i}=x_{i}$ for $i\neq m$ and $y_{i}=x_{i}-1$ for $i=m$.

Claim 6.2**.**

For any compact set $C\subset\Pi_{l}$ of positive measure $\lambda(C)$ and any $\varepsilon>0$ there exists $k\geq l$ such that for any $m\geq k$ the intersection $C\cap s_{m}(C)$ has measure $\lambda(C\cap s_{m}(C))>(1-\varepsilon)\lambda(C)$.

Proof.

By the regularity of the measure $\lambda$, the set $C$ has a neighborhood $O(C)\subset\Pi$ such that $\lambda(O(C)\setminus C)<\varepsilon\lambda(K)$. By the compactness of $C$, there exists $k\geq l$ such that for any $m\geq k$ the shift $s_{m}(C)$ is contained in $O(C)$. Hence $\lambda(s_{m}(C)\setminus C)\leq\lambda(O(C)\setminus C)<\varepsilon\lambda(C)$ and thus $\lambda(s_{m}(C)\cap C)=\lambda(s_{m}(C))-\lambda(s_{m}(C)\setminus C)>\lambda(C)-\varepsilon\lambda(C)=(1-\varepsilon)\lambda(C)$. ∎

Using Claim 6.2 we can choose an increasing number sequence $(m_{k})_{k\in\omega}$ such that $m_{0}>l$ and the set $K_{\infty}:=\bigcap_{k\in\omega}s_{m_{k}}(K)$ has positive measure and hence contains a point $\vec{b}:=(b_{i})_{i\in\omega}$. It follows that for every $k\in\omega$ the point $\vec{b}_{k}:=s_{m_{k}}^{-1}(\vec{b})$ belongs to $K\subset\Sigma^{-1}(z+A)$.

Observe that $\Sigma(\vec{b}_{k})=\Sigma(\vec{b})+a_{m_{k}}\in z+A$ and hence $-z+\Sigma(\vec{b})+a_{m_{k}}\in A$ for all $k\in\omega$, which contradicts the choice of the sequence $(a_{n})_{n\in\omega}$. ∎

Remark 6.3**.**

After writing the initial version of this paper we discovered that a result similar to Theorem 6.1 was independently found by Bingham and Ostaszewski [6, Theorem 3].

7. The $\sigma$-ideal generated by (closed) Borel null-finite sets

In this section we introduce two new invariant $\sigma$-ideals generated by (closed) Borel null-finite sets and study the relation of these new ideals to the $\sigma$-ideals of Haar-null and Haar-meager sets.

Namely, for a complete metric group $X$ let

• •

$\sigma\mathsf{NF}_{X}$ be the smallest $\sigma$-ideal containing all Borel null-finite sets in $X$;

• •

$\sigma\overline{\mathsf{NF}}_{X}$ be the smallest $\sigma$-ideal containing all closed null-finite sets in $X$;

• •

$\sigma\overline{\mathsf{HN}}_{X}$ be the smallest $\sigma$-ideal containing all closed Haar-null sets in $X$.

Theorems 5.1 and 6.1 imply that $\sigma\overline{\mathsf{NF}}_{X}\subset\sigma\overline{\mathsf{HN}}_{X}$ and $\sigma\mathsf{NF}_{X}\subset\mathsf{HN}_{X}\cap\mathsf{HM}_{X}$. So, we obtain the following diagram in which an arrow $\mathcal{A}\to\mathcal{B}$ indicates that $\mathcal{A}\subset\mathcal{B}$.

[TABLE]

In Examples 7.1 and 7.2 we show that the $\sigma$-ideal $\sigma\overline{\mathsf{NF}}_{X}$ is strictly smaller than $\sigma\overline{\mathsf{HN}}_{X}$ and the ideal $\sigma\mathsf{NF}_{X}$ is not contained in $\sigma\overline{\mathsf{HN}}_{X}$.

Example 7.1**.**

The closed set $A=\prod_{n\geq 2}C_{n}^{*}$ in the product $X=\prod_{n\geq 2}C_{n}$ of cyclic groups is Haar-null but cannot be covered by countably many closed null-finite sets. Consequently, $A\in\sigma\overline{\mathsf{HN}}_{X}\setminus\sigma\overline{\mathsf{NF}}_{X}$.

Proof.

The set $A=\prod_{n=2}^{\infty}C_{n}^{*}$ has Haar measure

[TABLE]

and hence is Haar-null in the compact Polish group $X$.

Next, we show that $A$ cannot be covered by countably many closed null-finite sets. To derive a contradiction, assume that $A=\bigcup_{n\in\omega}A_{n}$ where each $A_{n}$ is closed and null-finite in $X$. By the Baire Theorem, for some $n\in\omega$ the set $A_{n}$ has non-empty interior in $A$. Consequently, we can find $m>2$ and $a\in\prod_{n=2}^{m-1}C_{n}^{*}$ such that $\{a\}\times\prod_{n=m}^{\infty}C_{n}^{*}\subset A_{n}$. By Example 2.4, the set $\prod_{n=m}^{\infty}C_{n}^{*}$ is not null-finite in $\prod_{n=m}^{\infty}C_{n}$, which implies that the set $A_{n}\supset\{a\}\times\prod_{n=m}^{\infty}C_{n}^{*}$ is not null-finite in the group $X$. But this contradicts the choice of $A_{n}$. ∎

Our next example shows that $\sigma\mathsf{NF}_{X}\not\subset\sigma\overline{\mathsf{HN}}_{X}$ for some compact Polish group $X$.

Example 7.2**.**

For any function $f:\omega\to[2,\infty)$ with $\prod_{n\in\omega}\frac{f(n)-1}{f(n)}>0$, the compact metrizable group $X=\prod_{n\in\mathbb{N}}C_{f(n)}$ contains a null-finite $G_{\delta}$-set $A\subset X$ which cannot be covered by countably many closed Haar-null sets in $X$. Consequently, $A\in\mathsf{NF}_{X}\setminus\sigma\overline{\mathsf{HN}}_{X}$.

Proof.

For every $n\in\omega$ let $g_{n}$ be a generator of the cyclic group $C_{f(n)}$. In the compact metrizable group $X=\prod_{n\in\omega}C_{f(n)}$ consider the null-sequence $(x_{n})_{n\in\omega}$ defined by the formula

[TABLE]

Consider the closed subset $B=\prod_{n\in\omega}C_{f(n)}^{*}$ in $X$ and observe that it has positive Haar measure, equal to the infinite product

[TABLE]

It is easy to see that for every $n\in\omega$ the set $C_{f(n)}^{*}=C_{f(n)}\setminus\{\theta\}$ is not equal to $C_{f(n)}^{*}\cap(g_{n}+C_{f(n)}^{*})$, which implies that the intersection $B\cap(x_{n}+B)$ is nowhere dense in $B$. Consequently, the set $A=B\setminus\bigcup_{n\in\omega}(x_{n}+B)$ is a dense $G_{\delta}$-set in $B$.

We claim that the set $A$ is null-finite. Given any $a\in X$ we should prove that the set $\{n\in\omega:a+x_{n}\in A\}$ is finite. If $a\notin B$, then the open set $X\setminus B$ is a neighborhood of $a$ in $X$. Since the sequence $(a+x_{n})_{n\in\omega}$ converges to $a\in X\setminus B$, the set $\{n\in\omega:a+x_{n}\in B\}\supset\{n\in\omega:a+x_{n}\in A\}$ is finite.

If $a\in B$, then $\{n\in\omega:a+x_{n}\in A\}\subset\{n\in\omega:(B+x_{n})\cap A\neq\emptyset\}=\emptyset$ by the definition of the set $A$.

Next, we prove that the $G_{\delta}$-set $A$ cannot be covered by countably many closed Haar-null sets. To derive a contradiction, assume that $A\subset\bigcup_{n\in\omega}F_{n}$ for some closed Haar-null sets $F_{n}\subset X$. Since the space $A$ is Polish, we can apply the Baire Theorem and find $n\in\omega$ such that the set $A\cap F_{n}$ has non-empty interior in $A$ and hence its closure $\overline{A\cap F_{n}}$ has non-empty interior in $B=\prod_{n\in\omega}C_{f(n)}^{*}$. It is easy to see that each non-empty open subset of $B$ has positive Haar measure in $X$. Consequently, the set $\overline{A\cap F_{n}}$ has positive Haar measure, which is not possible as this set is contained in the Haar-null set $F_{n}$. ∎

Remark 7.3**.**

Answering a question posed in a preceding version of this paper, Adam Kwela [21] constructed two compact null-finite subsets $A,B$ on the real line, whose union $A\cup B$ is not null-finite. This means that the family of subsets of (closed) Borel null-finite subsets on the real line is not an ideal, and the ideal $\sigma\overline{\mathsf{NF}}_{\mathbb{R}}$ contains compact subsets of the real line, which fail to be null-finite.

8. Decomposing non-discrete metric groups into unions of two null-finite sets

By Theorem 6.1 and the countable additivity of the family of Borel Haar-null sets [8], the countable union of Borel null-finite sets in a complete metric group $X$ is Haar-null in $X$ and hence is not equal to $X$. So, a complete metric group cannot be covered by countably many Borel null-finite sets. This result dramatically fails for non-Borel null-finite sets.

Theorem 8.1**.**

Each non-discrete metric group $X$ can be written as the union $X=A\cup B$ of two null-finite subsets $A,B$ of $X$.

Proof.

Being non-discrete, the metric group $X$ contains a non-trivial null-sequence, which generates a non-discrete countable subgroup $Z$ in $X$.

Let $Z=\{z_{n}\}_{n\in\omega}$ be an enumeration of the countable infinite group $Z$ such that $z_{0}=\theta$ and $z_{n}\neq z_{m}$ for any distinct numbers $n,m\in\omega$. By induction we can construct sequences $(u_{n})_{n\in\omega}$ and $(v_{n})_{n\in\omega}$ in $Z$ such that $u_{0}=v_{0}=\theta$ and for every $n\in\mathbb{N}$ the following two conditions are satisfied:

1. (1)

$\|u_{n}\|\leq\frac{1}{n}$ and $u_{n}\notin\{-z_{i}+z_{j}+v_{k}:i,j\leq n,\;k<n\}\cup\{-z_{i}+z_{j}+u_{k}:i,j\leq n,\;k<n\}$; 2. (2)

$\|v_{n}\|\leq\frac{1}{n}$ and $v_{n}\notin\{-z_{i}+z_{j}+u_{k}:i,j\leq n,\;k\leq n\}\cup\{-z_{i}+z_{j}+v_{k}:i,j\leq n,\;k<n\}$.

At the $n$-th step of the inductive construction, the choice of the points $u_{n},v_{n}$ is always possible as the ball $\{z\in Z:\|z\|\leq\frac{1}{n}\}$ is infinite and $u_{n},v_{n}$ should avoid finite sets.

After completing the inductive construction, we obtain the null-sequences $(u_{n})_{n\in\omega}$ and $(v_{n})_{n\in\omega}$ of pairwise distinct points of $Z$ such that for any points $x,y\in Z$ the intersection $\{x+u_{n}\}_{n\in\omega}\cap\{y+u_{m}\}_{m\in\omega}$ is finite and for every distinct points $x,y\in Z$ the intersections $\{x+u_{n}\}_{n\in\omega}\cap\{y+u_{m}\}_{m\in\omega}$ and $\{x+v_{n}\}_{n\in\omega}\cap\{y+v_{m}\}_{m\in\omega}$ are finite.

Using these facts, for every $n\in\omega$ we can choose a number $i_{n}\in\omega$ such that the set $\{z_{n}+v_{k}:k\geq i_{n}\}$ is disjoint with the set $\{z_{i}+v_{m}:i<n,\;m\in\omega\}\cup\{z_{i}+u_{m}:i\leq n,\;m\in\omega\}$ and the set $\{z_{n}+u_{k}:k\geq i_{n}\}$ is disjoint with the set $\{z_{i}+u_{m}:i<n,\;m\in\omega\}\cup\{z_{i}+v_{m}:i\leq n,\;m\in\omega\}$.

We claim that the set $A:=\bigcup_{n\in\omega}\{z_{n}+u_{m}:m>i_{n}\}$ is null-finite in $Z$. This will follow as soon as we verify the condition:

1. (3)

for any $z\in Z$ the set $\{n\in\omega:z+v_{n}\in A\}$ is finite.

Find $j\in\omega$ such that $z=z_{j}$ and observe that for every $n\geq j$ the choice of the number $i_{n}$ guarantees that $\{z_{n}+u_{m}\}_{m>i_{n}}\cap\{z_{j}+v_{n}\}_{n\in\omega}=\emptyset$, so $\{n\in\omega:z+v_{n}\in A\}$ is contained in the finite set $\bigcup_{n<j}\{z_{n}+v_{m}\}_{m\in\omega}\cap\{z_{j}+u_{m}\}_{m\in\omega}$ and hence is finite.

Next, we show that the set $B:=Z\setminus A$ is null-finite in $Z$. This will follow as soon as we verify the condition

1. (4)

for any $z\in Z$ the set $\{m\in\omega:z+u_{m}\in B\}$ is finite.

Find a number $n\in\omega$ such that $z=z_{n}$ and observe that $\{m\in\omega:z+u_{m}\in B\}=\{m\in\omega:z_{n}+u_{m}\notin A\}\subset\{m\in\omega:m\leq i_{n}\}$ is finite. Therefore, the sets $A$ and $B=Z\setminus A$ are null-finite in $Z$.

Using Axiom of Choice, choose a subset $S\subset X$ such that for every $x\in X$ the intersection $S\cap(x+Z)$ is a singleton. It follows that $X=S+Z=(S+A)\cup(S+B)$. We claim that the sets $S+A$ and $S+B$ are null-finite in $X$. This will follow as soon as for any $x\in X$ we check that the sets $\{n\in\omega:x+v_{n}\in S+A\}$ and $\{n\in\omega:x+u_{n}\in S+B\}$ are finite. Since $X=S+Z$, there exist elements $s\in S$ and $z\in Z$ such that $x=s+z$. Observe that if for $n\in\omega$ we get $x+v_{n}\in S+A$, then $s+z+v_{n}=t+a$ for some $t\in S$ and $a\in A\subset Z$. Subtracting $a$ from both sides, we get $t=s+z+v_{n}-a\in s+Z$ and hence $t=s$ as $S\cap(s+Z)$ is a singleton containing both points $t$ and $s$. So, the inclusion $x+v_{n}=s+z+v_{n}\in S+A$ is equivalent to $z+v_{n}\in A$. By analogy we can prove that $x+u_{n}\in S+B$ is equivalent to $z+u_{n}\in B$. Then the sets

[TABLE]

are finite by the properties (3) and (4) of the null-finite sets $A,B$. ∎

Proposition 2.3 and Theorem 8.1 imply the following corollary.

Corollary 8.2**.**

Let $X$ be a non-discrete metric group.

1. (1)

Each null-finite subset of $X$ is contained in some invariant ideal on $X$; 2. (2)

For any ideal $\mathcal{I}$ on $X$ there exists a null-finite set $A\subset X$ such that $A\notin\mathcal{I}$.

Proof.

1. For any null-finite set $A$ in $X$ the family

[TABLE]

is an invariant ideal on $X$ whose elements have empty interiors in $X$ by Proposition 2.3. We recall that by $[X]^{<\omega}$ we denote the ideal of finite subsets of $X$.

2. By Theorem 8.1, the group $X$ can be written as the union $X=A\cup B$ of two null-finite sets. Then for any ideal $\mathcal{I}$ on $X$ one of the sets $A$ or $B$ does not belong to $\mathcal{I}$. ∎

9. Applying null-finite sets to additive functionals

In this section we apply null-finite sets to prove a criterion of continuity of additive functionals on metric groups.

A function $f:X\to Y$ between groups is called additive if $f(x+y)=f(x)+f(y)$ for every $x,y\in X$. An additive function into the real line is called an additive functional.

Theorem 9.1**.**

An additive functional $f:X\to\mathbb{R}$ on a metric group $X$ is continuous if it is upper bounded on a set $B\subset X$ which is not null-finite.

Proof.

Suppose that the functional $f$ is not continuous. Then there exists $\varepsilon>0$ such that $f(U)\not\subset(-\varepsilon,\varepsilon)$ for each neighborhood $U\subset X$ of zero. It follows that for every $n\in\omega$ there is a point $x_{n}\in X$ such that $\|x_{n}\|\leq\frac{1}{4^{n}}$ and $|f(x_{n})|>\varepsilon$. Observe that, $\|2^{n}x_{n}\|\leq 2^{n}\cdot\|x_{n}\|\leq\frac{1}{2^{n}}$ and $|f(2^{n}x_{n})|=2^{n}|f(x_{n})|>2^{n}\varepsilon$. Choose $\varepsilon_{n}\in\{1,-1\}$ such that $\varepsilon_{n}f(2^{n}x_{n})$ is positive and put $z_{n}:=\varepsilon_{n}2^{n}x_{n}$. Then $f(z_{n})>2^{n}\varepsilon$ and $\|z_{n}\|\leq\frac{1}{2^{n}}$.

We claim that the null-sequence $(z_{n})_{n\in\omega}$ witnesses that the set $B$ is null-finite. Given any point $x\in X$ we need to check that the set $\Omega:=\{n\in\omega:x+z_{n}\in B\}$ is finite.

Let $M:=\sup f(B)$ and observe that for every $n\in\Omega$ we have

[TABLE]

which implies $2^{n}\leq\frac{1}{\varepsilon}(M-f(x))$. Hence the set $\Omega$ is finite. Therefore, the null-sequence $(z_{n})_{n\in\omega}$ witnesses that the set $B$ is null-finite, which contradicts our assumption. ∎

Theorems 9.1 and 5.1 imply

Corollary 9.2**.**

An additive functional $f:X\to\mathbb{R}$ on a complete metric group is continuous if and only if it is upper bounded on some uBP set $B\subset X$ which is not Haar-meager.

An analogous result for Haar-null sets follows from Theorems 9.1 and 6.1.

Corollary 9.3**.**

An additive functional $f:X\to\mathbb{R}$ on a complete metric group is continuous if and only if it is upper bounded on some universally measurable set $B\subset X$ which is not Haar-null.

10. Applying null-finite sets to additive functions

In this section we prove some continuity criteria for additive functions with values in Banach spaces or locally convex spaces.

Corollary 10.1**.**

An additive function $f:X\to Y$ from a complete metric group $X$ to a Banach space $Y$ is continuous if for any linear continuous functional $y^{*}:Y\to\mathbb{R}$ the function $y^{*}\circ f:X\to\mathbb{R}$ is upper bounded on some set $B\subset X$ which is not null-finite in $X$.

Proof.

Assuming that $f$ is discontinuous, we can find $\varepsilon>0$ such that for each neighborhood $U\subset X$ of $\theta$ we get $f(U)\not\subset B_{\varepsilon}:=\{y\in Y:\|y\|<\varepsilon\}$. Then for every $n\in\omega$ we can find a point $x_{n}\in X$ such that $\|x_{n}\|\leq\frac{1}{4^{n}}$ and $\|f(x_{n})\|\geq\varepsilon$. Since $\|2^{n}x_{n}\|\leq 2^{n}\|x_{n}\|\leq\frac{1}{2^{n}}$, the sequence $(2^{n}x_{n})_{n\in\omega}$ is a null-sequence in $X$. On the other hand, $\|f(2^{n}x_{n})\|=2^{n}\|f(x_{n})\|\geq 2^{n}\varepsilon$ for all $n\in\omega$, which implies that the set $\{f(2^{n}x_{n})\}_{n\in\omega}$ is unbounded in the Banach space $Y$. By the Banach-Steinhaus Uniform Boundedness Principle [14, 3.15], there exists a linear continuous functional $y^{*}:Y\to\mathbb{R}$ such that the set $\{y^{*}\circ f(2^{n}x_{n}):n\in\omega\}$ is unbounded in $\mathbb{R}$. By our assumption, the additive functional $y^{*}\circ f:X\to\mathbb{R}$ is upper bounded on some set $B\subset X$ which is not null-finite in $X$. By Theorem 9.1, the additive functional $y^{*}\circ f$ is continuous and thus the sequence \big{(}y^{*}\circ f(2^{n}x_{n})\big{)}_{n\in\omega} converges to zero and hence cannot be unbounded in $\mathbb{R}$. This is a desired contradiction, completing the proof. ∎

Corollary 10.1 admits a self-generalization.

Theorem 10.2**.**

An additive function $f:X\to Y$ from a complete metric group $X$ to a locally convex space $Y$ is continuous if for any linear continuous functional $y^{*}:Y\to\mathbb{R}$ the function $y^{*}\circ f:X\to\mathbb{R}$ is upper bounded on some set $B\subset X$ which is not null-finite in $X$.

Proof.

This theorem follows from Corollary 10.1 and the well-known fact [27, p.54] that each locally convex space is topologically isomorphic to a linear subspace of a Tychonoff product of Banach spaces. ∎

Combining Theorem 10.2 with Theorems 5.1 and 6.1, we derive two corollaries.

Corollary 10.3**.**

An additive function $f:X\to Y$ from a complete metric group $X$ to a locally convex space $Y$ is continuous if for any linear continuous functional $y^{*}:Y\to\mathbb{R}$ the function $y^{*}\circ f:X\to\mathbb{R}$ is upper bounded on some uBP-set $B\subset X$ which is not Haar-meager in $X$.

Corollary 10.4**.**

An additive function $f:X\to Y$ from a complete metric group $X$ to a locally convex space $Y$ is continuous if for any linear continuous functional $y^{*}:Y\to\mathbb{R}$ the function $y^{*}\circ f:X\to\mathbb{R}$ is upper bounded on some universally measurable set $B\subset X$ which is not Haar-null in $X$.

11. Applying null-finite sets to mid-point convex functions

In this section we apply null-finite sets to establish a continuity criterion for mid-point convex functions on linear metric spaces.

A function $f:C\to\mathbb{R}$ on a convex subset $C$ of a linear space is called mid-point convex if

[TABLE]

for any $x,y\in C$. Mid-point convex functions are alternatively called Jensen convex.

Theorem 11.1**.**

A mid-point convex function $f:G\to\mathbb{R}$ defined on an open convex subset $G\subset X$ of a metric linear space $X$ is continuous if and only if $f$ is upper bounded on some set $B\subset G$ which is not null-finite in $X$ and whose closure $\overline{B}$ is contained in $G$.

Proof.

The “only if” part is trivial. To prove the “if” part, assume that $f$ is upper bounded on some set $B\subset G$ such that $\overline{B}\subset G$ and $B$ is not null-finite in $X$. We need to check the continuity of $f$ at any point $c\in G$. Shifting the set $G$ and the function $f$ by $-c$, we may assume that $c=\theta$. Also we can replace the function $f$ by $f-f(\theta)$ and assume that $f(\theta)=0$. In this case the mid-point convexity of $f$ implies that $f(\frac{1}{2^{n}}x)\leq\frac{1}{2^{n}}f(x)$ for every $x\in G$ and $n\in\mathbb{N}$.

To derive a contradiction, suppose that the function $f$ is not continuous at $\theta$. Then there exists $\varepsilon>0$ such that $f(U)\not\subset(-\varepsilon,\varepsilon)$ for any open neighborhood $U\subset G$. For every $n\in\omega$ consider the neighborhood $U_{n}:=\{x\in G:\|x\|\leq\frac{1}{4^{n}},\;x\in(\frac{1}{2^{n}}G)\cap(-\frac{1}{2^{n}}G)\}$ and find a point $x_{n}\in U_{n}$ such that $|f(x_{n})|\geq\varepsilon$. Replacing $x_{n}$ by $-x_{n}$, if necessary, we can assume that $f(x_{n})\geq\varepsilon$ (this follows from the inequality $\frac{1}{2}(f(x_{n})+f(-x_{n}))\geq f(\theta)=0$ ensured by the mid-point convexity of $f$).

Since $x_{n}\in\frac{1}{2^{n}}G$, for every $k\leq n$ the point $2^{k}x_{n}$ belongs to $G$ and has prenorm $\|2^{k}x_{n}\|\leq 2^{k}\|x_{n}\|\leq\frac{2^{n}}{4^{n}}=\frac{1}{2^{n}}$ in the metric linear space $(X,\|\cdot\|)$. This implies that $(2^{n}x_{n})_{n\in\omega}$ is a null-sequence in $X$. Since the set $B$ is not null-finite, there exists $a\in X$ such that the set $\Omega:=\{n\in\mathbb{N}:a+2^{n}x_{n}\in B\}$ is infinite. Then $a\in\overline{B}\subset G$. Choose $k\in\mathbb{N}$ so large that $2^{-k+1}a\in-G$ (such number $k$ exists as $-G$ is a neighborhood of $\theta$).

Observe that for every number $n\in\Omega$ with $n>k$ the mid-point convexity of $f$ ensures that

[TABLE]

and hence

[TABLE]

which contradicts the upper boundedness of $f$ on $B$. ∎

Theorem 11.1 implies the following generalization of the classical Bernstein-Doetsch Theorem [5], due to Mehdi [24]; see also a survey paper of Bingham and Ostaszewski [7].

Corollary 11.2**.**

A mid-point-convex function $f:G\to\mathbb{R}$ defined on an open convex subset $G$ of a metric linear space $X$ is continuous if and only if $f$ is upper-bounded on some non-empty open subset of $G$.

Combining Theorem 11.1 with Theorems 5.1 and 6.1, we obtain the following two continuity criteria, which answer the Problem of Baron and Ger [3, P239].

Corollary 11.3**.**

A mid-point convex function $f:G\to\mathbb{R}$ defined on a convex subset $G$ of a complete metric linear space $X$ is continuous if and only if it is upper bounded on some uBP-set $B\subset G$ which is not Haar-meager in $X$.

Proof.

Assume that $f$ is upper bounded on some uBP-set $B\subset G$ which is not Haar-meager in $X$.

Write the open set $G$ of $X$ as the union $U=\bigcup_{n\in\omega}F_{n}$ of closed subsets of $X$. By [9], the countable union of uBP Haar-meager sets is Haar-meager. Since the set $B=\bigcup_{n\in\omega}B\cap F_{n}$ is not Haar-meager, for some $n\in\mathbb{N}$ the subset $B\cap F_{n}$ of $B$ is not Haar-meager. By Theorem 5.1, $B\cap F_{n}$ is not null-finite. Since $\overline{B\cap F_{n}}\subset F_{n}\subset G$ and $f$ is upper bounded on $B\cap F_{n}$, we can apply Theorem 11.1 and conclude that the mid-point convex function $f$ is continuous. ∎

Corollary 11.4**.**

A mid-point convex function $f:G\to\mathbb{R}$ defined on a convex subset $G$ of a complete metric linear space $X$ is continuous if and only if it is upper bounded on some universally measurable set $B\subset G$ which is not Haar-null in $X$.

Proof.

The proof of this corollary runs in exactly the same way as the proof of Corollary 11.3 and uses the well-known fact [8] that the countable union of universally measurable Haar-null sets in a complete metric group is Haar-null. ∎

12. Some Open Problems

In this section we collect some open problems related to null-finite sets.

It is well-known that for a locally compact Polish group $X$, each Haar-meager set in $X$ can be enlarged to a Haar-meager $F_{\sigma}$-set and each Haar-null set in $X$ can be enlarged to a Haar-null $G_{\delta}$-set in $X$. Those “enlargement” results dramatically fail for non-locally compact Polish groups, see [13], [10]. We do not know what happens with null-finite sets in this respect.

Problem 12.1**.**

Is each Borel null-finite subset $A$ of a (compact) Polish group $X$ contained in a null-finite set $B\subset X$ of low Borel complexity?

By [28] (resp. [12, 4.1.6]), each analytic Haar-null (resp. Haar-meager) set in a Polish group is contained in a Borel Haar-null (resp. Haar-meager) set. On the other hand, each non-locally compact Polish group contains a coanalytic Haar-null (resp. Haar-meager) set which cannot be enlarged to a Borel Haar-null (resp. Haar-meager) set, see [13] (resp. [10]).

Problem 12.2**.**

Is each (co)analytic null-finite set $A$ in a Polish group $X$ contained in a Borel null-finite set?

Our next problem ask about the relation of the $\sigma$-ideal $\sigma\mathsf{NF}_{X}$ generated by null-finite sets to other known $\sigma$-ideals.

Problem 12.3**.**

Let $X$ be a Polish group.

1. (1)

Is $\sigma\overline{\mathsf{HN}}_{X}\subset\sigma\mathsf{NF}_{X}$? 2. (2)

Is $\sigma\mathsf{NF}_{X}=\mathsf{HN}_{X}\cap\mathsf{HM}_{X}$?

The negative answer to both parts of Problem 12.3 would follow from the negative answer to the following concrete question.

Problem 12.4**.**

Can the closed Haar-null set $\prod_{n=2}^{\infty}C_{n}^{*}$ in the group $X=\prod_{n=2}^{\infty}C_{n}$ be written as a countable union of Borel null-finite sets in $X$?

Problem 12.5**.**

For an infinite Polish group $X$ and an ideal $\mathcal{I}\in\{\sigma\overline{\mathsf{NF}}_{X},\sigma\mathsf{NF}_{X}\}$ evaluate the standard cardinal characteristics of $\mathcal{I}$:

• •

$\mathrm{add}(\mathcal{I})=\min\{|\mathcal{J}|:\mathcal{J}\subset\mathcal{I}$* and $\bigcup\mathcal{J}\notin\mathcal{I}\}$;*

• •

$\mathrm{non}(\mathcal{I})=\min\{|A|:A\subset X$* and $A\notin\mathcal{I}\}$;*

• •

$\mathrm{cov}(\mathcal{I})=\min\{|\mathcal{J}|:\mathcal{J}\subset\mathcal{I}$* and $\bigcup\mathcal{J}=X\}$;*

• •

$\mathrm{cof}(\mathcal{I})=\min\{|\mathcal{J}|:\forall I\in\mathcal{I}\;\exists J\in\mathcal{J}\;\;(I\subset J)\}$.

The cardinal characteristics of the $\sigma$-ideals $\mathsf{HN}_{X}$, $\mathsf{HM}_{X}$, $\sigma\overline{\mathsf{HN}}_{X}$ on Polish groups are well-studied [4], [2] and play an important role in Modern Set Theory.