Integration with respect to deficient topological measures on locally compact spaces

TL;DR

This paper explores integration with respect to deficient topological measures on locally compact spaces, establishing properties, new measures, and convergence theorems, thus extending measure theory to a broader class of non-linear functionals.

Contribution

It introduces a novel integration framework for deficient topological measures, characterizes resulting measures, and proves convergence results, expanding the theoretical understanding of non-linear measure-like functionals.

Findings

Integration over sets yields new deficient topological measures.

Such measures are absolutely continuous and Lipschitz continuous.

Monotone convergence theorem is established.

Abstract

Topological measures and deficient topological measures generalize Borel measures and correspond to certain non-linear functionals. We study integration with respect to deficient topological measures on locally compact spaces. Such an integration over sets yields a new deficient topological measure if we integrate a nonnegative vanishing at infinity function; and it produces a signed deficient topological measure if we use a continuous function on a compact space. We present many properties of these resulting deficient topological measures and of signed deficient topological measures. In particular, they are absolutely continuous with respect to the original deficient topological measure and Lipschitz continuous. Deficient topological measures obtained by integration over sets can also be obtained from non-linear functionals. We show that for a deficient topological measure $ \mu$ that assumes finitely many values, there is a function $ f $ such that $\int_X f \, d \mu = 0$, but $\int_X (-f )\, d \mu \neq 0$. We present different criteria for $\int_X f \, d \mu = 0$. We also prove some convergence results, including a Monotone convergence theorem.

Equations

$$|\nu |(U)=\sup \{\sum _{i=1}^n|\nu (K_i)|:\underset{i=1}{\overset{n}{⨆}}{K}_i\subseteq U,K_i\subseteq \mathcal{K}(X),n\in \mathbb{N}\};$$

$$|\nu |(F)=\inf \{|\nu |(U):F\subseteq U,U\in \mathcal{O}(X)\}.$$

$$R_1(t)=R_{1,\mu ,f}(t)=\mu (f^{-1}((t,\infty ))),$$

$$R_2(t)=R_{2,\mu ,f}(t)=\mu (f^{-1}([t,\infty ))).$$

$$\mathcal{R}(f)$$

$$\mathcal{R}(f)={\int }_0^{b}iddr={\int }_0^b{R}_1(t)dt={\int }_0^b{R}_2(t)dt.$$

$${\int }_{X}fd\mu =\mathcal{R}(f)={\mathcal{R}}_{\mu }(f)={\int }_{\mathbb{R}}iddr.$$

$$\underset{\alpha }{\lim }{\int }_X{f}_{\alpha }(x)l(dx)={\int }_{X}f(x)l(dx).$$

$$\underset{\alpha }{\lim }{\int }_a^b{g}_{\alpha }(x)l(dx)={\int }_a^{b}g(x)l(dx).$$

$${\mu }_V(U)=\mu (U\cap V),U\in \mathcal{O}(X),$$

$${\mu }_V(F)=\inf \{{\mu }_V(U):F\subseteq U,U\in \mathcal{O}(X)\},F\in \mathcal{C}(X).$$

$${\mu }_F(K)=\mu (F\cap K)\text{mbox}forK\in \mathcal{K}(X),$$

$${\mu }_F(U)=\sup \{{\mu }_F(K):K\subseteq U,K\in \mathcal{K}(X)\}\text{mbox}forU\in \mathcal{O}(X).$$

$$\mu (F\cap C)=\inf \{{\mu }_V(C):F\subseteq V,V\in \mathcal{O}(X)\},$$

$${\int }_{A}gd\mu ={\int }_{X}gd{\mu }_A={\mathcal{R}}_{{\mu }_A}(g).$$

$${\int }_{E}fd\mu =a\mu (E)+{\int }_a^b{R}_{1,{\mu }_E,f}(t)dt=a\mu (E)+{\int }_a^b{R}_{2,{\mu }_E,f}(t)dt,$$

$$R_{1,{\mu }_V,f}(t)={\mu }_V(f^{-1}((t,\infty )))=\nu (V\cap f^{-1}((t,\infty )))\text{mbox}forE=V\in \mathcal{O}(X),$$

$$R_{2,{\mu }_C,f}(t)={\mu }_C(f^{-1}([t,\infty )))=\mu (C\cap f^{-1}([t,\infty )))\text{mbox}forE=C\in \mathcal{K}(X).$$

$${\int }_{A}fd\mu ={\int }_0^b{R}_{1,{\mu }_A,f}(t)dt={\int }_0^b{R}_{2,{\mu }_A,f}(t)dt.$$

$${\int }_{E}fd\mu =a^{\prime }\mu (E)+{\int }_{a^{\prime }}^{b^{\prime }}{R}_{1,{\mu }_E,f}(t)dt=a^{\prime }\mu (E)+{\int }_{a^{\prime }}^{b^{\prime }}{R}_{2,{\mu }_E,f}(t)dt$$

$$\int_{V}g\,d\mu=\lim\Big{\{}\int_{K}g\,d\mu:\ K\subseteq U,\ K\in\mathscr{K}(X)\Big{\}}.$$

$$\int_{F}g\,d\mu=\lim\Big{\{}\int_{V}g\,d\mu:\ F\subseteq V,\ V\in\mathscr{O}(X)\Big{\}}.$$

$$\underset{K\subseteq V,K\in \mathcal{K}(X)}{\lim }{R}_{1,{\mu }_K,g}(t)=R_{1,{\mu }_V,g}(t).$$

$$\underset{V\in \mathcal{O}(X),F\subseteq V}{\lim }{R}_{2,{\mu }_V,g}(t)=R_{2,{\mu }_F,g}(t).$$

$${\mu }_g(K)={\mathcal{R}}_{{\mu }_K}(g)\le \Vert g\Vert {\mu }_K(X)=\Vert g\Vert \mu (K),$$

$$g^{-1}([t,\infty ))\cap F=(g^{-1}([t,\infty ))\cap E)\bigsqcup (g^{-1}([t,\infty ))\cap G),$$

$${\int }_0^b{R}_{2,{\mu }_F,g}(t)dt={\int }_0^b{R}_{2,{\mu }_E,g}(t)dt+{\int }_0^b{R}_{2,{\mu }_D,g}(t)dt,$$

$$\mu (A)\cdot \underset{A}{\inf }g\le {\lambda }_g(A)\le \mu (A)\cdot \underset{A}{\sup }g.$$

$${\mu }_g(A)\le \mu (A)\cdot \underset{A}{\sup }g\le \Vert g\Vert \mu (A),|{\nu }_g(A)|\le \mu (A)\underset{A}{\sup }|g|\le \Vert g\Vert \mu (A).$$

$${\nu }_{g+c}(K)=\mathcal{R}{\mu }_K(g+c)=\mathcal{R}{\mu }_K(g)+c{\mu }_K(X)={\nu }_g(K)+c\mu (K).$$

Full text

1. Introduction

Topological measures (initially called quasi-measures) were introduced by J. F. Aarnes in [1], [2], and [3]. These generalizations of measures are defined on open and closed subsets of a topological space. Despite the lack of an algebraic structure on the domain, absence of subadditivity, and unavailability of many standard techniques of measure theory and functional analysis, topological measures still possess many features of regular Borel measures. Via integration, topological measures correspond to functionals that are not linear, but are linear on singly generated subalgebras. Topological measures and corresponding quasi-linear functionals are connected to the problem of linearity of the expectational functional on the algebra of observables in quantum mechanics. Initial papers by Aarnes were followed by many papers devoted to the subject. Applications of topological measures and corresponding non-linear functionals to symplectic topology have been studied in numerous papers beginning with [13] (which has been cited over 100 times), and in a monograph [15].

Topological measures are a subclass of deficient topological measures, which also correspond via integration to certain non-linear functionals. (See, for example, [14], [17], [18], [7], and [9] for more information). Deficient topological measures are not only interesting by themselves, but also provide an essential framework for studying topological measures and various non-linear functionals. This provided a motivation for our study of integration with respect to deficient topological measures on locally compact spaces, especially given the fact that the vast majority of results so far has been proven for compact spaces (which has impeded the development of the area and its applications). We demonstrate that integration with respect to a deficient topological measure sometimes gives the same results as integration with respect to a measure would give; in other instances, the results are very different. Some of our results are new, and some are generalizations of known results about deficient topological measure on compact spaces to signed deficient topological measures on compact spaces and /or deficient topological measures on locally compact spaces. (See [10] and [11] for more information about signed deficient topological measures.)

We begin (Section 2) with the study of integration with respect to a deficient topological measure over a set. It is done by using ”restricted” deficient topological measures. Integration over sets yields a new deficient topological measure if we integrate a nonnegative vanishing at infinity function; and it produces a signed deficient topological measure if we use a continuous function on a compact space. We present many properties of these resulting deficient topological measures and of signed deficient topological measures. In particular, these resulting (signed) deficient topological measures are absolutely continuous with respect to the original deficient topological measure, and they are Lipschitz continuous. In Section 3 we show that the deficient topological measures obtained in Section 2 from finite deficient topological measures by integrating over sets can also be obtained from non-linear functionals. We present more properties of such deficient topological measures. In Section 4 we show that for a deficient topological measure $\mu$ that assumes finitely many values, there is $f\geq 0$ such that $\int_{X}f\,d\mu=0$, but $\int_{X}(-f)\,d\mu\neq 0$. Using integration over zero and cozero sets, we present different criteria for $\int_{X}f\,d\mu=0$. We conclude the paper (Section 5) with some convergence results, including a Monotone convergence theorem.

In this paper $X$ is a locally compact, connected space.

By $C(X)$ we denote the set of all real-valued continuous functions on $X$ with the uniform norm, by $C_{b}(X)$ the set of bounded continuous functions on $X$, by $C_{0}(X)$ the set of continuous functions on $X$ vanishing at infinity, and by $C_{c}(X)$ the set of continuous functions with compact support. By $C_{0}^{+}(X)$ and $C_{c}^{+}(X)$ we denote the collection of all nonnegative functions from $C_{0}(X)$ and $C_{c}(X)$, respectively. $Z(f)$ and $Coz(f)$ stand for the zero and the cozero sets of the function $f$, i.e. $Z(f)=\{x\in X:\,f(x)=0\},\ \ Coz(f)=X\setminus Z(f).$

When we consider maps into $[-\infty,\infty]$ we assume that any such map attains at most one of $\infty,-\infty$, and is not identically $\infty$ or $-\infty$.

We denote by $\overline{E}$ the closure of a set $E$, and by $\bigsqcup$ a union of disjoint sets. We denote by $1$ the constant function $1(x)=1$, by $id$ the identity function $id(x)=x$, and by $1_{K}$ the characteristic function of a set $K$. By $supp\,f$ we mean $\overline{\{x:f(x)\neq 0\}}$.

Several collections of sets are used often. They include: $\mathscr{O}(X)$; $\mathscr{C}(X)$; and $\mathscr{K}(X)$– the collection of open subsets of $X$; the collection of closed subsets of $X$; and the collection of compact subsets of $X$, respectively.

Definition 1**.**

Let $X$ be a topological space and $\nu$ be a set function on a family of subsets of $X$ that contains $\mathscr{O}(X)\cup\mathscr{C}(X)$. We say that

• •

$\nu$ is compact-finite if $|\nu(K)|<\infty$ for any $K\in\mathscr{K}(X)$;

• •

$\nu$ is simple if it only assumes values [math] and $1$;

• •

a nonnegative set-function $\nu$ is finite if $\nu(X)<\infty$.

Definition 2**.**

A Radon measure $m$ on $X$ is a Borel measure that is finite on all compact sets, outer regular on all Borel sets, and inner regular on all open sets, i.e. $m(E)=\inf\{m(U):\ \ E\subseteq U,\ U\text{ open }\}$ for every Borel set $E$, and $m(U)=\sup\{m(K):\ \ K\subseteq U,\ K\text{ compact }\}$ for every open set $U$.

Recall the following fact (see, for example, [12, Chapter XI, 6.2]):

Lemma 3**.**

Let $K\subseteq U,\ K\in\mathscr{K}(X),\ U\in\mathscr{O}(X)$ in a locally compact space $X$. Then there exists a set $V\in\mathscr{O}(X)$ with compact closure such that $K\subseteq V\subseteq\overline{V}\subseteq U.$

Definition 4**.**

A deficient topological measure on a locally compact space $X$ is a set function $\nu:\mathscr{C}(X)\cup\mathscr{O}(X)\longrightarrow[0,\infty]$ which is finitely additive on compact sets, inner compact regular, and outer regular, i.e. :

1. (DTM1)

if $C\cap K=\emptyset,\ C,K\in\mathscr{K}(X)$ then $\nu(C\sqcup K)=\nu(C)+\nu(K)$; 2. (DTM2)

$\nu(U)=\sup\{\nu(C):\ C\subseteq U,\ C\in\mathscr{K}(X)\}$ for $U\in\mathscr{O}(X)$; 3. (DTM3)

$\nu(F)=\inf\{\nu(U):\ F\subseteq U,\ U\in\mathscr{O}(X)\}$ for $F\in\mathscr{C}(X)$.

For a closed set $F$, $\nu(F)=\infty$ iff $\nu(U)=\infty$ for every open set $U$ containing $F$.

Definition 5**.**

A topological measure on $X$ is a set function $\mu:\mathscr{C}(X)\cup\mathscr{O}(X)\longrightarrow[0,\infty]$ 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)

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

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

The following two theorems from [7, Section 4] give criteria for a deficient topological measure to be a topological measure or a measure.

Theorem 6**.**

1. (I)

Let $X$ be compact, and $\nu$ a deficient topological measure. The following are equivalent:

1. (a)

$\nu$* is a topological measure* 2. (b)

$\nu(X)=\nu(C)+\nu(X\setminus C),\ \ \ C\in\mathscr{C}(X)$** 3. (c)

$\nu(X)\leq\nu(C)+\nu(X\setminus C),\ \ \ C\in\mathscr{C}(X)$** 2. (II)

Let $X$ be locally compact, and $\nu$ a deficient topological measure. The following are equivalent:

1. (a)

$\nu$* is a topological measure* 2. (b)

$\nu(U)=\nu(C)+\nu(U\setminus C),\ \ \ C\in\mathscr{K}(X),\ \ \ U\in\mathscr{O}(X)$** 3. (c)

$\nu(U)\leq\nu(C)+\nu(U\setminus C),\ \ \ C\in\mathscr{K}(X),\ \ \ U\in\mathscr{O}(X)$**

Theorem 7**.**

Let $\mu$ be a deficient 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.*

Remark 8**.**

In [7, Section 3] we show that a deficient topological measure $\nu$ is $\tau$-smooth on compact sets (i.e. if a net $K_{\alpha}\searrow K$ , where $K_{\alpha},K\in\mathscr{K}(X)$ then $\mu(K_{\alpha})\rightarrow\mu(K)$), and also $\tau$-smooth on open sets (i.e. if a net $U_{\alpha}\nearrow U$, where $U_{\alpha},U\in\mathscr{O}(X)$ then $\mu(U_{\alpha})\rightarrow\mu(U)$).

A deficient topological measure $\nu$ is also superadditive, i.e. if $\bigsqcup_{t\in T}A_{t}\subseteq A,$ where $A_{t},A\in\mathscr{O}(X)\cup\mathscr{C}(X)$, and at most one of the closed sets (if there are any) is not compact, then $\nu(A)\geq\sum_{t\in T}\nu(A_{t})$.

If $F\in\mathscr{C}(X)$ and $C\in\mathscr{K}(X)$ are disjoint, then $\nu(F)+\nu(C)=\nu(F\sqcup C)$.

One may consult [7] for more properties of deficient topological measures on locally compact spaces.

Definition 9**.**

A signed deficient topological measure on a locally compact space $X$ is a set function $\nu:\mathscr{C}(X)\cup\mathscr{O}(X)\longrightarrow[-\infty,\infty]$ that assumes at most one of $\infty,-\infty$ and that is finitely additive on compact sets, inner compact regular on open sets, and outer regular on closed sets, i.e.

1. (SDTM1)

If $C\cap K=\emptyset,\ C,K\in\mathscr{K}(X)$ then $\nu(C\sqcup K)=\nu(C)+\nu(K);$ 2. (SDTM2)

$\mu(U)=\lim\{\mu(K):K\in\mathscr{K}(X),\ K\subseteq U\}$ for $U\in\mathscr{O}(X)$; 3. (SDTM3)

$\mu(F)=\lim\{\mu(U):U\in\mathscr{O}(X),\ F\subseteq U\}$ for $F\in\mathscr{C}(X)$.

Remark 10**.**

In condition (SDTM2) we mean the limit of the net $\nu(C)$ with the index set $\{C\in\mathscr{K}(X):\ C\subseteq U\}$ ordered by inclusion. The limit exists and is equal to $\nu(U)$. Condition (SDTM3) is interpreted in a similar way, with the index set being $\{U\in\mathscr{O}(X):\ U\supseteq C\}$ ordered by reverse inclusion.

Remark 11**.**

Since we consider set-functions that are not identically $\infty$ or $-\infty$, we see that for a signed deficient topological measure $\nu(\emptyset)=0$. If $\nu$ and $\mu$ are (signed) deficient topological measures that agree on $\mathscr{K}(X)$ (or on $\mathscr{O}(X)$) then $\nu=\mu$; if $\nu\leq\mu$ on $\mathscr{K}(X)$ (or on $\mathscr{O}(X)$) then $\nu\leq\mu$.

Remark 12**.**

For a signed deficient topological measure $\nu$ we may define its total variation, a deficient topological measure $|\nu|$, by

[TABLE]

for an open set $U$, and for a closed subset $F\subseteq X$

[TABLE]

See [7, Sections 2,3] for detail.

Definition 13**.**

We define $\|\nu\|=\sup\{|\nu(K)|:K\in\mathscr{K}(X)\}$ for a signed deficient topological measure $\nu$.

If $\mu$ is a deficient topological measure then $\|\mu\|=\mu(X)$.

Definition 14**.**

A signed deficient topological measure $\nu$ is called proper if from $m\leq|\nu|$, where $m$ is a Radon measure and $|\nu|$ is a total variation of $\nu$, it follows that $m=0$.

Let $\mu$ be a deficient topological measure, and $\nu$ be a signed deficient topological measure. We say that $\nu$ is absolutely continuous with respect to $\mu$ (and we write $\nu\ll\mu$) if $\mu(A)=0$ implies $|\nu|(A)=0$.

Definition 15**.**

A signed topological measure on a locally compact space $X$ is a set function $\mu:\mathscr{O}(X)\cup\mathscr{C}(X)\longrightarrow[-\infty,\infty]$ that assumes at most one of $\infty,-\infty$ and satisfies the following conditions:

1. (STM1)

if $A,B,A\sqcup B\in\mathscr{K}(X)\cup\mathscr{O}(X)$ then $\mu(A\sqcup B)=\mu(A)+\mu(B);$ 2. (STM2)

$\mu(U)=\lim\{\mu(K):K\in\mathscr{K}(X),\ K\subseteq U\}$ for $U\in\mathscr{O}(X)$; 3. (STM3)

$\mu(F)=\lim\{\mu(U):U\in\mathscr{O}(X),\ F\subseteq U\}$ for $F\in\mathscr{C}(X)$.

Remark 16**.**

Definition 9 was first introduced in [10]. Any deficient topological measure (topological measure, signed topological measure) is a signed deficient topological measure.

Remark 17**.**

There is a correspondence between deficient topological measures and certain non-linear functionals, see [9, Section 8]. In particular, 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 [8] and Theorem 3.9 in [16] for the first version of the representation theorem.) We outline the correspondence.

1. (I)

Given a finite deficient topological measure $\mu$ on a locally compact space $X$ and $f\in C_{b}(X)$, define functions on $\mathbb{R}$:

[TABLE]

[TABLE]

Let $r$ be the Lebesque-Stieltjes measure associated with $-R_{1}$, a regular Borel measure on $\mathbb{R}$. We define a functional on $C_{0}(X)$:

[TABLE]

where $[a,b]$ is any interval containing $f(X)$. If $f$ is nonnegative with $f(X)\subseteq[0,b]$ we have:

[TABLE]

We call the functional $\mathcal{R}$ a quasi-integral (with respect to a deficient topological measure $\mu$) and write:

[TABLE] 2. (II)

Functional $\mathcal{R}$ is non-linear. By [9, Lemma 7.7, Lemma 7.10, Lemma 3.6, Lemma 7.12] we have:

1. (a)

$\mathcal{R}(f)$ is positive-homogeneous, i.e. $\mathcal{R}(cf)=c\mathcal{R}(f)$ for $c\geq 0$, and $\mathcal{R}(0)=0$. 2. (b)

if $g\,h=0$, where $g,h\geq 0$ or $g\geq 0,h\leq 0$, then $\mathcal{R}(gh)=\mathcal{R}(g)+\mathcal{R}(h)$. 3. (c)

$\mathcal{R}$ is monotone, i.e. if $f\leq g$ then $\mathcal{R}(f)\leq\mathcal{R}(g)$. 4. (d)

$\mu(X)\cdot\inf_{x\in X}f(x)\leq\mathcal{R}(f)\leq\mu(X)\cdot\sup_{x\in X}f(x)$ for any $f\in C_{0}(X)$. 3. (III)

A functional $\rho$ with values in $[-\infty,\infty]$ (assuming at most one of $\infty,-\infty$) and $|\rho(0)|<\infty$ is called a d-functional if on nonnegative functions it is positive-homogeneous, monotone, and orthogonally additive, i.e. for $f,g\in D(\rho)$ (the domain of $\rho$) we have: (d1) $f\geq 0,\ a>0\Longrightarrow\rho(af)=a\rho(f)$; (d2) $0\leq g\leq f\Longrightarrow\rho(g)\leq\rho(f)$; (d3) $f\cdot g=0,f,g\geq 0\Longrightarrow\rho(f+g)=\rho(f)+\rho(g)$.

Let $\rho$ be a d-functional with $C_{c}^{+}(X)\subseteq D(\rho)\subseteq C_{b}(X)$. In particular, we may take functional $\mathcal{R}$ on $C_{0}^{+}(X)$. The corresponding deficient 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, $\mu_{\rho}(K)=\inf\{\rho(g):\ g\in C_{c}(X),g\geq 1_{K}\}=\inf\{\rho(g):\ g\in C_{c}(X),1_{K}\leq g\leq 1\}.$ (See Definition 33 and Lemma 35 in Section 4 of [8].)

If given a finite deficient topological measure $\mu$, we obtain $\mathcal{R}$, and then $\mu_{\mathcal{R}}$, then $\mu=\mu_{\mathcal{R}}$.

Definition 18**.**

Let $X$ be locally compact. A functional $\rho$ on $C_{0}(X)$ is Lipschitz continuous if for every compact $K\subseteq X$ there exists a number $N_{K}$ such that $|\rho(f)-\rho(g)|\leq N_{K}\|f-g\|$ for all $f,g\in C_{c}(X)$ with support in $K$.

The proof of the following theorem can be found, for example, in [5], §7.2.6.

Theorem 19**.**

Let $l$ be a regular $\tau$-smooth measure on a topological space $X$ and let $\{f_{\alpha}\}$ be an increasing net of nonnegative lower semicontinuous functions such that the function $f=\lim_{\alpha}f_{\alpha}$ is bounded. Then

[TABLE]

Remark 20**.**

It is easy to see that a left-continuous nondecreasing function (that does not assume $-\infty$) or right-continuous nonincreasing function (that does not assume $-\infty$) is lower semicontinuous.

Corollary 21**.**

Let $l$ be a regular $\tau$-smooth measure on a topological space $X$. Let $\{g_{\alpha}\}$ be a decreasing net of nonnegative nonincreasing left-continuous functions converging to $g$. Suppose one of the functions $\{g_{\alpha}\}$ is bounded. Then

[TABLE]

Proof.

Suppose $g_{\gamma}$ is bounded above by $M>0$. Then $g_{\alpha}\leq M$ for all $\alpha\geq\gamma$. Consider functions $f_{\alpha}=M-g_{\alpha}$ and $f=M-g$. By Remark 20, functions $f_{\alpha}$ are lower semi-continuous. Apply Theorem 19 to $f_{\alpha}$. ∎

2. Integrals over a set via ”restricted” deficient topological measures

We would like to begin with integration with respect to a deficient topological measure over a set. Integration of a continuous bounded function with respect to a deficient topological measure is described in part (I) of Remark 17, but it will not work for a function restricted to a set, as such a function need not be continuous anymore. Another approach to obtain integration over a set would be to integrate the function with respect to a deficient measure that is restricted to a set. But if $\mu$ is a deficient topological measure on $X$, one can not obtain a deficient topological measure on $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ by simply restricting $\mu$ to $A$, i.e., considering $\mu(A\cap B),\ B\in\mathscr{O}(X)\cup\mathscr{C}(X)$. One simple reason for this is that the intersection of two arbitrary sets from $\mathscr{O}(X)\cup\mathscr{C}(X)$ does not in general belong to $\mathscr{O}(X)\cup\mathscr{C}(X)$. However, there is still a way to obtain new deficient topological measures by ”restriction”. The next two results are from [7, Section 5].

Theorem 22**.**

Let $\mu$ be a deficient topological measure on a locally compact space $X,\ V\in\mathscr{O}(X)$. Define a set function $\mu_{V}$ on $\mathscr{O}(X)\cup\mathscr{C}(X)$ by letting

[TABLE]

[TABLE]

Then $\mu_{V}$ is a deficient topological measure on $X$.

Theorem 23**.**

Let $X$ be locally compact, and let $F\subseteq\mathscr{C}(X)$. There exists a deficient topological measure $\mu_{F}$ on $X$ such that

[TABLE]

[TABLE]

If $F\in\mathscr{K}(X)$ then $\mu_{F}(C)=\mu(F\cap C)$ for every $C\in\mathscr{C}(X)$.

The next theorem states some properties of ”restricted” deficient topological measures $\mu_{V}$ and $\mu_{F}$, given by Theorem 22 and Theorem 23.

Theorem 24**.**

Let $X$ be locally compact, $V\in\mathscr{O}(X),F\in\mathscr{C}(X)$, and $\mu,\nu$ be deficient topological measures on $X$. Let $\mu_{V}$ and $\mu_{F}$ be deficient topological measures given by Theorem 22 and Theorem 23. Then

1. (v1)

$\mu_{V}(U)=\sup\{\mu_{K}(U):\ K\subseteq U,\ K\in\mathscr{K}(X)\}$* for every $U\in\mathscr{O}(X)$.* 2. (v2)

$\mu_{F}(C)=\inf\{\mu_{V}(C):\ F\subseteq V,\ V\in\mathscr{O}(X)\}$* for every $C\in\mathscr{K}(X)$. If $F$ is compact then the equality holds for every $C\in\mathscr{C}(X)$.* 3. (v3)

If $A,B$ are disjoint compact sets, or disjoint open sets, then $\mu_{A\sqcup B}=\mu_{A}+\mu_{B}$. 4. (v4)

If $\mu\leq\nu$ then $\mu_{A}\leq\nu_{A}$ for any set $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. 5. (v5)

If $A,B\subseteq\mathscr{O}(X)\cup\mathscr{C}(X),\ A\subseteq B$ then $\mu_{A}\leq\mu_{B}$. 6. (v6)

If $a,b\geq 0$ then $(a\mu+b\nu)_{A}=a\mu_{A}+b\nu_{A}$ for any set $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. 7. (v7)

If $A\in\mathscr{O}(X)\cup\mathscr{K}(X)$ then $\mu_{A}(X)=\mu(A)$. If $F\in\mathscr{C}(X)$ then $\mu_{F}(X)\leq\mu(F)$.

Proof.

1. (v1)

Assume first that $\mu_{V}(U)=\mu(U\cap V)<\infty$. Given $\epsilon>0$, let $K\in\mathscr{K}(X)$ be such that $K\subseteq U\cap V,\ \mu(U\cap V)-\mu(K)<\epsilon.$ Note that for any $C\subseteq U$ we have $\mu_{K}(C)=\mu(K\cap C)\leq\mu(U\cap V)=\mu_{V}(U)$, which gives $\mu_{K}(U)\leq\mu_{V}(U)$. Since $\mu_{V}(U)-\mu_{K}(U)\leq\mu_{V}(U)-\mu_{K}(K)=\mu(U\cap V)-\mu(K)<\epsilon$, we have: $\mu_{V}(U)=\sup\{\mu_{K}(U):\ K\subseteq U,\ K\in\mathscr{K}(X)\}.$

Now assume that $\mu_{V}(U)=\mu(U\cap V)=\infty$. For each $n\in\mathbb{N}$ choose $K_{n}\in\mathscr{K}(X)$ such that $K_{n}\subseteq U\cap V,\ \mu(K_{n})\geq n$. For any compact $C$ such that $K_{n}\subseteq C\subseteq U$ we have $\mu_{K_{n}}(C)=\mu(K_{n}\cap C)=\mu(K_{n})\geq n$, and we see that $\mu_{K_{n}}(U)\geq n$. Then $\sup\{\mu_{K}(U):\ K\subseteq U,\ K\in\mathscr{K}(X)\}=\infty=\mu_{V}(U)$. 2. (v2)

Follows from Theorem 23 and the following result from [7, Section 5].

Lemma 25**.**

Let $X$ be locally compact, $\mu$ be a deficient topological measure on $X$, and $F\in\mathscr{C}(X)$. Then for any $C\in\mathscr{K}(X)$

[TABLE]

where $\mu_{V}$ is a deficient topological measure on $X$ from Theorem 22. If $F\in\mathscr{K}(X)$ then (4) holds for any $C\in\mathscr{C}(X)$. 3. (v3)

By Remark 11 it is enough to check the equality on compact sets (or open sets). The argument is easy and left to the reader. 4. (v4)

Take $A\in\mathscr{C}(X)$ (respectively, $A\in\mathscr{O}(X)$). Then $\mu_{A}(K)\leq\nu_{A}(K)$ for every $K\in\mathscr{K}(X)$ (respectively, $\mu_{A}(U)\leq\nu_{A}(U)$ for every $U\in\mathscr{O}(X)$). The statement then follows from Remark 11. 5. (v5)

Arguing as in part (v4) we see that $\mu_{A}\leq\mu_{B}$ if $A,B\in\mathscr{C}(X)$ or if $A,B\in\mathscr{O}(X)$. If $F\subseteq U,F\subseteq\mathscr{C}(X),\ U\in\mathscr{O}(X)$, then by part (v2) $\mu_{F}(K)\leq\mu_{U}(K)$ for every $K\in\mathscr{K}(X)$. By Remark 11 $\mu_{F}\leq\mu_{U}$. Consider the last case: $U\subseteq F,\ U\in\mathscr{O}(X),F\in\mathscr{C}(X)$. Let $V\in\mathscr{O}(X)$ be such that $F\subseteq V$. So $U\subseteq F\subseteq V$. Then $\mu_{U}(K)\leq\mu_{V}(K)$, and by part (v2) $\mu_{U}(K)\leq\mu_{F}(K)$ for every $K\in\mathscr{K}(X)$. By Remark 11 $\mu_{U}\leq\mu_{F}$. 6. (v6)

The argument is similar to one for part (v4). 7. (v7)

Easy to see from Theorem 22 and Theorem 23.

∎

Definition 26**.**

Suppose $X$ is locally compact, $\mu$ is a deficient topological measure on $X$, and $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ is such that $\mu(A)<\infty$. In particular, for a finite deficient topological measure $\mu$ on $X$, we take $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. Let $\mu_{A}$ be a deficient topological measure given by Theorem 22 or Theorem 23. We define the the quasi-integral (with respect to a deficient topological measure $\mu$) of a function $g\in C_{b}(X)$ over a set $A$ with $\mu(A)<\infty$ to be the functional $\mathcal{R}_{\mu_{A}}(g)$ as in formula (3) (where we use a deficient topological measure $\mu_{A}$) and we write:

[TABLE]

Remark 27**.**

If $A=X$ we have the usual $\mathcal{R}_{\mu}(g)$. If $\mu$ is a measure, we obtain the standard integral $\int_{A}g\,d\mu$ over a set $A$ . To evaluate $\mathcal{R}_{\mu_{A}}(g)$ we may use formulas (1), (2), and part (v7) of Theorem 24. For example, if $f\in C_{b}(X),f(X)\subseteq[a,b],E\in\mathscr{O}(X)\cup\mathscr{K}(X)$, and $\mu(E)<\infty$, we may use:

[TABLE]

where, for example,

[TABLE]

[TABLE]

If $f\in C_{0}^{+}(X)$ and $f(X)\subseteq[0,b]$ then for $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ with $\mu(A)<\infty$ we may use:

[TABLE]

Let $a^{\prime}=\sup_{E}f,\ b^{\prime}=\inf_{E}f$. Since $R_{i,\mu_{E},f}(t)=\mu(E)$ for $t\in[a,a^{\prime})$, and $R_{i,\mu_{E},f}(t)=\emptyset$ for $t\in(b^{\prime},b]$ (where $i=1,2$), we may rewrite ( 6 ) as:

[TABLE]

Lemma 28**.**

Let $X$ be locally compact, $\mu$ be a deficient topological measure on $X$, and $g\in C_{b}(X)$. We have:

1. (i)

If $V\in\mathscr{O}(X),\mu(V)<\infty$ then

[TABLE] 2. (ii)

Suppose $g\in C(X)$ if $X$ is compact and $g\in C_{0}^{+}(X)$ if $X$ is locally compact. If $F\in\mathscr{C}(X),\mu(F)<\infty$ then

[TABLE]

Proof.

Let $g(X)\subseteq[a,b]$. If $g\in C_{0}^{+}(X)$ we take $a=0$.

1. (i)

For any $K\subseteq V,K\in\mathscr{K}(X)$ the function $R_{1,\mu_{K},g}(t)$ is right-continuous (see [9, Lemma 6.3]), nonincreasing, so by Remark 20 it is lower-semicontinuous. For each $t\in[a,b]$ by part (v5) of Theorem 24 the net $\{R_{1,\mu_{K},g}(t):\,K\subseteq V,K\in\mathscr{K}(X)\}$ (ordered by inclusion) is nondecreasing, and by part (v1) of Theorem 24

[TABLE]

Note that $0\leq R_{1,\mu_{V},g}(t)\leq\mu(V)$. Apply formula (6) and Theorem 19 to obtain the statement. 2. (ii)

Since $\mu(F)<\infty$, there exists $U\in\mathscr{O}(X)$ such that $F\subseteq U,\mu(U)<\infty$, and then $\mu(V)<\infty$ for all open sets $V$ such that $F\subseteq V\subseteq U$. For any $V\in\mathscr{O}(X)$ containing $F$ the function $R_{2,\mu_{V},g}$ is left-continuous on $[a,b]$ (use [9, Lemma 6.3]), nonincreasing, and bounded. The assumptions assure that each set $g^{-1}([t,\infty))$ is compact for $t\in(a,b]$. For $t\in[a,b]$ by part (v5) of Theorem 24 the net $\{R_{2,\mu_{V},g}(t):\,V\in\mathscr{O}(X),F\subseteq V\}$ (ordered by reverse inclusion) is nonincreasing, and by part (v2) of Theorem 24

[TABLE]

Applying formula (6) and Corollary 21 we obtain the statement.

∎

Theorem 29**.**

Let $X$ be locally compact, $\mu$ be a compact-finite deficient topological measure on $X$.

1. (a)

Let $g\in C_{0}^{+}(X)$. Then there exists a compact-finite deficient topological measure $\mu_{g}$ on $X$ such that $\mu_{g}(A)=\int_{A}g\,d\mu$ for every $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ with $\mu(A)<\infty$, and $\mu_{g}(K)\leq\|g\|\mu(K)$ for any $K\in\mathscr{K}(X)$. Also, if $E\sqcup D=F$, where $E,D,F\in\mathscr{C}(X)$ and $\mu(F)<\infty$, then $\mu_{g}(E)+\mu_{g}(D)=\mu_{g}(F)$. 2. (b)

Let $X$ be compact, $g\in C(X)$. Then a set function $\nu_{g}$ on $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ given by $\nu_{g}(A)=\int_{A}g\,d\mu$ is a signed deficient topological measure with finite norm $\|\nu_{g}\|\leq\|g\|\mu(X)$. If $g\geq 0$ then $\nu_{g}$ is a deficient topological measure, and $\nu_{g}=\mu_{g}$.

Proof.

1. (a)

Consider a set function $\lambda$ on $\mathscr{K}(X)$ given by $\lambda(K)=\int_{K}g\,d\mu$. By part (v3) of Lemma 24 $\lambda$ is finitely additive on $\mathscr{K}(X)$. We take $\mu_{g}$ to be the positive variation $\lambda^{+}$ of $\lambda$, so $\mu_{g}$ is a deficient topological measure (see [7, Section 3]). By definition of $\lambda^{+}$ and Lemma 28 we see that $\mu_{g}(A)=\int_{A}g\,d\mu$ for every $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ with $\mu(A)<\infty$. For any $K\in\mathscr{K}(X)$ using part (II) of Remark 17 and part (v7) of Theorem 24 we have:

[TABLE]

and we see that $\mu_{g}$ is compact finite.

Now suppose $E\sqcup D=F$, where $E,D,F\in\mathscr{C}(X)$ and $\mu(F)<\infty$. Let $g(X)\subseteq[0,b]$. For each $t>0$

[TABLE]

and all sets are compact. Since $\mu$ is finitely additive on compact sets, we see that

[TABLE]

which (since $\mu(F),\mu(E),\mu(D)<\infty$) means that $\mu_{g}(F)=\mu_{g}(E)+\mu_{g}(D)$. 2. (b)

From part (v3) of Theorem 24 and Lemma 28 we see that $\nu_{g}$ is a signed deficient topological measure. As in (10), $|\nu_{g}(K)|\leq\|g\|\mu(K)\leq\|g\|\mu(X)$ for each $K\in\mathscr{K}(X)$, so by Definition 13 $\|\nu_{g}\|\leq\|g\|\mu(X)$. If $g\geq 0$ then $\nu_{g}=\mu_{g}$ on $\mathscr{K}(X)$, so by Remark 11 $\nu_{g}=\mu_{g}$.

∎

The next example shows that by Theorem 29 we can obtain a signed deficient topological measure.

Example 30**.**

In this example we shall use a ”parliamentry” topological measure $\mu$ on $X=[-4,0]\times[0,4]$: given an odd number of distinct points of a compact space $X$, we assign a closed or open set which is also solid (i.e. which is connected and has a connected complement) $\mu$-value 1 if the set contains more than half of the points, and $\mu$-value 0 otherwise. (For more information, see [2, Remark 6.1] or [4, Introduction, Example b].) Let $\mu$ be the simple topological measure based on points $p=(-4,0),q=(-4,4)$, and $s=(0,4)$. Let $g(x,y)=x$, so $g(X)=[-4,0]$. Let $K\in\mathscr{K}(X)$ be the closed triangle with vertices $(-4,0),(-4,4)$, and $(0,0)$, so $K$ is a closed solid set with $\mu(K)=1$. Note that for any $t>-4$ the set $K\cap g^{-1}([t,\infty))$ is a closed solid set with $\mu(K\cap g^{-1}([t,\infty)))=0$, and so $R_{2,\mu_{K},g}(t)=\mu(K\cap g^{-1}([t,\infty)))=0.$ By formula (6) we see that $\nu_{g}(K)=-4\mu(K)=-4.$ Thus, $\nu_{g}$ is a signed deficient topological measure. Note also that $\nu_{g}(K)=-4=\inf_{K}g\cdot\mu(K)$.

Now we shall show that $\nu_{g}$ is not a signed topological measure. Let $C\in\mathscr{K}(X)$ be the closed triangle with vertices $(-4,0),(-2,4)$, and $(0,0)$, and let $U,V$ be the connected components of $X\setminus C$. Triangles $U,V$ are open solid sets. Since $\mu(U)=\mu(V)=0,\mu(X)=1$, the sets $U\cap g^{-1}((t,\infty)),V\cap g^{-1}((t,\infty))$, and $g^{-1}((t,\infty))$ are open solid sets with at most one of points $p,q,s$ for each $t>0$, by formula (6) we calculate $\nu_{g}(U)=\nu_{g}(V)=0,\nu_{g}(X)=-4$. For each $t>-4$ the set $C\cap g^{-1}([t,\infty))$ is not solid, but its complement consists of two disjoint open solid sets which we call $U_{t}$ and $V_{t}$, and $\mu(U_{t})=1,\mu(V_{t})=0$. Then $R_{2,\mu_{C},g}(t)=\mu(C\cap g^{-1}([t,\infty)))=\mu(X)-\mu(U_{t})-\mu(V_{t})=0$ for each $t>-4$. Since $\mu(C)=0$, we obtain $\nu_{g}(C)=0$. Since $\nu_{g}(X)=-4$, while $\nu_{g}(V)=\nu_{g}(W)=\nu_{g}(C)=0$, we see that $\nu_{g}$ is not a signed topological measure.

Remark 31**.**

If in Theorem 29 $\mu$ is a measure, then it is easy to see that $\nu_{g}$ is a signed measure, which is a measure if $g\geq 0$. However, the next example shows that if $\mu$ is a topological measure, then $\mu_{g}$ may not be a a topological measure.

Example 32**.**

Let $X=\mathbb{R}^{2}$, and $D$ be the disk of radius 2 centered at the origin. Let $g\in C_{c}(X)$ be the function with $\operatorname{supp}g=D$, whose graph is the cone with the vertex $(0,2)$ and the base $D$. Thus, $g(X)=[0,2]$. Let $\mu$ be the simple topological measure based on points $(0,0),(1,0)$ and $(4,0)$ as in [6, Example 2]. Since $g^{-1}((t,\infty))$ is a bounded open solid set, by [6, Remark 4] we see that $R_{1,\mu,g}(t)=1$ for $0<t<1$, and $R_{1,\mu,g}(t)=0$ for $1\leq t\leq 2$. Then $\mu_{g}(X)=\int_{X}g\,d\mu=\int_{0}^{4}R_{1,\mu,g}(t)\,dt=1$. Let compact $K=\{(x,0):1\leq x\leq 4\}$ and $V=X\setminus K$. For each $t\geq 0$ we see that $K\cap g^{-1}([t,\infty))$ is a compact solid set, and $V\cap g^{-1}((t,\infty))$ is a bounded open solid set. Note that $R_{2,\mu_{K},g}(t)=\mu(K\cap g^{-1}([t,\infty)))=0$, and $R_{1,\mu_{V},g}(t)=0$ for each $t>0$. Thus, $\mu_{g}(K)=\mu_{g}(V)=0$. It follows that $\mu_{g}$ is not a topological measure.

When $\mu$ is a topological measure, $\mu_{g}$ may not be a topological measure even for a strictly positive function $g$. Let $X=[1,3]\times[0,1]$. Let $\mu$ be the ”parliamentry” simple topological measure based on points $p=(2,0),q=(1,1)$, and $s=(3,1)$. Let $g(x,y)=x$, so $g>0$ and $g(X)=[1,3]$. Let $K$ be the triangle with vertices $(1,1),(1,0),(3,0)$ and let $V=X\setminus K$. So $K\in\mathscr{K}(X),V\in\mathscr{O}(X),X=K\sqcup V$, and $\mu(K)=1,\mu(V)=0,\mu(X)=1.$ Note that for $t\in(1,3]$ the set $K\cap g^{-1}([t,\infty))$ is a closed solid set containing at most one of the points $p,q,s$, and so $R_{2,\mu_{K},g}(t)=\mu(K\cap g^{-1}([t,\infty)))=0$. Similarly, $R_{1,\mu_{V},g}(t)=0$ for $t\in(1,3].$ Using formula (6) we see that $\mu_{g}(V)=0$ and $\mu_{g}(K)=1$. By formula (1) we also have $\mu_{g}(X)=2$, since $R_{2,\mu,g}(t)=1$ if $t\in[1,2]$, and $R_{2,\mu,g}(t)=0$ if $t>2$. Thus, $\mu_{g}$ can not be a topological measure.

Theorem 33**.**

*Let $X$ be locally compact, $\mu$ be a compact-finite deficient topological measure on $X$. Let $\mu_{g}$ (for $g\in C_{0}^{+}(X)$) and $\nu_{g}$ (for $g\in C(X)$ and $X$ compact) be as in Theorem 29. Then *

1. (b1)

If $a\geq 0$ then $\mu_{ag}=a\mu_{g}$. If $g=0$ then $\mu_{g}=0$. The same holds for $\nu_{g}$. 2. (b2)

If $g\,h=0$, where $g,h\geq 0$ then $\mu_{g+h}=\mu_{g}+\mu_{h}$. If $g\,h=0,\,g\geq 0,h\leq 0$ then $\nu_{g+h}=\nu_{g}+\nu_{h}$. 3. (b3)

If $g\leq h$ then $\mu_{g}\leq\mu_{h}$ (respectively, $\nu_{g}\leq\nu_{h}$). 4. (b4)

If $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ and $\mu(A)<\infty$ then for $\lambda_{g}=\mu_{g}$ or $\lambda_{g}=\nu_{g}$

[TABLE]

In particular,

[TABLE] 5. (b5)

If $A\in\mathscr{O}(X)\cup\mathscr{C}(X),\,\mu(A)<\infty$ and $g=h$ on $A$ then $\mu_{g}(A)=\mu_{h}(A)$ (respectively, $\nu_{g}(A)=\nu_{h}(A)$). 6. (b6)

(Lipschitz continuity) If $K\in\mathscr{K}(X),f,g\in C_{c}^{+}(X)$ and $\operatorname{supp}f,\operatorname{supp}g\subseteq K$ then $|\mu_{g}(K)-\mu_{f}(K)=|\mathcal{R}_{\mu_{K}}(g)-\mathcal{R}_{\mu_{K}}(f)|\leq\|f-g\|\mu(K)$. 7. (b7)

If $A\in\mathscr{O}(X)\cup\mathscr{K}(X)$ and $g=1$ on $A$ then $\mu_{g}(A)=\mu(A)$ (respectively, $\nu_{g}(A)=\mu(A)$). 8. (b8)

For any open set $U\subseteq X$ we have $\mu(U)=\sup\{\mu_{g}(K):K\subseteq U,K\in\mathscr{K}(X),g=1\mbox{ on }K\}$ (respectively, $\mu(U)=\lim\{\nu_{g}(K):K\subseteq U,K\in\mathscr{K}(X),g=1\mbox{ on }K\}$). 9. (b9)

If $\mu=c\lambda+d\delta$, where $c,d\geq 0$ and $\lambda,\delta$ are compact-finite deficient topological measures on $X$, then $\mu_{g}=c\lambda_{g}+d\delta_{g}$. (Here $\lambda_{g},\delta_{g}$ are deficient topological measures obtained from $\lambda,\delta$ by Theorem 29.) Respectively, $\nu_{g}=c\lambda_{g}+d\delta_{g}$, where $\lambda_{g},\delta_{g}$ are signed deficient topological measures obtained from $\lambda,\delta$ by Theorem 29. 10. (b10)

If $\mu\leq\lambda$ where $\lambda$ is a compact-finite deficient topological measures on $X$, then $\mu_{g}\leq\lambda_{g}$ (respectively, $\nu_{g}\leq\lambda_{g}$) where $\lambda_{g}$ is a deficient topological measure (respectively, a signed deficient topological measure) obtained from $\lambda$ by Theorem 29. 11. (b11)

If $\mu$ is a proper deficient topological measure, then so is $\mu_{g}$ (respectivey, $\nu_{g}$). If $g>0$ then $\mu$ is a proper deficient topological measure iff $\mu_{g}$ is a proper deficient topological measure. 12. (b12)

If $\mu$ is a measure then $\nu_{g}$ is a signed measure, which is a measure if $g\geq 0$. If $g>0$ then $\mu$ is a Radon measure iff $\mu_{g}$ is a Radon measure. 13. (b13)

If $\mu$ is finite and $g\geq 0$ then $\mu_{g}(X)=\mu_{g}(Coz(g))$. 14. (b14)

(Absolute continuity) For any $\epsilon>0$ there exists $\delta>0$ such that if $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ and $\mu(A)<\delta$, then $\mu_{g}(A)<\epsilon$ (respectively, $|\nu_{g}(A)|<\epsilon$). Hence, $\mu_{g}\ll\mu$ (respectively, $\nu_{g}\ll\mu$).

If $X$ is compact, $g,h\in C(X)$ we also have:

1. (i)

If $c$ is a constant then $\nu_{c}=c\mu$ and $\nu_{g+c}=\nu_{g}+c\mu$. 2. (ii)

If $g\leq c,\ g=c$ on $\operatorname{supp}h,\ h\in C(X)$ then $\nu_{g+h}=\nu_{g}+\nu_{h}$. 3. (iii)

*(Lipschitz continuity): $|\nu_{g}(A)-\nu_{h}(A)|=|\mathcal{R}_{\mu_{K}}(g)-\mathcal{R}_{\mu_{K}}(h)|\leq\mu(A)\|g-h\|$ for $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. *

Proof.

1. (b1)

Since $ag\in C_{0}^{+}(X)$, we may define the deficient topological measure $\mu_{ag}$. By Remark 11 it is enough to show that $a\mu_{g}=\mu_{ag}$ on $\mathscr{K}(X)$. Let $K\in\mathscr{K}(X)$. By formula (5) and part (II) of Remark 17 we have: $\mu_{ag}(K)=\mathcal{R}_{\mu_{K}}(ag)=a\mathcal{R}_{\mu_{K}}(g)=a\mu_{g}(K).$ Thus, $\mu_{ag}=a\mu_{g}.$

If $g=0$ then by part (II) of Remark 17$\mu_{g}(K)=\mathcal{R}_{\mu_{K}}(0)=0$ for each $K\in\mathscr{K}(X)$. Thus, $\mu_{g}=0$. The argument for $\nu_{g}$ is the same. 2. (b2)

We need to check that $\mu_{g+h}=\mu_{g}+\mu_{h}$ on $\mathscr{K}(X)$. For $K\in\mathscr{K}(X)$ by part (II) of Remark 17 $\mu_{g+h}(K)=\mathcal{R}_{\mu_{K}}(g+h)=\mathcal{R}_{\mu_{K}}(g)+\mathcal{R}_{\mu_{K}}(h)=\mu_{g}(K)+\mu_{h}(K).$ The proof for $\nu_{g}$ is the same. 3. (b3)

By Remark 11 it is enough to check that $\mu_{g}\leq\mu_{h}$ (or $\nu_{g}\leq\nu_{h}$) on $\mathscr{K}(X)$, which easily follows from part (II) of Remark 17. 4. (b4)

Let $A\in\mathscr{O}(X)\cup\mathscr{C}(X),\,\mu(A)<\infty$. By formula (9) we have $a^{\prime}\mu(A)\leq\int_{A}f\,d\mu\leq a^{\prime}\mu(A)+\int_{a^{\prime}}^{b^{\prime}}R_{2,\mu_{A},f}(t)\,dt\leq a^{\prime}\mu(A)+\int_{a^{\prime}}^{b^{\prime}}\mu(A)\,dt=b^{\prime}\mu(A)$, and the statement follows. 5. (b5)

If $F\in\mathscr{C}(X)$ and $g=h$ on $F$ then $R_{2,\mu_{F},g}(t)=R_{2,\mu_{F},h}(t)$ for every $t>0$. If $U\in\mathscr{O}(X)$ and $g=h$ on $U$ then $R_{1,\mu_{U},g}=R_{1,\mu_{U},h}$ for every $t\geq 0$. From formula (8) (applied on $[a,b]$ that contains $f(X)$ and $g(X)$) we see that $\mu_{g}(A)=\mu_{h}(A)$ for $A=F$ or $A=U$. For $\nu_{g}$ we apply a similar argument using formula (6). 6. (b6)

Let $K\in\mathscr{K}(X)$. Using [9, Lemma 7.12(z8)] and part (v7) of Theorem 24 we have: $|\mu_{g}(K)-\mu_{f}(K)|=|\mathcal{R}_{\mu_{K}}(g)-\mathcal{R}_{\mu_{K}}(f)|\leq\mu_{K}(X)\|g-f\|=\mu(K)\|g-f\|$. 7. (b7)

Follows from formula (11). 8. (b8)

Follows from part (b7) and regularity of $\mu$. 9. (b9)

It is enough to check that $\mu_{g}(K)=c\lambda_{g}(K)+d\delta_{g}(K)$ (respectively, $\nu_{g}(K)=c\lambda_{g}(K)+d\delta_{g}(K)$) for each $K\in\mathscr{K}(X)$, which follows easily from formula (8) (respectively, from (6)). 10. (b10)

Let $K\in\mathscr{K}(X)$. By part (v4) of Theorem 24 $\mu_{K}\leq\lambda_{K}$. By [9, Theorem 8.7 (II)] $\mu_{g}(K)=\mathcal{R}_{\mu_{K}}(g)\leq\mathcal{R}_{\lambda_{K}}(g)=\lambda_{g}(K)$. The statement follows. 11. (b11)

Follows from formula (11) and [11, Theorem 4.5(I), Section 4]. 12. (b12)

See first Remark 31. Now let $g>0$. Assume that $\mu_{g}$ is a Radon measure. By [11, Theorem 4.3, Section 4] we may write $\mu=m+\lambda$ where $m$ is a Radon measure and $\lambda$ is a proper deficient topological measure. By part (b9) $\mu_{g}=m_{g}+\lambda_{g}$. But since $\mu_{g}$ and $m_{g}$ are both Radon measures, by uniqueness of decomposition in [11, Theorem 4.3, Section 4] we see that $\lambda_{g}=0$. Then from part (b4) it follows that $\lambda(K)=0$ for every $K\in\mathscr{K}(X)$. Then $\lambda=0$ and $\mu=m$ is a Radon measure. 13. (b13)

Let $P=Cozg$. Since $R_{1,\mu_{P},g}(t)=R_{1,\mu_{X},g}(t)$ for all $t\geq 0$, by formula (8) we have $\mu_{g}(P)=\mathcal{R}_{\mu_{P}}(g)=\mathcal{R}_{\mu_{X}}(g)=\mu_{g}(X)$. 14. (b14)

Follows from (12).

1. (i)

It is enough to show that $\nu_{g+c}=\nu_{g}+c\mu$ on $\mathscr{K}(X)$; the other statement then will follow from part (b1). For $K\in\mathscr{K}(X)$ by [9, Lemma 7.12(i)] and part (v7) of Theorem 24 we have:

[TABLE] 2. (ii)

Suppose first $0\leq g\leq c,\,g=c$ on $\operatorname{supp}h$. We need to check that $\nu_{g+h}=\nu_{g}+\nu_{h}$ on $\mathscr{K}(X)$. Let $K\in\mathscr{K}(X)$. From [9, Theorem 7.10, Lemma 4.4] it follows that $\mathcal{R}_{\mu_{K}}$ satisfies the $c$-level condition (see [9, Definition 4.3], so

[TABLE]

For the general case $g\leq c,\,g=c$ on $\operatorname{supp}h$ choose constant $b\geq 0$ such that $0\leq g+b\leq c+b$ on $\operatorname{supp}h$. For any $K\in\mathscr{K}(X)$ we have: $\nu_{g+b+h}(K)=\nu_{g+b}(K)+\nu_{h}(K)$, which by (13) gives $\nu_{g+h}(K)+b\mu(K)=\nu_{g}(K)+b\mu(K)+\nu_{h}(K)$, i.e $\nu_{g+h}(K)=\nu_{g}(K)+\nu_{h}(K)$. Thus, $\nu_{g+h}=\nu_{g}+\nu_{h}$. 3. (iii)

Using [9, Lemma 7.12(ii), ] and part (v7) of Theorem 24, as in part (b6) we obtain the statement.

∎

Remark 34**.**

Example 30 and similar examples show that inequalities in part (b4) of Theorem 33 can be realized as equalities.

Theorem 35**.**

Suppose $X$ is locally compact and $g\in C_{0}^{+}(X)$, or $X$ is compact and $g\in C(X)$. Suppose $\mu$ is a compact-finite topological measure on $X$. Suppose $A,B,A\sqcup B\in\mathscr{O}(X)\cup\mathscr{C}(X)$, and $\mu(A\sqcup B)<\infty$. Then

[TABLE]

where $\omega(f,A)=\sup_{x\in A}f(x)-\inf_{x\in A}f(x)$.

Proof.

Let $g(X)\subseteq[a,b]$ where $a\leq 0$. Suppose first that $D\sqcup V=U$, where $D\subseteq\mathscr{C}(X),V,U\in\mathscr{O}(X)$, and $\mu(U)<\infty$. Let $p=\inf_{x\in D}g(x),\,q=\sup_{x\in D}g(x)$.

If $a<t<p$ then $D\subseteq g^{-1}((t,\infty))$, and so $g^{-1}((t,\infty))\cap U=(g^{-1}((t,\infty))\cap V)\sqcup D$. Thus, $R_{1,\mu_{U},g}(t)=R_{1,\mu_{V},g}(t)+\mu(D)$.

If $t>q$ then $g^{-1}((t,\infty))\cap D=\emptyset$, so $g^{-1}((t,\infty))\cap U=g^{-1}((t,\infty))\cap V$, i.e. $R_{1,\mu_{U},g}(t)=R_{1,\mu_{V},g}(t)$. Also,

[TABLE]

By superadditivity $\mu(U)-\mu(V)-\mu(D)\geq 0$. Using also (6) and (11) we obtain:

[TABLE]

The statement now follows.

All other possible cases are proved similarly. For example, if $D\sqcup V=C$, where $D,C\subseteq\mathscr{C}(X),\,V\in\mathscr{O}(X)$ and $\mu(C)<\infty$, we use the fact that $R_{2,\mu_{C},g}(t)=R_{2,\mu_{D},g}(t)+\mu(V)$ for $t<p=\inf_{x\in V}g(x)$, and $R_{2,\mu_{C},g}(t)=R_{2,\mu_{D},g}(t)$ for $t>q=\sup_{x\in V}g(x)$. ∎

Remark 36**.**

Theorem 35 says that for a deficient topological measure $\mu_{g}$ or a signed deficient topological measures $\nu_{g}$ obtained in Theorem 29 we have

[TABLE]

[TABLE]

whenever $A,B,A\sqcup B\in\mathscr{O}(X)\cup\mathscr{C}(X)$, and $\mu(A\sqcup B)<\infty$. When $X$ is compact, in [19, Theorem 5.10] it was shown that $\mu_{g}(A\sqcup B)\leq\mu_{g}(A)+\mu_{g}(B)+\mu(X)\,\max\{\omega(g,A),\omega(g,B)\}$, and it inspired Theorem 35.

By Theorem 6, $\mu_{g}$ is a topological measure iff $\mu_{g}(U)\leq\mu_{g}(K)+\mu_{g}(U\setminus K)$ for any open $U$ and any compact $K\subseteq U$. Example 32 shows that even when $\mu$ is a finitely defined simple topological measure and $g$ is strictly positive, $\mu_{g}$ may not be a topological measure. This is a contrast to the situation with measures (see part (b12) of Theorem 33). Inequality (14) is what we can say in general. Using Theorem 7, we can also say, that, unless $m_{g}$ is a measure, any inequality that estimates $\mu_{g}(A\sqcup B)$ above using $\mu_{g}(A)$ and $\mu_{g}(B)$ (like (inequality 14)) must have the form $\mu_{g}(A\sqcup B)\leq\mu_{g}(A)+\mu_{g}(B)+\mu(A\sqcup B)+\delta(A,B)$ with $\delta(A,B)>0$ for some $A,B\in\mathscr{O}(X)$ or $A,B\in\mathscr{K}(X)$.

3. Integrals over a set via functionals

In this section we present another interpretation of the deficient topological measure $\mu_{g}$ given by Theorem 29 in the case where $\mu$ is finite.

Let $X$ be locally compact. Let $\mu$ be a finite deficient topological measure on $X$ with the corresponding functional $\rho=\mathcal{R}_{\mu}$ according to part (I) of Remark 17. Let $g\in C_{0}^{+}(X)$. Let $\rho_{g}$ be a functional on $C_{0}(X)$ defined by $\rho_{g}(f)=\rho(fg)$. Since $\rho$ is a d-functional, so is $\rho_{g}$.

Definition 37**.**

Let ${\widehat{\mu_{g}}}$ be the deficient topological measure corresponding to $\rho_{g}$ by part (III) of Remark 17.

Remark 38**.**

By part (III) of Remark 17

[TABLE]

for $K\in\mathscr{K}(X)$, and for $U\in\mathscr{O}(X)$

[TABLE]

Remark 39**.**

If $\mu$ is a regular Borel measure, it is easy to see that

[TABLE]

for every $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. In the next theorem we will show that the equality ${\widehat{\mu_{g}}}(A)=\int_{A}g\,d\mu$ also holds when $\mu$ is a finite deficient topological measure.

Theorem 40**.**

Let $X$ be locally compact. Let $\mu$ be a finite deficient topological measure on $X$. Let $g\in C_{0}^{+}(X)$. Let ${\widehat{\mu_{g}}}$ be as in Definition 37, and let $\mu_{g}$ be the deficient topological measure given by Theorem 29. Then ${\widehat{\mu_{g}}}=\mu_{g}$.

Proof.

It is enough to show that ${\widehat{\mu_{g}}}=\mu_{g}$ on $\mathscr{K}(X)$. Let $K\in\mathscr{K}(X)$. Let $g(x)\subseteq[0,b],\,b\geq 1.$ By formula (8)

[TABLE]

where $R_{2,\mu_{K},g}(t)$ is given by formula (7). Take any $f\in C_{c}(X)$ such that $1_{K}\leq f\leq 1$. Then $[0,b]$ contains the ranges of functions $g$ and $gf$. By formula (2)

[TABLE]

Note that $g^{-1}([t,\infty))\cap K\subseteq(fg)^{-1}([t,\infty)),$ so $R_{2,\mu_{K},g}(t)\leq R_{2,\mu,fg}(t).$ Then by (15) $\mu_{g}(K)\leq\rho(fg)=\rho_{g}(f)$. From Remark 38 we see that $\mu_{g}(K)\leq{\widehat{\mu_{g}}}(K)$.

Now we shall show the opposite inequality. By Lemma 3 choose $U,V\in\mathscr{O}(X)$ and $C=\overline{V}\in\mathscr{K}(X)$ such that $K\subseteq V\subseteq C\subseteq U$. Let ${{\mathcal{K}}}=\{C\in\mathscr{K}(X):\exists V\mbox{ with }C=\overline{V},K\subseteq V\subseteq C\subseteq U\}$. ${{\mathcal{K}}}$ is a directed set with respect to inverse inclusion and $\bigcap_{C\in{{\mathcal{K}}}}C=K$. Let $C\in{{\mathcal{K}}}$. Let $f$ be the Urysohn function such that $1_{K}\leq f\leq 1_{V}$. For every $t>0$ we have $(fg)^{-1}([t,\infty))\subseteq g^{-1}([\frac{t}{f(t)},\infty))\cap f^{-1}((0,\infty))\subseteq g^{-1}([t,\infty))\cap C$, so $R_{2,\mu,fg}(t)\leq R_{2,\mu_{C},g}(t)$. Then

[TABLE]

Each function $R_{2,\mu_{C},g}$ is bounded above by $\mu(U)<\infty$. Also, $R_{2,\mu_{C},g}$ is left-continuous on $[0,b]$ (where left-continuity of $R_{2,\mu_{C},g}(t)$ for $t>0$ follows from $\tau$-smoothness on compact sets in Remark 8). By Corollary 21 $\int_{0}^{b}R_{2,\mu_{C},g}(t)\,dt\rightarrow\int_{0}^{b}R_{2,\mu_{K},g},$ and so ${\widehat{\mu_{g}}}(K)\leq\int_{0}^{b}R_{2,\mu_{K},g}(t)\,dt=\mu_{g}(K).$ ∎

We can say more about the values of the deficient topological measure $\mu_{g}$ using the measure $n_{r}$ on $X$ given by [9, Definition 7.15]:

Theorem 41**.**

*Suppose $\mu$ is a finite deficient topological measure on a locally compact space $X$, $g\in C_{0}^{+}(X)$. Then *

1. (i)

For $E=g^{-1}(A)$, where $A\subseteq\mathbb{R}$ is open or closed, $\mu_{g}(E)\leq\int_{E}g\,dn_{r}$. 2. (ii)

$\mu_{g}(Kerg)=\int_{Kerg}g\,dn_{r}=0$, and $\mu_{g}(E)=\int_{E}g\,dn_{r}=0$ for $E=g^{-1}(A)$, where $A\subseteq(-\infty,0)$ is open or closed, 3. (iii)

If $\mu$ is a topological measure then $\mu_{g}(E)=\int_{E}g\,dn_{r}$ for $E=g^{-1}(A)$, where $A\subseteq\mathbb{R}\setminus\{0\}$ is closed or $A\subseteq\mathbb{R}$ is open. 4. (iv)

If $X$ is compact and $\mu$ is a topological measure, then $\nu_{g}(E)=\int_{E}g\,dn_{r}$ for $E=g^{-1}(A)$, where $A\subseteq\mathbb{R}$ is open or closed. 5. (v)

If $X$ is compact then $\mu$ is a topological measure on $X$ iff $\mu_{g}(E)=\int_{E}g\,dn_{r}$ for any $g\in C^{+}(X)$ and $E=g^{-1}(C)$ for any closed $C\subseteq\mathbb{R}$.

Proof.

1. (i)

First, let $D=g^{-1}(C)$, where $C\subseteq\mathbb{R}$ is closed. Let $g(X)\subseteq[0,b]$. Set the measure $p$ to be the restriction of the measure $n_{r}$ to $D$. Then $\int_{D}g\,dn_{r}=\int_{X}g\,dp=\int_{0}^{b}F(t)dt,$ where $F$ is the distribution function for the measure $p\circ g^{-1}$ defined by

[TABLE]

For $t>0$ using [9, Lemma 7.17] we have:

[TABLE]

so

[TABLE]

The case $U=g^{-1}(W)$, where $W\subseteq\mathbb{R}$ is open, is similar. We use intervals $(t,\infty)$ instead of $[t,\infty)$ in the distribution function $F$. Then as above, $F(t)\geq R_{1,\mu_{U},g}(t)$ and $\int_{U}g\,dn_{r}=\int_{0}^{b}F(t)dt\geq\int_{0}^{b}R_{1,\mu_{U},g}(t)dt=\mu_{g}(U)$. 2. (ii)

Using part (i), $0\leq\mu_{g}(Kerg)\leq\int_{Kerg}g\,dn_{r}=0$. If $A\subseteq\mathbb{R}$ is an open or closed subset of $(-\infty,0)$ then $E=g^{-1}(A)=\emptyset$. So $\mu_{g}(E)=\int_{E}g\,dn_{r}=0.$ 3. (iii)

Let $A\subseteq\mathbb{R}\setminus\{0\}$ be closed or open and $E=g^{-1}(A)$. The argument as in the proof of part (i), where by [9, Lemma 7.17(iii)] inequalities become equalities, shows that $\mu_{g}(E)=\int_{E}g\,dn_{r}.$ We are only left to consider the case $\mu_{r}(U)$, where $U=g^{-1}(W)$ for an open set $W\subseteq\mathbb{R}$ such that $0\in W$. Write $W=W_{1}\sqcup\{0\},\,U_{1}=g^{-1}(W_{1})$. Then $U=U_{1}\sqcup Kerg$ and

[TABLE]

Together with part (i) this gives $\mu_{g}(U)=\int_{U}g\,dn_{r}.$ 4. (iv)

The argument is essentially the same as the one for part (i), where by [9, Lemma 7.17(iv)] inequalities become equalities. For example, for $D$ we have $F(t)=R_{2,\mu_{D},g}(t)$ for any $t\in[a,b]=g(X)$, and then

[TABLE]

(Note that when $g\geq 0$, one may also use the argument from [18, Theorem 20].) 5. (v)

The proof is basically one from [18, Theorem 20] and is given here for completeness. Let $Z\subseteq X$ be a zero set. Say, $Z=f^{-1}(0)$. We may assume that $0\leq f\leq 1$. Let $g=1-\displaystyle{\frac{1}{n}}f$. Then $g\in C^{+}(X),\ 1-\displaystyle{\frac{1}{n}}\leq g\leq 1$, and $g^{-1}(1)=Z$. Let $U=X\setminus Z$. Using formula (11) and part (iv) we have:

[TABLE]

It follows that $\mu(X)\leq\mu(Z)+\mu(U)=\mu(Z)+\mu(X\setminus Z)$. Every locally compact space is completely regular, and so the family of all zero sets is a base for closed sets. From Remark 8 it follows that $\mu(X)\leq\mu(C)+\mu(X\setminus C)$ for any closed set $C$, and by Theorem 6 $\mu$ is a topological measure.

∎

4. Integration over zero and cozero sets

Integration with respect to a deficient topological measure sometimes gives the same results as integration with respect to a measure would give. For example, we have the following lemma:

Lemma 42**.**

Supppose $X$ is locally compact, $\mu$ is a deficient topological measure on $X,\,f\in C_{b}(X)$. If $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$ is such that $\mu(A)=0$ then $\int_{A}fd\mu=0$.

Proof.

Let $f(X)\subseteq[a,b]$. Consider first $V\in\mathscr{O}(X)$ such that $\mu(V)=0$. Since $R_{1,\mu_{V},f}(t)=0$ for any $t\in[a,b]$, using formula (6) we see that $\int_{V}fd\mu=0$. For $C\in\mathscr{C}(X),\mu(C)=0$ we may apply a similar argument using formula (6) and the fact that $R_{2,\mu_{C},f}(t)=0$. ∎

Integration with respect to a deficient topological measure sometimes is also very different from integration with respect to a measure. If $\mu$ is a measure and $\mu(Coz(f))=0$, then $\int_{X}fd\mu=0$. Example 43 below shows that this is generally false if $\mu$ is a deficient topological measure. (On the other hand, part (y1) of Theorem 44 below shows that this is true for nonnegative $f$, assuming that $\mu(Cozf)<\infty$ (in particular, for finite $\mu$); this theorem also generalizes Lemma 42).

Example 43**.**

Let $X=\mathbb{R},D=[1,4]$ and $\mu$ be a deficient topological measure such that $\mu(A)=1$ if $D\subseteq A$ and $\mu(A)=0$ otherwise, for any $A\in\mathscr{O}(X)\cup\mathscr{C}(X)$. (See [7, Section 6]). Let $f=(|x|-2)\wedge 0$, so $f(X)=[-2,0]$, $Coz(f)=(-2,2)$ and $\mu(Coz(f))=0$. Note that $R_{2,\mu,f}(t)=1$ if $t\in[-2,-1]$ and $R_{2,\mu,f}(t)=0$ if $t\in(-1,0]$. By formula (3) we have:

[TABLE]

Now let $g=-f$, so $g(X)=[0,2]$. Since $R_{2,\mu,g}(t)=0$ for every $t>0$, we have

[TABLE]

Thus,

[TABLE]

Theorem 44**.**

Suppose $X$ is locally compact, $\mu$ is a deficient topological measure on $X$, $f\in C_{b}(X)$. Then

1. (y1)

If $f\geq 0$, then for any open set $V$ with $\mu(V)<\infty$

[TABLE]

and

[TABLE]

In particular, for a finite $\mu$ we have $\int_{X}fd\mu=\int_{Coz(f)}fd\mu,$ and

[TABLE] 2. (y2)

If $V\in\mathscr{O}(X),\mu(V)<\infty$, then

[TABLE]

In particular, if $\mu$ is finite and $\mu(Z(f))=\mu(X)$, then $\int_{X}fd\mu=0$. 3. (y3)

If $X$ is compact, $V\in\mathscr{O}(X)$, and $f\leq 0$ then

[TABLE]

In particular,

[TABLE]

Proof.

1. (y1)

Let $f\in C_{b}(X),f\geq 0$. Take any $V\in\mathscr{O}(X)$ with $\mu(V)<\infty$, and let $U=V\cap Coz(f)$. For $t>0$ we have

[TABLE]

so from (7) and ( 8) we see that $R_{1,\mu_{U},f}(t)=R_{1,\mu_{V},f}(t),$ and $\int_{V}f\,d\mu=\int_{V\cap Coz(f)}f\,d\mu$. Now suppose $\int_{V}fd\mu=\int_{0}^{b}R_{1,\mu_{V},f}(t)=0.$ Since the integrand in the second integral is nonnegative, using the right-continuity of $R_{1,\mu_{V},f}(t)$ (see [9, Lemma 6.3]), we must have $\mu(V\cap Coz(f))=R_{1,\mu_{V},f}(0)=0$. On the other hand, if $\mu(V\cap Coz(f))=0$, then $R_{1,\mu_{V},f}(t)=0$ for every $t>0$, and by formula (6) $\int_{V}fd\mu=0$. 2. (y2)

Suppose $f(X)\subseteq[a,b],\,a\leq 0,b\geq 0$, and $\mu_{V}(Z(f))=\mu(V)<\infty$. Using part (v7) of Theorem 24 and superadditivity (see Remark 8), we have $\mu(V)=\mu_{V}(X)\geq\mu_{V}(Z(f))+\mu_{V}(Coz(f)),$ and so we must have $\mu_{V}(Coz(f))=\mu(V\cap Coz(f))=0$. With $W=(f^{-1}(t,\infty))\cap Coz(f)$, we have: $\mu_{V}(W)=0$. Since $f^{-1}((t,\infty))\cap Z(f)=\emptyset$ for $t\geq 0$ and $(f^{-1}(t,\infty))\cap Z(f)=Z(f)$ for $t<0$, we see that $f^{-1}((t,\infty))=W$ if $t\geq 0$, and $f^{-1}((t,\infty))=Z(f)\sqcup W$ if $t<0$. Then $R_{1,\mu_{V},f}(t)=\mu_{V}(W)=0$ for $t\geq 0$ and $R_{1,\mu_{V},f}(t)=\mu_{V}(f^{-1}((t,\infty)))\geq\mu_{V}(Z(f))+\mu_{V}(W)=\mu_{V}(Z(f))=\mu(V)$ for $t<0$. Since also $R_{1,\mu_{V},f}(t)\leq\mu(V)$, we have $R_{1,\mu_{V},f}(t)=\mu(V)$ for $t<0$. Then

[TABLE] 3. (y3)

Suppose that $X$ is compact. ($\Longrightarrow$) is given by part (y2). ($\Longleftarrow$) Assume that $\int_{V}fd\mu=0$, i.e. by formula (6)

[TABLE]

The integrand function in the last integral is nonpositive. Since $X$ is compact, by [9, Lemma 6.3(V)] $R_{2,\mu_{V},f}(t)$ is left-continuous at [math]. From left continuity it follows that $R_{2,\mu_{V},f}(0)-\mu(V)=0$. But $R_{2,\mu_{V},f}(0)=\mu_{V}(Z(f))$.

∎

Lemma 45**.**

If $\mu$ assumes only finitely many values and is not a topological measures then there exists $f\geq 0$ such that $\int_{X}fd\mu=0,\,\int_{X}(-f)d\mu<0$.

Proof.

Suppose that $\mu$ is a deficient topological measure, but not a topological measure and that $\mu$ assumes finitely many values. By Theorem 6 there are $U\in\mathscr{O}(X)$ and $C\in\mathscr{K}(X),C\subseteq U$ such that $\mu(C)+\mu(U\setminus C)<\mu(U)$. We may assume that $\overline{U}$ is compact. (Otherwise, choose by regularity compact $D\subseteq U$ with $\mu(D)=\mu(U)$, then using regularity and Lemma 3 choose $U_{1}\in\mathscr{O}(X)$ such that $\overline{U_{1}}\in\mathscr{K}(X),\,D\subseteq U_{1}\subseteq U,$ and $\mu(U_{1})=\mu(D)=\mu(U)$. Then $\mu(C)+\mu(U_{1}\setminus C)\leq\mu(C)+\mu(U\setminus C)<\mu(U)=\mu(U_{1})$, and we replace $U$ by $U_{1}$.) We have

[TABLE]

Pick a compact $B\subseteq U\setminus C$ with $\mu(B)=\mu(U\setminus C)$. Let $V=(U\setminus C)\setminus B$. By superadditivity $\mu(V)=0$. Set a closed set $G=(X\setminus U)\sqcup B\sqcup C$. By Remark 8 $\mu(G)=\mu(X\setminus U)+\mu(B)+\mu(C)$, which is equal to the left-hand side of (16), so $\mu(G)<\mu(X)$. Now choose $W\in\mathscr{O}(X)$ such that $G\subseteq W,\,\mu(W)=\mu(G)$. Let $\delta=\mu(X)-\mu(W)>0$.

Let $K=X\setminus W$. Then $K\subseteq X\setminus G=V\subseteq U$, so $K$ is compact. Let $f\in C_{c}(X)$ be a Urysohn function such that $f=1$ on $K$ and $\operatorname{supp}f\subseteq V$. Then $Coz(f)\subseteq V$, so $\mu(Coz(f)=0$, and by part (y1) of Theorem 44 $\int_{X}f\,d\mu=0$.

Now let $g=-f$, so $g(X)\subseteq[-1,0]$. For $t\in(-1,0)$ we have $g^{-1}((t,\infty))\subseteq g^{-1}((-1,\infty))\subseteq X\setminus K=W$, so $R_{1,\mu,g}(t)\leq\mu(W)$. Then by formula (1)

[TABLE]

∎

Remark 46**.**

Example 43 illustrates Lemma 45. Also, in Example 43 $\int_{X}fd\mu=-1$, while $\mu(Coz(f))=0$. This shows that part (y1) of Theorem 44 may not hold if the condition $f\geq 0$ is relaxed.

Remark 47**.**

Parts (v1) - (v6) of Theorem 24 for the compact space are stated without proof in [19, Proposition 5.2]. Some statements from Theorem 29 and Lemma 28 are related to [19, Theorem 5.4]. Parts (b1) - (b4), (b9) - (b14) of Theorem 33 are generalizations of results presented in [18, Theorem 22] and [19, Theorem 5.7, Proposition 5.9]; for part (i) of Theorem 33 see also [19, Theorem 5.7]. Theorem 40 is inspired by [18, Theorem 19]. Statements ”In particular…” in Theorem 44 are generalizations of [19, Theorem 3.6, 1-3], and Lemma 45 generalizes [19, Theorem 3.6, 4].

5. Convergence theorems

Theorem 48**.**

Let $X$ be a locally compact space, $\mu$ a finite deficient topological measure on $X$.

1. (I)

Suppose $f_{\alpha}\in C_{b}(X)$ and $f_{\alpha}\nearrow f$ in the topology of uniform convergence. Then

[TABLE] 2. (II)

Suppose $f_{\alpha}\in C_{0}^{+}(X)$ and $f_{\alpha}\searrow f$ in the the topology of uniform convergence. Then

[TABLE]

Proof.

1. (I)

Since $f\in C_{b}(X)$, we may consider $\int f\,d\mu$. For any $t$, $f_{\alpha}^{-1}((t,\infty))\nearrow f^{-1}((t,\infty))$, so by Remark 8 $\mu(f_{\alpha}^{-1}((t,\infty)))\rightarrow\mu(f^{-1}((t,\infty)))$. i.e. $R_{1,\mu,f_{\alpha}}(t)\rightarrow R_{1,\mu,f}(t)$. Applying formula (1) on $[a,b]$ containing $f(X)$ we see that

[TABLE] 2. (II)

The proof is similar. Note that for any $t>0$ the set $f_{\alpha}^{-1}([t,\infty))$ is compact, $f_{\alpha}^{-1}([t,\infty))\searrow f^{-1}([t,\infty))$, and we may again apply Remark 8.

∎

Theorem 49**.**

Let $\mu$ be a compact-finite deficient topological measure on a locally compact space $X$. Suppose ${f_{\alpha}}$ converges uniformly to $f$, $f_{\alpha},f\in C_{c}(X),f_{\alpha}\geq 0$. If $\operatorname{supp}f_{\alpha},\operatorname{supp}f\subseteq K$ for some compact $K$, then $\int_{X}f_{\alpha}\,d\mu\longrightarrow\int_{X}f\,d\mu$. If $X$ is compact, $f_{\alpha}\in C(X)$ and ${f_{\alpha}}$ converges uniformly to $f$ then $\int_{X}f_{\alpha}\,d\mu\longrightarrow\int_{X}f\,d\mu$.

Proof.

Follows from parts (b6) and (iii) of Theorem 33. ∎

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