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%%        Monotone set-valued functions defined by         %%
%%             set-valued Choquet integrals                %%
%%                                                         %%
%%               Sun Hongxia, Li Jun, M. Yasuda            %%
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%%--- 2005/09/28(Sun)                                      %%
%%--- 2006/03/18(Li)                                       %%
%%--- 2007/08/30(Li)                                       %%
%%--- 2007/09/02(Li)                                       %%
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\documentclass[11pt]{article}
\usepackage{latexsym,amssymb,theorem}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\newtheorem{prop}{Proposition}[section]
\newtheorem{df}{Definition}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}{Corollary}
\newtheorem{lem}{Lemma}
\newtheorem{rem}{Remark}[section]
\newtheorem{exa}{Example}[section]

\begin{document}

%---------------------------the text-----------------------------
\title{ Monotone set-valued functions defined by \\
              set-valued Choquet integrals}
\vspace{4mm}
\author{    Sun Hongxia \\
\small{\it Department of  Mathematics, Dezhou University,} \\
\small{\it Dezhou, Shandong 253023, China } \\[2mm]
            Li Jun{\thanks{
            {\it E-mail address}: lijun@cuc.edu.cn }} \\
\small{\it School of Science, Communication University of China,} \\
\small{\it Beijing, 100024, China } \\[2mm]
            Masami Yasuda \\
\small{\it Department of Mathematics \& Informatics,
                                Faculty of Science,} \\
\small{\it Chiba University, Chiba 263-8522, Japan }  }
%
\date{}
\maketitle
%
\begin{abstract}
In this paper, some properties of
the monotone set-valued function defined by the set-valued Choquet
integral are discussed. It is shown that several important
structural characteristics of the original set function, such as
null-additivity, strong order continuity, property(S) and
pseudometric generating property, etc., are preserved by the new
set-valued function.  It is also shown that integrable assumption
is inevitable for the preservation of strong order continuous and
pseudometric generating property. Several kind of absolute continuity of
set-valued function with respect to set function are also discussed.

 {\it Keywords}: Monotone set-valued function; set-valued
Choquet integrals; Choquet integral
\end{abstract}
%
%---------- 1. Introduction ----------------
%
\section{Introduction}
\hspace{6mm}
Given a measurable space $(X,\cal A)$, a nonnegative
monotone set function $\mu$ on $\cal A$ with $\mu(\emptyset)=0$,
and a nonnegative measurable function f, then the set function
$v_f$ defined by the  Chouqet integral
$$v_f(A)=(C)\displaystyle\int_Afd\mu \   \ ( \forall A\in \cal A)$$
 is also nonnegative
and monotone on $\cal A$ with $v_f(\emptyset)=0 \cite{WK}$.
 Wang and Klir (\cite{WK}) have discussed that $v_f(A)$ preserved some
important structure characteristics of the original function
$\mu$, such as null-additivit, subadditivity, autocontinuity,
e.t.c. Ouyang and Li (\cite{OL}) further discussed that $v_f$
preserved other important structure characteristics of $\mu$,
 such as strong order continuity, order
continuity, property(S),  pseudometric generating property.

Similarly to the single-valued fuzzy measure, we can define the
structural characteristics for a set-valued fuzzy measure, whether
these characteristics can be preserved by the set-valued set functions
defined by set-valued Choquet integral?

In this paper, we give the definition of these characteristics of
set-valued set function, and we prove that a new monotone
set-valued function defined by set-valued Choquet integral also
preserved the characteristics of original set-valued set function.
We also discuss several kind of absolute continuity of
set-valued function with respect to set function.
%
%------------2. Preliminaries-----------------
\section{Preliminaries}
\hspace{6mm}
Throughout this paper, we suppose that
$(X,\cal A)$ is a measurable space, $f$ is a nonnegative
measurable function on $(X, \cal A)$, and $\mu$ is a monotone set
function with $\mu(\emptyset)=0$.
$R^+$ denotes the
interval [0,$\infty$],  ${\cal P}_0(R^+)({\cal P}_f(R^+)$, resp.)
denotes the class of all nonempty subsets(closed, resp.) of $R^+$.
%
\begin{df} {\rm(\cite{MS})} \
  The Choquet integral  of f on A with respect to $\mu$,
  denoted by $(C)\int_Afd\mu$, is defined as
$$(C)\int_Afd\mu={\int_0}^{\infty}\mu(F_\alpha \cap A)d\alpha$$ where
$F_\alpha=\{f\geq\alpha\}=\{x|f(x)\geq\alpha\}$ for any
$\alpha\geq0$ and the right-hand side is the Lebesgue integral.
\end{df}
%
\begin{df}{\rm(\cite{ZW} ) } \
  A set-valued function $F: X\rightarrow{\cal P}_0(R^+)$
    is said to be measurable if its
graph is measurable, that is ,
$$
G_r(F)=\{(x,r)\in X\times
R^+:r\in F(x)\}\in {\cal A}\otimes Borel (R^+)
$$
where $ Borel (R^+)$ is the Borel field  of $R^+$.
F is said to be closed-valued if it values in ${\cal P}_f(R^+)$
\end{df}
%
\begin{df}{\rm(\cite{JK})}
 A set-valued function F is said to be Choquet
integrably  bounded if there is a Choquet integrable function g
such that
$$\|F(x)\|=\sup_{r\in F(x)}|r|\leq g(x)$$
\end{df}
\begin{df}{\rm(\cite{ZW})}
  Let F be a set-valued function and $A\in\cal A$. The  Choquet integral
  of F on A with
respect to $\mu$, denote by $(C)\int_AFd\mu$, is defined as
$$(C)\int_AFd\mu=\{(C)\int_Afd\mu:f\in S(F)\}$$
where S(F) is the family of $\mu$-a.e. measurable selection of F.
\end{df}
\par Instead of $(C) \int_XFd\mu $, we will write $(C) \int Fd\mu $.
Obviously, $(C)\int Fd\mu$ may be empty. A set-valued function F
is said to be integrable existing if $(C)\int Fd\mu\neq\emptyset$,
and F is said to be integrable if $(C) \int Fd\mu$ exists and does
not include $\infty$.
\begin{df}{\rm(\cite{ZW1})}
 Let A, B $\in {\cal P}_0(R^+)$, then $A\leq B$   means that
   \par{$(1)$} \  For each $x_0\in A$,there exists $y_0\in B$, such that
   $x_0\leq y_0$;
   \par{$(2)$} \  For each $y_0\in A$,there exists $x_0\in B$, such that
   $x_0\leq y_0$.
 \end{df}
\begin{df}
A set-valued function $\pi:{\cal A}\rightarrow {\cal P}_0(R^+)$
 is said to be a monotone set-valued function if
it satisfies \\
(1) $\pi(\emptyset)={0}$.\\
(2) $A\subset B$ implies $\pi(A)\leq\pi(B)$
\end{df}
\begin{df}
 The set-valued function  $(C)\pi_F$ is
defined as
$$(C)\pi_F(A)=(C)\int_AFd\mu (A\in \cal A)$$
where F is a set-valued function with $S(F)\neq\emptyset, \mu$ is
a monotone set function.
\end{df}
\begin{rem}
By proposition $3.5 \cite{ZW}$ , we know that $(C)\pi_F$ is a
monotone set-valued function.
\end{rem}
\begin{df}{\rm(\cite{ZW1})}
 Let $\{A_n\}\subset {\cal P}_0(R^+)$, we
write
$$\limsup_{n\rightarrow\infty}{A_n}=\{x\in R^+ :
x=\lim\limits_{k\rightarrow\infty}x_{n{_k}}, x_{n{_k}}\in A_{n{_k}}
(k\geq 1) \}$$
$$\liminf_{n\rightarrow\infty}{A_n}=\{x\in R^+ :
x=\lim\limits_{k\rightarrow\infty}x_{n}, x_{n}\in A_{n} \ (n\geq 1)
\}$$ If
$\lim\limits_{n\rightarrow\infty}\sup{A_n}
   =\lim\limits_{n\rightarrow\infty}\inf{A_n}=A$,
then $\{A_n\}$ is said to be convergent to A and it is simply
 noted with $A_n\rightarrow A$.
\end{df}

Similarly to the single-valued function, we will defined
 several important structural characteristics for a set-valued function.
%
\begin{df} {\rm} \ Let $\pi$ is a set-valued function.

{$(1)$} $\pi$ is called null-additive if for any A, B
$\in\cal A$,
 $\pi(A)=\{0\}$ implies $\pi(A \cup B)=\pi(B)$.
\par{$(2)$} $\pi$ is called weakly null-additive if for any A, B $\in\cal
A$,$\pi(A)=\pi(B)=\{0\}$ implies $\pi(A \cup B)= \{0\}$.
\par{$(3)$}$\pi$ is said to be strongly order continuous if for any
  ${A_n}\in\cal A$,  $A_n\downarrow A  $ and
$\pi(A)=\{0\}$ implies $\lim\limits_{n \to \infty}\pi (A_n)=\{0\}$
\par{$(4)$} $\pi$ is said to be order continuous if for any
$\{A_n\}\subset\cal A$, $A_n\downarrow \emptyset$ implies
$\lim\limits_{n \to\infty}\pi (A_n)=\{0\}$
\par{$(5)$} $\pi$ is said to  have the property (S) if for any
${A_n}\in\cal A$ that satisfies $\lim\limits_{n \to \infty}\pi
(A_n)=\{0\}$ there exists a subsequence $\{A_{n_k}\}\subset \{A_n\}$
such that $\pi(\lim\limits_{k \to\infty}\sup A_{n_k})=\{0\}$.
\par{$(6)$}  $\pi$  is said to have the pseudometric generating property,
abbreviated as p.g.p. if for any $\epsilon>0$, there
 exists $\delta>0$ such that $\pi(A \cup B)\subset[0,\epsilon]$
whenever $A, B\in \cal A$, $\pi(A)\subset [0,\delta]$ and
 $\pi(B)\subset [0,\delta]$.
 \end{df}
 It is easy to prove that the above definition is the generalization
 of single-valued case.
%
%----------------------------
%
\section{Preservation of structural characteristics}
\hspace{6mm}
In this section, we suppose that F is a measurable
closed-valued function.
%
\begin{thm}
If $\mu$ is null-additive, then so is $(C)\pi_F$.
\end{thm}
{\bf  Proof.}\ For  any A, B$\in\cal A$ with
  $(C)\pi_F(B)=\{0\}$, i.e. $(C)\int_BFd\mu=\{0\}$, then for any $f\in S(F)$,
  we have $(C) \int_Bfd\mu=0.$ Hence by theorem 7 [1]
, we have $$(C) \int_{A\cup B} fd\mu=(C)\int_Afd\mu.$$  Let
$a_0\in (C)\pi_F(A\cup B)$, then exists $f\in S(F)$, such that
$$a_0=(C)\int_{A\cup B} fd\mu=(C)\int_Afd\mu\in(C)\pi_F(A).$$
\indent Similarly, for $b_0\in (C)\pi_F(A)$, we can prove that
$b_0\in (C)\pi_F(A\cup B)$.

The proof is completed.
%
\begin{thm} \
If $\mu$ is weakly null-additive , then so is $(C)\pi_F$.
\end{thm}
{\bf Proof.}\ It is similar to the proof of Theorem 3.1.
%
\begin{thm} \
Let  F is  integrable. If $\mu$ is strongly order continuous,
then so is $(C)\pi_F$.
\end{thm}
%
{\bf Proof.}\ For any $\{A_n\}\subset\cal A$ with $A_n  \downarrow
A$ and $(c)\pi_F(A)=\{0\}$, then for any $f\in S(F)$, $
v_f(A)=(C)\displaystyle\int_Afd\mu=0$  , hence by Theorem
3.(2) (\cite{OL}), we have
$$
\lim\limits_{n\rightarrow\infty}v_f(A_n)=0 , \ \ for\ \ any \ \ f\in S(F).
$$
 Hence
 $$
 \limsup_{n\rightarrow\infty}(C)\pi(F)(A_n)
 =\liminf_{n\rightarrow\infty}(C)\pi(F)(A_n)=\{0\}.
 $$
 So
 $$\lim\limits_{n\rightarrow\infty}(C)\pi_F(A_n)=\{0\}.$$
 %
\begin{thm} \
 Let F is integrable. If $\mu$ is order continuous,
   then so is $(C)\pi_F$.
\end{thm}
%
{\bf Proof.}\ Similar to Theorem 3.3.
%
\begin{thm}
 Let F is  integrable. If $\mu$ has p.g.p, then so is $(C)\pi_F$.
\end{thm}
%
{\bf Proof.}\  Since F is integrable, hence $ (C)\int fd\mu<\infty
$. Since $\mu$ have p.g.p., there exists $\delta >0$, such that
$\mu(E\cup F)<\epsilon $ whenever $E,F\in \cal A$ and $\mu(E)\vee
\mu(F)<\delta$.

We will prove that $(C)\pi_F(A\cup B)\subset[0,\epsilon]$
whenever  $A, B\in\cal A$ ,
$(c)\pi_F(A)\subset [0,\delta]$ and $(c)\pi_F(B)\subset [0,\delta]$.
In fact, for any $f\in S(F)$, we have
$(C)\int_Afd\mu\vee(C) \int_Bfd\mu<\delta$, by Theorem 5 (\cite{OL}),
we have $(C) \int_{A\cup B}fd\mu<\epsilon$. Hence$(C)\pi_F(A\cup
B)\subset[0,\epsilon]$.
%
\begin{rem}
 {\rm Observe that Theorem $3.3$ to Theorem $3.5$ are based on the
assumption that F is integrable. The following examples show that
the conclusions in Theorem 3.3 to Theorem 3.5 may not be true when
the assumption is abandoned.
}
\end{rem}
%
\begin{exa}
{\rm
  Let $X = \{1,2,..., \}$, ${\cal A=\cal P} (X)$. $\mu$
is defined as \\
$$\mu(E)= \left\{
\begin{array}{ll}
0, & {\rm }\ \ E=\emptyset,\\
\ &\ \\
max\{1/i|i\in E\}, & {\rm  } \ \ E\neq\emptyset.
\end{array}\right.
$$
\par It is easy to see that $\mu$ is strongly order continuous \ $($so
$\mu$ is order continuous$)$ monotone set function. Let
$F(X)=\{x\}(x\in X)$, then we have $(C)\pi_F(X)=\{\infty \}$,
hence F is not Choquet integrable. we will show that $(C)\pi_F$ is
not order continuous\ $($so $(C)\pi_F$ is not strongly order
continuous$)$.
\par  In fact, let $E_n= \{n,n+1,..., \}$, then
$E_n\downarrow\emptyset$. For any $f\in S(F)$, $f(x)=x$.
$$(C)\int_{E_n}fd\mu=\int_0^\infty\mu(E_n\cap
F_\alpha)d\alpha\geq \int_0^n\mu(E_n\cap F_\alpha)d\alpha=1.$$
Hence $(C)\pi_F$ is not order continuous.
 }
\end{exa}
%
\begin{exa} \
{\rm
Let $X_1 = \{1,3,..., \}, X_2=\{2,4,...\}, X=X_1\bigcup
X_2$, ${\cal A=\cal P }(X)$. $\mu$
is defined as \\
$$\mu(E)= \left\{
\begin{array}{ll}
0,& {\rm if}\ \ E=\emptyset,\\
\ &  \ \\
 max\{1/i^2|i\in E\},& {\rm if } \ \ E\in X_1 or X_2\\
\ & \  \\
 max\{1/i|i\in E\},& {\rm  } \ \  otherwise
\end{array}\right.
$$
\par It is easy to see that $\mu$ have p.g.p.. Let $F(X)=\{x\} (x\in
X)$, then we have $(C)\pi_F(X)=\{\infty \}$, hence F is not
Choquet integrable. we will show that $(C)\pi_F$ do not have
p.g.p..
\par Let $A_n=\{2n\}, B_n=\{2n+1\}$,then
$$(C)\int_{A_n}fd\mu= \int_0^\infty\mu(A_n\cap F_\alpha)d\alpha=
\int_0^{2n} 1/(2n)^2d\alpha=1/2n\rightarrow
0(n\rightarrow\infty)$$
 \par Similarly,
$$ (C)\int_{B_n}fd\mu\rightarrow 0 \ \ \
(n\rightarrow\infty)$$
 But
\begin{eqnarray*}
 (C) \int_{A_n\cup B_n}fd\mu&=&\int_0^\infty\mu(({A_n\cup B_n})\cap
F_\alpha)d\alpha\\[2mm]
&=&\int_0^{2n}\mu(({A_n\cup B_n})\cap F_\alpha)d\alpha+\int_{2n+1}^{2n+2}\mu(({A_n\cup B_n})\cap F_\alpha)d\alpha \\[2mm]
&=&\int_0^{2n}1/2nd\alpha+\int_{2n+1}^{2n+2}1/(2n+1)^2d\alpha\\[2mm]
&=&1+1/(2n+1)^2\rightarrow 1  (n\rightarrow\infty).
\end{eqnarray*}
\indent Hence  $(C)\pi_F$ do not have p.g.p..
 }
\end{exa}
%
\begin{thm} \
Let F is Choquet bounded and $\mu$ is continuous . If $\mu$ have
property (S), then so is $(C)\pi_F$.
\end{thm}
%
{\bf Proof.}\ For any For any $\{A_n\}\subset\cal A$ with
$\lim\limits_{n\rightarrow\infty}(C)\pi_F(A_n)=\{0\}$, we have
$$\lim\limits_{n\rightarrow\infty}d(0,(C)\pi_F(A_n)) =0$$ i.e. $$ \liminf
_{n\rightarrow\infty}{_{f\in
S(F)}}\mid(C)\displaystyle\int_{A_n}fd\mu\mid=0$$ Then from the
definition of limit inferior, we have that
$$\lim\limits_{m\rightarrow\infty}(C)\int_{A_n}f_{n_{(m)}}d\mu=\inf_{f\in
 S(F)}\mid(C)\int_{A_n}fd\mu\mid  \ \ \ f_{n_{(m)}}\in S(F)$$
   Since F is integrably bounded , so there exists a Choquet
integrable function g , such that $\mid f(x)\mid\leq g(x)$ for any
$f\in S(F)$. Hence we have
$$
\lim\limits_{m\rightarrow\infty}(C)\int_{A_n}f_{n_{(m)}}d\mu=
 (C)\int_{A_n}\lim\limits_{m\rightarrow\infty}f_{n_{(m)}}d\mu.
 $$
 Denote
 $f_0=\lim\limits_{m\rightarrow\infty}f_{n_{(m)}}d\mu$, then
 $\lim\limits_{n\rightarrow\infty}(C)\int_{A_n}f_0d\mu=0$.
  By Theorem 2 in \cite{OL}, there exists $\{A_{n{_k}}\}\subset{A_n}$,
  such that
 $v_{f{_0}}(\lim\limits_{k}sup A_{n{_k}})=0$.
  i.e.
  $$
  (C)\int_{{ \lim\limits_{k}sup}
 A_{n{_k}}}f_0d\mu={\int_0}^{\infty} \mu(\limsup_{k}
 A_{n{_k}}
 \cap{\{f_0\geq \alpha}\}))=0.
 $$
  Hence for any $a_0>0$, we have
  $$
  \mu(\lim\limits_{k} sup
 A_{n{_k}}\cap{\{f_0\geq \alpha}\})
 =0 \ \ \  a.e.
 $$
 Denote
 $B_{n_k}=A_{n_k}\cap\{f_0\geq\alpha_0\}$, then
 $$
 \mu(\limsup_{k}B_{n{_k}})={0}\ \ \  a.e.
 $$
 Therefore, for any $ f\in S(F)$
 $$
 (C)\int_{\lim\limits_{k }\sup B_{n{_k}}}f d\mu
 ={\int_0}^{\infty} \mu(\limsup_{k}
 A_{n{_k}}\cap{\{f\geq \alpha}\}))=0.
 $$
 Hence
$$
(C)\pi_F(\limsup_{k}B_{n{_k}})=\{0\}.
$$
%
 \begin{df}{\rm(\cite{MS})}
 A set $N\in \cal A$ is called a null set $($with
 respect to $\mu$$)$ if $\mu(A\cup N)=\mu(A)$, for all $A\in \cal
 A$.
 \end{df}
%
 \begin{df}{\rm(\cite{MS})}
   A fuzzy measure $\mu$ is said to be
 m-continuous if there exists a complete and finite measure m,
 such that $\mu\ll m,$ i.e. m(A)=0 implies A is a null set (with
 respect to $\mu$).
 \end{df}
 %
\begin{lem} \ $(C)\int_A Fd\mu= (C)\int {X_A}\circ
Fd\mu$  where $X_A$ is the characteristic function of A, and
$$
{X_A}\circ F(x)=\left\{
\begin{array}{ll}
F(x), & {\rm }\ \ x\in A,\\
\ &  \  \\
 {0}, & {\rm  } \ \  otherwise.
\end{array}\right.
$$
\end{lem}
{\bf Proof.}\ It is easy to be proved.
%
\begin{thm} \ Let $\mu$ is m-continuous and F is Choquet integrable bounded. If
$\mu$ is continuous, then so is $(C)\pi_F$.
\end{thm}
{\bf Proof.}\ It is similar to the proof of Proposition 3 in \cite{GZ}.

\section{ Absolutely continuity  }
\hspace{6mm}Absolutely plays an important role in measure theorem,
we expect that most of the classical results, such as
Radon-Nikodym theorem, etc., remain valid for monotone set-valued
function. In this section, we will introduce several absolutely
continuous.
\begin{df} {\rm } \ Let $\mu$ is a set function and $\pi$ is a  set-valued
function , we say that
 \par{$(1)$}\ $\pi$ is absolutely continuous
of type I with respect to $\mu$, denote by $\pi\ll_I\mu$, if and
only if $\pi(A)=\{0\}$ whenever $\mu(A)=0$.
\par{$(2)$}\ $\pi$ is absolutely continuous of type II with respect to
$\mu$, denote by $\pi\ll_{II_\alpha}\mu$, if and only if
$\pi(A)=\pi(B)$ whenever B$\subset$A and $\mu(A\setminus B)=0$.
\par{$(3)$}\ $\pi$ is absolutely continuous of type III with respect to
$\mu$, denote by $\pi\ll_{III_\alpha}\mu$, if and only if
$\lim\limits_{n \to \infty}\pi (A_n)=\{0\}$ whenever $A_n\downarrow
\emptyset $ and $ \lim\limits_{n \to\infty}\mu(A_n)={0}$
\par{$(4)$}\ $\pi$ is absolutely continuous of type IV with respect to
$\mu$, denote by $\pi\ll_{IV}\mu$, if and only if for any
$\epsilon>0$, there exists $\delta>0$, such that
$\pi(A)\subset[0,\epsilon]$ whenever $\mu(A)<\delta$.
 \end{df}
\begin{thm}
 $(c)\pi_F\ll_I\mu$.
\end{thm}
 {\bf Proof.}\ For any $A\in \cal A$ with
$\mu(A)=0$ . Then for  any $f\in S(F)$ $$(C) \int_Afd\mu=
\int_0^\infty \mu(A\cap F_\alpha)d\alpha=0  $$ Hence
$(C)\pi_F(A)={0}$.
\begin{thm}
 If $\mu$ is null additive,then $(C)\pi_F\ll_{II_a}\mu$;
\end{thm}
{\bf Proof.}\ For any $A,B\in \cal A$ with $B\subset A$ and
$\mu(A-B)=0$. Let $a_0\in(C)\pi_F(A)$, there exists $f\in S(F)$
such that
$$a_0=(C)\int _Afd\mu= e{\int_0}^{\infty}\mu(F_\alpha\cap A)d\alpha$$
 From the monotonicity of $\mu$, we have
$\mu((A-B)\cap F_\alpha)=0$. Since $\mu$ is null additive , hence
$$\mu(A\cap F_\alpha)=\mu[((A-B)\cap F_\alpha)\cup(B\cap
F_\alpha)]=\mu(B\cap F_\alpha)$$
So
$$a_0={\int_0}^{\infty}\mu(F_\alpha \cap B)d\alpha=(C)\int_Bfd\mu\in\pi_F(B)$$
\begin{thm}
  Let F is Choquet integrable  .$(c)\pi_F\ll_{III_a}\mu$;
\end{thm}
{\bf Proof.}  For any $\{A_n\}\subset\cal A$,
 $A_n\downarrow\emptyset$. since$\lim\limits_{n\rightarrow
\infty}\mu(A_n)=0$ for any $f\in S(F)$, so
$$\lim\limits_{n\rightarrow\infty}\int fd\mu=\lim\limits_{n\rightarrow\infty}\int\mu(F_\alpha\cap A_n)d\alpha
=\int\lim\limits_{n\rightarrow\infty}\mu(F_\alpha\cap A_n)d\alpha$$
 hence
$$\limsup_{n\rightarrow\infty}(C)\pi(F)(A_n)=\liminf_{n\rightarrow\infty}(C)\pi(F)(A_n)=\{0\}$$
i.e. $$\lim\limits_{n\rightarrow\infty}(C)\pi_F(A_n)=\{0\}.$$

\begin{thm}
  Let F is Choquet integrable. $(C)\pi_F\ll_{IV}\mu$.
\end{thm}
{\bf Proof.}  Since F is Choquet integrable,  so  $(C)\int
fd\mu<\infty$ for any $f\in S(F)$, i.e.
$$\int_0^{\infty}\mu(F_\alpha)d\alpha<\infty$$
Hence for any $\epsilon>0$, there exits $0<a<b$, such that
$$(C)\int_0^{a}\mu(F_\alpha)d\alpha +\int_b^{\infty}\mu(F_\alpha)d\alpha<\frac{\epsilon}{2}$$
Let$\delta=\frac{\epsilon}{2(b-a)}$ and $\mu(A)<\delta$ , then for
any $f\in S(F)$
\begin{eqnarray*}
(C)\int fd\mu&=&(C)\int_0^{\infty}\mu(F_\alpha\cap A)d\alpha\\
&=&(C)\int_0^{a}\mu(F_\alpha\cap
A)d\alpha+\int_b^{\infty}\mu(F_\alpha\cap A)d\alpha
+\int_a^{b}\mu(F_\alpha\cap A)d\alpha\\
&<&\frac{\epsilon}{2}+\mu(F_\alpha\cap A)(b-a)\\
&<&\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{eqnarray*}
Hence $\pi(A)\subset[0, \epsilon].$

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 {\it Fuzzy Sets and Systems}, \  56  (1993)237-241.
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 Guo C., Zhang D. \  On set-valued fuzzy measures,\
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\newtheorem{df}{Definition}[section]
\newtheorem{th}{Theorem}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{rem}{Remark}[section]
\newtheorem{ex}{Example}[section] 