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%--- MDAI2007                                                 %
%--- Paper ID: 088                                            %
%                                                             %
%    "A Version of Lebesgue decomposition theorems of         %
%     non-additive measure"                                   %
%     by  J.Li, M.Yasuda, JZ.Li                               %
%                                                             %
%---2007/01/30 (Li Jun)                                       %
%---Review Version (2007/04/11)                               %
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\begin{document}
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\mainmatter              % start of the contributions
%\addtocmark{Hamiltonian Mechanics} % additional mark in the TOC
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\title{  A Version of Lebesgue decomposition theorem for
                  non-additive measure
      \thanks{This work was partially supported by
                   the NSFC Grant No.70471063, as well as by
                   the Subject Building Foundation of
      Communication University of China (No. XK060106). } }
\titlerunning{Lebesgue decomposition theorem}
%
\author{
  Jun LI\inst{1}, Masami YASUDA\inst{2} and Jianzeng LI\inst{3}}
\institute{
  %Department of Applied Mathematics,
  School of Science, Communication University of China,
       Beijing 100024, China,
       \email{lijun@cuc.edu.cn}
\and
  Faculty of Science, Chiba University
       Chiba 263-8522 Japan,
       \email{yasuda@math.s.chiba-u.ac.jp}
\and
  School of Information Engineering,
       Communication University of China, \\
       Beijing 100024, China, \\
       \email{jzli@cuc.edu.cn}
}% end of institute
%
\date{}
\maketitle
\begin{abstract}
 In this paper, Lebesgue decomposition type theorems
 for non-additive measure are shown under the conditions of
 null-additivity, converse null-additivity, weak null-additivity
 and $\sigma$-null-additivity, etc..
 In our discussion, the monotone continuity of set function is not
 required.
\end{abstract}
%
{\it Keywords}: Non-additive measure; null-additivity;
absolute continuity; Lebesgue decomposition theorem;
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%----------------- Section 1 -------------------------
\section{Introduction}
%\hspace{6mm}
Lebesgue decomposition of a set function $\mu$ is
stated as: for another given set function $\nu$, $\mu$ is
represented as $\mu = \mu_c + \mu_s$, where $\mu_c$ and $\mu_s$
are absolutely continuous and singular with respect to $\nu$,
respectively. In measure theory this decomposition is a well-known
fact, which is referred to as 
Lebesgue decomposition theorem \cite{Halm68}.

For the case of non-additive measure theory, the situation is not so
simple.  There are many discussions on Lebesgue decomposition type
theorems such as, a version on submeasure (cf.~\cite{Dobra1974}),
a version on $\bot$-decomposition measure (cf.~\cite{Pap1990a}),
a version on $\sigma$-finite fuzzy measure (cf.~\cite{Jiang1995}),
and
a version on signed fuzzy measure (cf.~\cite{Zhang1999}), and so on.
In those discussion, the monotone continuity or autocontinuity 
of set function are required. 
However we will try to weaken this condition.

In this paper, we shall show several versions of 
Lebesgue decomposition theorems of non-additive measure $\mu$ 
for another given non-additive measure $\nu$, where
 $\mu$ is converse null-additive or exhaustive, 
 superadditive or order continuous,
 and
 $\nu$ is either weakly null-additive and continuous from below or
 $\sigma$-null-additive.
 In our discussion the monotone continuity
 of set function is not required, so Lebesgue decomposition
 theorem is formulated in generality.
%
%----------------- Section 2 -------------------------
%
\section{Preliminaries}                         \label{sec:2}
Let $X$ be a non-empty set, ${\cal R}$ a
$\sigma$-ring of subsets of $X$, and $(X, {\cal R})$
denotes the measurable space.

\begin{df} \                            \label{df:2.1}
{\rm
 A non-additive measure on ${\cal R}$
is an real valued set function $\mu:{\cal R} \to [0, +\infty)$
satisfying the following two conditions:
\begin{itemize}
 \item[{\rm (1)}] \ \ $\mu(\emptyset)=0;$
 \item[{\rm (2)}] \ \ $A \subset B$ and $A,
   B \in {\cal R}$ \ $\Rightarrow$ \  $\mu(A) \leq \mu(B)$.   
   \hfill  (monotonicity)
\end{itemize}
} %\rm
\end{df}

When a non-additive measure $\mu$ is continuous from below,
it is called a lower semicontinuous fuzzy measure (cf.~\cite{Wang92}).
%
In some literature, a set function $\mu$ satisfying
the conditions (1) and (2) of Definition \ref{df:2.1} 
is called a fuzzy measure. 


A set function $\mu:{\cal R} \to [0, +\infty)$ is said to be
 (i) {\it exhaustive} (cf.~\cite{Pap95}),
 if $\displaystyle{\lim_{n \to +\infty}} \mu(E_n) = 0$ for any
 infinite disjoint sequence $\{E_n \}_n$;
 (ii) {\it order-continuous} (cf.~\cite{Halm68}),
 if $\lim_{n \to \infty}\mu(A_n) = 0$ whenever $A_n \in {\cal R}$ and
   $A_n \searrow \emptyset \ (n \to \infty)$;
 (iii) {\it strongly order-continuous} (cf.~\cite{Li03d}),
 if $\lim_{n \to \infty}\mu(A_n) = 0$
     whenever $A_n, A \in {\cal R}$, $\searrow A$ and $\mu(A) = 0$;
 (iv) to have {\it property $(S)$} (cf.~\cite{Sun94}),
  if for any $\{A_n\}_{n=1,2,\cdots}  \subset {\cal R}$
  with $\textstyle{\lim_{n \to +\infty}} \mu(A_n) = 0$,
there exists a subsequence $\{A_{n_i} \}_{i=1,2,\cdots}$ of $\{A_n \}_{n=1,2,\cdots}$ such that
$\mu(\limsup A_{n_i}) = 0$.
%
\begin{df} {\rm (\cite{Wang96})} \               \label{df:2.2}
{\rm
 Let $\mu$ and $\nu$ be two non-additive measures.
We say that
\vspace{-2.5ex}
\begin{itemize}
   \item[(1)] \, $\mu$ is absolutely continuous of Type $I$
     with respect to $\nu$,
     denoted by $\mu \ll_{I} \nu$, if $\mu(A) = 0$
     whenever $A \in  {\cal R}$, $\nu(A) = 0$;
   \item[(2)] \, $\mu$ is absolutely
     continuous of Type $V\!I$  with respect to $\nu$,
     denoted by $\mu \ll_{V\!I} \nu$, if for any $\epsilon > 0$,
     there exists $\delta > 0$ such that $\mu(A) < \epsilon$
     whenever $A \in  {\cal R}$, $\nu(A) < \delta$.
\end{itemize}
}
\end{df}
%
\begin{df} {\rm(\cite{Dobra1974})} \              \label{df:2.3}
{\rm
 We say that $\mu$ is singular with respect to $\nu$,
and denote $\mu \perp \nu$, if there is a set $Q \in  {\cal R}$
 such that
$\nu(Q)=0$ and $\mu(E-Q)=0$ for any $E \in {\cal R}$.
}
\end{df}
%

For non-additive measures $\mu$ and $\nu$, if $\mu \ll_{V\!I} \nu$,
then $\mu \ll_{I} \nu$. The inverse statement may not be true.
%

\begin{prop} \                           \label{Prop-soc}
Let $\mu$ and $\nu$ be two non-additive measures.
If $\mu$ is strongly order-continuous, $\nu$
have property (S), then
$\mu \ll_{V\!I} \nu$ if and only if $\mu \ll_{I} \nu$.
\end{prop}




%----------------- Section 3 -------------------------
%
\section{Null-Additivity of Set Function}     \label{sec:3}
The several kinds of null-additivity of non-additive
 measure play important role in establishing
 Lebesgue decomposition type theorems
of non-additive measure. We recall them in the following.
\begin{df} {\rm (\cite{Wang92,Wang96})} \     \label{df:3.1}
{\rm
A set function $\mu:{\cal R} \to [0, +\infty)$ is said to be
  (i) {\it null-additive}, if $\mu(E
\cup F) = \mu(E)$ whenever $E, F \in {\cal R}$ and $\mu(F) = 0$;
  (ii) {\it weakly null-additive},
if $\mu(E \cup F) = 0$ whenever $E, F \in {\cal R}$, $\mu(E) = \mu(F) = 0$;
  (iii) {\it converse null-additive}, if
$\mu(A - B) = 0$ whenever $A \in {\cal R}, \
 B \in A \cap {\cal R}, \ \mu(B) = \mu(A)$;
  (iv) {\it pseudo-null-additive}, if
$\mu(B \cup C) = \mu(C)$, whenever
  $A \in {\cal R}, B \in A \cap {\cal R},
   C \in A \cap {\cal R}, \mu(A - B) = \mu(A)$.
}
\end{df}
%
\begin{prop}                                   \label{Prop-null}
If $\mu$ is null-additive, then it is weakly null-additive.
If $\mu$ is pseudo-null-additive, then it is converse null-additive.
\end{prop}
%
\begin{df} {\rm (\cite{Pap95})} \              \label{def:3.2}
{\rm
A non-additive measure $\mu$ is said to be $\sigma$-{\it null-additive},
if for every sequence $\{B_i \}_{i=1,2,\cdots }$ of pairwise 
disjoint sets from ${\cal R}$ and $\mu(B_i) = 0$ we have
\[
  \mu\left(A \cup \bigcup_{i=1}^{+\infty}B_i \right) = \mu(A)
  \quad \forall A \in {\cal R}.
\]
}
\end{df}
%
\begin{prop}                                  \label{Prop-sigema}
   $\mu$ is $\sigma$-null-additive if and only if $\mu$ is 
   null-additive
   and $\mu(B_i) = 0 \ (i=1,2,\cdots)$ implies
   $\mu(\displaystyle{\cup_{i=k}^{+\infty}B_i}) = 0$
   for every sequence $\{B_i \}_{i=1,2,\cdots }$ of pairwise 
   disjoint sets from ${\cal R}$.
\end{prop}
%
%----------------- Section 4 -------------------------
%
\section{Lebesgue Decomposition Theorems of Non-Additive Measure}
                                                   \label{sec:4}
In this section we show the Lebesgue decomposition type theorems
of non-additive measure. Unless stated, in the following we always 
assume that all set functions are non-additive measure.
%
\begin{lem}  \                                     \label{Lemma 1}
Let $\mu$ be converse null-additive non-additive measure
on ${\cal R}$. Then, there is a set $Q \in {\cal R}$ such that
$\mu(E - Q) = 0$; and further, if $\mu$ is null-additive, then
$\mu(E) = \mu(Q \cap E)$ for any $E \in {\cal R}$.
\end{lem}
%
{\it Proof.} \ Let $\alpha = \sup \{\mu(E) \,\, : \, \, 
E \in {\cal R} \}$.
By the definition of $\alpha$, we can choose a sequence
$\{E_{n} ^{(1)} \}_{n=1,2,\cdots}$ from ${\cal R}$ such that
for every $n=1,2,\cdots$,
\[
  \alpha - \frac{1}{n} < \mu(E_{n} ^{(1)}) \leq \alpha.
\]
Denote
$Q_1 = \displaystyle{\bigcup_{n=1}^{+\infty}} E_{n} ^{(1)}$,
then $Q_1 \in {\cal R}$. Thus we have
\[
   \alpha - \frac{1}{n} < \mu(E_{n} ^{(1)}) \leq \mu(Q_1) \leq \alpha,
\]
$n=1,2,\cdots$. Let $n \to +\infty$, we have $\mu(Q_1) = \alpha$.

Similarly, there exists a sequence $\{E_{n} ^{(2)} \}_{n=1,2,\cdots}$
from ${\cal R}$
such that
\[
   E_{n} ^{(2)} \subset X - Q_1, \quad \forall n \geq 1,
\]
 and
\[
  \mu(Q_2) =
  \sup \{\mu(E) \,\, : \,\, E \in {\cal R}, \quad E \subset X - Q_1 \}
\]
where $Q_2 = \displaystyle{\bigcup_{n=1}^{+\infty}} E_{n} ^{(2)}$.

Let us denote $Q = Q_{1} \cup Q_{2}$. Then $Q \in {\cal R}$, and
\[
  \alpha = \mu(Q_1) \leq \mu(Q) \leq
              \sup \{\mu(E) \,\, : \, \, E \in {\cal R} \}.
\]
Therefore $\mu(Q) = \mu(Q_1) = \alpha$.
By the converse null-additivity of $\mu$, and noting
  $Q_{1} \cap Q_{2} = \emptyset$, we have
  $\mu(Q_{2}) = \mu(Q - Q_{1}) = 0$.
Therefore, for any $E \in {\cal R}$, it is follows from
\[
  E - Q \subset E - Q_{1} \subset X - Q_{1}
\]
that
\[
  \mu(E - Q) \leq \mu(Q_2) = 0.
\]
Thus, for any $E \in {\cal R}$, we have $\mu(E - Q) = 0$.
When $\mu$ is null-additive, then
$\mu(E) = \mu((E - Q)\cup (E\cap Q) = \mu(Q \cap E)$ for any 
$E \in {\cal R}$.
The proof of the lemma is now complete. $\Box$
%
\begin{lem}  \                                 \label{Lemma 2}
Let $\mu$ be exhaustive non-additive measure
on ${\cal R}$. Then, there is a set $Q \in {\cal R}$ such that
$\mu(E - Q) = 0$; and further, if $\mu$ is null-additive, then
$\mu(E) = \mu(Q \cap E)$ for any $E \in {\cal R}$.
\end{lem}
{\it Proof.} \ It is similar to the proof of Theorem 2 in 
\cite{Jiang1995} and Lemma~\ref{Lemma 1} above. \\
%
%---------Lebesgue decomposition theorems--------------

The following theorems with their corollaries are several versions of
Lebesgue decomposition theorem for non-additive measure.
%
\begin{thm} {\rm (Lebesgue decomposition theorem) }
                                           \label{Lebes Thm 1}
Let $\mu$ and $\nu$ be non-additive measures on ${\cal R}$.
If $\mu$ is either converse null-additive or exhaustive,
$\nu$ is weakly null-additive and continuous from below, then there
exists a set $Q \in {\cal R}$ such that those non-additive measures
$\mu_c$ and $\mu_s$ defined by
\[
  \mu_c(E)=\mu(E-Q) \ \ \mbox{and} \ \ \mu_s(E)=\mu(E \cap Q), 
  \quad \forall E \in {\cal R}
\]
satisfy $\mu_c \ll_{I} \nu$ \ and \ $\mu_s \perp \nu$, respectively.
%
\end{thm}
%
{\bf Proof.} \ Put ${\cal R}_1 = \{A \in {\cal R} \ : \ \nu(A) = 0 \}$.
Then, ${\cal R}_1$ is a $\sigma$-subring of ${\cal R}$ by the
weak null-additivity and continuity from below of $\nu$.
By using Lemma~\ref{Lemma 1} and \ref{Lemma 2}, we may take a set $Q \in {\cal R}_1$
 such that
\[
\mu(Q)=\sup\{\mu(A) \ : \ A \in {\cal R}_1\} \quad \mbox{and} \quad
\mu(E-Q)=0, \quad \forall E \in {\cal R}_1.
\]
Now we take that $\mu_c(E)=\mu(E-Q)$ and $\mu_s(E)=\mu(E \cap Q)$, \ $E \in
{\cal R}$, then $\mu_c$ and $\mu_s$ satisfy the required properties:
\[
  \mu_c(E)=0 \ \ \mbox{if} \ \ \nu(E)=0
\]
and
\[
  \mu_s(E - Q) = \mu((E-Q) \cap Q) = \mu(\emptyset) = 0
\]
for any $E \in {\cal R}$. \ $\Box$\\

From Proposition~\ref{Prop-null} and Theorem~\ref{Lebes Thm 1},
we have the following corollary.
\begin{cor} \                                     \label{cor-1}
If $\mu$ is pseudo-null-additive, $\nu$ is null-additive
and continuous from below, then there exists a set $Q \in {\cal R}$ 
such that those non-additive measures
$\mu_c$ and $\mu_s$ defined by
\[
  \mu_c(E)=\mu(E-Q) \ \ \mbox{and} \ \ \mu_s(E)=\mu(E \cap Q),
   \quad \forall E
   \in {\cal R}
\]
satisfy $\mu_c \ll_{I} \nu$ \ and \ $\mu_s \perp \nu$, respectively.
%
\end{cor}
%
\begin{df} \ {\rm (\cite{Pap95})}                \label{df:4.1}
{\rm
A set function $\mu:{\cal R} \to [0, +\infty)$ is called
{\it superadditive},
 if for every $A, B \in {\cal R}$ and
 $A \cap B = \emptyset$,
\[
  \mu(A \cup B) \geq  \mu(A) + \mu(B).
\]
}
\end{df}
%
\begin{prop}                                   \label{Prop-supp}
If $\mu$ is superadditive, then it is converse null-additive and
exhaustive.
\end{prop}

Note that a condition of superadditivity do not imply the
order continuity. There is a counterexample:
A superadditive non-additive measure is not necessarily order continuous.
 Counterexample:
 Let $\mu$ be the non-additive measure on $2^N$, 
  the power set of the set $N$ of all the positive integers,
 defined by
 $\mu(E)=1$ if $E$ is cofinite (i.e., 
  the complement of a finite set), 
 $0$ otherwise.
%
 Then obviously $\mu$ is superadditive non-additive measure.
 However $A_n=\{n, n+1, \cdots\}$ for $n=1, 2, \cdots$
 decreasingly converges to the empty set,
 but the limit of $\{\mu(A_n)\}$ is equal to $1$, not zero.
This counterexample is noticed by the anonymous referee.


As a direct result of Proposition~\ref{Prop-supp} and
Theorem~\ref{Lebes Thm 1},
we can  obtain the following theorem immediately.
\begin{thm} \                     \label{Lebes Thm 2}
Let $\mu$ and $\nu$ be non-additive measures on ${\cal R}$.
If $\mu$ is superadditive,
   $\nu$ is weakly null-additive and continuous from below,
then there exist non-additive measures
$\mu_c$ and $\mu_s$ on ${\cal R}$ such that
 $\mu_c \ll_{I} \nu$, \ $\mu_s \perp \nu$, and
 $\mu \geq \mu_c + \mu_s$.
\end{thm}
%

By using Proposition~\ref{Prop-sigema}, similar to the proof
of Theorem~\ref{Lebes Thm 1}, we can prove the following 
theorem.
\begin{thm} \                         \label{Lebes Thm 3}
  Let $\mu$ and $\nu$ be non-additive measures on ${\cal R}$.
  If $\mu$ is either converse null-additive or exhaustive,
  $\nu$ is $\sigma$-null-additive,
  then there exist non-additive measures
  $\mu_c$ and $\mu_s$ on ${\cal R}$ such that
  $\mu_c \ll_{I} \nu$ and $\mu_s \perp \nu$.
\end{thm}
%
\begin{cor} \                          \label{cor-3}
  Let $\mu$ and $\nu$ be non-additive measures on ${\cal R}$.
  If $\mu$ is superadditive, $\nu$ is $\sigma$-null-additive,
  then there exist non-additive measures
  $\mu_c$ and $\mu_s$ on ${\cal R}$ such that
  $\mu_c \ll_{I} \nu$, \ $\mu_s \perp \nu$, and
  $\mu \geq \mu_c + \mu_s$.
\end{cor}

\noindent
Note that
it is not required that $\nu$ has the continuity from below
in Theorem~\ref{Lebes Thm 3} and Corollary~\ref{cor-3}. 

Combining Theorem~\ref{Lebes Thm 1} and
Proposition~\ref{Prop-soc}, we can obtain the following result.
%
\begin{thm} \                         \label{Lebes Thm 4}
  Let $\mu$ and $\nu$ be non-additive measures on ${\cal R}$.
  If $\mu$ is strongly order continuous,
  $\nu$ is weakly null-additive
  and continuous from below, and have property (S), then there
  exists a set $Q \in {\cal R}$ such that those non-additive measures
  $\mu_c$ and $\mu_s$ defined by
  \[
    \mu_c(E)=\mu(E-Q) \ \ \mbox{and} \ \ \mu_s(E)=\mu(E \cap Q), 
    \quad \forall E \in {\cal R}
  \]
  satisfy $\mu_c \ll_{V\!I} \nu$ \ and \ $\mu_s \perp \nu$, 
  respectively.
\end{thm}
%

\noindent
{\bf Conclusions:}
There are several versions of Lebesgue decomposition theorem 
as noted in the first section. Here we proved that 
Lebesgue decomposition type theorems
for non-additive measure are shown under the conditions of
null-additivity, converse null-additivity, weak null-additivity
and $\sigma$-null-additivity.
It should be clarified these relations and also considered
applications to the various fields.


\vspace{1.5ex}
\noindent
{\bf Acknowledgements:}
The authors would like to thank anonymous referees a useful 
comment. We impose the exhaustivity for Lebesgue decomposition 
theorem of Theorem \ref{Lebes Thm 2} in stead of the order 
continuity. 
By this comment, the assertion of Theorem \ref{Lebes Thm 2} is 
corrected.
%
%         This work was partially supported by
%                   the NSFC Grant No.70471063.
The research of the first author was supported by the Postdoctoral 
Research Finance of Communication University of China.

%----------------References---------------------
%
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%----------
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%-----------------
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%
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%
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%-------
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%-----------
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%
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%
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%
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%
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%
\end{thebibliography}
\end{document}
%%%%%%%%%%%%%%%%%%%%%%%
