Subrearrangement-invariant function spaces

TL;DR

This paper introduces the concept of subrearrangement-invariance in function spaces, explores its relationship with rearrangement-invariance, and constructs new examples of such spaces with specific properties.

Contribution

It defines subrearrangement-invariance as a generalization of 1-subsymmetry, proves that all rearrangement-invariant spaces are subrearrangement-invariant, and constructs new spaces inspired by Garling.

Findings

Not all function spaces admit an equivalent subrearrangement-invariant norm.

Not all subrearrangement-invariant spaces admit an equivalent rearrangement-invariant norm.

Constructed spaces contain copies of ll_p.

Abstract

Rearrangement-invariance in function spaces can be viewed as a kind of generalization of 1-symmetry for Schauder bases. We define subrearrangement-invariance in function spaces as an analogous generalization of 1-subsymmetry. It is then shown that every rearrangement-invariant function space is also subrearrangement-invariant. Examples are given to demonstrate that not every function space on $(0,\infty)$ admits an equivalent subrearrangement-invariant norm, and that not every subrearrangement-invariant function space on $(0,\infty)$ admits an equivalent rearrangement-invariant norm. The latter involves constructing a new family of function spaces inspired by D.J.H.\ Garling, and we further study them by showing that they are Banach spaces containing copies of $\ell_p$.

Equations

$$\varphi (y)-\epsilon <\theta (x)<\varphi (y)+\epsilon .$$

$$\rho_{X}(f)=\left\{\begin{array}[]{ll}\displaystyle\left\|\sum_{i=1}^{\infty}f(i)e_{i}\right\|_{X}&\text{ if }\sum_{i=1}^{\infty}f(i)e_{i}\text{ converges, and}\\ \\ \infty&\text{ otherwise.}\end{array}\right.$$

$${\text{dist}}_f(s)=\mu \left\{x\in \Omega :|f(x)|>s\right\}.$$

$$\begin{gathered} \Vert f\circ m{\Vert }_X={\Vert \sum _{i=1}^{\infty }(f)(m(i))e_i\Vert }_X={\Vert \sum _{i\in F}f(i)e_{m^{-1}(i)}\Vert }_X \\ ={\Vert \sum _{i\in F}f(i){\text{1}}_F(i)e_{m^{-1}(i)}\Vert }_X={\Vert \sum _{i=1}^{\infty }f(i){\text{1}}_F(i)e_i\Vert }_X=\Vert f{\text{1}}_F{\Vert }_X. \end{gathered}$$

$$\begin{gathered} {\Vert \sum _{k=1}^{\infty }f(k)e_{i_k}\Vert }_X={\Vert \sum _{k=1}^{\infty }(f\circ m^{-1})(i_k)e_{i_k}\Vert }_X={\Vert \sum _{i\in F}(f\circ m^{-1})(i)e_i\Vert }_X \\ ={\Vert \sum _{i=1}^{\infty }g(i){\text{1}}_F(i)e_i\Vert }_X=\Vert g{\text{1}}_F{\Vert }_X=\Vert g\circ m{\Vert }_X \\ =\Vert f\circ m^{-1}\circ m{\Vert }_X=\Vert f{\Vert }_X={\Vert \sum _{i=1}^{\infty }f(i)e_i\Vert }_X. \end{gathered}$$

$${\rho }_Y(f)=\underset{A\in \mathcal{A}}{\sup }{\int }_{A}f(t)dt.$$

$${\int }_A{1}_{(0,b]}(t)dt⩽\lambda (A)⩽\sqrt{\inf A}⩽\sqrt{b}$$

$$\frac{\Vert 1_{(c,c+b]}{\Vert }_Y}{\Vert 1_{(c,c+b]}\circ m{\Vert }_Y}=\frac{\Vert 1_{(c,c+b]}{\Vert }_Y}{\Vert 1_{(0,b]}{\Vert }_Y}⩾\frac{b}{\sqrt{b}}=\sqrt{b}.$$

$$\Vert f{\Vert }_g:=\underset{(i(k){)}_{k=1}^{\infty }\in {\mathbb{N}}^{\uparrow }}{\sup }{\left(\sum _{k=1}^{\infty }|f(i(k)){|}^{p}w(k)\right)}^{1/p}<\infty ,$$

$${\rho }_g(f)=\underset{\begin{array}{c}E,F\subseteq \mathbb{N}\\ m\in \mathbb{MO}(E,F)\end{array}}{\sup }{\left(\sum _{k\in E}(f\circ m)(k{)}^{p}w(k)\right)}^{1/p}.$$

$$\begin{gathered} \sum _{k\in E}(f\circ m)(k{)}^{p}w(k)=\sum _{j=1}^{\infty }(f\circ m\circ n)(j{)}^{p}w(n(j)) \\ ⩽\sum _{j=1}^{\infty }(f\circ m\circ n)(j{)}^{p}w(j)⩽\Vert f{\Vert }_g^p. \end{gathered}$$

$${\rho }_G(f)=\underset{\begin{array}{c}E,F\in \Lambda \\ m\in \mathbb{MO}(E,F)\end{array}}{\sup }{\left({\int }_E(f\circ m)(t{)}^{p}W(t)dt\right)}^{1/p},$$

$$\begin{gathered} \Vert f{\text{1}}_D{\Vert }_G^p{\approx }_{\epsilon }{\int }_E((f{\text{1}}_D)\circ m)(t{)}^{p}W(t)dt={\int }_E(f\circ m)(t{)}^p({\text{1}}_D\circ m)(t)W(t)dt \\ ={\int }_{m^{-1}(D)\cap E}(f\circ m)(t{)}^{p}W(t)dt⩽{\int }_{m^{-1}(D)}(f\circ m)(t{)}^{p}W(t)dt \\ ={\int }_{m^{-1}(D)}((f\circ n)\circ (n^{-1}\circ m))(t{)}^{p}W(t)dt⩽\Vert f\circ n{\Vert }_G^p, \end{gathered}$$

$$\begin{gathered} \Vert f\circ n{\Vert }_G^p{\approx }_{\epsilon }{\int }_A(f\circ n\circ \ell )(t{)}^{p}W(t)dt={\int }_A(f\circ n\circ \ell )(t{)}^p({\text{1}}_D\circ n\circ \ell )(t)W(t)dt \\ ={\int }_A((f{\text{1}}_D)\circ (n\circ \ell ))(t{)}^{p}W(t)dt⩽\Vert f{\text{1}}_D{\Vert }_G^p, \end{gathered}$$

$$\Vert f{\Vert }_G={\left({\int }_0^{s}f(t+r-s{)}^{p}W(t)dt\right)}^{1/p}$$

$${\int }_{\Omega }|f|d\mu ={\int }_0^{\infty }{\text{dist}}_f(t)dt.$$

$${\int }_E(f\circ m)(t)dt={\int }_{F}f(t)dt.$$

$${\int }_E(f\circ m)(t)dt={\int }_0^{\infty }{\text{dist}}_{f\circ m}(t)dt={\int }_0^{\infty }{\text{dist}}_f(t)dt={\int }_{F}f(t)dt.$$

$$b\mapsto {\int }_0^{b}f(t+r-b{)}^{p}W(t)dt$$

$${\int }_0^{s}f(t+r-s{)}^{p}W(t)dt=\underset{b\in [0,r]}{\sup }{\int }_0^{b}f(t+r-b{)}^{p}W(t)dt.$$

$$\Vert f{\Vert }_G^p{\approx }_{\epsilon }{\int }_E(f\circ m)(t{)}^{p}W(t)dt.$$

$$n:(0,b)\setminus D_0\to E\setminus E_0.$$

$$(f\circ m\circ n)(t{)}^p⩽f(t+r-b{)}^p,$$

$$(m\circ n)((t,c)\setminus D_0)\subseteq [(m\circ n)(t),(m\circ n)(c)]$$

$$\begin{gathered} c-t=\lambda (t,c)=\lambda ((t,c)\setminus D_0)=\lambda ((m\circ n)((t,c)\setminus D_0)) \\ ⩽\lambda [(m\circ n)(t),(m\circ n)(c)]=(m\circ n)(c)-(m\circ n)(t). \end{gathered}$$

$$b-t<c-t+\epsilon ⩽(m\circ n)(c)-(m\circ n)(t)+\epsilon ⩽r-(m\circ n)(t)+\epsilon .$$

$$(W\circ n)(t)⩽W(t),$$

$$n((\delta ,t)\setminus D_0)\subseteq [n(\delta ),n(t)]$$

$$t-\delta =\lambda (n((\delta ,t)\setminus D_0))⩽\lambda [n(\delta ),n(t)]=n(t)-n(\delta )⩽n(t).$$

$$(f\circ m\circ n)(t{)}^p(W\circ n)(t)⩽f(t-r+b{)}^{p}W(t)$$

Full text

1. Introduction

A 1-symmetric basis can be viewed as a rearrangement-invariant function space over $\mathbb{N}$ (equipped with the counting measure). Some bases, though, are only 1-subsymmetric, and we would like to explore an analogous property for certain nonatomic function spaces corresponding to this generalization. For instance, in the case where we have a function space $X$ on $(0,\infty)$, we say that it is subrearrangement-invariant whenever the following holds:

For every function $f\in X$, every measurable $F\subseteq(0,\infty)$, and every strictly increasing bijection $m:(0,\infty)\to F$ such that $m$ and $m^{-1}$ are both measure-preserving, we have $\|f\circ m\|_{X}=\|f\textbf{1}_{F}\|_{X}$.

Later, in §2, we give a broader definition for other nonatomic function spaces besides $(0,\infty)$.

D.J.H. Garling was the first to prove that not every subsymmetric basis is symmetric, by publishing a counterexample in 1968 ([Ga68, §5]). Garling’s sequence space, then, already furnishes us with a subrearrangement-invariant function space on $\mathbb{N}$ which is not rearrangement-invariant under any equivalent norm. It is also well-known that there exist 1-unconditional bases which are not subsymmetric, and hence function spaces on $\mathbb{N}$ which fail to admit an equivalent subrearrangement-invariant norm. In §3, we extend these results to a purely nonatomic case, exhibiting an example of a function space on $(0,\infty)$ which is not essentially subrearrangement-invariant, and another example which is subrearrangement-invariant but not essentially rearrangement-invariant. Some geometric properties of these “Garling function spaces” are then explored in §4.

All Banach spaces and function spaces are taken over the real field $\mathbb{R}$, and all measure spaces we assume to be countably additive and $\sigma$-finite. If $\theta$ and $\phi$ are real-valued functions, we use the symbolism $\theta(x)\approx_{\varepsilon}\phi(y)$ to mean that for any $\varepsilon>0$, the arguments $x$ and $y$ can be chosen such that

[TABLE]

Beyond that, all notation and terminology is either standard (such as appears, for instance, in [LT77]) or defined as encountered. In §5 we give an appendix with proofs of some measure-theoretic results we need for the main results, that are probably already known, but which we couldn’t locate in the published literature.

2. Subrearrangement-invariant function spaces

For the following definition, $\beta$ denotes the Borel measure.

Definition 2.1**.**

Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, and let $\mathcal{M}_{0}^{+}(\Omega)$ denote the cone of (nonnegative) $(\mu,\beta)$-measurable functions $f:\Omega\to[0,\infty]$. Suppose $\rho:\mathcal{M}_{0}(\Omega)\to[0,\infty]$ satisfies the following properties for all $a\in(0,\infty)$ and all $f,g\in\mathcal{M}_{0}^{+}(\Omega)$:

• (i)

$\rho(f+g)\leqslant\rho(f)+\rho(g)$;

• (ii)

$\rho(af)=a\rho(f)$; and

• (iii)

$\rho(f)=0$ if and only if $f\equiv 0$ almost everywhere.

We can then define a normed linear space $(X,\|\cdot\|_{X})$ consisting the a.e.-equivalence classes of measurable functions $f:\Omega\to[-\infty,\infty]$ satisfying $\|f\|_{X}:=\rho(|f|)<\infty$. In this case we say that $\rho$ is a function norm on $\Omega$, and $X$ is a function space on $\Omega$ with respect to $\rho$.

Note that our definition differs from other classes of function spaces such as Banach function spaces defined in [BS88, §1] or Köthe function spaces. It is suitable for the present purposes, however.

Remark 2.2*.*

If $(e_{i})_{i=1}^{\infty}$ is a 1-unconditional basis for a Banach space $X$, we can define a function norm $\rho_{X}$ by setting, for all $f:\mathbb{N}\to[0,\infty]$,

[TABLE]

In this way, $X$ can be viewed as a function space on $\mathbb{N}$, with respect to $(e_{i})_{i=1}^{\infty}$.

Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, and $f:\Omega\to[-\infty,\infty]$ a $(\mu,\beta)$-measurable function (where $\beta$ is the Borel measure). The distribution function $\text{dist}_{f}:[0,\infty]\to[0,\infty]$ of $f$ is given by the rule

[TABLE]

Two such measurable functions $f$ and $g$ are said to be equimeasurable whenever $\text{dist}_{f}=\text{dist}_{g}$. In this case we write $f\sim g$.

Definition 2.3**.**

Let $(\Omega,\mu)$ be a $\sigma$-finite measure space. A function space $X$ on $\Omega$ is called rearrangement-invariant iff $\|f\|_{X}=\|g\|_{X}$ for all equimeasurable functions $f,g\in X$. It is essentially rearrangement-invariant provided it admits an equivalent rearrangement-invariant norm.

Remark 2.4*.*

The above definition follows [BS88] rather than the somewhat more stilted definition of rearrangement-invariance found, for instance, in [LT79].

Let $(E,\mu_{E})$ and $(F,\mu_{F})$ be measure spaces. A map $m:E\to F$ is called $\boldsymbol{(\mu_{E},\mu_{F})}$-measurable (or, when $\mu_{E}$ and $\mu_{F}$ are clear from context, simply, measurable) if whenever $A$ is a measurable subset of $F$, the set $m^{-1}(A)$ is measurable in $E$. The map $m$ is called a measure-preserving transformation if whenever $A$ is a measurable subset of $F$, the set $m^{-1}(A)$ is measurable with $\mu_{E}(m^{-1}(A))=\mu_{F}(A)$. If furthermore $m$ is bijective with $m^{-1}$ also measure-preserving, we say that it is a measure-isomorphism.

Definition 2.5**.**

If $E$ and $F$ are totally-ordered measure spaces, we denote by $\mathbb{MO}(E,F)$ the set of all maps $m:E\to F$ such that $m$ is both a stictly increasing measure-isomorphism. Any such $m\in\mathbb{MO}(E,F)$ is called an $\boldsymbol{\mathbb{MO}}$-isomorphism between $E$ and $F$.

Let us now introduce the main subject under study.

Definition 2.6**.**

Let $(\Omega,\mu)$ be a totally-ordered $\sigma$-finite measure space satisfying $\mu(\Omega)=\infty$. We say that a function space $X$ on $\Omega$ is subrearrangement-invariant if for every measurable $F\subseteq\Omega$, every $m\in\mathbb{MO}(\Omega,F)$, and every $f\in X$, we have $\|f\circ m\|_{X}=\|f$1${}_{F}\|_{X}$. We say that $X$ is essentially subrearrangement-invariant whenever it admits an equivalent subrearrangement-invariant norm.

Here, the restriction $\mu(\Omega)=\infty$ has been included since $\mathbb{MO}(\Omega,F)$ would be empty otherwise, whenever $\mu(F)\neq\mu(\Omega)$, and that would make every function space on $\Omega$ trivially subrearrangement-invariant.

The following well-known fact is proved in the appendix.

Proposition 2.7**.**

A 1-unconditional basis $(e_{i})_{i=1}^{\infty}$ for a real Banach space $X$ is 1-symmetric if and only if $X$ is rearrangement-invariant as a function space on $\mathbb{N}$ with respect to $(e_{i})_{i=1}^{\infty}$. It is symmetric if and only if $X$ is essentially rearrangement-invariant.

Next, we give a result which in some sense justifies our definition of subrearrangement-invariance.

Proposition 2.8**.**

A 1-unconditional basis $(e_{i})_{i=1}^{\infty}$ for a real Banach space $X$ is 1-subsymmetric if and only if $X$ is subrearrangement-invariant as a function space on $\mathbb{N}$ with respect to $(e_{i})_{i=1}^{\infty}$. It is subsymmetric if and only if $X$ is essentially subrearrangement-invariant.

Proof.

($\Rightarrow$): Suppose $(e_{i})_{i=1}^{\infty}$ is 1-subsymmetric. Let $F\subseteq\mathbb{N}$ and $m\in\mathbb{MO}(\mathbb{N},F)$, and select any $f\in X$. By 1-subsymmetry of $(e_{i})_{i=1}^{\infty}$ we have

[TABLE]

Hence, $X$ is subrearrangement-invariant with respect to $(e_{i})_{i=1}^{\infty}$.

($\Leftarrow$): Suppose now that $X$ is subrearrangement-invariant with respect to $(e_{i})_{i=1}^{\infty}$. Let $(e_{i_{k}})_{k=1}^{\infty}$ be a subsequence and $f\in X$. Define $m(k)=i_{k}$ for $k\in\mathbb{N}$, and $F:=(i_{k})_{k=1}^{\infty}$. Clearly, $m\in\mathbb{MO}(\mathbb{N},F)$. Define $g:\mathbb{N}\to[0,\infty]$ by letting $g(i)=(|f|\circ m^{-1})(i)$ if $i\in F$ and $g(i)=0$ otherwise. We will need to check that $g\in X$, but this follows from the facts below, together with the identity $g\boldsymbol{1}_{F}=g$. Now, by subrearrangement-invariance and 1-unconditionality we have

[TABLE]

That $(e_{n})_{n=1}^{\infty}$ is subsymmetric if and only if it is essentially subrearrangement-invariant follows easily by considering the equivalent norm $|||x|||=\sup_{(n_{k})\in\mathbb{N}^{\uparrow}}\|\sum_{k=1}^{\infty}x_{k}^{*}(x)x_{n_{k}}\|$. ∎

It is well-known that every 1-symmetric basis is 1-subsymmetric. Similarly, it is easy to show that rearrangement-invariance implies subrearrangement-invariance. We just need a quick preliminary fact before we do.

Proposition 2.9** ([BS88, Proposition 2.7.2]).**

Let $m:E\to F$ be a measure preserving transformation between $\sigma$-finite measure spaces $(E,\mu_{E})$ and $(F,\mu_{F})$. If $f:F\to[0,\infty]$ is a $(\mu_{F},\beta)$-measurable function on $F$, then $f\circ m:E\to[0,\infty]$ is a $(\mu_{E},\beta)$-measurable function on $E$, and $f$ and $f\circ m$ are equimeasurable.

Proposition 2.10**.**

Let $(\Omega,\mu)$ be a totally-ordered $\sigma$-finite measure space satisfying $\mu(\Omega)=\infty$. If $X$ is a rearrangement-invariant function space on $\Omega$, then it is also subrearrangement-invariant.

Proof.

Select any $f\in X$, measurable $F\subseteq\Omega$, and $m\in\mathbb{MO}(\Omega,F)$. Notice that $f|_{F}\circ m=f\circ m$ so that, by Proposition 2.9, $f\circ m\sim f|_{F}$. We also clearly have $f\textbf{1}_{F}\sim f|_{F}$, and hence $f\circ m\sim f\textbf{1}_{F}$. By rearrangment-invariance this means $\|f\circ m\|_{X}=\|f\textbf{1}_{F}\|_{X}$. ∎

Let us close this section by discussing the nontriviality of essential-subrearrangement invariance. There are, after all, well-known examples of 1-unconditional bases which are not subsymmetric under any renorming, for instance the basis for the Tsirelson space. This furnishes us with examples of function spaces on $\mathbb{N}$ which are not essentially subrearrangement-invariant. The following example—a simple modification of the Schreier sequence space—gives us a function space on the purely nonatomic measure space $(0,\infty)$ which fails to be essentially subrearrangement-invariant.

Example 2.11**.**

Denote by $\mathcal{A}$ the family of all subsets $A$ of $(0,\infty)$ satisfying $\lambda(A)\leqslant\sqrt{\inf A}$. For a nonnegative $(\lambda,\beta)$-measurable function $f:(0,\infty)\to[0,\infty]$, we set

[TABLE]

Then $\rho_{Y}$ is a function norm, and we can denote by $Y$ the function space it generates. Furthermore, $Y$ is a Banach space which fails to be essentially subrearrangement-invariant.

Proof.

That $\rho_{Y}$ is a function norm is clear from the definition, and it’s routine (via an argument such as in Proposition 4.1) to show completeness. So we need only prove that it fails to be essentially subrearrangement-invariant. Select any $b\in(0,\infty)$. When selecting $A\in\mathcal{A}$ to estimate $\|\boldsymbol{1}_{(0,b]}\|_{Y}$, we may assume without loss of generality that $\inf A\leqslant b$, else $\int_{A}\boldsymbol{1}_{(0,b]}(t)\;dt=0$. Hence,

[TABLE]

so that $\|\boldsymbol{1}_{(0,b]}\|_{Y}\leqslant\sqrt{b}$. On the other hand, if $c\geqslant\sqrt{b}$ then $\|\boldsymbol{1}_{(c,c+b]}\|_{Y}=b.$ It is clear that $\boldsymbol{1}_{(c,c+b]}\circ m=\boldsymbol{1}_{(0,b]}$ for the shift map $m\in\mathbb{MO}((0,\infty),(c,\infty))$ defined by $m(t)=t+c$. Hence,

[TABLE]

As $b\in(0,\infty)$ was arbitrary, it follows that $Y$ is not essentially subrearrangement-invariant. ∎

3. Garling function spaces

The converse of Proposition 2.10 fails to hold in general, as can be seen from the following example. If $1\leqslant p<\infty$ and $w=(w(k))_{k=1}^{\infty}$ is a nonincreasing sequence of positive real numbers satisfying $w\in c_{0}\setminus\ell_{1}$, then we can define the Garling sequence space $g(w,p)$ as the space of all scalar sequences $f:\mathbb{N}\to[-\infty,\infty]$ satisfying

[TABLE]

where $\mathbb{N}^{\uparrow}$ denotes the family of all increasing sequences in $\mathbb{N}$. (We usually also impose the condition that $w(1)=1$ but this is not always necessary.) It is known from [AAW18, Proposition 2.4] and [AALW18, Lemma 3.1] that the unit vectors in $g(w,p)$ form a 1-unconditional basis which is 1-subsymmetric but not symmetric. In particular, thusly viewed as a function space on $\mathbb{N}$, by Propositions 2.7 and 2.8, it is subrearrangement-invariant but fails to be rearrangement-invariant, or even just essentially rearrangement-invariant. Nevertheless, it remains to be seen whether essential subrearrangement-invariance is a strictly weaker condition than essential rearrangement-invariance in the nonatomic setting. We devote this section, therefore, to exhibiting a function space on $(0,\infty)$ which is subrearrangement-invariant but fails to be essentially rearrangement-invariant.

To accomplish this, we shall simply generalize Garling’s construction. In fact, we will use the very same “split into two sums” trick that Garling did in his original paper [Ga68, §5]. However, in order for this strategy to work, we need to make some adaptations. Part of that will involve the using the measure-theoretic results from §2 of the present paper. Also, we need to characterize Garling sequence spaces slightly differently.

Proposition 3.1**.**

Fix a nonincreasing function $w:\mathbb{N}\to(0,\infty)$ with $w\in c_{0}\setminus\ell_{1}$. For each function $f:\mathbb{N}\to[0,\infty]$ we define

[TABLE]

Then $\rho_{g}$ is a function norm generating the space $g(w,p)$.

Proof.

Let $(i(k))_{k=1}^{\infty}\in\mathbb{N}^{\uparrow}$. By taking $E=\mathbb{N}$, $F=(i(k))_{k=1}^{\infty}$, and $m(k)=i(k)$, it is clear that $\rho_{g}(f)\geqslant\|f\|_{g}$. For the reverse inequality, let $E,F\subseteq\mathbb{N}$ and $m\in\mathbb{MO}(E,F)$. We may assume without loss of generality that $E$ and $F$ are both infinite. Thus, there is a unique $n\in\mathbb{MO}(\mathbb{N},E)$, and this satisfies $m\circ n\in\mathbb{MO}(\mathbb{N},F)$. Since $w$ is nonincreasing, we have

[TABLE]

That $\rho_{g}$ is a function norm generating $g(w,p)$ follows immediately. ∎

Definition 3.2**.**

Let $\mathbb{W}$ denote the set of all nonincreasing $(\lambda,\beta)$-measurable functions $W:(0,\infty)\to(0,\infty)$ satisfying the following conditions:

• (W1)

$\displaystyle\lim_{t\to\infty}W(t)=0$,

• (W2)

$\int_{0}^{\infty}W(t)\;dt=\infty$, and

• (W3)

$\int_{0}^{1}W(t)\;dt<\infty$.

We also denote by $\lambda$ the Lebesgue measure and $\Lambda$ the family of Lebesgue-measurable subsets of $(0,\infty)$. (Recall that $\beta$ is the Borel measure.) For each $(\lambda,\beta)$-measurable $f:(0,\infty)\to[0,\infty]$, set

[TABLE]

where $W\in\mathbb{W}$ and $1\leqslant p<\infty$. We then define a Garling function space, denoted $G_{W,p}(0,\infty)$, as the space of all a.e.-equivalence classes of measurable functions $f:(0,\infty)\to[-\infty,\infty]$ satisfying $\|f\|_{G}:=\rho_{G}(|f|)<\infty$.

Remark 3.3*.*

Conditions (W1) and (W2) are the only ones we use in §3 and the proof of Proposition 4.1. However, for the other results in §4, condition (W3) is needed.

It is clear that $\rho_{G}$ is a function norm, and hence $G_{W,p}(0,\infty)$ is a function space on $(0,\infty)$. We will show later in §4 that it is in fact a Banach space, i.e. that it is complete.

Proposition 3.4**.**

Fix $1\leqslant p<\infty$, and let $W\in\mathbb{W}$. Then $G_{W,p}(0,\infty)$ is subrearrangement-invariant.

Proof.

Fix $D\in\Lambda$ and $n\in\mathbb{MO}((0,\infty),D)$, and $f\in G_{W,p}(0,\infty)$. Observe that there are $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ such that

[TABLE]

where the last inequality follows from the fact that $n^{-1}\circ m$ is an $\mathbb{MO}$-isomorphism from $m^{-1}(D)$ onto its image. On the other hand, there are $A,B\in\Lambda$ and $\ell\in\mathbb{MO}(A,B)$ such that

[TABLE]

where the first equality follows due to the fact that $\textbf{1}_{D}\circ n\circ\ell$ is the identity function on $A$, and the final inequality follows from the fact that $n\circ\ell$ is an $\mathbb{MO}$-isomorphism from $A$ onto its image. ∎

To show that a Garling function space fails to admit an equivalent rearrangement-invariant norm, we need the following intuitively obvious lemma.

Lemma 3.5**.**

Fix $p\in[1,\infty)$ and $r\in(0,\infty)$. Let $W\in\mathbb{W}$ and $f:(0,\infty)\to[0,\infty]$ a measurable function which is nondecreasing on $(0,r)$ and zero everywhere else. Then there is $s\in[0,r]$ so that

[TABLE]

Unfortunately, it requires a somewhat technical proof. We begin with some preliminaries.

Proposition 3.6** ([Bo07, Theorem 2.9.3]).**

Let $(\Omega,\mu)$ be a measure space and $f:\Omega\to[-\infty,\infty]$ a $(\mu,\beta)$-measurable function. Then the $\mu$-integrability of $f$ is equivalent to the Lebesgue integrability of the function $t\mapsto\text{dist}_{f}(t)$, and

[TABLE]

Corollary 3.7**.**

If $E$ and $F$ are measurable subsets of $(0,\infty)$, and $f:(0,\infty)\to[0,\infty]$ is a (nonnegative) $(\lambda,\beta)$-measurable function, then for any measure-preserving transformation $m:E\to F$ we have

[TABLE]

Proof.

By Proposition 2.9, $f$ and $f\circ m$ are equimeasurable, which is to say that $\text{dist}_{f}=\text{dist}_{f\circ m}$. Now by Proposition 3.6 we have

[TABLE]

∎

The proof of the following is given in the appendix.

Proposition 3.8**.**

Let $E$ be a measurable subset of $[-\infty,\infty]$ with $\lambda(E)<\infty$. Then there is a measure-zero subset $E_{0}$ of $E$, a measure-zero subset $D_{0}$ of $[0,\lambda(E)]$, and an $\mathbb{MO}$-isomorphism between $E\setminus E_{0}$ and $[0,\lambda(E)]\setminus D_{0}$.

Proof of Lemma 3.5.

First, observe that since the map

[TABLE]

is continuous on the compact set $[0,r]$, we can find $s\in[0,r]$ so that

[TABLE]

Let $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ be such that

[TABLE]

Without loss of generality we may assume that $F\subseteq(0,r)$, and set $b:=\lambda(F)\leqslant r$. By Proposition 3.8 we can find measure-zero subsets $E_{0}$ of $E$ and $D_{0}$ of $(0,r)$, and an $\mathbb{MO}$-isomorphism

[TABLE]

We claim that

[TABLE]

or, equivalently, $b-t\leqslant r-(m\circ n)(t)$, for each $t\in(0,b)\setminus D_{0}$. Indeed, as $m\circ n$ is order-preserving, we have, for $c\in(t,b)\setminus D_{0}$,

[TABLE]

and since $m\circ n$ is a measure isomorphism from $(0,b)\setminus D_{0}$ onto its image, we also have

[TABLE]

For $\varepsilon>0$ we are free to choose $c\in(t,b)\setminus D_{0}$ so that $b-c<\varepsilon$. Hence,

[TABLE]

As $\varepsilon>0$ was arbitrary, this means $b-t\leqslant r-(m\circ n)(t)$ as desired.

Next we claim that

[TABLE]

or, equivalently, $t\leqslant n(t)$, for each $t\in(0,b)\setminus D_{0}$. Indeed, for $\delta\in(0,t)\setminus D_{0}$ we have

[TABLE]

and hence

[TABLE]

As $\delta\in(0,t)\setminus D_{0}$ can be chosen arbitrarily close to zero, this means $t\leqslant n(t)$ as claimed.

From (3.3) and (3.4) we obtain that

[TABLE]

for all $t\in(0,b)\setminus D_{0}$, and hence, by the above together with (3.1), (3.2), and Corollary 3.7, we have

[TABLE]

∎

We are now set to prove the main result of this section.

Theorem 3.9**.**

If $W(t)=(t+1)^{-1/2}$ then $G_{W,1}(0,\infty)$ fails to admit an equivalent rearrangement-invariant norm.

Proof.

Fix $r\in(0,\infty)$ and let $f_{r}:(0,\infty)\to[0,\infty]$ and $f_{r}^{*}:(0,\infty)\to[0,\infty]$ be defined by

[TABLE]

and

[TABLE]

We claim that $f_{r}$ and $f_{r}^{*}$ are equimeasurable. Indeed, it is clear that $\text{dist}_{f_{r}}(s)=\text{dist}_{f_{r}^{*}}(s)=r$ for all $0\leqslant s\leqslant(1+r)^{-1/2}$ and $\text{dist}_{f_{r}}(s)=\text{dist}_{f_{r}^{*}}(s)=0$ for all $1\leqslant s\leqslant\infty$. Now select $(1+r)^{-1/2}<s<1$. We have $f_{r}(t)>s$ if and only if both $0<t<r$ and $(r+1-t)^{-1/2}>s$, or, equivalently, $r+1-s^{-2}<t<r$. In this case we have

[TABLE]

Similarly, $f_{r}^{*}(t)>s$ if and only if both $0<t<r$ and $(t+1)^{-1/2}>s$, or, equivalently, $0<t<s^{-2}-1$. This gives us

[TABLE]

so that $f_{r}$ and $f_{r}^{*}$ are equimeasurable as claimed.

Note that

[TABLE]

as $r\to\infty$. Thus, to complete the proof, it is enough to show that $\|f_{r}\|_{G}$ is bounded by a number not depending on $r$.

Now we apply Garling’s own “split into two sums” trick, except in our case the “sums” are actually integrals. Since $f_{r}$ is increasing on its support $(0,r)$, and $W\in\mathbb{W}$, by Lemma 3.5 we must have $s\in[0,r]$ so that

[TABLE]

Hence, it suffices to show that each of these pieces is bounded by a number not depending on $s$. For the first piece, note that if $t\in(0,s/2]$ then $(1-t+s)^{-1/2}\leqslant(s/2+1)^{-1/2}$. Hence,

[TABLE]

For the second piece, note that if $t\in[s/2,s]$ then $(t+1)^{-1/2}\leqslant(s/2+1)^{-1/2}$, so that

[TABLE]

∎

4. Geometric properties of Garling function spaces

In this section we show that Garling function spaces are Banach spaces containing $(1+\varepsilon)$-isomorphic copies of $\ell_{p}$. As a consequence, the space $G_{W,1}(0,\infty)$ is nonreflexive. It remains an open question as to whether $G_{W,p}(0,\infty)$ is reflexive when $1<p<\infty$.

Proposition 4.1**.**

Fix $1\leqslant p<\infty$ and $W\in\mathbb{W}$. Then space $G_{W,p}(0,\infty)$ is complete.

Proof.

Let $(f_{i})_{i=1}^{\infty}$ be a Cauchy sequence in $G_{W,p}$. Let $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$. Observe that

[TABLE]

so that $(|f_{i}|W^{1/p})_{i=1}^{\infty}$ is Cauchy in $L_{p}(0,\infty)$. As such, it converges a.e.-pointwise to $g\in L_{p}(0,\infty)$. Similarly,

[TABLE]

so that $(|f_{i}\circ m|W^{1/p})_{i=1}^{\infty}$ converges both in $L_{p}(E)$ and a.e.-pointwise to some $g_{E}\in L_{p}(E)$. Set $f:=gW^{-1/p}$ so that $(|f_{i}|)_{i=1}^{\infty}$ converges a.e.-pointwise to $f$. As $(|f_{i}\circ m|W^{1/p})_{i=1}^{\infty}$ now converges a.e.-pointwise to $|f\circ m|W^{1/p}$, it follows that $|f\circ m|W^{1/p}$ and $g_{E}$ are a.e.-identical.

Since $(f_{i})_{i=1}^{\infty}$ is Cauchy, we can find $M\in(0,\infty)$ so that $\|f_{i}\|_{G}^{p}\leqslant M$ for all $i\in\mathbb{N}$. Furthermore, we can find $i_{0}\in\mathbb{N}$ so that $\|g_{E}-|f_{i_{0}}\circ m|W^{1/p}\|_{L_{p}(E)}^{p}\leqslant 1$. Then

[TABLE]

As $E,F,m$ were arbitrary, we have $\rho_{G}(|f|)\leqslant(1+M)^{1/p}<\infty$ so that $f\in G_{W,p}(0,\infty)$.

Next, select $\varepsilon>0$ and find $N\in\mathbb{N}$ so that $\|f_{i}-f_{j}\|_{G}<\varepsilon/2$ for all $i,j\geqslant N$. Select $j_{0}\geqslant N$ so that $\|g_{E}-|f_{j_{0}}\circ m|W^{1/p}\|_{L_{p}(E)}^{p}<\varepsilon/2$. Then for $i\geqslant N$ we have

[TABLE]

Again as $E,F,m$ were arbitrary and independent of $N$, it follows that $\|f-f_{i}\|_{G}^{p}<\varepsilon$ for all $i\geqslant N$. As $\varepsilon>0$ was also arbitrary, $f_{i}\to f$ in $G_{W,p}(0,\infty)$. ∎

To close, we will show that when $1\leqslant p<\infty$ and $W\in\mathbb{W}$, the space $G_{W,p}(0,\infty)$ contains a copy of $\ell_{p}$. To do this, we will use a basic sequence of characteristic functions as an auxiliary structure. Let us gather some facts about it in the next lemma. In what follows, we denote $\boldsymbol{1}_{i}=\boldsymbol{1}_{(i-1,i]}$ for each $i\in\mathbb{N}$.

Lemma 4.2**.**

Fix $1\leqslant p<\infty$ and $W\in\mathbb{W}$, and set $K=\int_{0}^{1}W(t)\;dt$. Then the sequence $(\boldsymbol{1}_{i}/K)_{i=1}^{\infty}$ is a normalized, monotone, 1-unconditional and 1-subsymmetric basic sequence in $G_{W,p}(0,\infty)$ which 1-dominates the unit vector basis $(g_{i})_{i=1}^{\infty}$ of the Garling sequence space $g(w,p)$, where $w=(w(i))_{i=1}^{\infty}$ is formed by letting $w(i)=K^{-1}\int_{i-1}^{i}W(t)\;dt$ for each $i\in\mathbb{N}$. Furthermore, they are isometrically equivalent for constant coefficients.

Proof.

By replacing $W$ with $K^{-1}W$ if necessary, we may assume without loss of generality that $K=1$.

It’s clear that $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ is normalized. It is also clear that if $M<N\in\mathbb{N}$ and $(a_{i})_{i=1}^{\infty}$ is any sequence of scalars then we have

[TABLE]

which is precisely the criterion for forming a monotone basic sequence.

Next we show that it is 1-unconditional. Let $(a_{i})_{i=1}^{\infty},(b_{i})_{i=1}^{\infty}\in c_{00}$ and satisfy $|a_{i}|\leqslant|b_{i}|$ for all $i\in\mathbb{N}$. Then we can find $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ such that, setting $U_{j}=m^{-1}(F\cap(j-1,j])$ for each $j\in\mathbb{N}$ so that $U_{1}<U_{2}<\cdots$ with $E=\bigcup_{j=1}^{\infty}U_{j}$ and each $\boldsymbol{1}_{j}\circ m=\boldsymbol{1}_{U_{j}}|_{E}$,

[TABLE]

By an analogous argument we have

[TABLE]

whence

[TABLE]

so that $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ is 1-unconditional.

Let us show that it is 1-subsymmetric. Indeed, if $(a_{i})_{i=1}^{\infty}\in c_{00}$ and $(\boldsymbol{1}_{i_{k}})_{k=1}^{\infty}$ is some subsequence, then we can find $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ such that

[TABLE]

Set $E^{\prime}=\bigcup_{k=1}^{\infty}m^{-1}(F\cap(i_{k}-1,i_{k}])$, and define an $\mathbb{MO}$-isomorphism $\ell:(0,\infty)\to\bigcup_{k=1}^{\infty}(i_{k}-1,i_{k}]$ by gluing together the shift maps $(k-1,k]\mapsto(i_{k}-1,i_{k}]$. Then $\ell^{-1}\circ m$ is an $\mathbb{MO}$-isomorphism between $E^{\prime}$ and its image, and for each $k\in\mathbb{N}$ and $t\in E^{\prime}$ we have $\boldsymbol{1}_{i_{k}}(m(t))=\boldsymbol{1}_{k}(\ell^{-1}(m(t)))$. Furthermore, $\boldsymbol{1}_{i_{k}}(m(t))=0$ for each $k\in\mathbb{N}$ and $t\in E\setminus E^{\prime}$. Hence,

[TABLE]

To show the reverse inequality, we instead choose $E,F,m$ so that

[TABLE]

Define $\ell$ as before so that $\ell\circ m$ is an $\mathbb{MO}$-isomorphism between $E$ and its image, and $\boldsymbol{1}_{k}(m(t))=\boldsymbol{1}_{i_{k}}(\ell(m(t)))$ for each $k\in\mathbb{N}$ and $t\in E$. Then

[TABLE]

It follows that $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ is 1-subsymmetric.

To show that it 1-dominates $g(w,p)$, we again let $(a_{i})_{i=1}^{\infty}\in c_{00}$. Select any subsequence $(a_{i_{k}})_{k=1}^{\infty}$. As before, there is an $\mathbb{MO}$-isomorphism $\ell:(0,\infty)\to\bigcup_{k=1}^{\infty}(i_{k}-1,i_{k}]$ defined by gluing together the shift maps $(k-1,k]\mapsto(i_{k}-1,i_{k}]$. Note that $\boldsymbol{1}_{i_{k}}\circ\ell=\boldsymbol{1}_{k}$ for each $k\in\mathbb{N}$. We now have

[TABLE]

By taking the supremum over all subsequences we obtain

[TABLE]

Finally, we consider the last part of the lemma, about being isometrically equivalent for constant coefficients to $(g_{i})_{i=1}^{\infty}$. Indeed, as $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ already 1-dominates it as shown above, we need only show the reverse inequality, i.e. that $(g_{i})_{i=1}^{\infty}$ 1-dominates $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ for constant coefficients. To that end, fix $N\in\mathbb{N}$ and let $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ be such that

[TABLE]

For each $i=1,\cdots,N$, define $A_{i}:=m^{-1}(F\cap(i-1,i])$, and then set $A:=\bigcup_{i=1}^{N}A_{i}$. It is clear that $\lambda(A)<\infty$, so by Proposition 3.8 we can find measure-zero subsets $D_{0}$ of $[0,\lambda(A)]$ and $A_{0}$ of $A$, and an $\mathbb{MO}$-isomorphism $n$ from $D:=[0,\lambda(A)]\setminus D_{0}$ onto $A\setminus A_{0}$. We claim that $t\leqslant n(t)$ for all $t\in D$. Indeed, if we set $b:=\inf n(D)$ then since $b\geqslant 0$ and $n(D_{\leqslant t})\subseteq[b,n(t)]$ we have

[TABLE]

As $W$ is nonincreasing it follows that $W(n(t))\leqslant W(t)$ for all $t\in D$. Note also that $\lambda(A)\leqslant N$ so that $D\subseteq[0,N]$. Furthermore, it is clear that $\boldsymbol{1}_{i}(m(t))=0$ for all $i=1,\cdots,N$ and all $t\in E\setminus A$. Together with Corollary 3.7 we now obtain

[TABLE]

As $(g_{i})_{i=1}^{\infty}$ and $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ are both 1-subsymmetric, we are done. ∎

Theorem 4.3**.**

Fix $1\leqslant p<\infty$ and let $W\in\mathbb{W}$. Then for any $\varepsilon>0$ the basic sequence $(\boldsymbol{1}_{i})_{i=1}^{\infty}$ admits a normalized constant coefficient block basic sequence which is $(1+\varepsilon)$-equivalent to $\ell_{p}$, and which is 2-complemented in $[\boldsymbol{1}_{i}]_{i=1}^{\infty}$.

Proof.

Let $g(w,p)$, $(g_{i})_{i=1}^{\infty}$, and $K$ be as in Lemma 4.2, so that $(\boldsymbol{1}_{i}/K)_{i=1}^{\infty}$ is isometrically equivalent to $(g_{i})_{i=1}^{\infty}$ for constant coefficients. It was shown in [AAW18, §3] that there exists a constant coefficient block basic sequence of $(g_{n})_{n=1}^{\infty}$ which is $(1+\varepsilon)$-equivalent to $\ell_{p}$, for any $\varepsilon>0$. In particular, we can select

[TABLE]

where $1=k_{1}<k_{2}<k_{3}<\cdots\in\mathbb{N}$, so that $(y_{i})_{i=1}^{\infty}$ is $(1+\varepsilon)$-equivalent to $\ell_{p}$.

Next, write

[TABLE]

where $1=k_{1}<k_{2}<k_{3}<\cdots\in\mathbb{N}$.

We claim that $(x_{i})_{i=1}^{\infty}$ is 1-dominated by the unit vector basis of $\ell_{p}$. Indeed, if $(a_{i})_{i=1}^{\infty}\in c_{00}$ then we can find $E,F\in\Lambda$ and $m\in\mathbb{MO}(E,F)$ such that

[TABLE]

so that $(x_{i})_{i=1}^{\infty}\lesssim_{1}\ell_{p}$ as claimed.

By Lemma 4.2, $(\boldsymbol{1}_{i}/K)_{i=1}^{\infty}$ is isometrically equivalent to $(g_{i})_{i=1}^{\infty}$ for constant coefficients, and so $\|y^{\prime}_{i}\|_{g}=\|x^{\prime}_{i}\|_{G}$ for each $i\in\mathbb{N}$. Again from Lemma 4.2, we know that $(g_{i})_{i=1}^{\infty}$ is 1-dominated by $(\boldsymbol{1}_{i}/K)_{i=1}^{\infty}$. It follows that

[TABLE]

That $(x_{i})_{i=1}^{\infty}$ spans a 2-complemented subspace of $[\boldsymbol{1}_{i}]_{i=1}^{\infty}$ follows from the fact that constant-coefficient block basic sequences of a 1-subsymmetric basis are always 2-complemented (see, for instance, [LT77, Proposition 3.a.4]). ∎

Remark 4.4*.*

Although $\ell_{p}$ is complemented in $[\boldsymbol{1}_{i}]_{i=1}^{\infty}$, we do not yet know if it is complemented in $G_{W,p}(0,\infty)$.

Corollary 4.5**.**

Fix $1\leqslant p<\infty$ and $W\in\mathbb{W}$. Then for every $\varepsilon>0$, the space $G_{W,p}(0,\infty)$ contains a subspace which is $(1+\varepsilon)$-isomorphic to $\ell_{p}$. Hence, in particular, the space $G(W,1)$ is nonreflexive.

5. Appendix

Proposition 5.1**.**

Let $E$ and $F$ be Lebesgue-measurable subsets of $\mathbb{R}$, and let $m:E\to F$ be a bijection which is both order-preserving and measure-preserving. Then $m^{-1}$ is also order-preserving and measure-preserving, i.e. $m\in\mathbb{MO}(E,F)$.

Proof.

Clearly, it is enough to show that $m^{-1}$ is measurable. To that end, let us fix a measurable set $A\subseteq E$; we claim that $m(A)$ is also measurable, which will complete the proof.

Denote by $\mathcal{B}=\sigma(\tau)$ the Borel $\sigma$-algebra on $\mathbb{R}$, where $\tau$ denotes the usual metric topology $\mathbb{R}$. Let $\tau_{E}$ be the subspace topology on $E$, i.e. the topology defined by

[TABLE]

Similarly, we denote by $\tau_{F}$ the subspace topology for $F$. It is well-known (and easy to see) that the set

[TABLE]

is a $\sigma$-algebra on $E$, called the trace $\sigma$-algebra. Since $E\cap\tau\subset E\cap\mathcal{B}$, we obtain

[TABLE]

For the reverse inclusion, define

[TABLE]

It is routine to verify that $\Sigma$ is a $\sigma$-algebra on $\mathbb{R}$. Also, it is clear that $\tau\subseteq\Sigma$, since for $U\in\tau$ we have $E\cap U\in\tau_{E}\subseteq\sigma(\tau_{E})$. It follows that $\sigma(\tau)\subseteq\Sigma$, whence also by definition of $\Sigma$ we obtain $E\cap\sigma(\tau)\subseteq\sigma(\tau_{E})$. This gives us the reverse inclusion as desired. We now have the identity $\sigma(\tau_{E})=E\cap\mathcal{B},$ and an identical argument shows that $\sigma(\tau_{F})=F\cap\mathcal{B}.$

It’s a well-known fact in real analysis that we can find $C\in\mathcal{B}$ such that $A\subseteq C$ and $\lambda(C\setminus A)=0$. Now set $C^{\prime}=E\cap C\in\sigma(\tau_{E})$. Since $\lambda(C\setminus A)=0$ there is a measure-zero set $D\in\mathcal{B}$ with $C^{\prime}\setminus A\subseteq C\setminus A\subseteq D$. Set $D^{\prime}:=E\cap D\in\sigma(\tau_{E})$ so that $C^{\prime}\setminus A\subseteq D^{\prime}$ and $\lambda(D^{\prime})=0$.

By a standard argument found, for instance, in the proof of [Bo07, Theorem 2.1.2], it follows that $m(B)$ is Lebesgue-measurable whenever $B\in\sigma(\tau_{E})$. Thus we have $\lambda[m(D^{\prime})]=0$. Observe $m(C^{\prime})\setminus m(A)=m(C^{\prime}\setminus A)\subseteq m(D^{\prime})$ so that (since subsets of measure-zero sets are themselves measure-zero) $\lambda[m(C^{\prime})\setminus m(A)]=0$ as well. Note also that since $C^{\prime}\in\sigma(\tau_{E})$ we have $m(C^{\prime})$ measurable. Since $A\subseteq C^{\prime}$, we obtain $m(A)=m(C^{\prime})\setminus[m(C^{\prime})\setminus m(A)]$, which shows that $m(A)$ is measurable. ∎

Proposition 5.2**.**

Let $E$ and $F$ be Lebesgue-measurable subspaces of $[-\infty,\infty]$, and let $m:F\to E$ be a surjective measure-preserving transformation which is also order-preserving. Then there is a measure-zero subset $F_{0}$ of $F$ such that $m$ is a bijection between $F\setminus F_{0}$ and $E$.

Proof.

For each $x\in E$, let $I_{x}$ be an interval containing $m^{-1}\{x\}$ which is minimal under the relation $\subseteq$. Since $m$ is order-preserving, the $I_{x}$’s are all disjoint, which means only countably many of them have positive measure. In particular, $m^{-1}\{x\}$ is a singleton for all but countably many $x\in E$. Set $E_{0}:=\{x\in E:m^{-1}\{x\}\text{ is not a singleton}\}.$ For each $x\in E_{0}$, select some $f_{x}\in m^{-1}\{x\}$. Now set $F_{0}:=\bigcup_{x\in E_{0}}\left(m^{-1}\{x\}\setminus\{f_{x}\}\right).$ Clearly, $m$ is a bijection between $F\setminus F_{0}$ and $E$. Observe that each $m^{-1}\{x\}\setminus\{f_{x}\}$ has measure zero and that $E_{0}$ is countable. It follows that $F_{0}$ has measure zero. ∎

Let $E$ be a totally-ordered set. An initial segment of $E$ is any subset of $E^{\prime}$ of $E$ such that $E^{\prime}<E\setminus E^{\prime}$.

Proposition 5.3**.**

Let $E$ be a Lebesgue-measurable subset of $[-\infty,\infty]$ with $\lambda(E)<\infty$. Then for each $t\in[0,\lambda(E)]$ there is an initial segment $E_{t}$ of $E$ such that $\lambda(E_{t})=t$.

Proof.

Note that if $E_{t}$ is an initial segment of $E\setminus\{-\infty,\infty\}$ then $E_{t}\cup\{-\infty\}$ is an initial segment of $E$ with the same measure as $E_{t}$. Hence, without loss of generality, we may assume $E\subset\mathbb{R}$. We may also assume that $E$ is bounded, since if the result holds in that case then it can be extended to the unbounded case by considering the union of sets

[TABLE]

Say $E\subseteq[a,b]$ for $-\infty<a<b<\infty$. Define $f:[a,b]\to[0,\lambda(E)]$ by the rule

[TABLE]

Observe that if $y<x\in[a,b]$ then

[TABLE]

so that $f$ is Lipschitz, in particular, continuous. As $f(a)=0$ and $f(b)=\lambda(E)$, we may now apply the Intermediate Value Theorem. ∎

Lemma 5.4**.**

Let $E$ be a measurable subset of $\mathbb{R}$ with $\lambda(E)<\infty$. For each $t\in[0,\lambda(E)]$, let $E_{t}$ be an initial segment of $E$ (whose existence is guaranteed by Proposition 5.3). Define the map $m:E\to[0,\lambda(E)]$ by the rule

[TABLE]

Then $m$ is both measure-preserving and order-preserving. Furthermore, $m$ can be extended to a map $m:\mathbb{R}\to[0,\lambda(E)]$ defined by

[TABLE]

Proof.

It is obvious that $m$ is order-preserving, and it is explicitly proved in [BS88, Proposition 2.7.4] that it is also measure-preserving. For $x\in E$ we set $E_{\leqslant x}:=(-\infty,x]\cap E$, and observe that if $t>\lambda(E_{\leqslant x})$ then $x\in E_{t}$ and if $t<\lambda(E_{\leqslant x})$ then $x\notin E_{t}$. It follows that $m(x)=\lambda(E_{\leqslant x})$ for all $x\in E$. Thus, we can extend $m$ continuously to the function $M:\mathbb{R}\to[0,\lambda(E)]$ via the rule $M(x)=\lambda((-\infty,x]\cap E).$ ∎

Proof of Proposition 3.8.

Since $\{-\infty,\infty\}$ has measure zero, we may assume without loss of generality that $E\subset\mathbb{R}$. Let $m:\mathbb{R}\to E$ be as in Lemma 5.4. It is clear (as in, for instance, the proof of Proposition 5.3) that $m$ is Lipschitz, and hence continuous in the usual sense as well. Since $\lambda$ is inner-regular, we can find a sequence $(K_{n})_{n=1}^{\infty}$ of compact sets and a measure-zero set $L$ such that $E=L\cup\bigcup_{n=1}^{\infty}K_{n}.$ It is known that the image of a bounded measure-zero set under a Lipschitz function is again measure-zero. Furthermore, the continuous image of a compact set is again compact, and in particular measurable. We now have $m(E)=m\left(L\cup\bigcup_{n=1}^{\infty}K_{n}\right)=m(L)\cup\bigcup_{n=1}^{\infty}m(K_{n}).$ It follows that $m(E)$ is measurable. We can now apply Proposition 5.2 to find a subset $E_{0}$ of measure zero such that $m$ is a bijection between $E\setminus E_{0}$ and $m(E)$. Set $D_{0}=[0,\lambda(E)]\setminus m(E)$. We have $\lambda(D_{0})=0$, and by Proposition 5.1, $m$ is an $\mathbb{MO}$-isomorphism between $E\setminus E_{0}$ and $[0,\lambda(E)]\setminus D_{0}$. ∎

Proposition 5.5**.**

If $f,g:\mathbb{N}\to[0,\infty)$ are equimeasurable with

[TABLE]

then either they are both identically zero or else there is a measure-isomorphism $m:\text{supp}(f)\to\text{supp}(g)$ such that $g\circ m=f$ on $\text{supp}(f)$.

Proof.

Obviously, if one of $f$ and $g$ is identically zero then, since they are equimeasurable, so is the other. So let us assume that neither is identically zero. Let $f^{*}(n)=\inf\{\lambda:\text{dist}_{f}(\lambda)\leqslant n\}$ denote the “decreasing rearrangement” of $f$. Since $\lim_{n\to\infty}f(n)=0$ we have also $\lim_{n\to\infty}f^{*}(n)=0$. Now [BS88, Corollary 7.6] gives us a measure-preserving transformation $m_{f}:\text{supp}(f)\to\text{supp}(f^{*})$ such that $f=f^{*}\circ m_{f}$ on $\text{supp}(f)$, and analogously we get $m_{g}:\text{supp}(g)\to\text{supp}(g^{*})$ with $g=g^{*}\circ m_{g}$. Since $f$ and $g$ are equimeasurable, $f^{*}=g^{*}$, and hence $g=f^{*}\circ m_{g}$. This means $g\circ m_{g}^{-1}=f^{*}$ and hence, setting $m=m_{g}^{-1}\circ m_{f}$, we obtain $g\circ m=g\circ m_{g}^{-1}\circ m_{f}=f^{*}\circ m_{f}=f$. ∎

Proof of Proposition 2.7.

($\Rightarrow$): Let $(e_{i})_{i=1}^{\infty}$ be 1-symmetric, and suppose $f$ and $g$ are equimeasurable sequences in $X$. Then so are $|f|$ and $|g|$. If $f$ and $g$ are identically zero then $\|f\|_{X}=0=\|g\|_{X}$ and we are done. Otherwise by Proposition 5.5 there is a bijection $m:\text{supp}(f)\to\text{supp}(g)$ with $f=g\circ m$ on $\text{supp}(f)$. Now we have, by 1-symmetry and 1-unconditionality

[TABLE]

($\Leftarrow$): Suppose that $X$ is rearrangement-invariant with respect to $(e_{i})_{i=1}^{\infty}$, and select a permutation $\pi$ of $\mathbb{N}$. Then its inverse $\pi^{-1}$ exists and is a measure-preserving transformation. Select any $f\in X$, and note that $|f(i)|<\infty$ for all $i\in\mathbb{N}$. By Proposition 2.9, $|f|$ and $|f|\circ\pi^{-1}$ are equimeasurable. Now we have, by 1-unconditionality and rearrangement-invariance

[TABLE]

That $(e_{n})_{n=1}^{\infty}$ is symmetric if and only if it is essentially rearrangement-invariant is clear from considering the equivalent norm $|||x|||=\sup_{\sigma\in\Pi_{\mathbb{N}}}\|\sum_{n=1}^{\infty}e_{n}^{*}(x)e_{\sigma(n)}\|$. ∎

Acknowledgments. Thanks to Lukas Geyer and Ramiro Affonso de Tadeu Guerreiro for assisting in the proofs of Propositions 5.1 and 3.8, respectively.