Repeated quasi-integration on locally compact spaces

TL;DR

This paper investigates the properties of quasi-integrals on locally compact spaces, providing criteria for their repeated integration to produce quasi-linear functionals and characterizing when different orders of integration yield the same results.

Contribution

It introduces criteria for repeated quasi-integration to produce quasi-linear functionals and characterizes when different integration orders are equivalent in this context.

Findings

Criteria for repeated quasi-integration to yield quasi-linear functionals

Conditions under which double quasi-integrals are simple

Characterization of when order of integration does not affect the result

Abstract

When $X$ is locally compact, a quasi-integral (also called a quasi-linear functional) on $ C_c(X)$ is a homogeneous, positive functional that is only assumed to be linear on singly-generated subalgebras. We study simple and almost simple quasi-integrals, i.e., quasi-integrals whose corresponding compact-finite topological measures assume exactly two values. We present a criterion for repeated quasi-integration (i.e., iterated integration with respect to topological measures) to yield a quasi-linear functional. We find a criterion for a double quasi-integral to be simple. We describe how a product of topological measures acts on open and compact sets. We show that different orders of integration in repeated quasi-integrals give the same quasi-integral if and only if the corresponding topological measures are both measures or one of the corresponding topological measures is a positive scalar multiple of a point mass.

Equations

$$B(f)=\{\varphi \circ f:\varphi (0)=0,\varphi \in C(f(X))\}.$$

$$A(f)=\{\varphi \circ f:\varphi \in C(f(X))\}.$$

$$m_f(W)=\mu (f^{-1}(W)\setminus \{0\})\text{ for every open set }W\in \mathbb{R},$$

$$m_f(W)=\mu (f^{-1}(W))\text{ for every open set }W\in \mathbb{R}\setminus \{0\}.$$

$$m_f(W)=\mu (f^{-1}(W))\text{ for every open set }W\in \mathbb{R}.$$

$${\rho }_{\mu }(f)={\int }_{\mathbb{R}}idd{m}_f.$$

$${\rho }_{\mu }(\varphi \circ f)={\int }_{\mathbb{R}}\varphi d{m}_f={\int }_{[a,b]}\varphi d{m}_f,$$

$$|\rho (f)-\rho (g)|\le \parallel f-g\parallel \mu (K),$$

$$|\rho (f)|\le \parallel f\parallel \mu (suppf).$$

$$|\rho (f)-\rho (g)|\le 2\parallel f-g\parallel \mu (K).$$

$$\rho (\varphi \circ f)={\int }_{[a,b]}\varphi d{m}_f=\varphi (y)=\varphi (\rho (f)).$$

$$T_{\rho }(f)(y)=\rho (f_y),S_{\eta }(f)(x)=\eta (f_x).$$

$$|T_{\rho }(f)(y)-T_{\rho }(f)(y^{\prime })|=|\rho (f_y)-\rho (f_{y^{\prime }}|\le 2\parallel f_y-f_{y^{\prime }}\parallel \mu (C)<2ϵ\mu (C),$$

$$\parallel T_{\rho }(f)\parallel \le \parallel f\parallel \mu (C)=\parallel f\parallel \mu ({\pi }_1(suppf)).$$

$$(\eta \times \rho )(f)=\eta (T_{\rho }(f))={\int }_Y{T}_{\rho }(f)d\nu ,$$

$$(\rho \times \eta )(f)=\rho (S_{\eta }(f))={\int }_X{S}_{\eta }(f)d\mu .$$

$$\mu (X)\nu (Y)\le (\nu \times \mu )(X\times Y),\mu (X)\nu (Y)\le (\mu \times \nu )(X\times Y).$$

$$(\nu \times \mu )(X\times Y)=\mu (X)\nu (Y),(\mu \times \nu )(X\times Y)=\mu (X)\nu (Y).$$

$$T_{\rho }(f_1\otimes g$$

$$=\rho (g(y)f_1)+\rho (h(y)f_2)=g(y)\rho (f_1)+h(y)\rho (f_2),$$

$$T_{\rho }(f_1\otimes g+f_2\otimes h)=\rho (f_1)g+\rho (f_2)h=g+h.$$

$$\eta (g)$$

$$=(\eta \times \rho )(f_1\otimes g)+(\eta \times \rho )(f_2\otimes h)=(\eta \times \rho )(f_1\otimes g+f_2\otimes h)$$

$$=\eta (T_{\rho }(f_1\otimes g+f_2\otimes h))=\eta (g+h).$$

$$(\eta \times \rho )(\varphi \circ f)=\eta (T_{\rho }(\varphi \circ f)=\eta (\varphi \circ T_{\rho }(f))=\varphi (\eta \circ T_{\rho }(f))=\varphi ((\eta \times \rho )f).$$

$$(\eta \times \rho )(g^2\otimes h_1{h}_2)$$

$$=((\eta \times \rho )(g\otimes h_1))\cdot ((\eta \times \rho )(g\otimes h_2))$$

$$=\rho (g)\eta (h_1)\rho (g)\eta (h_2)=1.$$

$$(\eta \times \rho )(g^2\otimes h_1{h}_2)=\rho (g^2)\eta (h_1{h}_2)=\eta (h_1{h}_2)=0,$$

$$(\eta \times \rho )(g^2\otimes h_1{h}_2)=\rho (g)\eta (h_1)\rho (g)\eta (h_2)=(\rho (g){)}^2>0.$$

$$(\eta \times \rho )(g^2\otimes h_1{h}_2)=\rho (g^2)\eta (h_1{h}_2)=0.$$

$$(\nu \times \mu )(U)={\int }_Y\mu (U_y)d\nu (y).$$

$$(\nu \times \mu )(K)={\int }_Y\mu (K_y)d\nu (y).$$

Full text

1. Introduction

Quasi-integrals (also called quasi-linear functionals) generalize linear functionals. They were first introduced on a compact space by J. Aarnes in [1], where Aarnes also studied their properties, and proved the representation theorem of quasi-integrals in terms of topological measures (formerly called quasi-measures). For uses of quasi-linear functionals in symplectic topology one may consult numerous papers beginning with [10] (which has been cited over 100 times), and a monograph [11].

In [9] D. Grubb gives a nice treatment of repeated quasi-integration on a product of compact spaces. In this paper we generalize all results from [9] to a locally compact setting, and also obtain new results previously not known in the compact case. In proving results about quasi-linear functionals and topological measures in a locally compact setting one loses certain convenient features present in the compact case: quasi-integrals and topological measures are not finite in general, subalgebras do not contain constants, quasi-integrals are no longer states, etc. On another level, to tackle the problems related to repeated integration on locally compact spaces, one needs a variety of results about quasi-liner functionals and topological measures. These include a Representation Theorem showing that quasi-linear functionals are obtained by integration with respect to compact-finite topological measures, continuity of quasi-integrals with respect to the topology of uniform convergence on compacta, and others. These results are obtained in [12], [4], [7], and [6].

In this paper $X$ is a Hausdorff, locally compact, connected space. In Section 2 we give necessary definitions and facts. We define quasi-integrals and topological measures on a locally compact space and outline the correspondence between them. In Section 3 we study simple and almost simple quasi-integrals and topological measures, which are necessary for investigating repeated quasi-integration. In Section 4 we define and study repeated quasi-integrals. In particular, we give a criterion for repeated quasi-integration to yield a quasi-linear functional and a criterion for a double quasi-integral to be simple. In Section 5 we give formulas that describe how a product of compact-finite topological measures acts on open and compact sets. We show that different orders of integration in repeated quasi-integrals give the same quasi-linear functional if and only if corresponding compact-finite topological measures are both measures or one of the corresponding topological measures is a positive scalar multiple of a point mass.

2. Preliminaries

We use continuous functions with the uniform norm; $C_{c}(X)$ is the set of real-valued continuous functions on $X$ with compact support, and $C_{c}^{+}(X)$ is the set of nonnegative functions from $C_{c}(X)$. By $supp\ f$ we mean $\overline{\{x:f(x)\neq 0\}}$. We denote by $1$ the constant function $1(x)=1$, by $id$ the identity function $id(x)=x$, and by $1_{E}$ the indicator function of the set $E$. When we consider maps into extended real numbers, they are not identically $\infty$. We denote by $\overline{E}$ the closure of a set $E$, and by $\bigsqcup$ a union of disjoint sets. $\mathscr{O}(X)$, $\mathscr{C}(X)$, and $\mathscr{K}(X)$ stand, respectively, for the collections of open, closed, and compact subsets of $X$.

We are interested in repeated quasi-integrals, i.e. repeated quasi-integration with respect to topological measures on locally compact spaces. Since $X\times Y$ is locally compact iff $X$ and $Y$ are locally compact, we assume that $X$ and $Y$ are locally compact.

We will use the following definitions and facts.

Definition 2.1**.**

A topological measure on a locally compact space $X$ is a set function $\mu:\mathscr{C}(X)\cup\mathscr{O}(X)\to[0,\infty]$ satisfying the following conditions:

1. (TM1)

If $A,B,A\sqcup B\in\mathscr{K}(X)\cup\mathscr{O}(X)$ then $\mu(A\sqcup B)=\mu(A)+\mu(B).$ 2. (TM2)

If $U\in\mathscr{O}(X)$ then $\mu(U)=\sup\{\mu(K):K\in\mathscr{K}(X),\ K\subseteq U\}.$ 3. (TM3)

If $F\in\mathscr{C}(X)$ then $\mu(F)=\inf\{\mu(U):U\in\mathscr{O}(X),\ F\subseteq U\}.$

A topological measure $\mu$ on $X$ is called compact-finite if $\mu(K)<\infty$ for $K\in\mathscr{K}(X)$; $\mu$ is finite if $\mu(X)<\infty$.

Theorem 2.2**.**

Let $\mu$ be a topological measure on a locally compact space $X$. The following are equivalent:

• (a)

If $C,K$ are compact subsets of $X$, then $\mu(C\cup K)\leq\mu(C)+\mu(K)$.

• (b)

If $U,V$ are open subsets of $X$, then $\mu(U\cup V)\leq\mu(U)+\mu(V)$.

• (c)

$\mu$* admits a unique extension to an inner regular on open sets, outer regular Borel measure $m$ on the Borel $\sigma$-algebra of subsets of $X$. $m$ is a Radon measure iff $\mu$ is compact-finite. If $\mu$ is finite then $m$ is a regular Borel measure.*

(See Theorem 34 in Section 4 of [6].)

Remark 2.3**.**

Let $X$ be locally compact. For $f\in C_{c}(X)$ we have $0\in f(X)=[a,b]$. By a singly generated subalgebra of $C_{c}(X)$ generated by $f$ we mean the smallest closed subalgebra of $C_{c}(X)$ containing $f$; it has the form

[TABLE]

When $X$ is compact, by a singly generated subalgebra of $C(X)$ generated by $f$ we mean the smallest closed subalgebra of $C(X)$ containing $f$ and $1$; it has the form:

[TABLE]

(See Remarks 4 and 2 in [7]).

Definition 2.4**.**

Let $X$ be locally compact. A quasi-integral (or a quasi-linear functional) on $C_{c}(X)$ is a map $\rho:C_{c}(X)\longrightarrow\mathbb{R}$ such that:

1. (QI1)

$\rho$* is homogeneous, i.e. $\rho(af)=a\rho(f)$ for $a\in\mathbb{R}.$* 2. (QI2)

For each $f\in C_{c}(X)$ we have: $\rho(g+h)=\rho(g)+\rho(h)$ for $g,h$ in the singly generated subalgebra $B(f)$. 3. (QI3)

$\rho$* is positive, i.e. $f\geq 0\Longrightarrow\rho(f)\geq 0.$*

When $X$ is compact, we call $\rho$ a quasi-state if $\rho(1)=1$.

In this paper we are interested in quasi-integrals on $C_{c}(X)$ and compact-finite topological measures for the reason given in the next remark.

Remark 2.5**.**

There is an order-preserving isomorphism between compact-finite topological measures on $X$ and quasi-integrals on $C_{c}(X)$, and $\mu$ is a measure iff the corresponding functional is linear. See Theorem 42 in Section 4 of [7] for this result and Theorem 3.9 in [12] for the first version of the representation theorem. We outline the correspondence.

1. (I)

Suppose $\mu$ is a compact-finite topological measure on a locally compact space $X$, $f\in C_{c}(X)$. Then there exists a measure $m_{f}$ on $\mathbb{R}$ with $supp\ m_{f}\subseteq f(X)$ such that

[TABLE]

thus,

[TABLE]

If $\mu$ is finite then

[TABLE]

The measure $m_{f}$ is the Stieltjes measure associated with the function $F(t)=\mu(f^{-1}((t,\infty)\setminus\{0\})$ (respectively, $F(t)=\mu(f^{-1}((t,\infty))$ if $\mu$ is finite.) (See Lemma 25 in Section 3 of **[7*]**.) *

Define a quasi-integlral $\rho=\rho_{\mu}$ on $C_{c}(X)$ by:

[TABLE]

We also write $\rho_{\mu}(f)=\int_{X}f\ \,d\mu$. If $\mu$ is a measure then $\rho_{\mu}(f)=\int_{X}f\ \,d\mu$ in the usual sense. On singly generated subagerbras $\rho_{\mu}$ acts as follows: for every $\phi\in C([a,b])$ (with $\phi(0)=0$ if $X$ is locally compact but not compact)

[TABLE]

where $[a,b]$ is any interval containing $f(X)$. (See Theorem 30 and Proposition 29 in Section 3 of **[7]**.) 2. (II)

*Let $\rho$ be a quasi-integral on $C_{c}(X)$. The corresponding compact-finite topological measure $\mu=\mu_{\rho}$ is given as follows: *

*If $U$ is open, *$\mu_{\rho}(U)=\sup\{\rho(f):\ f\in C_{c}(X),0\leq f\leq 1,supp\ f\subseteq U\},$

*if $F$ is closed, $\mu_{\rho}(F)=\inf\{\mu_{\rho}(U):\ F\subseteq U,U\in\mathscr{O}(X)\}$. *

If $K$ is compact, $\mu_{\rho}(K)=\inf\{\rho(g):\ g\in C_{c}(X),g\geq 1_{K}\}=\inf\{\rho(g):\ g\in C_{c}(X),0\leq g\leq 1,g=1\mbox{ on }K\}.$ (See Definition 33 and Lemma 35 in Section 4 of **[7]**.)

For examples of topological measures and quasi-integrals on locally compact spaces see [5] and the last sections of [4] and [7].

Remark 2.6**.**

If $f\cdot g=0,f,g\geq 0$ then $f,g$ are in the same singly generated subalgebra. If $\rho$ is a quasi-integral on a locally compact space, then $\rho(f)=\rho(f^{+})-\rho(f^{-})$ for any $f\in C_{c}(X)$. If $f\cdot g=0$ then $\rho(f+g)=\rho(f)+\rho(g).$ Also, $\rho(0)=0$. (See Lemma 19 and Lemma 20 in Section 3 of [7].)

The following is a part of Theorem 47 in Section 5 of [7].

Theorem 2.7**.**

Suppose $X$ is locally compact and $\rho$ is af quasi-linear functional on $C_{c}(X)$. If $f,g\in C_{c}(X),\,f,g\geq 0,supp\ f,supp\ g\subseteq K$, where $K$ is compact, then

[TABLE]

where $\mu$ is the compact-finite topological measure corresponding to $\rho$. In particular, for any $f\in C_{c}(X)$

[TABLE]

If $f,g\in C_{c}(X),\,supp\ f,supp\ g\subseteq K$, where $K$ is compact, then

[TABLE]

3. Almost simple quasi-integrals

A nontrivial topological measure assumes at least two values. Topological measures that assume exactly two values are important for proving results about repeated quasi-integration.

Definition 3.1**.**

Let $X$ be locally compact. A topological measure is called simple if it only assumes values [math] and $1$. A topological measure is almost simple if it is a positive scalar multiple of a simple topological measure. A quasi-linear functional is simple (almost simple) if the corresponding topological measure is simple (almost simple).

Lemma 3.2**.**

Suppose $X$ is locally compact and $\mu$ is a compact-finite topological measure that assumes more than two values. Let $\rho$ be the corresponding quasi-integral. Then there are functions $f_{1},f_{2}\in C_{c}^{+}(X)$ such that $f_{1}f_{2}=0,\ \rho(f_{1})=\rho(f_{2})=1$. Functions $f_{1},f_{2}$ belong to the same subalgebra.

Proof.

Choose compact $C_{1}$ such that $0<\mu(C_{1})<\mu(X)$. If $\mu(X)<\infty$ by inner regularity of $\mu$ on $X\setminus C_{1}$ choose compact $C_{2}\subseteq X\setminus C_{1}$ such that $0<\mu(C_{2})<\mu(X)$. If $\mu(X)=\infty$, i.e. $\mu(X\setminus C_{1})=\infty$, choose a compact $C_{2}\subseteq X\setminus C_{1}$ with $\mu(C_{2})>n$. Let $U_{1},U_{2}$ be open disjoint sets containing $C_{1},C_{2}$. For $\epsilon>0$ let $f_{i}\in C_{c}(X)$ be such that $f_{i}=1$ on $C_{i},\ supp\ f_{i}\subseteq U_{i}$ and $0<\mu(C_{i})\leq\rho(f_{i})\leq\mu(C_{i})+\epsilon$ for $i=1,2$. Then $f_{1}f_{2}=0$ and by calibrating $f_{i}$ we may assume that $\rho(f_{i})=1$ for $i=1,2$. (This proof is adapted from part of an argument in Theorem 1 in [9].) The last statement follows from Remark 2.6. ∎

The next theorem extends results for a simple quasi-state on $C(X)$ where $X$ is compact, given in Section 2 of [2].

Theorem 3.3**.**

Let $X$ be locally compact. The following are equivalent for a quasi-integral $\rho$ on $C_{c}(X)$:

1. (i)

$\rho$* is simple.* 2. (ii)

$m_{f}$* is a point mass at $y=\rho(f)\in f(X)$.* 3. (iii)

$\rho(\phi\circ f)=\phi(\rho(f))$* for any $\phi\in C([a,b])$ (with $\phi(0)=0$ if $X$ is locally compact but not compact), where $f(X)\subseteq[a,b]$.* 4. (iv)

$\rho$* is multiplicative on each singly generated subalgebra, i.e. for each $f\in C_{c}(X)$ we have: $\rho(gh)=\rho(g)\,\rho(h)$ for $g,h$ in the singly generated subalgebra $B(f)$.*

Proof.

(i) $\Longrightarrow$ (ii). If $\rho$ is simple, i.e. the corresponding topological measure $\mu$ is simple, then the measure $m_{f}$ in part (I) of Remark 2.5 is a point mass. From formula (2.2) we see that $m_{f}$ is a point mass at $y=\rho(f)$. To show that $y\in f(X)$, suppose the opposite, and choose an open set $W$ such that $y\in W,\ W\cap f(X)=\emptyset$. Then $m_{f}(W)=1$, while $\mu(f^{-1}(W))=\mu(\emptyset)=0$, which contradicts formula (2.1).

(ii) $\Longrightarrow$ (iii). Since $m_{f}$ is a point mass at $y=\rho(f)\in f(X)$, by formula (2.3)

[TABLE]

(iii) $\Longrightarrow$ (iv). Let $\phi\circ f,\,\psi\circ f\in B(f)$. Then $\rho((\phi\circ f)\,(\psi\circ f))=\rho((\phi\,\psi)\circ f))=(\phi\,\psi)(\rho(f))=\phi(\rho(f))\,\psi(\rho(f))=\rho(\phi\circ f)\,\rho(\psi\circ f)$.

(iv) $\Longrightarrow$ (i) Suppose to the contrary that $\rho$ is not simple, i.e. the corresponding compact-finite topological measure $\mu$ is not simple. Then there is a compact set $K\subseteq X$ with $a=\mu(K)\in\mathbb{R},\,a\neq 0,1$. Suppose first $a<1$. Let $\epsilon>0$ be such that $a>(a+\epsilon)^{2}$. By part (II) of Remark 2.5 choose $g\in C_{c}(X),g\geq 1_{K}$ with $\rho(g)<a+\epsilon$. Note that $g^{2}\in B(g)$ and $g^{2}\geq 1_{K}$. Then $\rho(g^{2})\geq\mu(K)=a>(a+\epsilon)^{2}>(\rho(g))^{2}$, and $\rho$ is not multiplicative on $B(g)$.

Assume now that $a>1$. There is an open set $U$ (containing $K$) such that $\mu(U)=b\in\mathbb{R}$ and $b>1$. Let $\epsilon>0$ be such that $b<(b-\epsilon)^{2}$. By part (II) of Remark 2.5 let $f\in C_{c}(X)$ be such that $0\leq f\leq 1,supp\ f\subseteq U,$ and $b-\epsilon<\rho(f)$. Since $f^{2}\in B(f)$ and $supp\ f^{2}\subseteq U$ we have $\rho(f^{2})\leq b<(b-\epsilon)^{2}<(\rho(f))^{2}$, and $\rho$ is not multiplicative on $B(f)$. ∎

Theorem 3.4**.**

Let $X$ be locally compact. The following are equivalent for a quasi-integral $\rho$ on $C_{c}(X)$:

1. (i)

$\rho$* is almost simple.* 2. (ii)

If $fg=0$ then $\rho(f)\rho(g)=0$. 3. (iii)

If $fg=0,\,f,g\geq 0$ then $\rho(f)\rho(g)=0$. 4. (iv)

If $fg=0,\,f,g\leq 0$ then $\rho(f)\rho(g)=0$.

Proof.

(i) $\Longrightarrow$ (ii). If $\rho$ is almost simple, the topological measure $\mu$ corresponding to $\rho$ assumes two values, and the measure $m_{f}$ supported on $f(X)$ as in Remark 2.5 has the form $m_{f}=c\delta_{y}$, where $c>0,y\in f(X)$. For $g$ the measure $m_{g}=k\delta_{z}$, where $k>0,z\in g(X)$. Then $\rho(f)\rho(g)=ckf(y)g(z)=0$. (ii) $\Longrightarrow$ (i). Lemma 3.2 shows that the topological measure corresponding to $\rho$ can not assume more than 2 values. Thus, $\rho$ is almost simple. The equivalence of (iii) and (ii) follows from $\rho(f)=\rho(f^{+})-\rho(f^{-}),\rho(g)=\rho(g^{+})-\rho(g^{-})$ (see Remark 2.6) and the fact that the product of any two of $f^{+},f^{-},g^{+},g^{-}$ is [math]. Equivalence of (iii) and (iv) is obvious. ∎

Remark 3.5**.**

If a compact-finite topological measure $\mu$ assumes only two values, but not $0,1$, then $\mu$ is almost simple. Write $\mu=c\mu^{\prime},$ where $\mu^{\prime}$ is simple and $c>0$. Then $\rho=c\rho^{\prime}$, where quasi-integrals $\rho$ and $\rho^{\prime}$ correspond to $\mu$ and $\mu^{\prime}$. Quasi-integral $\rho$ is no longer multiplicative on singly generated subalgebras.

4. Repeated quasi-integrals

Let $\mu$ be a compact-finite topological measure on $X$ with corresponding quasi-integral $\rho$, and $\nu$ be a compact-finite topological measure on $Y$ with corresponding quasi-integral $\eta$.

Definition 4.1**.**

For a set $A$ in $X\times Y$ let $A_{y}=\{x:\ (x,y)\in A\}$, and let $A_{x}=\{y:\ (x,y)\in A\}$.

Remark 4.2**.**

If the set $A$ is closed/compact/open then so is the set $A_{y}$. Note also that $(A\setminus B)_{y}=A_{y}\setminus B_{y}$.

Let $f\in C_{c}(X\times Y)$. Define continuous functions $f_{y}$ on $X$ and $f_{x}$ on $Y$ by $f_{y}(x)=f(x,y)=f_{x}(y)$. Compact $C=\pi_{1}(supp\ f)$, where $\pi_{1}:X\times Y\rightarrow X$ is the canonical projection, contains $supp\ f_{y}$ for any $y$. We have $f_{y}\in C_{c}(X)$, and $f_{x}\in C_{c}(Y)$.

Definition 4.3**.**

Define real-valued functions $T_{\rho}(f)$ on $Y$ and $S_{\eta}(f)$ on $X$ by:

[TABLE]

Proposition 4.4**.**

Suppose $X\times Y$ is locally compact and $f\in C_{c}(X\times Y)$. If $\rho$ is a quasi-integral on $C_{c}(X)$ then $T_{\rho}(f)\in C_{c}(Y)$ and $\parallel T_{\rho}(f)\parallel\leq\parallel f\parallel\mu(\pi_{1}(supp\ f))$. Similarly, if $\eta$ is a quasi-integral on $C_{c}(Y)$ then $S_{\eta}(f)\in C_{c}(X)$ and $\parallel S_{\eta}(f)\parallel\leq\parallel f\parallel\nu(\pi_{2}(supp\ f))$. Here $\pi_{1}:X\times Y\rightarrow X$ and $\pi_{2}:X\times Y\rightarrow Y$ are canonical projections.

Proof.

Let $y\in Y$. We shall show that $T_{\rho}(f)$ is a continuous function at $y$. Compact $C=\pi_{1}(supp\ f)$ contains $supp\ f_{y}$. Let $\epsilon>0$. For each $x\in C$ let $U_{x}$ be a neighborhood of $x$ and $V_{x,y}$ be a neighborhood of $y$ such that $|f(x,y)-f(x^{\prime},y^{\prime})|<\epsilon$ whenever $(x^{\prime},y^{\prime})\in U_{x}\times V_{x,y}$. Open sets $U_{x}$ cover $C$, so let $U_{x_{1}},\ldots,U_{x_{n}}$ be a finite subcover, and let $V_{y}=\bigcap_{i=1}^{n}V_{x_{i},y}$. Take $y^{\prime}\in V_{y}$. Then for each $x\in C$ there is $i$ such that $x\in U_{x_{i}}$, so $(x,y^{\prime})\in U_{x_{i}}\times V_{y}\subseteq U_{x_{i}}\times V_{x_{i},y}$, and then $|f_{y}(x)-f_{y^{\prime}}(x)|=|f(x,y)-f(x,y^{\prime})|<\epsilon$. For any $x\notin C$ we have $f_{y}(x)=f_{y^{\prime}}(x)=0$. Therefore, $\parallel f_{y}-f_{y^{\prime}}\parallel<\epsilon$ (for any $y^{\prime}\in V_{y}$). Since $supp\ f_{y},supp\ f_{y^{\prime}}\subseteq C$, by Theorem 2.7 for any $y^{\prime}\in V_{y}$ we have:

[TABLE]

and the continuity of $T_{\rho}(f)$ at $y$ follows.

From Theorem 2.7 for any $y$ we have $|T_{\rho}(f)(y)|=|\rho(f_{y})|\leq\parallel f_{y}\parallel\mu(C)\leq\parallel f\parallel\mu(C).$ Thus,

[TABLE]

Since $X$ and $Y$ are locally compact, for $x\in X$ let $U(x)$ be a neighborhood of $x$ such that $\overline{U(x)}$ is compact in $X$, and let $V(y)$ be a neighborhood of $y\in Y$ such that $\overline{V(y)}$ is compact in $Y$. Open sets $U(x)\times V(y)$ cover $supp\ f$, so let $U_{1}\times V_{1},\ldots,U_{n}\times V_{n}$ be a finite subcover of $supp\ f$. Let $G=\overline{V}_{1}\cup\ldots\cup\overline{V}_{n}$, a compact in $Y$. For each $x\in X$ and each $y\notin G$ we have$(x,y)\notin supp\ f$, so $f(x,y)=0$. This means that $f_{y}=0$ for each $y\notin G$. Then $\rho(f_{y})=0$, i.e. $T_{\rho}(f)(y)=\rho(f_{y})=0$ for each $y\notin G$. Hence, $T_{\rho}(f)\in C_{c}(Y)$. ∎

We are now ready to define repeated quasi-integrals.

Definition 4.5**.**

Using Definition 4.3 and Proposition 4.4 we define the following functionals on $C_{c}(X\times Y)$:

[TABLE]

[TABLE]

Remark 4.6**.**

Functionals $\eta\times\rho$ and $\rho\times\eta$ are real-valued since $\eta$ and $\rho$ are quasi-integrals.

Lemma 4.7**.**

Suppose $\eta$ and $\rho$ are quasi-integrals. Let $\rho=c\rho^{\prime}$, where $c>0$. Then $\eta\times\rho$ is a quasi-integral iff $\eta\times\rho^{\prime}$ is a quasi-integral, and $\eta\times\rho=c(\eta\times\rho^{\prime})$. Similarly, if $\eta=k\eta^{\prime},\,k>0$ then $\eta\times\rho$ is a quasi-integral iff $\eta^{\prime}\times\rho$ is a quasi-integral, and $\eta\times\rho=k(\eta^{\prime}\times\rho)$.

Proof.

We have $T_{\rho}(f)(y)=cT_{\rho^{\prime}}(f)(y)$ for every $y$, i.e. $T_{\rho}(f)=cT_{\rho^{\prime}}$. Since $\eta$ is homogeneous, we see that $\eta\times\rho=c(\eta\times\rho^{\prime})$. Since $\eta\times\rho$ and $\eta\times\rho^{\prime}$ satisfy conditions (QI1), (QI2), and (QI3) of Definition 2.4 simultaneously, $\eta\times\rho$ is a quasi-integral iff $\eta\times\rho^{\prime}$ is. The proof of the last part is similar but easier. ∎

We would like to know whether $\eta\times\rho=\rho\times\eta$. We shall see later that, unlike the case of linear functionals, this is not usually the case.

For $g\in C_{c}(X)$ and $h\in C_{c}(Y)$ let $(g\otimes h)(x,y)=g(x)h(y)$. Then $(g\otimes h)\in C_{c}(X\times Y)$.

Proposition 4.8**.**

1. (1)

$T_{\rho}(g\otimes h)=\rho(g)\ h$. 2. (2)

$(\eta\times\rho)(g\otimes h)=(\rho\times\eta)(g\otimes h)=\rho(g)\eta(h).$**

Proof.

The proof is as in Proposition 1 in [9]. ∎

Remark 4.9**.**

If $\mu$ is the compact-finite topological measure corresponding to quasi-linear functional $\rho$, $\nu$ is the compact-finite topological measure corresponding to $\eta$, let $\nu\times\mu$ be the compact-finite topological measure corresponding to $\eta\times\rho$, and $\mu\times\nu$ be the compact-finite topological measure corresponding to $\rho\times\eta$. Using part (2) of Proposition 4.8 and part (II) of Remark 2.5 it is easy to see that

[TABLE]

By Definition 4.5, Theorem 2.7, and Proposition 4.4 the opposite inequalities also hold, so

[TABLE]

Remark 4.10**.**

It is easy to see that if $\eta$ and $\rho$ are homogeneous, then so are $\eta\times\rho$ and $\rho\times\eta$; and that if $\eta$ and $\rho$ are positive, then so are $\eta\times\rho$ and $\rho\times\eta$. Yet, $\rho\times\eta$ and $\eta\times\rho$ are not always quasi-integrals. The criterion for $\eta\times\rho$ to be a quasi-integral is given in Theorem 4.12 below.

Theorem 4.11**.**

1. (1)

If the compact-finite topological measure corresponding to $\rho$ assumes more than two values and $\eta\times\rho$ is a quasi-integral, then $\eta$ is linear. 2. (2)

If $\eta$ is linear or $\rho$ is simple then $\eta\times\rho$ is a quasi-linear functional.

Proof.

1. (1)

Our proof follows part of Theorem 1 in [9]. It is given for completeness and because intermediate results from the proof are needed elsewhere in the paper. Suppose that $\eta\times\rho$ is a quasi-integral and the compact-finite topological measure $\mu$ corresponding to $\rho$ assumes more than two values. By Lemma 3.2 choose $f_{1},f_{2}\in C_{c}(X)$ such that $\rho(f_{1})=\rho(f_{2})=1$ and $f_{1}f_{2}=0$. Take any $g,h\in C_{c}(Y)$. For any $y\in Y$, $\ (g(y)f_{1})(h(y)f_{2})=0,$ so using Remark 2.6 we have:

[TABLE]

i.e.

[TABLE]

Since $\eta\times\rho$ is a quasi-integral and $(f_{1}\otimes g)(f_{2}\otimes h)=0$, using Proposition 4.8 and Remark 2.6 we see that

[TABLE]

Thus, $\eta$ is linear on $C(Y)$. 2. (2)

To show that $\eta\times\rho$ is a quasi-integral, by Remark 4.10 we need to check condition (QI2) of Definition 2.4. Let $\phi,\psi\in C([a,b])$ where $[a,b]$ contains the ranges of $f$ and $T_{\rho}(f)$. Recall that $f_{y}\in C_{c}(X)$ and note that $(\phi\circ f)_{y}=\phi\circ f_{y}$. In particular, $\phi\circ f_{y}$ and $\psi\circ f_{y}$ are in the same subalgebra generated by $f_{y}$. One may apply the argument from Theorem 1 in [9] to prove the statement.

∎

Theorem 4.12**.**

Suppose $\eta$ and $\rho$ are quasi-integrals. $\eta\times\rho$ is a quasi-linear functional iff $\eta$ is linear or $\rho$ is almost simple.

Proof.

($\Longrightarrow$). Suppose $\eta\times\rho$ is a quasi-integral. If $\rho$ is not almost simple then $\rho$ assumes more then two values, and by Theorem 4.11 $\eta$ must be linear.

($\Longleftarrow$). If $\eta$ is linear then $\eta\times\rho$ is a quasi-linear functional by Theorem 4.11. Suppose $\rho=c\rho^{\prime}$ , where $\rho^{\prime}$ is simple, $c>0$. Then by Theorem 4.11 $\eta\times\rho^{\prime}$ is a quasi-integral, and then by Lemma 4.7 so is $\eta\times\rho$. ∎

Corollary 4.13**.**

Suppose $\mu(X)=1$. The functional $\eta\times\rho$ is a quasi-integral iff $\eta$ is linear or $\rho$ is simple.

Remark 4.14**.**

We may phrase the results of Theorem 4.12 it terms of topological measures. If $\mu$ is a compact-finite topological measure on $X$ and $\nu$ is a compact-finite topological measure on $Y$ we may define a compact-finite product topological measure $\nu\times\mu$ on $X\times Y$ if either $\nu$ is a measure or $\mu$ is an almost simple topological measure. Similar results hold for topological measure $\mu\times\nu$ corresponding to quasi-integral $\rho\times\eta$.

Theorem 4.15**.**

Suppose $\rho,\ \eta$, and $\eta\times\rho$ are quasi-integrals. If any two of them are simple, then so is the third one.

Proof.

Assume that $\rho$ and $\eta$ are simple. Let $f\in C_{c}(X\times Y)$ and $\phi\in C([a,b])$ (with $\phi(0)=0$ if $X$ is locally compact but not compact), where $[a,b]$ contains the ranges of $f$ and $T_{\rho}(f)$. Since $\rho$ is simple, by part (iii) of Theorem 3.3 we have $T_{\rho}(\phi\circ f)(y)=\phi\circ T_{\rho}(f)(y)$ for all $y$, so $T_{\rho}(\phi\circ f)=\phi\circ T_{\rho}(f)$. Since $\eta$ is simple, we have

[TABLE]

By part (iii) of Theorem 3.3 $\eta\times\rho$ is simple. (Cf. Corollary 1 in [9].)

Now assume that $\rho$ and $\eta\times\rho$ are simple. Take $g\in C_{c}^{+}(X)$ such that $\rho(g)=1$. Since $\rho$ is multiplicative on singly generated sublagebras, $\rho(g^{2})=1$. Suppose that $\eta$ is not simple. If $\eta$ assumes two values then $\eta$ is almost simple, but not simple; by Lemma 4.7 $\eta\times\rho$ is almost simple, but not simple, which contradicts our assumption. So $\eta$ assumes more than 2 values. By Lemma 3.2 let $h_{1},\ h_{2}\in C_{c}^{+}(Y)$ be such that $\eta(h_{1})=\eta(h_{2})=1$, $h_{1}h_{2}=0$, so $\eta(h_{1}h_{2})=0$. Since $(g\otimes h_{1})\cdot(g\otimes h_{2})=0$, by Remark 2.6 functions $g\otimes h_{1},\,g\otimes h_{2}$ belong to the same singly generated subalgebra. Since $\eta\times\rho$ is simple, it is multiplicative on this subalgebra and then using Proposition 4.8 we have:

[TABLE]

On the other hand, by Proposition 4.8

[TABLE]

and we arrive at contradiction. Thus, $\eta$ must be simple.

The proof that $\rho$ is simple when $\eta$ and $\eta\times\rho$ are is similar. ∎

From Theorem 4.15 and Lemma 4.7 we obtain

Corollary 4.16**.**

If $\rho$ and $\eta$ are almost simple, then so is $\eta\times\rho$.

Theorem 4.17**.**

Suppose $\rho,\ \eta,\ \eta\times\rho$ are quasi-integrals. Then $\eta\times\rho$ is simple iff $\eta\times\rho=\eta^{\prime}\times\rho^{\prime}$ for some simple quasi-integrals $\rho^{\prime}$ and $\eta^{\prime}$.

Proof.

If $\rho$ and $\eta$ are simple, then $\eta\times\rho$ is simple by Theorem 4.15.

Now assume that $\eta\times\rho$ is simple. First, we shall show that $\eta$ assumes exactly two values. If $\eta$ attains more than two values, by Lemma 3.2 let $h_{1},\ h_{2}\in C_{c}^{+}(Y)$ be such that $\eta(h_{1})=\eta(h_{2})=1$, $h_{1}h_{2}=0$, so $\eta(h_{1}h_{2})=0$. Choose $g\in C_{c}(X)$ such that $\rho(g)>0$. Since $(g\otimes h_{1})\cdot(g\otimes h_{2})=0$, by Remark 2.6 functions $g\otimes h_{1},\,g\otimes h_{2}$ belong to the same singly generated subalgebra. Since $\eta\times\rho$ is simple, as in the proof of Theorem 4.15 we have:

[TABLE]

On the other hand,

[TABLE]

This contradiction shows that $\eta$ assumes only two values.

Write $\eta=k\eta^{\prime}$ for some $k$ and some simple quasi-integral $\eta^{\prime}$. Then $\eta\times\rho=(k\eta^{\prime})\times\rho=\eta^{\prime}\times(k\rho)$. Since $\eta\times\rho$ and $\eta^{\prime}$ are simple, by Theorem 4.15 so is $\rho^{\prime}=k\rho$, and the assertion follows. ∎

5. Products of topological measures and Fubini’s theorem

The next two theorems describe how $\nu\times\mu$ acts on sets.

Theorem 5.1**.**

Let $\mu$ be a compact-finite topological measure on $X$ and $\nu$ a finite measure on $Y$. Then for $U$ open in $X\times Y$ we have:

[TABLE]

If $\mu$ is finite, for a compact set $K$ in $X\times Y$ we also have

[TABLE]

Proof.

Part (II) of Remark 2.5 allows us to use the same argument as in Claim 2 and Claim 3 of Theorem 2 in [9] to prove that for any open $U\subseteq X\times Y$ we have $(\nu\times\mu)(U)=\int_{Y}\mu(U_{y})d\nu(y).$

If $\mu$ is finite, then by Remark 4.9 so is $\nu\times\mu$ and $(\nu\times\mu)(X\times Y)=\mu(X)\nu(Y)$. If $K\subseteq X\times Y$ is compact, by Definition 2.1 we have:

[TABLE]

∎

Theorem 5.2**.**

Let $\mu$ be a simple topological measure on $X$ and $\nu$ be a compact-finite topological measure on $Y$. Then for $U$ open in $X\times Y$ we have:

[TABLE]

If $\nu$ is finite, for $K$ compact in $X\times Y$ we also have:

[TABLE]

Proof.

For $A$ in $X\times Y$ let $B(A)=\{y:\mu(A_{y})=1\}$. Let $U$ be open in $X\times Y$. It is not hard to show (for example, by applying the argument from Claim 1 of Theorem 3 in [9]) that $B(U)$ is also open. Since $\mu$ is simple, for any compact $D\subseteq X\times Y$ we have $(B(D))^{c}=B(D^{c})$, and $B(D)$ is closed.

Suppose that $K\subseteq B(U)$ is compact. For each $y\in K,\ \mu(U_{y})=1$, so there is a compact set $C(y)\subseteq U_{y}$ with $\mu(C(y))=1$. Then $C(y)\times\{y\}\subseteq U$, so there are open sets $V(y),\,W(y)$ such that $C(y)\subseteq V(y)\subseteq X,\ y\in W(y)\subseteq Y$, and $V(y)\times\overline{W(y)}\subseteq U$. Finitely many $W(y_{1}),\ldots,W(y_{n})$ cover $K$. Set $D=\bigcup_{i=1}^{n}C(y_{i})\times\overline{W(y_{i})}$. Then $D$ is compact and $D\subseteq U$. Choose $f\in C_{c}(X\times Y)$ such that $1_{D}\leq f\leq 1,\ supp\ f\subseteq U$. Then for $y\in K$, say, $y\in W(y_{i})$, we have $f_{y}=1$ on $C(y_{i})$, so $1=\mu(C(y_{i}))\leq\rho(f_{y})=T_{\rho}(f)(y)$. Thus $T_{\rho}(f)=1$ on $K$. Then by part (II) of Remark 2.5

[TABLE]

Taking the supremum over $K\subseteq B(U)$ shows that $\nu(B(U))\leq(\nu\times\mu)(U).$

Now we shall show that $(\nu\times\mu)(U)\leq\nu(B(U))$. Since $(\nu\times\mu)(U)=\sup\{(\eta\times\rho)(f):\,f\in C_{c}(X\times Y),supp\ f\subseteq U,0\leq f\leq 1\}$, it is enough to show that $(\eta\times\rho)(f)\leq\nu(B(U))$ for $f\in C_{c}(X\times Y)$ such that $supp\ f\subseteq U,0\leq f\leq 1$. Let $C=supp\ f$. Pick an open set $V$ with compact closure such that $C\subseteq V\subseteq\overline{V}\subseteq U$ (see, for example, [8], Chapter XI, 6.2). Note that $f_{y}\in C_{c}(X),supp\ f_{y}\subseteq V_{y},0\leq f_{y}\leq 1$. Then

[TABLE]

Since $B(\overline{V})$ is closed, we have $supp\ T_{\rho}(f)\subseteq B(\overline{V})\subseteq B(U)$. By Proposition 4.4 $\parallel T_{\rho}(f)\parallel\leq 1$. By part (II) of Remark 2.5 $(\eta\times\rho)(f)=\eta(T_{\rho}(f))\leq\nu(B(U))$.

Now we have: $(\nu\times\mu)(U)=\nu(\{y:\mu(U_{y})=1\})$ for any open set $U\subseteq X\times Y$. When $\nu$ is finite, the formula for compact $K$ can be proved as in Theorem 5.1. ∎

Corollary 5.3**.**

Suppose $\mu=c\mu^{\prime}$ is an almost simple topological measure on $X$, where $\mu^{\prime}$ is simple and $c>0$, and $\nu$ is a topological measure on $Y$. Then for $A$ open in $X\times Y$ we have:

[TABLE]

When $\nu$ is finite, the same also holds for compact sets in $X\times Y$.

Proof.

Using Lemma 4.7 and Theorem 5.2 we have: $(\nu\times\mu)(A)=c(\nu\times\mu^{\prime})(A)=c\nu(\{y:\mu^{\prime}(A_{y})=1\})=c\nu(\{y:\mu(A_{y})=c\})$. ∎

Corollary 5.4**.**

Suppose $\mu$ is a compact-finite topological measure on $X$, $\nu$ is a finite topological measure on $Y$, and $\nu\times\mu$ is a compact-finite topological measure. If $A\subseteq X$ and $B\subseteq Y$ are both open then $(\nu\times\mu)(A\times B)=\mu(A)\nu(B)$. When $\mu$ is finite, the same also holds when $A\subseteq X$ and $B\subseteq Y$ are compact sets.

Proof.

By Theorem 4.12 either $\mu$ is almost simple or $\nu$ is a finite measure. The corollary now follows from Theorem 5.1 and Corollary 5.3, and generalizes Remark 4.9. ∎

Lemma 5.5**.**

If $\mu$ and $\nu$ are almost simple but not measures then $\eta\times\rho\neq\rho\times\eta$.

Proof.

We first prove (as in Corollary 3 in [9]) that if $\mu$ and $\nu$ are simple but not measures then $\eta\times\rho\neq\rho\times\eta$. If $\mu$ and $\nu$ are not measures, by Theorem 2.2 they are not subadditive, and we may find open sets $U,V\subseteq X$ with $\mu(U)=\mu(V)=0,\ \mu(U\cup V)=1$, and compact sets $C,K\subseteq Y$ with $\nu(C)=\nu(K)=0,\ \nu(C\cup K)=1$. Taking complements of $C$ and $K$ we get open sets $W,E\subseteq Y$ such that $\nu(W)=\nu(E)=1,\ \nu(W\cap E)=0$. Let $A=(U\times W)\cup(V\times E)$. We shall show that $(\nu\times\mu)(A)=0$ using Theorem 5.2. If $y\in E\cap W$ then $A_{y}=U\cup V$, so $\mu(A_{y})=1$. For the cases $y\in W\setminus E,\ y\in E\setminus W,\ y\notin E\cup W$ we have $\mu(A_{y})=0$. Then $(\nu\times\mu)(A)=\nu(E\cap W)=0$. A similar argument shows that $(\mu\times\nu)(A)=1$. Since $(\nu\times\mu)(A)\neq(\mu\times\nu)(A)$, we have $\eta\times\rho\neq\rho\times\eta$.

If $\mu$ and $\nu$ are almost simple but not measures, write $\mu=c\mu^{\prime},\nu=k\nu^{\prime}$, where $c,k>0$ and $\mu^{\prime},\nu^{\prime}$ are simple, but not measures. With simple quasi-integrals $\rho^{\prime},\eta^{\prime}$ corresponding to $\mu^{\prime},\nu^{\prime}$ we have $\eta\times\rho=ck(\eta^{\prime}\times\rho^{\prime})\neq ck(\rho^{\prime}\times\eta^{\prime})=\rho\times\eta$. ∎

Lemma 5.6**.**

Suppose one of compact-finite topological measures $\mu,\nu$ is a positive scalar multiple of a point mass. Then $(\nu\times\mu)=(\mu\times\nu)$.

Proof.

If we prove the statement in the case when one of $\mu,\nu$ is a point mass, the lemma easily follows. So suppose $\mu=\delta_{x_{0}}$. It is enough to show that $(\nu\times\mu)=(\mu\times\nu)$ for open sets. Let $U\subseteq X\times Y$ be open. Then $\mu(U_{y})=1$ iff $(x_{0},y)\in U$, and by Theorem 5.2 we have:

[TABLE]

By Theorem 5.1 we have:

[TABLE]

Thus, $(\nu\times\mu)=(\mu\times\nu)$. ∎

Now we are ready to answer the question of when a version of Fubini’s theorem holds for repeated quasi-integrals, in other words, when $\eta\times\rho=\rho\times\eta$.

Theorem 5.7**.**

Let $\rho$ be a quasi-integral with corresponding compact-finite topological measure $\mu$ on $Y$ and let $\eta$ be a quasi-integral with corresponding compact-finite topological measure $\nu$ on $Y$. Then $\eta\times\rho=\rho\times\eta$ if and only if $\mu,\ \nu$ are both measures or one of $\mu,\ \nu$ is a positive scalar multiple of a point mass.

Proof.

(${\Longleftarrow}$) If both $\mu,\ \nu$ are measures, then $\eta\times\rho=\rho\times\eta$ by Fubini’s theorem. If one of $\mu,\ \nu$ is a positive scalar multiple of a point mass, then $\eta\times\rho=\rho\times\eta$ by Lemma 5.6.

(${\Longrightarrow}$) First we shall show that $\eta\times\rho$ and $\rho\times\eta$ must be quasi-integrals. Suppose $\eta\times\rho$ is not a quasi-integral, that is, $\rho$ is not almost simple and $\eta$ is not linear by Theorem 4.12. Let $f_{1},f_{2}\in C_{c}(X)$ be functions given by Lemma 3.2, so $f_{1}f_{2}=0$, $\rho(f_{1})=\rho(f_{2})=1$. Also choose $g,h\in C_{c}(Y)$ such that $\eta(g+h)\neq\eta(g)+\eta(h)$. Let $f=f_{1}\otimes g+f_{2}\otimes h$. As in formula (4.1), we have $T_{\rho}(f)=g+h$. Then

[TABLE]

Now we shall look at $(\rho\times\eta)(f)=\rho(S_{\eta}(f)).$ For each $x$, $S_{\eta}(f)(x)=\eta(f_{x})=\eta(f_{1}(x)g+f_{2}(x)h).$ Since $(f_{1}(x)g)(f_{2}(x)h)=0$, we have:

[TABLE]

Thus, $S_{\eta}(f)=f_{1}\eta(g)+f_{2}\eta(h)$. Since $f_{1}\eta(g)\cdot f_{2}\eta(h)=0$, we have

[TABLE]

and we see that $\eta\times\rho\neq\rho\times\eta$. Thus, $\eta\times\rho$ must be a quasi-integral, and so must $\rho\times\eta$.

Both $\eta\times\rho$ and $\rho\times\eta$ are quasi-integrals. From Theorem 4.12 we see that this happens only when (a) both $\mu$ and $\nu$ are measures, or (b) at least one of $\mu$ or $\nu$ is a positive scalar multiple of a point mass, or (c) both $\mu$ and $\nu$ are almost simple, but not measures. The first two cases produce $\eta\times\rho=\rho\times\eta$, by Fubini’s Theorem or Lemma 5.6. In the last case (c), by Lemma 5.5 $\eta\times\rho\neq\rho\times\eta$. This finishes the proof. ∎

Remark 5.8**.**

As in the compact case (see [9]), we have the following interesting phenomenon: if $\mu$ and $\nu$ are almost simple, but not measures, then $\nu\times\mu$ and $\mu\times\nu$ are different, even though they agree on rectangles. This holds even when $X=Y$ and $\mu=\nu$. Taking $a(\mu\times\nu)+(m-a)(\nu\times\mu)$, where $m=\mu(X)^{2}$, we obtain uncountably many topological measures that agree on rectangles, but are distinct. This is impossible for measures on product spaces, as they are determined by values on rectangles.

Acknowledgments: The author would like to thank the Department of Mathematics at the University of California Santa Barbara for its hospitality and supportive environment.