Validity space of Dunford-Schwartz pointwise ergodic theorem

TL;DR

This paper characterizes the conditions under which the Dunford-Schwartz pointwise ergodic theorem holds in quasi-non-atomic infinite measure spaces, linking it to the finiteness of measure of level sets of functions.

Contribution

It provides a necessary and sufficient condition for the validity of the ergodic theorem in certain infinite measure spaces.

Findings

Ergodic theorem holds iff measure of {f ≥ λ} is finite for all λ>0

Characterizes the validity of the theorem in quasi-non-atomic spaces

Links measure properties of functions to ergodic convergence

Abstract

We show that if a $\sigma-$finite infinite measure space $(\Omega,\mu)$ is quasi-non-atomic, then the Dunford-Schwartz pointwise ergodic theorem holds for $f\in \mathcal L^1(\Omega)+\mathcal L^{\infty}(\Omega)$ if and only if $\mu\{f\ge \lambda\}<\infty$ for all $\lambda>0$.

Equations

$$A_n(f)=A_n(T,f)=\frac{1}{n}\sum _{k=0}^{n-1}{T}^k(f)$$

$${\mu }_t(f)=\inf \{\lambda >0:\mu \{|f|>\lambda \}\le t\},t>0$$

$$f\in E,g\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty },{\mu }_t(g)\le {\mu }_t(f)\forall t>0$$

$$f\in E,g\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty },\underset{0}{\overset{s}{\int }}{\mu }_t(g)dt\le \underset{0}{\overset{s}{\int }}{\mu }_t(f)dt\forall s>0(\text{writing }g\prec \prec f))$$

$$\Vert f{\Vert }_{{\mathcal{L}}^1\cap {\mathcal{L}}^{\infty }}=\max \left\{\Vert f{\Vert }_1,\Vert f{\Vert }_{\infty }\right\}$$

$$\Vert f{\Vert }_{{\mathcal{L}}^1+{\mathcal{L}}^{\infty }}=\inf \left\{\Vert g{\Vert }_1+\Vert h{\Vert }_{\infty }:f=g+h,g\in {\mathcal{L}}^1,h\in {\mathcal{L}}^{\infty }\right\}={\int }_0^1{\mu }_t(f)dt$$

$${\mathcal{R}}_{\mu }=\{f\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty }:{\mu }_t(f)\to 0\text{ as }t\to \infty \}.$$

$${\mathcal{R}}_{\mu }=\{f\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty }:\mu \{f>\lambda \}<\infty \text{ for all }\lambda >0\}.$$

$$f\in {\mathcal{R}}_{\mu },g\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty },\text{ and }g\prec \prec f,$$

$$\Vert T(f){\Vert }_1\le \Vert f{\Vert }_1\forall f\in {\mathcal{L}}^1\text{ and }\Vert \mathcal{T}(f){\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }\forall f\in {\mathcal{L}}^{\infty }.$$

$$\Vert T{\Vert }_{{\mathcal{L}}^1+{\mathcal{L}}^{\infty }\to {\mathcal{L}}^1+{\mathcal{L}}^{\infty }}\le 1$$

$$\Vert T{\Vert }_{E\to E}\le 1$$

$${\mathcal{C}}_{\mu }=\left\{f\in {\mathcal{L}}^1+{\mathcal{L}}^{\infty }:\{A_n(T,f)\}\text{ converges a.e. for all}T\in DS\right\}.$$

$${\mathcal{C}}_{\mu }={\mathcal{R}}_{\mu }.$$

$${\sigma }^{-1}(E)\in {\mathcal{A}}_1\text{ and }{\mu }_1\left({\sigma }^{-1}(E)\right)={\mu }_2(E)\text{ for every }E\in {\mathcal{A}}_2.$$

$$f={\mu }_t(f)\circ \sigma$$

$$\Vert f{\Vert }_{\infty }=\underset{t\in Q(\nabla )}{\max }|f(t)|.$$

$${\Phi }_0(f)=\sum _{i=1}^k{\lambda }_i\phi (e_i).$$

$${\Phi }_0(f{e}_i)\phi (e_i)=g\phi (e_i)\text{ for all }i.$$

$$g\phi (p_{ij})=g\phi (e_i)\phi (q_j)={\Phi }_0(f{e}_i)\phi (e_i)\phi (q_j)={\Phi }_0(f{e}_i{q}_j)=$$

$$={\Phi }_0(f{q}_j)\phi (q_j)\phi (e_i)=h\phi (q_j)\phi (e_i)=h\phi (p_{ij})$$

$$\underset{i,j}{\sup }\phi (p_{ij})=\left(\underset{i}{\sup }\phi (e_i)\right)\left(\underset{j}{\sup }\phi (q_j)\right)=1_{{\nabla }_2},$$

$$\hat{\mu }(e)=\mu (E)$$

$${\hat{\mu }}_2(\phi (e))={\hat{\mu }}_1(e),e\in {\nabla }_{{\mu }_1}.$$

$$\Vert \Phi (f){\Vert }_1={\Vert \sum _{i=1}^k{\lambda }_i\phi (e_i)\Vert }_1=\sum _{i=1}^k|{\lambda }_i|{\hat{\mu }}_2(\phi (e_i))=\sum _{i=1}^k|{\lambda }_i|{\hat{\mu }}_1(e_i)=\Vert f{\Vert }_1.$$

$$\Phi (f_n)\stackrel{(o)}{⟶}\Phi (f)⟺f_n\stackrel{(o)}{⟶}f,$$

$$U(f)=\sum _{n=1}^{\infty }{U}_n(f{\chi }_{G_n})$$

$$\sum _{n=1}^k{U}_n(f{\chi }_{G_n})=S\left(\sum _{n=1}^{k}f{\chi }_{G_n}\right)\uparrow S(f).$$

$${\Vert U\left(\sum _{n=1}^{k}f{\chi }_{G_n}\right)\Vert }_{\infty }\le {\Vert \sum _{n=1}^{k}f{\chi }_{G_n}\Vert }_{\infty }\le \Vert f{\Vert }_{\infty },$$

$$\hat{T}(g)=T(g)\text{and}{A}_n(\hat{T},g)=A_n(T,g)$$

Full text

1. Introduction

Let $(\Omega,\mathcal{A},\mu)$ be a complete measure space. Denote by $\cal L^{0}=\cal L^{0}(\Omega)$ the algebra of equivalence classes of almost everywhere (a.e.) finite real-valued measurable functions on $\Omega$. Let $\mathcal{L}^{p}\subset\mathcal{L}^{0}$, $1\leq p\leq\infty$, be the $L^{p}-$space equipped with the standard norm $\|\cdot\|_{p}$.

Let $T:\cal L^{1}+\cal L^{\infty}\to\cal L^{1}+\cal L^{\infty}$ be a Dunford-Schwartz operator (see Definition 2.1). Dunford-Schwartz theorem on a.e. convergence of the ergodic averages

[TABLE]

for a Dunford-Schwartz operator $T$ acting in the $L^{p}-$space, $1\leq p<\infty$, of real valued functions of an arbitrary measure space was established in [3]; see also [4, Theorem VIII.6.6].

If $\mu(\Omega)=\infty$, then there is the problem of describing the largest subspace of $\cal L^{1}+\cal L^{\infty}$ for which Dunford-Schwartz theorem is valid. It can be noticed that the averages $A_{n}(f)$ converge a.e. for all $f\in\cal R_{\mu}$, that is, when $f\in\cal L^{1}+\cal L^{\infty}$ is such that $\mu\{f\geq\lambda\}<\infty$ for all $\lambda>0$ (see, for example, [2]). We show that if a $\sigma-$finite infinite measure space $(\Omega,\mu)$ has finitely many atoms or its atoms have equal measures, then $\cal R_{\mu}$ is the largest subspace of $\cal L^{1}+\cal L^{\infty}$ for which the convergence takes place: if $f\in(\cal L^{1}+\cal L^{\infty})\setminus\cal R_{\mu}$, then there exists a Dunford-Schwartz operator $T$ such that the sequence $\{A_{n}(T,f)\}$ does not converge a.e. (Theorem 3.4).

As a corollary, we derive that a fully symmetric space $E\subset\cal L^{1}+\cal L^{\infty}$ of a $\sigma-$finite quasi-non-atomic measure space (see Definition 2.3) possesses the individual ergodic theorem property (see Definition 2.2) if and only if the characteristic function of $\Omega$ does not belong to $E$. In conclusion, we outline some (classes of) fully symmetric spaces that do or do not possess the individual ergodic theorem property.

2. Preliminaries

If $f\in\cal L^{1}+\cal L^{\infty}$, then a non-increasing rearrangement of $f$ is defined as

[TABLE]

(see [6, Ch.II, §2]).

A Banach space $(E,\|\cdot\|_{E})\subset\cal L^{1}+\cal L^{\infty}$ is called symmetric (fully symmetric) if

[TABLE]

(respectively,

[TABLE]

implies that $g\in E$ and $\|g\|_{E}\leq\|f\|_{E}$.

Immediate examples of fully symmetric spaces are $\cal L^{1}\cap\cal L^{\infty}$ with the norm

[TABLE]

and $\cal L^{1}+\cal L^{\infty}$ with the norm

[TABLE]

(see [6, Ch. II, §4]).

Define

[TABLE]

It is clear that $\mathcal{R}_{\mu}$ admits a more direct description:

[TABLE]

Note that if $\mu(\Omega)<\infty$, then $\cal R_{\mu}$ is simply $\cal L^{1}$. Therefore, we will be concerned with infinite measure spaces.

By [6, Ch.II, §4, Lemma 4.4], $(\cal R_{\mu},\|\cdot\|_{\cal L^{1}+\cal L^{\infty}})$ is a symmetric space. In addition, $\cal R_{\mu}$ is the closure of $\cal L^{1}\cap\cal L^{\infty}$ in $\cal L^{1}+\cal L^{\infty}$ (see [6, Ch.II, §3, Section 1]). Furthermore, it follows from definitions of $\cal R_{\mu}$ and $\|\cdot\|_{\cal L^{1}+\cal L^{\infty}}$ that if

[TABLE]

then $g\in\cal R_{\mu}$ and $\|g\|_{\cal L^{1}+\cal L^{\infty}}\leq\|f\|_{\cal L^{1}+\cal L^{\infty}}$. Therefore $(\cal R_{\mu},\|\cdot\|_{\cal L^{1}+\cal L^{\infty}})$ is a fully symmetric space.

Definition 2.1**.**

A linear operator $T:\cal L^{1}+\cal L^{\infty}\to\cal L^{1}+\cal L^{\infty}$ is called a Dunford-Schwartz operator if

[TABLE]

In what follows, we will write $T\in DS$ to indicate that $T$ is a Dunford-Schwartz operator. If $T$ is, in addition, positive, we shall write $T\in DS^{+}$. It is clear that

[TABLE]

for all $T\in DS$ and $T(f)\prec\prec f$ for all $f\in\cal L^{1}+\cal L^{\infty}$ [6, Ch.II, §3, Sec.4]. Therefore $T(E)\subset E$ for every fully symmetric space $E$ and

[TABLE]

(see [6, Ch.II, §4, Sec.2]). In particular, $T(\cal R_{\mu})\subset\cal R_{\mu}$, and the restriction of $T$ on $\cal R_{\mu}$ is a linear contraction.

Definition 2.2**.**

We say that a fully symmetric space $E\subset\mathcal{L}^{1}(\Omega)+\mathcal{L}^{\infty}(\Omega)$ possesses the individual ergodic theorem property, writing $E\in$ IET$\,(\Omega)$, if for every $f\in E$ and $T\in DS$ the averages (1) converge a.e. to some $\widehat{f}\in E$.

Let $\chi_{E}$ be the characteristic function of a set $E\in\mathcal{A}$. Denote $\mathbf{1}=\chi_{\Omega}$. The following fact was noticed in [2, Proposition 2.1].

Proposition 2.1**.**

If $\mu(\Omega)=\infty$, then a symmetric space $E\subset\cal L^{1}+\cal L^{\infty}$ is contained in $\cal R_{\mu}$ if and only if $\mathbf{1}\notin E$.

Here is a Dunford-Schwartz pointwise ergodic theorem in a fully symmetric space $E\subset\cal R_{\mu}$ [2, Theorem 4.3]:

Theorem 2.1**.**

Let $(\Omega,\mu)$ be an infinite measure space. If $E\subset\mathcal{L}^{1}+\mathcal{L}^{\infty}$ is a fully symmetric space with $\mathbf{1}\notin E$, then $E\in IET(\Omega)$. In particular, $\mathcal{R}_{\mu}\in IET(\Omega)$.

Definition 2.3**.**

We say that a measure space $(\Omega,\mu)$ is quasi-non-atomic if it has finitely many atoms or its atoms have the same measure.

Denote

[TABLE]

By Theorem 2.1, $\mathcal{R}_{\mu}\subset\mathcal{C}_{\mu}$.

Our main result is Theorem 3.4 below stating that if a $\sigma-$finite measure space $(\Omega,\mu)$ is quasi-non-atomic, then the fully symmetric space $\mathcal{R}_{\mu}$ is maximal relative to Dunford-Schwartz pointwise ergodic theorem, that is,

[TABLE]

We will begin with the case when $\Omega=(0,\infty)$ equipped with Lebesgue measure; see Theorem 3.1.

In order to establish Theorem 3.1 for a quasi-non-atomic measure space, we will need some properties of non-atomic $\sigma-$finite measure spaces.

Let $(\Omega_{1},\mathcal{A}_{1},\mu_{1})$ and $(\Omega_{2},\mathcal{A}_{2},\mu_{2})$ be measure spaces. A mapping $\sigma:\Omega_{1}\to\Omega_{2}$ is said to be a measure-preserving transformation (m.p.t.) if

[TABLE]

If $f\in\cal L^{0}(\Omega)$, denote $\mathop{\rm supp}(f)=\{f\neq 0\}$. The following property of non-atomic measure spaces can be found in [1, Ch.2, Corollary 7.6].

Theorem 2.2**.**

Let $(\Omega,\mu)$ be a non-atomic $\sigma-$finite measure space. Then, given $0\leq f\in\mathcal{R}_{\mu}$, there is a surjective m.p.t. $\sigma:\mathop{\rm supp}(f)\to\mathop{\rm supp}(\mu_{t}(f))$ such that

[TABLE]

on $\mathop{\rm supp}(f)$.

Let $\nabla$ be a complete Boolean algebra. Denote by $Q(\nabla)$ the Stone compact of $\nabla$, and let $C_{\infty}(Q(\nabla))$ be the algebra of continuous functions $f:Q(\nabla)\rightarrow[-\infty,+\infty]$ assuming possibly the values $\pm\infty$ on nowhere dense subsets of $Q(\nabla)$ [7, 1.4.2]. Let $C(Q(\nabla))$ be the Banach algebra of continuous functions $f:Q(\nabla)\rightarrow\mathbb{R}$ relative to the norm

[TABLE]

We will employ the following extension result.

Theorem 2.3**.**

If $\varphi:\nabla_{1}\to\nabla_{2}$ is an isomorphism of complete Boolean algebras, then there exists a unique isomorphism $\Phi:C_{\infty}(Q(\nabla_{1}))\rightarrow C_{\infty}(Q(\nabla_{2}))$ such that $\Phi(e)=\varphi(e)$ for all $e\in\nabla_{1}$.

Proof.

Show first that the isomorphism $\varphi$ extends up to an isomorphism $\Phi_{0}:C(Q(\nabla_{1}))\rightarrow C(Q(\nabla_{2}))$. Let $\mathbb{R}(\nabla_{1})$ be the subalgebra of step-elements of the algebra $C_{\infty}(Q(\nabla_{1}))$, that is, elements of the form $f=\sum\limits_{i=1}^{k}\lambda_{i}e_{i}$, where $e_{i}\in\nabla_{1}$, $e_{i}e_{j}=0$ if $i\neq j$, and $\lambda_{i}\in\mathbb{R}$, $i=1,...,k$. It is clear that the subalgebra $\mathbb{R}(\nabla_{1})$ is dense in the Banach algebra $(C(Q(\nabla_{1})),\|\cdot\|_{\infty})$.

Put

[TABLE]

Since $\varphi$ is an isomorphism of Boolean algebras, it follows that $\Phi_{0}:\mathbb{R}(\nabla_{1})\to\mathbb{R}(\nabla_{2})$ is an isomorphism and an isometry with respect to the norm $\|\cdot\|_{\infty}$. Due to the density of the subalgebra $\mathbb{R}(\nabla_{i})$ in the Banach algebra $(C(Q(\nabla_{i})),\|\cdot\|_{\infty})$, $i=1,2$, $\Phi_{0}$ uniquely extends to an isomorphism $\Phi_{0}:C(Q(\nabla_{1}))\to C(Q(\nabla_{2}))$. It is clear that $\Phi_{0}(e)=\varphi(e)$ for all $e\in\nabla_{1}$.

If $f\in C_{\infty}(Q(\nabla_{1}))$, then there exists a partition $\{e_{i}\}$ of unity $\mathbf{1}_{\nabla_{1}}$, such that $fe_{i}\in C(Q(\nabla_{1}))$ for each $i$. Since $\varphi$ is an isomorphism of Boolean algebras, it follows that $\{\varphi(e_{i})\}$ is a partition of unity $\mathbf{1}_{\nabla_{2}}$. Therefore, there is a unique $g\in C_{\infty}(Q(\nabla_{2}))$, such that

[TABLE]

If $\{q_{j}\}$ is another partition of unity $\mathbf{1}_{\nabla_{1}}$ such that $fq_{j}\in C(Q(\nabla_{1}))$ for all $j$, then $\{p_{ij}=e_{i}q_{j}\}$ is also a partition of unity $\mathbf{1}_{\nabla_{1}}$ and $fp_{ij}\in C(Q(\nabla_{1}))$ for any $i,j$. If $h\in C_{\infty}(Q(\nabla_{2}))$ is such that $h\varphi(q_{j})=\Phi_{0}(fq_{j})\varphi(q_{j})$ for all $j$, then

[TABLE]

[TABLE]

for any $i,j$. Since

[TABLE]

we have $h=g$. Therefore, we can define the mapping $\Phi:C_{\infty}(Q(\nabla_{1}))\rightarrow C_{\infty}(Q(\nabla_{2}))$ by $\Phi(f)=g$. Clearly, $\Phi:C_{\infty}(Q(\nabla_{1}))\rightarrow C_{\infty}(Q(\nabla_{2}))$ is a unique isomorphism such that $\Phi(e)=\varphi(e)$ for each $e\in\nabla_{1}$. ∎

If $\nabla_{\mu}$ is the complete Boolean algebra of the classes $e=[E]$ of $\mu-$a.e. equal sets in $\mathcal{A}$, then

[TABLE]

is a strictly positive measure on $\nabla_{\mu}$. The following is a refinement of Theorem 2.3 in the case of measure spaces.

Theorem 2.4**.**

Let $(\Omega_{i},\mu_{i})$ be a $\sigma-$finite measure space, $i=1,2$, and let $\varphi:\nabla_{\mu_{1}}\to\nabla_{\mu_{2}}$ be an isomorphism such that

[TABLE]

Then there exists a unique isomorphism $\Phi:\cal L^{0}(\Omega_{1})\rightarrow\cal L^{0}(\Omega_{2})$ such that

1. (1)

$\Phi(e)=\varphi(e)$* for all $e\in\nabla_{\mu_{1}}$;* 2. (2)

$\Phi:\cal L^{1}(\Omega_{1})\to\cal L^{1}(\Omega_{2})$* and $\Phi:\cal L^{\infty}(\Omega_{1})\to\cal L^{\infty}(\Omega_{2})$ are bijective linear isometries;* 3. (3)

If $f_{n},f\in\cal L^{0}(\Omega_{1})$ and $g_{n},g\in\cal L^{0}(\Omega_{2})$, then $\Phi(f_{n})\rightarrow\Phi(f)$ ($\mu_{2}-$a.e.) if and only if $f_{n}\to f$ ($\mu_{1}-$a.e.) and $\Phi^{-1}(g_{n})\to\Phi^{-1}(g)$ ($\mu_{1}-$a.e.) if and only if $g_{n}\to g$ ($\mu_{2}-$a.e.).

Proof.

The existence of a unique isomorphism $\Phi:\cal L^{0}(\Omega_{1})\to\cal L^{0}(\Omega_{2})$ with property (1) follows directly from Theorem 2.3.

(2) It is clear that $\Phi$ is the bijection from the algebra $\mathbb{R}(\nabla_{\mu_{1}})$ of step-elements in $\cal L^{0}(\Omega_{1})$ onto the algebra $\mathbb{R}(\nabla_{\mu_{2}})$. By (2), for every $f=\sum\limits_{i=1}^{k}\lambda_{i}e_{i}\in\mathbb{R}(\nabla_{\mu_{1}})$, we have

[TABLE]

Since the algebra $\mathbb{R}(\nabla_{\mu_{i}})$ is dense in $(\cal L^{1}(\mu_{i}),\|\cdot\|_{1}),i=1,2$, it follows that there exists a bijective linear isometry $U$ from $(\cal L^{1}(\Omega_{1}),\|\cdot\|_{1})$ onto $(\cal L^{1}(\Omega_{2}),\|\cdot\|_{1})$ such that $U(f)=\Phi(f)$ for every $f\in\mathbb{R}(\nabla_{\mu_{1}})$. If $f\in\cal L^{1}_{+}(\Omega_{1})$, then there exists a sequence $\{f_{n}\}\subset\mathbb{R}(\nabla_{\mu_{1}})$ such that $0\leq f_{n}\uparrow f$. Then we have $\|f_{n}-f\|_{1}\to 0$, which implies that $\|U(f_{n})-U(f)\|_{1}\to 0$. Therefore $\Phi(f_{n})=U(f_{n})\uparrow U(f)$. Since $\Phi:\cal L^{0}(\Omega_{1})\to\cal L^{0}(\Omega_{2})$ is an isomorphism, it follows that $\Phi(f_{n})\uparrow\Phi(f)$, and we conclude that $\Phi(f)=U(f)$ for every $f\in\cal L^{1}_{+}(\Omega_{1})$. Now, since $\cal L^{1}(\Omega_{1})=\cal L_{+}^{1}(\Omega_{1})-\cal L^{1}_{+}(\Omega_{1})$, we obtain $\Phi(x)=U(x)$ for all $\in\cal L^{1}(\Omega_{1})$, hence $\Phi:(\cal L^{1}(\Omega_{1}),\|\cdot\|_{1})\to(\cal L^{1}(\Omega_{2}),\|\cdot\|_{1})$ is a bijective linear isometry.

If $0\leq f\in\cal L^{\infty}_{+}(\Omega_{1})$, we choose a sequence $\{f_{n}\}\subset\mathbb{R}(\nabla_{\mu_{1}})$ such that $0\leq f_{n}\uparrow f$ and $\|f_{n}-f\|_{\infty}\to 0$ and repeat the preceding argument.

(3) Since $\Phi:\cal L^{0}(\Omega_{1})\to\cal L^{0}(\Omega_{2})$ is an isomorphism, it follows that, given $f_{n},f\in\cal L^{0}(\Omega_{1})$,

[TABLE]

where $f_{n}\stackrel{{\scriptstyle(o)}}{{\longrightarrow}}f$ is the $(o)-$convergence, that is, there exist $g_{n},h_{n}\in\cal L^{0}(\Omega_{1})$ such that $g_{n}\leq f_{n}\leq h_{n}$ for all $n$ and $g_{n}\uparrow f,\ h_{n}\downarrow f$. This completes the argument because $(o)-$convergence and convergence a.e. coincide (see, for example, [10, Ch.VI, §3]). ∎

We will also need some properties of conditional expectations.

In the case of finite measure the following is known (see, for example, [9, Ch.IV, §IV.3]).

Theorem 2.5**.**

Let $(\Omega,\mathcal{A},\mu)$ be a finite measure space, and let $\mathcal{B}$ be a $\sigma-$subalgebra of $\mathcal{A}$. Then there exists a positive linear contraction $U:\cal L^{1}(\Omega,\cal A,\mu)\rightarrow\cal L^{1}(\Omega,\cal B,\mu)$ such that

1. (1)

$U(\cal L^{1}(\Omega,\cal A,\mu))=\cal L^{1}(\Omega,\cal B,\mu)$* and $U(\cal L^{\infty}(\Omega,\cal A,\mu))=\cal L^{\infty}(\Omega,\cal B,\mu)$;* 2. (2)

$U(\mathbf{1})=\mathbf{1}$* and $\|U(f)\|_{\infty}\leq\|f\|_{\infty}$ for all $f\in\cal L^{\infty}(\Omega,\mathcal{A},\mu)$;* 3. (3)

$U^{2}=U$* and $\int_{\Omega}U(f)d\mu=\int_{\Omega}fd\mu$ for all $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)$.*

The next theorem is a version of Theorem 2.5 for $\sigma-$finite measure.

Theorem 2.6**.**

Let $(\Omega,\mathcal{A},\mu)$ be a $\sigma-$finite measure space, and let $\cal B$ be a $\sigma-$subalgebra of $\cal A$. Then there exists $S\in DS^{+}(\Omega,\cal A,\mu)$ such that

1. (1)

$S(\cal L^{1}(\Omega,\mathcal{A},\mu))=\cal L^{1}(\Omega,\mathcal{B},\mu)$* and $S(\cal L^{\infty}(\Omega,\mathcal{A},\mu))=\cal L^{\infty}(\Omega,\mathcal{B},\mu)$;* 2. (2)

$S(\mathbf{1})=\mathbf{1}$; 3. (3)

$S^{2}=S$* and $\int_{\Omega}S(f)d\mu=\int_{\Omega}fd\mu$ for all $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)$.*

Proof.

If $\mu(\Omega)<\infty$, then Theorem 2.6 follows from Theorem 2.5. So, let $\mu(\Omega)=\infty$. Then there exists the sequence $\{G_{n}\}_{n=1}^{\infty}\subset\cal B$ such that $\mu(G_{n})<\infty$ for all $n$, $G_{n}\cap G_{k}=\varnothing$ if $n\neq k$, and $\bigcup_{n=1}^{\infty}G_{n}=\Omega$. Let $\cal A_{G_{n}}=\{G_{n}\cap E:E\in\cal A\}$ and $\cal B_{G_{n}}=\{G_{n}\cap E:E\in\cal B\},\ n\in\mathbb{N}$. By Theorem 2.5, there exists a conditional expectation $U_{n}:\cal L^{1}(\Omega,\mathcal{A}_{G_{n}},\mu)\rightarrow\cal L^{1}(\Omega,\mathcal{B}_{G_{n}},\mu)$ satisfying conditions (i)-(iii) of Theorem 2.5.

Since $\int_{\Omega}|f|d\mu=\sum_{n=1}^{\infty}\int_{G_{n}}|f|d\mu$, $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)$, the map $U:\cal L^{1}(\Omega,\mathcal{A},\mu)\rightarrow\cal L^{1}(\Omega,\mathcal{B},\mu)$ given by

[TABLE]

is well-defined. It clear that $U$ is a positive linear $\|\cdot\|_{1}-$contraction such that $\|U(f)\|_{\infty}\leq\|f\|_{\infty}$ for all $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)\cap\cal L^{\infty}(\Omega,\mathcal{A},\mu)$ and $U(g)=g$ for all $g\in L^{1}(\Omega,\mathcal{B},\mu)$.

By [2, Theorem 3.1], there is $S\in DS^{+}(\Omega,\cal A,\mu)$ such that $S(f)=U(f)$ for all $f\in\cal L^{1}(\Omega,\cal A,\mu)$; in particular, $S(g)=g$ for all $g\in L^{1}(\Omega,\mathcal{B},\mu)$.

If $0\leq f\in\cal L^{\infty}(\Omega,\mathcal{A},\mu)$, then $0\leq f\chi_{G_{n}}\in\cal L^{\infty}(\Omega,\mathcal{A}_{G_{n}},\mu)$ and $\sum_{n=1}^{k}f\chi_{G_{n}}\uparrow f$ as $k\rightarrow\infty$. Since, by [2, Theorem 3.1], $S|\cal L^{\infty}(\Omega,\mathcal{A},\mu)$ is $\sigma(\cal L^{\infty},\cal L^{1})-$continuous, it follows that

[TABLE]

as $k\to\infty$. Besides, for any $k$ we have

[TABLE]

hence $S(\cal L^{\infty}(\Omega,\cal A,\mu))=\cal L^{\infty}(\Omega,\mathcal{B},\mu)$ and $\|S(f)\|_{\infty}\leq\|f\|_{\infty}$ if $f\in\cal L^{\infty}(\Omega,\cal A,\mu)$.

From the definition of $U$ and $\int_{\Omega}|f|d\mu=\sum_{n=1}^{\infty}\int_{G_{n}}|f|d\mu$, $f\in\cal L^{1}(\Omega,\cal A,\mu)$, it follows that $S(\cal L^{1}(\Omega,\mathcal{A},\mu))=\cal L^{1}(\Omega,\mathcal{B},\mu)$. It is also clear that $S(\mathbf{1})=\mathbf{1}$, $S^{2}=S$, and $\int_{\Omega}S(f)d\mu=\int_{\Omega}fd\mu$ for all $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)$. ∎

If $(\Omega,\cal A,\mu)$ be a $\sigma-$finite measure space, $E\in\mathcal{A}$ and $\cal A_{E}=\{E\cap G:G\in\cal A\}$, then $(E,\mathcal{A}_{E},\mu)$ is also a $\sigma-$finite measure space. We will need the next corollary of Theorem 2.6.

Corollary 2.1**.**

Let $(\Omega,\mathcal{A},\mu)$ be a $\sigma-$finite measure space, and let $E\in\cal A$. If $\cal B$ is a $\sigma-$subalgebra of $\mathcal{A}_{E}$ and $T\in DS(E,\cal B,\mu)$, then there exists $\widehat{T}\in DS(\Omega,\cal A,\mu)$ such that

[TABLE]

for all $g\in\cal L^{1}(E,\cal B,\mu)+\cal L^{\infty}(E,\cal B,\mu)$ and $n$.

Proof.

By Theorem 2.6, there exists $S\in DS^{+}(E,\cal A_{E},\mu)$ such that $S(\cal L^{1}(E,\cal A_{E},\mu))=\cal L^{1}(E,\cal B,\mu)$, $S(\cal L^{\infty}(E,\cal A_{E},\mu))=\cal L^{\infty}(E,\cal B,\mu)$, and $S^{2}=S$.

If we define $\widehat{T}(f)=T(S(f\chi_{E}))$, $f\in\cal L^{1}(\Omega,\mathcal{A},\mu)+\cal L^{\infty}(\Omega,\mathcal{A},\mu)$, then it follows that $\widehat{T}\in DS(\Omega,\cal A,\mu)$, and $\widehat{T}(g)=T(g)$ and $A_{n}(\widehat{T},g)=A_{n}(T,g)$ for all $g\in\cal L^{1}(E,\cal B,\mu)+\cal L^{\infty}(E,\cal B,\mu)$ and $n$. ∎

3. Maximality of the space $\mathcal{R}_{\mu}$

Theorem 3.1**.**

Let $\nu$ be Lebesgue measure on the interval $(0,\infty)$. Then, given $f\in\left(\mathcal{L}^{1}(0,\infty)+\mathcal{L}^{\infty}(0,\infty)\right)\setminus\mathcal{R}_{\nu}$, there exists $T\in DS$ such that the averages (1) do not converge a.e., hence $\mathcal{C}_{\nu}=\mathcal{R}_{\nu}$.

Proof.

Let $f\in(\mathcal{L}^{1}+\mathcal{L}^{\infty})\setminus\mathcal{R}_{\nu}$. If $f=f_{+}-f_{-}$, then either $f_{+}\in(\mathcal{L}^{1}+\mathcal{L}^{\infty})\setminus\mathcal{R}_{\nu}$ or $f_{-}\in(\mathcal{L}^{1}+\mathcal{L}^{\infty})\setminus\mathcal{R}_{\nu}$, so let us assume the former. Then we have

[TABLE]

which implies that there exist $0<a<b<\infty$ such that

[TABLE]

If we denote $G=\{a\leq f_{+}\leq b\}$, then $\mu(G)=\infty$. Since $\mu$ is non-atomic, one can construct a sequence $\{G_{m}\}_{m=0}^{\infty}$ of pairwise disjoint subsets of $G$ such that

[TABLE]

Let $0=n_{0},n_{1},n_{2}\dots$ be an increasing sequence of integers. Define the function $\varphi$ on $(0,\infty)$ by

[TABLE]

Note that $\varphi(t)=0$ when $t\in(0,\infty)\setminus G$.

Next, since $\mu(G_{m})=\mu(G_{m+1})$, there exists a measure preserving isomorphism $\tau_{m}:G_{m}\to G_{m+1}$, $m=0,1,2,\dots$ ([1, Ch.2, Proposition 7.4]). Therefore, we have a measure preserving isomorphism $\tau:G\to G$ such that $\tau(G_{m})=G_{m+1}$ for all $m$. Expand $\tau$ to $(0,\infty)$ by letting $\tau(t)=t$ if $t\in(0,\infty)\setminus G$.

Let us now define $T\in DS(0,\infty)$ as

[TABLE]

Then it follows that

[TABLE]

We have

[TABLE]

Further, since

[TABLE]

there exists such $n_{2}$ that

[TABLE]

As

[TABLE]

one can find $n_{3}$ for which

[TABLE]

Continuing this procedure, we choose $n_{1}<n_{2}<n_{3}<\dots$ to satisfy the inequalities

[TABLE]

for every $t\in G_{0}$, implying that the sequence $\{A_{n}(f_{+})(t)\}$ diverges whenever $t\in G_{0}$.

Finally, note that $T(f_{-})=0$, which implies that

[TABLE]

hence the sequence $\{A_{n}(f)(t)\}$ does not converge almost everywhere on $(0,\infty)$. ∎

Now we shall extend Theorem 3.1 to the class of $\sigma-$finite non-atomic measure spaces:

Theorem 3.2**.**

If $(\Omega,\cal A,\mu)$ is a non-atomic $\sigma-$finite measure space, then $\cal C_{\mu}=\cal R_{\mu}$.

Proof.

Take $f\in(\cal L^{1}(\Omega)+\cal L^{\infty}(\Omega))\setminus\mathcal{R}_{\mu}$. We need to show that $f\notin\cal C_{\mu}$. Without loss of generality, $f\geq 0$ (see the proof of Theorem 3.1). Assume that $\displaystyle\lim_{t\to\infty}\mu_{t}(f)=1$: if $\displaystyle\lim_{t\to\infty}\mu_{t}(f)=\alpha>0$, then we can take $\frac{f}{\alpha}$ instead of $f$. Then we have

[TABLE]

Assume first that $\mu(E)=\infty$, where $E=\{f>1\}$. In this case the functions $f$ and $h=f\chi_{E}$ are equimeasurable, so $\mu_{t}(f)=\mu_{t}(h)$. If $g=h-\chi_{E}$, then $\mathop{\rm supp}(g)=E$ and, given $t>0$,

[TABLE]

[TABLE]

hence $\displaystyle\lim_{t\rightarrow\infty}\mu_{t}(g)=0$.

By Theorem 2.2, there exists a surjective m.p.t.

[TABLE]

such that $g=\mu_{t}(g)\circ\sigma$ on $E$. Then we have

[TABLE]

Since $\mu_{t}(g)+1\in(\mathcal{L}^{1}(0,\infty)+\mathcal{L}^{\infty}(0,\infty))\setminus\mathcal{R}_{\nu}$, Theorem 3.1 entails that there is $T\in DS(0,\infty)$ such that the sequence

[TABLE]

If $\cal B$ is the $\sigma-$algebra of Lebesgue measurable sets in $(0,\infty)$, let us denote

[TABLE]

and let $\varphi([\sigma^{-1}(B)])=[B]$, $B\in\cal B$. It is clear that $\varphi:\nabla_{\sigma}\to\nabla_{\nu}(0,\infty)$ is an isomorphism of Boolean algebras and $\widehat{\nu}(\varphi(p))=\widehat{\mu}(p)$ for all $p\in\nabla_{\sigma}$. Besides, if $B\in\cal B$, $e=\chi_{B}$, and $e\circ\sigma=\chi_{\sigma^{-1}(B)}$, then $\chi_{B}\circ\sigma=\chi_{\sigma^{-1}(B)}$, hence

[TABLE]

By Theorem 2.4, there exists an isomorphism $\Phi:\cal L^{0}(E,\cal A_{\sigma},\mu)\rightarrow\cal L^{0}(0,\infty)$ such that $\Phi(e)=\varphi(e)$ for all $e\in\nabla_{\sigma}$. Show that

[TABLE]

If $u=\sum\limits_{i=1}^{k}\lambda_{i}e_{i}\in\mathbb{R}(\nabla_{\nu}(0,\infty))$, where $\lambda_{i}\in\mathbb{R}$ and $e_{i}\in\nabla_{\nu}(0,\infty)$ with $e_{i}e_{j}=0$ for $i\neq j$, then

[TABLE]

Let now $u\in\cal L_{+}^{0}(0,\infty)$. Then there exists a sequence $0\leq u_{n}\in\mathbb{R}(\nabla_{\nu}(0,\infty))$ such that $u_{n}\uparrow u$, hence $(u_{n}\circ\sigma)\uparrow(u\circ\sigma)$. Since $\Phi$ is an isomorphism, it follows that $\Phi(u_{n}\circ\sigma)\uparrow\Phi(u\circ\sigma)$. Therefore

[TABLE]

If $u\in\cal L^{0}(0,\infty)$, then $u=u_{+}-u_{-}$ with $u_{+},u_{-}\in\cal L_{+}^{0}(0,\infty)$, and (5) follows.

Let $\widehat{T}=\Phi^{-1}T\Phi$. In view of Theorem 2.4 (2), $\widehat{T}\in DS(E,\cal A_{\sigma},\mu)$. Note that, since $\sigma$ is a m.p.t. and $g=\mu_{t}(g)\circ\sigma$, it follows that $f,g,h\in\cal L^{0}(E,\cal A_{\sigma},\mu)$. Then, utilizing (5), we obtain

[TABLE]

[TABLE]

Combining now (4) with Theorem 2.4 (3), we conclude that the averages $A_{n}(\widehat{T},f)$ do not converge a.e. on $E$.

By Corollary 2.1, there exists $\widetilde{T}\in DS(\Omega,\cal A,\mu)$ such that $\widetilde{T}(u)=\widehat{T}(u)$ and $A_{n}(\widetilde{T},u)=A_{n}(\widehat{T},u)$ for all $u\in\cal L^{0}(E,\cal A_{\sigma},\mu)$. Therefore, $A_{n}(\widetilde{T},f)$ do not converge a.e. on $E$, hence $f\notin\cal C_{\mu}$.

If $\mu(E)<\infty$, let $E_{1}=\{f=1\}$. Then, in view of (3), we have

[TABLE]

In particular, if $g_{1}=f\cdot\chi_{E_{1}}-\chi_{E_{1}}=0$, Theorem 2.2 implies that there exists a surjective m.p.t. $\sigma_{1}:\Omega=\mathop{\rm supp}(g_{1})\to(0,\infty)$. Repeating the argument as in the case $\mu(E)=\infty$, we complete the proof. ∎

Next we need a version of Theorem 3.1 for a totally atomic infinite measure space with the atoms of equal measures. If $(\Omega,\mu)$ is such a measure space, then $\Omega=\{\omega_{n}\}_{n=1}^{\infty}$, where $\omega_{n}$ is an atom with $0<\mu(\omega_{n})=\mu(\omega_{n+1})<\infty$ for all $n$. In this case, the algebra $\cal L^{0}(\Omega)$ is isomorphic to the algebra of sequences of real numbers, the algebra $\cal L^{\infty}(\Omega)$ is isomorphic to the algebra

[TABLE]

and the algebra $\cal L^{1}(\Omega)$ is isomorphic to the algebra

[TABLE]

As $l^{1}\subset l^{\infty}$, we have $l^{1}+l^{\infty}=l^{\infty}$. Note also that if $f_{k}=\{\alpha^{(k)}_{n}\}_{n=1}^{\infty}\in l^{\infty}$ and $f=\{\alpha_{n}\}_{n=1}^{\infty}\in l^{\infty}$, then

[TABLE]

Theorem 3.3**.**

If $(\Omega,\mu)$ is a totally atomic infinite measure space with the atoms of equal measures, then $\cal C_{\mu}=\cal R_{\mu}$.

Proof.

We need to show that if $f=\{\alpha_{n}\}_{n=1}^{\infty}\in l^{\infty}\setminus\mathcal{R}_{\mu}$, then $f\notin\cal C_{\mu}$. As in the proof of Theorem 3.2, we may assume that $\alpha_{n}\geq 0$ for every $n$, and $\displaystyle\lim_{t\to\infty}\mu_{t}(f)=1$. Identifying $\Omega=\{\omega_{n}\}_{n=1}^{\infty}$ with $\mathbb{N}$, it is clear that

[TABLE]

Besides, there exists $\alpha>1$ such that the set

[TABLE]

is infinite.

Let $1=n_{0},n_{1},n_{2}\dots$ be an increasing sequence of integers. Define the function $\varphi:\Omega\to\mathbb{R}$ $(E,\mathcal{A}_{E},\mu)$ by

[TABLE]

[TABLE]

[TABLE]

Let $\tau:\Omega\to\Omega$ be defined as

[TABLE]

Finally, define the linear operator $T:l^{\infty}\rightarrow l^{\infty}$ by

[TABLE]

It is clear that $T(l^{1})\subset l^{1}$, $\|T\|_{l^{1}\rightarrow l^{1}}\leq 1$, and $\|T\|_{l^{\infty}\rightarrow l^{\infty}}\leq 1$, that is, $T\in DS(\Omega)$.

We have

[TABLE]

Further, since

[TABLE]

there exists such $n_{2}$ that

[TABLE]

Repeating remaining steps of the proof of Theorem 3.1, we see that the sequence $\{A_{n}(T,f)(m_{1})\}$ does not converge, that is, $f\notin\cal C_{\mu}$. ∎

Finally, we establish Theorem 3.1 for a quasi-non-atomic measure space:

Theorem 3.4**.**

If an infinite measure space $(\Omega,\cal A,\mu)$ is quasi-non-atomic, then $\cal C_{\mu}=\cal R_{\mu}$.

Proof.

We need to show that if $f\in(\cal L^{1}+\cal L^{\infty})\setminus\cal R_{\mu}$, then $f\notin\cal C_{\mu}$. Consider the complete Boolean algebra $\nabla_{\mu}=\{[E]:E\in\mathcal{A}\}$. There exists $e\in\nabla_{\mu}$ such that $e\nabla_{\mu}$ is a non-atomic and $(\mathbf{1}-e)\nabla_{\mu}$ is a totally atomic Boolean algebra. Since $\displaystyle\lim_{t\rightarrow\infty}\mu_{t}(f)>0$ and $f=ef+(\mathbf{1}-e)f$, it follows from the inequality

[TABLE]

(see [1, Ch.2, Proposition 1.3]) that

[TABLE]

Assume that the number of atoms in $(\Omega,\mu)$ is finite. Then $\displaystyle\lim_{t\rightarrow\infty}\mu_{t}((\mathbf{1}-e)f)=0$, implying that $\displaystyle\lim_{t\rightarrow\infty}\mu_{t}(ef)>0$, hence

[TABLE]

By Theorem 3.2, there exists $T\in DS(e\nabla_{\mu})$ such that the averages $A_{n}(T,ef)$ do not converge a.e.

Besides, Corollary 2.1 implies that there is $\widehat{T}\in DS(\Omega,\mu)$ such that

[TABLE]

Next, defining the operator $\widetilde{T}\in DS(\Omega,\mu)$ by

[TABLE]

we obtain

[TABLE]

hence $f\notin\cal C_{\mu}$.

Let now $(\Omega,\mu)$ have infinitely many items of the same measure. Then there exists $e\in\nabla_{\mu}$ such that either $e\nabla_{\mu}\neq\{0\}$ is a non-atomic Boolean algebra and (6) holds or $(\mathbf{1}-e)\nabla_{\mu}\neq\{0\}$ is a totally atomic Boolean algebra with the atoms of equal measures and, with $e^{\perp}=\mathbf{1}-e$,

[TABLE]

In the first case, we find $T$ and define $\widetilde{T}\in DS(\Omega,\mu)$ as above and conclude that the averages $A_{n}(\widetilde{T},f)$ do not converge $\mu-$a.e., that is, $f\notin\cal C_{\mu}$.

In the second case, we use Theorem 3.3 and repeat the preceding argument. ∎

4. Applications

Combining Proposition 2.1 with Theorems 2.1 and 3.4, we arrive at the following characterization of the fully symmetric spaces possessing individual ergodic theorem property.

Theorem 4.1**.**

Let an infinite measure space $(\Omega,\mu)$ be quasi-non-atomic. Given a fully symmetric space $E\subset\cal L^{1}(\Omega)+\cal L^{\infty}(\Omega)$, the following conditions are equivalent:

1. (1)

$E\in IET(\Omega)$; 2. (2)

$E\subset\cal R_{\mu}$; 3. (3)

$\mathbf{1}\notin E$.

To outline the scope of applications of Theorem 4.1, we present some examples of fully symmetric spaces $E$ such that $\mathbf{1}\notin E$ or $\mathbf{1}\in E$.

1. Let $\Phi$ be an Orlicz function, that is, $\Phi:[0,\infty)\to[0,\infty)$ is left-continuous, convex, increasing function such that $\Phi(0)=0$ and $\Phi(u)>0$ for some $u\neq 0$ (see, for example [5, Chapter 2, §2.1]). Let

[TABLE]

be the corresponding Orlicz space, and let

[TABLE]

be the Luxemburg norm in $\cal L^{\Phi}$.  It is well-known that $(\cal L^{\Phi},\|\cdot\|_{\Phi})$ is a fully symmetric space.

Since $\mu(\Omega)=\infty$, if $\Phi(u)>0$ for all $u\neq 0$, then $\int_{\Omega}\left(\Phi\left(\frac{\mathbf{1}}{a}\right)\right)d\mu=\infty$ for each $a>0$, hence $\mathbf{1}\notin\cal L^{\Phi}$. If $\Phi(u)=0$ for all $0\leq u<u_{0}$, then $\mathbf{1}\in\cal L^{\Phi}$. Therefore, Theorem 4.1 implies the following.

Theorem 4.2**.**

Let $(\Omega,\mu)$ be an infinite measure space, and let $\Phi$ be an Orlicz function. If $\Phi(u)>0$ for all $u\neq 0$, then $(\cal L^{\Phi},\|\cdot\|_{\Phi})\in IET(\Omega)$. If $(\Omega,\mu)$ is quasi-non-atomic and $(\cal L^{\Phi},\|\cdot\|_{\Phi})\in IET(\Omega)$, then $\Phi(u)>0$ for all $u\neq 0$.

2. A symmetric space $(E,\|\cdot\|_{E})$ is said to have an order continuous norm if

[TABLE]

If $E$ is a symmetric space with order continuous norm, then $\mu\{|f|>\lambda\}<\infty$ for all $f\in E$ and $\lambda>0$, so $E\subset\cal R_{\mu}$; in particular, $\mathbf{1}\notin E$.

3. Let $\varphi$ be an increasing concave function on $[0,\infty)$ with $\varphi(0)=0$ and $\varphi(t)>0$ for some $t>0$, and let

[TABLE]

the corresponding Lorentz space.

It is well-known that $(\Lambda_{\varphi},\|\cdot\|_{\Lambda_{\varphi}})$ is a fully symmetric space; in addition, if $\varphi(\infty)=\infty$, then $\mathbf{1}\notin\Lambda_{\varphi}$ and if $\varphi(\infty)<\infty$, then $\mathbf{1}\in\Lambda_{\varphi}$. Another application of Theorem 4.1 yields the following.

Theorem 4.3**.**

Let $(\Omega,\mu)$ be an infinite measure space, and let $\varphi$ be an increasing concave function on $[0,\infty)$ with $\varphi(0)=0$ and $\varphi(t)>0$ for some $t>0$. If $\varphi(\infty)=\infty$, then $\Lambda_{\varphi}\in IET(\Omega)$. If $(\Omega,\mu)$ is quasi-non-atomic and $\Lambda_{\varphi}\in IET(\Omega)$, then $\varphi(\infty)=\infty$.

4. Let $E=E(0,\infty)$ be a fully symmetric space. If $s>0$, let the bounded linear operator $D_{s}$ in $E$ be given by $D_{s}(f)(t)=f(t/s),\ t>0$. The Boyd index $q_{E}$ is defined as

[TABLE]

It is known that $1\leq q_{E}\leq\infty$ [8, Vol.II, Ch.II, §2b, Proposition 2.b.2]. Since $\|D_{s}\|\leq\max\{1,s\}$ [8, Vol.II, Ch.II, §2b], $\mathbf{1}\in E$ would imply $D_{s}(\mathbf{1})=\mathbf{1}$ and  $\|D_{s}\|=1$ for all $s\in(0,1)$, hence $q_{E}=\infty$. Thus, if $q_{E}<\infty$, we have $\mathbf{1}\notin E$.