# Some familiar graphs on the rings of measurable functions

Pratip Nandi Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata 700019, West Bengal, India
<EMAIL_ADDRESS>,  Atasi Deb Ray Department of Pure Mathematics,
University of Calcutta, 35, Ballygunge Circular Road, Kolkata 700019, West
Bengal, India<EMAIL_ADDRESS>and  Sudip Kumar Acharyya Department of
Pure Mathematics, University of Calcutta, 35, Ballygunge Circular Road,
Kolkata 700019, West Bengal, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, replacing ‘equality’ by ’equality almost everywhere’ we modify
several terms associated with the ring of measurable functions defined on a
measure space $(X,\mathcal{A},\mu)$ and thereby study the graph theoretic
features of the modified comaximal graph, annihilator graph and the weakly
zero-divisor graph of the said ring. The study reveals a structural analogy
between the modified versions of the comaximal and the zero-divisor graphs,
which prompted us to investigate whether these two graphs are isomorphic.
Introducing a quotient-like concept, we find certain subgraphs of the
comaximal graph and the zero-divisor graph of $\mathcal{M}(X,\mathcal{A})$ and
show that these two subgraphs are always isomorphic. Choosing $\mu$ as a
counting measure, we prove that even if these two induced graphs are
isomorphic, the parent graphs may not be so. However, in case of Lebesgue
measure space on $\mathbb{R}$, we establish that the comaximal and the zero-
divisor graphs are isomorphic. Observing that both of the comaximal and the
zero-divisor graphs of the ring $\mathcal{M}(X,\mathcal{A})$ are subgraphs of
the annihilator graph of the said ring, we find equivalent conditions for
their equalities in terms of the partitioning of $X$ into two atoms. Moreover,
the non-atomicity of the underlying measure space $X$ is characterized through
graph theoretic phenomena of the comaximal and the annihilator graph of
$\mathcal{M}(X,\mathcal{A})$.

###### Key words and phrases:

Rings of measurable functions, Zero-divisor graph, Co-maximal graph,
Annihilator graph, Complemented graph, Bipartite graph, dominating number,
Atomic measure, Lebesgue measure

###### 2020 Mathematics Subject Classification:

Primary 13A70; Secondary 05C60, 05C63, 05C90

The first author thanks the CSIR, New Delhi – 110001, India, for financial
support.

##  1\. Introduction

Our starting point is a triplet $(X,\mathcal{A},\mu)$, where $X$ is a non-
empty set, $\mathcal{A}$, a $\sigma$-algebra of subsets of $X$ and
$\mu:\mathcal{A}\to[0,\infty]$ is a measure. Such a triplet is often called a
measure space and members of $\mathcal{A}$ are called $\mathcal{A}$-measurable
sets or simply measurable sets, if there is no chance of confusion. To make
matters non-trivial, we further assume that $\mu(X)>0$. A function
$f:X\to\mathbb{R}$ is called measurable or $\mathcal{A}$-measurable if for any
$r\in\mathbb{R}$, $\\{x\in X:f(x)<r\\}$ is a measurable set. The family
$\mathcal{M}(X,\mathcal{A})$ of all measurable functions is a commutative
lattice ordered ring with identity if the relevant operations are defined
pointwise on $X$. Rings of measurable functions are being investigated in the
recent times. The articles [1], [2], [3] are referred to in this connection.
Study on several aspects of different well-known graph structures associated
with the ring $\mathcal{M}(X,\mathcal{A})$ sounds quite fascinating. In [9],
the authors take up the study on some relevant properties linked to the zero-
divisor graph of the ring $\mathcal{M}(X,\mathcal{A})$, where the zero-divisor
graph has been redefined via the measure $\mu$. Incidentally, only one paper
[9] appears in the literature, addressing problems of this kind. In that
paper, the authors determined conditions under which this graph becomes
triangulated and also realise that it is never hypertriangulated. Being
inspired by these findings, in the present article we endeavour to study the
comaximal graph, the annihilator graph and the weakly zero-divisor graph of
the ring $\mathcal{M}(X,\mathcal{A})$, redefining each of the graphs via the
measure $\mu$. These three graphs are generalized versions of the comaximal
graph, annihilator graph and the weakly zero-divisor respectively on
$\mathcal{M}(X,\mathcal{A})$.
To make the article self-contained and reader friendly, we recall in section 2
a few technical terms related to the graphs in general. A few of the ring
theoretic terms related to the parent ring $\mathcal{M}(X,\mathcal{A})$ demand
natural modifications, when ‘equality’ is replaced by ‘equality almost
everywhere’ in the presence of a measure $\mu$ on $(X,\mathcal{A})$. In the
last part of this section, we incorporate these changes thereby preparing our
measure theoretic preliminaries, necessary for any study on the contemplated
graphs on the ring $\mathcal{M}(X,\mathcal{A})$ mentioned above in the
subsequent sections.
In section 3, we take up the actual problem of studying the comaximal graph
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ of the ring
$\mathcal{M}(X,\mathcal{A})$. We compute the familiar graph parameters, viz.
the eccentricities, diameter and girth of this graph. This leads to a
necessary and sufficient condition for the graph
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ to be a complete bipartite
one [Theorem 3.10]. Incidentally it is also established that
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a triangulated graph if
and only if the measure ‘$\mu$’ is non-atomic in the sense that each
measurable subset of $X$ with positive measure is expressible as a disjoint
union of two measurable sets each with positive measure. Thus in particular if
$X=\mathbb{R}$, $\mu\equiv$ Lebesgue or Borel measure, then
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is triangulated. It is
further realized that $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
never hypertriangulated [Corollary 3.19]. Thus outwardly a lot of similarities
are observed between the properties of the zero-divisor graph
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ as established in [9] and the
corresponding properties of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
as realized in the present paper. It is therefore natural to ask, whether
these two graphs, viz. $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, become isomorphic in the
sense that there is a bijection between the sets of vertices of these two
graphs which preserves the adjacency relation. Being motivated by this query,
we construct an induced subgraph $G_{2}$ of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ by identifying some vertices
and making a quotient like construction. Analogously, we take into our
consideration the induced subgraph of $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
denote it by $G$. It is realized that the graphs $G_{2}$ and $G$ are
isomorphic. From this it is deduced by imposing certain condition on the
nature of the characteristic function of measurable sets in $(X,\mathcal{A})$
that $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ are isomorphic [Theorem 3.34].
In section 4, we take up our study on the annihilator graph
$AG(\mathcal{M}(X,\mathcal{A}))$ of the ring $\mathcal{M}(X,\mathcal{A})$. The
vertices of this later graph are the same as the vertices of
$\Gamma(\mathcal{M}(X,\mathcal{A}))$. Amongst several properties enjoyed by
$AG(\mathcal{M}(X,\mathcal{A}))$ we would like to mention the one which
characterizes the non-atomicity of the measure $\mu$. Indeed this reads:
$AG(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated if and only if $\mu$ is
non-atomic. This in turn leads to the fact that
$AG(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated if and only if
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is triangulated.
We make our formal study on the weakly zero-divisor graph
$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ of the ring $\mathcal{M}(X,\mathcal{A})$
in Section 5. The set of vertices of this graph are the zero-divisors $f$ so
that $Z(f)$ is an atom.
In the concluding section 6 of this article, we illustrate various concepts
and facts that we achieve in the preceding sections about three types of
graphs mentioned above with the aid of two well-known specific example of
measure spaces. The first case is so called Counting measure on the measurable
space $(X,\mathcal{P}(X))$. The second one is the Lebesgue measure on the set
$\mathbb{R}$ equipped with the $\sigma$-algebra of Lebesgue measurable sets in
$\mathbb{R}$. It is realized that the zero-divisor graph and the comaximal
graph of the ring of all Lebesgue measurable functions on $\mathbb{R}$ become
isomorphic as graphs [Theorem 6.2.7]. However it is not known to us, whether
more generally for a non-atomic measure $\mu$ on $(X,\mathcal{A})$,
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ are isomorphic.

##  2\. Preliminaries

Let $G$ be a graph with $V$ as the set of vertices of $G$. $G$ is said to be a
simple graph if $G$ contains no self-loops and no parallel edges. $G$ is a
connected graph if every pair of vertices is connected by a path in $G$. A
stable set in $G$ is a subset of $V$ in which no two vertices are adjacent.
For a cardinal number $\alpha$, $G$ is called $\alpha$-partite graph if $G$
contains $\alpha$-many disjoint stable sets whose union is $V$. An
$\alpha$-partite graph is called a complete $\alpha$-partite graph if any two
vertices from two different stable sets are adjacent. If $\alpha=2$, we call
these graphs bipartite and complete bipartite respectively. The distance
between two vertices $u,v$ in $G$ is the length of the shortest path joining
$u,v$, denoted by $d(u,v)$. The eccentricity of a vertex $v$ is defined by
$ecc(v)=min\\{d(u,v):u\in V\\}$. The diameter of $G$ is
$diam(G)=max\\{d(u,v):u,v\in V\\}$ and the girth $gr(G)$ of $G$ is the length
of the smallest cycle in $G$. $G$ is triangulated or hypertriangulated
according as every vertex is a vertex of a triangle or every edge is an edge
of a triangle in $G$. For two vertices $u,v\in V$, $c(u,v)$ denotes the length
of the smallest cycle in $G$ containing $u,v$. Two vertices $u,v$ are said to
be orthogonal in $G$, denoted by $u\perp v$, if $u,v$ are adjacent and
$c(u,v)>3$. $G$ is called a complemented graph if for every $u\in V$, there
exists $v\in V$ such that $u\perp v$. A complemented graph is called uniquely
complemented if whenever $u\perp v$ and $u\perp w$ in $G$ for $u,v,w\in V$,
then $v,w$ are adjacent to the same set of vertices in $G$. The clique number
$cl(G)$ of $G$ is the maximum cardinality of complete subgraphs of $G$. A
subset $V_{1}$ of $V$ is a dominating set in $G$ if for every $u\in V\setminus
V_{1}$, $u,v$ are adjacent in $G$ for some $v\in V_{1}$ and the dominating
number $dt(G)=min\\{|V_{1}|:V_{1}\text{ is a dominating set in }G\\}$. A
subset $V_{1}$ of $V$ is a total dominating set in $G$ if for every $u\in V$,
$u,v$ are adjacent in $G$ for some $v\in V_{1}$ and the total dominating
number $dt_{t}(G)=min\\{|V_{1}|:V_{1}\text{ is a total dominating set in
}G\\}$. For a cardinal number $\alpha$, $G$ is said to be $\alpha$-colorable
if there is a map $\psi:V\to[0,\alpha]$ such that $\psi(u)\neq\psi(v)$
whenever two $u,v$ are adjacent in $G$. The chromatic number $\chi(G)$ of $G$
is $min\\{\alpha:G\text{ is }\alpha\text{-colorable}\\}$. For any subset
$V^{\prime}$ of $V$, the induced subgraph $G^{\prime}$ of $G$ induced by
$V^{\prime}$ is the graph whose vertex set is $V^{\prime}$ and two vertices in
$G^{\prime}$ are adjacent if they are adjacent in $G$.
Let $G_{1},G_{2}$ be two graphs having $V_{1},V_{2}$ as set of vertices
respectively. A bijection map $\psi:V_{1}\to V_{2}$ is called a graph
isomorphism between two graphs $G_{1},G_{2}$, if it preserves the adjacency
relations; i.e., $a,b\in V_{1}$ are adjacent in $G_{1}$ if and only if
$\psi(a),\psi(b)$ are adjacent in $G_{2}$. For more graph related terms we
refer to [8].
Let $(X,\mathcal{A},\mu)$ be a measure space and $\mathcal{M}(X,\mathcal{A})$
be the corresponding ring of real-valued measurable functions on $X$. It is
clear that $Z(f),X\setminus Z(f)\in\mathcal{A}$ for all
$f\in\mathcal{M}(X,\mathcal{A})$, here $Z(f)=\\{x\in X:f(x)=0\\}$ is the zero
set of $f$. For any subset $A$ of $X$, the characteristic function of $A$ is
denoted by $1_{A}$ and is given by

$1_{A}(x)=\begin{cases}1&\text{ if }x\in A\\\ 0&\text{ otherwise}\end{cases}$

Therefore, $1_{A}\in\mathcal{M}(X,\mathcal{A})$ if and only if
$A\in\mathcal{A}$. An element $A\in\mathcal{A}$ is called an atom in $X$ if
$\mu(A)>0$ and $A$ can not be written as a union of two disjoint measurable
sets, each with positive measure, i.e., $A$ can not be written as $B\sqcup C$,
where $B,C\in\mathcal{A}$ with $\mu(B),\mu(C)>0$, here ‘$\sqcup$’ denotes the
disjoint union of two sets. Suppose that $A\in\mathcal{A}$ is an atom in $X$.
Then for any $B\in\mathcal{A}$ with $\mu(B)=0$, it can easily be proved that,
$A\setminus B$ and $A\cup B$ are both atoms in $X$. The measure $\mu$ is said
to be an atomic measure if every measurable set with positive measure contains
an atom. On the other hand, $\mu$ is said to be non-atomic if $\mathcal{A}$
does not contain any atom.
Let $R$ be a commutative ring. There are several graphs, viz. the zero-divisor
graph, the comaximal graph, etc., defined on $R$ to study the interaction
between the properties of the ring and the properties of the respective
graphs. The zero-divisor graph $\Gamma(R)$ of $R$ has its vertices as the set
of non-zero zero-divisors of $R$ with two distinct vertices $x,y$ declared
adjacent if and only if $x.y=0$ [[4], [7]]. The comaximal graph of $R$,
defined in [12], with vertices as elements of $R$, where two distinct vertices
$x$ and $y$ are adjacent if and only if the sum of the principal ideals
generated by $x,y$ is $R$. Later in [6], the vertex set of the comaximal graph
of $R$ was redefined as $R\setminus[U(R)\cup J(R)]$, where $U(R)$ and $J(R)$
denote the set of all units in $R$ and the Jacobson radical of $R$
respectively. This graph is denoted by $\Gamma^{\prime}_{2}(R)$. The
annihilator graph $AG(R)$ of $R$, introduced in [5], is a supergraph of
$\Gamma(R)$ having the same set of vertices and two distinct vertices $x,y$
are adjacent if and only if $ann(x)\cup ann(y)\subsetneqq ann(x.y)$, where
$ann(a)=\\{r\in R:a.r=0\\}$ is the annihilator ideal of $a$ in $R$. The weakly
zero-divisor graph $W\Gamma(R)$ of $R$ is also a supergraph of $\Gamma(R)$
with the same set of vertices where two distinct vertices $x,y$ are adjacent
if and only if there exists $a\in ann(x)\setminus\\{0\\}$ and $b\in
ann(y)\setminus\\{0\\}$ such that $a.b=0$, introduced in [11].
As proposed in the introduction of this paper, we study the behaviour of the
well-known graphs, viz. comaximal, annihilator and weakly zero-divisor graphs,
redefining them via $\mu$, of the ring of measurable functions defined over a
measure space $(X,\mathcal{A},\mu)$. In this context, a few of the ring
theoretic terms also demand natural modifications when ‘equality’ is replaced
by ‘equality almost everywhere’, in the presence of a measure $\mu$ on
$(X,\mathcal{A})$. We now describe this modification and redefine the terms,
retaining their nomenclature in most of the cases.
Two functions $f,g\in\mathcal{M}(X,\mathcal{A})$ are said to be equal almost
everywhere on $X$ with respect to the measure $\mu$ if $\mu(\\{x\in X:f(x)\neq
g(x)\\})=0$ and in this case we write $f\equiv g$ a.e. on $X$. If
$f\in\mathcal{M}(X,\mathcal{A})$ is such that $\mu(Z(f))=0$, then $f$ is said
to be non-zero almost everywhere on $X$ (in short, non-zero a.e. on $X$).

######  Definition 2.1.

  1. (1)

A function $f\in\mathcal{M}(X,\mathcal{A})$ which is non-zero a.e. on $X$, is
called a zero-divisor in $\mathcal{M}(X,\mathcal{A})$ if there exists
$g\in\mathcal{M}(X,\mathcal{A})$ such that $g$ is non-zero a.e. on $X$ and
$f.g\equiv 0$ a.e. on $X$. The set of all zero-divisors in
$\mathcal{M}(X,\mathcal{A})$ is denoted by
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ (in short, $\mathscr{D}$).

  2. (2)

A function $f\in\mathcal{M}(X,\mathcal{A})$ is called a $\mu$-unit in
$\mathcal{M}(X,\mathcal{A})$ if $\mu(Z(f))=0$. $U(\mathcal{M}(X,\mathcal{A}))$
stands for the set of all $\mu$-units in $\mathcal{M}(X,\mathcal{A})$.

  3. (3)

An ideal $I$ in $\mathcal{M}(X,\mathcal{A})$ is said to be almost
$\mathcal{M}(X,\mathcal{A})$ if $I$ is a principal ideal generated by a
$\mu$-unit in $\mathcal{M}(X,\mathcal{A})$.

Since the Jacobson radical of $\mathcal{M}(X,\mathcal{A})$ is $(0)$, we
redefine
$J(\mathcal{M}(X,\mathcal{A}))=\\{f\in\mathcal{M}(X,\mathcal{A}):f\equiv
0\text{ a.e. on }X\\}$. For each $f\in\mathcal{M}(X,\mathcal{A})$, the
annihilator ideal of $f$ is also redefined as
$ann(f)=\\{g\in\mathcal{M}(X,\mathcal{A}):f.g\equiv 0\text{ a.e. on }X\\}$. We
can also write $ann(f)$ as $\\{g\in\mathcal{M}(X,\mathcal{A}):\mu(X\setminus
Z(f)\cap X\setminus Z(g))=0\\}$, where $f\in\mathcal{M}(X,\mathcal{A})$.

The following theorem completely describes the zero-divisors of
$\mathcal{M}(X,\mathcal{A})$ :

######  Theorem 2.2.

$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))=\\{f\in\mathcal{M}(X,\mathcal{A}):\mu(Z(f))>0\text{
and }\mu(X\setminus Z(f))>0\\}$.

###### Proof.

Let $f\in\mathcal{M}(X,\mathcal{A})$ be such that $\mu(Z(f))>0$ and
$\mu(X\setminus Z(f))>0$. Then $f$ is non-zero a.e. on $X$ and there exists
$g=1_{Z(f)}$, such that $f.g=0$ on $X$. Clearly
$g\in\mathcal{M}(X,\mathcal{A})$ and $\mu(X\setminus Z(g))=\mu(Z(f))>0$, i.e.,
$g$ is non-zero a.e. on $X$. So, $f\in\mathscr{D}$.
Conversely let $f\in\mathscr{D}$. Then by definition of a zero-divisor,
$\mu(X\setminus Z(f))>0$ and there exists $g\in\mathcal{M}(X,\mathcal{A})$
with $\mu(X\setminus Z(g))>0$ such that $f.g\equiv 0$ a.e., i.e., there exists
$A\in\mathcal{A}$ with $\mu(A)=0$ such that $f.g=0$ on $X\setminus A$. So,
$X\setminus Z(g)\cap X\setminus A\subset Z(f)$. Now, $X\setminus
Z(g)=(X\setminus Z(g)\cap A)\sqcup(X\setminus Z(g)\cap X\setminus A)$ and
$\mu(X\setminus Z(g)\cap A)=0$. This implies that $\mu(X\setminus Z(g)\cap
X\setminus A)=\mu(X\setminus Z(g))>0$. Therefore, $\mu(Z(f))>0$. ∎

We observe in the next theorem that the ring $\mathcal{M}(X,\mathcal{A})$ has
no divisor of zero is equivalent to a purely measure theoretic phenomenon of
the measure space $(X,\mathcal{A},\mu)$.

######  Theorem 2.3.

The following statements are equivalent:

  1. (1)

Every measurable set in $X$ with positive measure is an atom.

  2. (2)

$X$ is an atom.

  3. (3)

$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ is an empty set.

###### Proof.

$(1)\implies(2)$ is trivial, as $\mu(X)>0$ and $(2)\implies(3)$ follows
directly from the definition of an atom.
$(3)\implies(1)$ : Let $(1)$ be false. Then some $A\in\mathcal{A}$ exists such
that $\mu(A)>0$ but $A$ is not an atom. So $A$ is expressible as the disjoint
union of some $A_{1},A_{2}\in\mathcal{A}$ with $\mu(A_{1})>0$ and
$\mu(A_{2})>0$. In such case, $1_{A_{1}},1_{A_{2}}\in\mathscr{D}$, proving
$\mathscr{D}\neq\emptyset$. ∎

Following are a couple of examples of measure spaces where every measurable
set with positive measure is an atom and therefore by Theorem 2.3, their
corresponding rings of measurable functions possess no divisor of zero.

######  Example 2.4.

$(1)$ Let $X$ be a non-empty set, $\mathcal{A}=\mathscr{P}(X)$ (= the power
set of $X$) and $x_{0}\in X$. Consider the Dirac measure at $x_{0}$ given by

$\delta_{x_{0}}(A)=\begin{cases}0&\text{ if }x_{0}\notin A\\\ 1&\text{ if
}x_{0}\in A.\end{cases}$

$(2)$ Let $X$ be an uncountable set, $\mathcal{A}=\\{A\subseteqq X:\text{
either }A\text{ or }X\setminus A\text{ is countable}\\}$. Consider the co-
countable measure given by

$\rho(A)=\begin{cases}0&\text{ if }A\text{ is countable}\\\ \infty&\text{ if
}X\setminus A\text{ is countable.}\end{cases}$

######  Theorem 2.5.

For any $f,g\in\mathcal{M}(X,\mathcal{A})$, $ann(f)\subset ann(g)$ if and only
if $\mu(Z(f)\setminus Z(g))=0$.

###### Proof.

Let $E=Z(f)\setminus Z(g)$. If $\mu(E)=0$, then $Z(f)\cap X\setminus E\subset
Z(g)$. Now let $h\in ann(f)$, i.e., $h.f=0$ on $X\setminus F$ for some
$F\in\mathcal{A}$ with $\mu(F)=0$. Therefore $X\setminus Z(h)\cap X\setminus
F\subset Z(f)\implies X\setminus Z(h)\cap X\setminus(E\cup F)\subset
Z(g)\implies h.g\equiv 0$ on $X\setminus(E\cup F)$. Since $\mu(E\cup F)=0$,
$h.g\equiv 0$ a.e. on $X$ i.e., $h\in ann(g)$.
Conversely let $\mu(E)>0$. Let $h=1_{E}$. Then
$h\in\mathcal{M}(X,\mathcal{A})$ and $X\setminus Z(h)=E=Z(f)\setminus Z(g)$.
Therefore $f.h\equiv 0$ on $X$ and $X\setminus Z(h)\cap X\setminus
Z(g)=E\implies\mu(X\setminus Z(h)\cap X\setminus Z(g))>0$. Thus $h\in
ann(f)\setminus ann(g)$. i.e., $ann(f)\not\subset ann(g)$. ∎

######  Corollary 2.6.

For $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, $\mu(Z(f)\triangle
Z(g))=0$ if and only if $ann(f)=ann(g)$.

In view of Theorem 2.3, throughout this paper, we assume that $X$ is not an
atom and therefore, for every atom $A\in\mathcal{A}$, $\mu(X\setminus A)>0$.

##  3\. The comaximal graph $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
of $\mathcal{M}(X,\mathcal{A})$

We redefine the comaximal graph of $\mathcal{M}(X,\mathcal{A})$, taking into
account the measure defined on the measurable space $(X,\mathcal{A})$ and
investigate in this section its graph features via the properties of the
underlying measure space $(X,\mathcal{A},\mu)$ and the ring
$\mathcal{M}(X,\mathcal{A})$.

######  Definition 3.1.

The comaximal graph $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ (in
short, $\Gamma^{\prime}_{2}$) of $\mathcal{M}(X,\mathcal{A})$ is a graph on
the vertex set
$\mathcal{M}(X,\mathcal{A})\setminus[U(\mathcal{M}(X,\mathcal{A}))\cup
J(\mathcal{M}(X,\mathcal{A}))]$ and the adjacency relation for two vertices
$f,g$ is given by : $f,g$ are adjacent if and only if the sum of the principal
ideals generated by $f,g$ is almost $\mathcal{M}(X,\mathcal{A})$. In notation,
$<f>+<g>=<u>$, where $u$ is a $\mu$-unit in $\mathcal{M}(X,\mathcal{A})$.

Theorem 2.3 suggests that the vertex set of the comaximal graph
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ and hence it is a non-empty graph
under the assumption that $X$ is not an atom. In [9], the author defined the
zero-divisor graph $\Gamma(\mathcal{M}(X,\mathcal{A}))$ (in short, $\Gamma$)
of $\mathcal{M}(X,\mathcal{A})$ whose vertex set is
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ and the adjacency relation is given
by: $f$ is adjacent to $g$ if and only if $\mu(X\setminus Z(f)\cap X\setminus
Z(g))=0$. So, the sets of vertices of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ coincide.
The adjacency relation in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
can be interpreted through the almost disjointness of the corresponding zero
sets of the vertices in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, as
seen in the next result.

######  Theorem 3.2.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then $f,g$ are adjacent
in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ if and only if
$\mu(Z(f)\cap Z(g))=0$.

###### Proof.

Let $f,g$ be adjacent in $\Gamma^{\prime}_{2}$. Then there exists a $\mu$-unit
$u$ in $\mathcal{M}(X,\mathcal{A})$ such that $<f>+<g>=<u>$. i.e.,
$f.f_{1}+g.g_{1}=u$ for some $f_{1},g_{1}\in\mathcal{M}(X,\mathcal{A})$.
Therefore, $Z(f)\cap Z(g)\subset Z(u)$. Since $u$ is a $\mu$-unit
$\mu(Z(u))=0\implies\mu(Z(f)\cap Z(g))=0$.
Conversely, let $\mu(Z(f)\cap Z(g))=0$. We claim that $<f>+<g>=<u>$, where
$u=f^{2}+g^{2}$, a $\mu$-unit in $\mathcal{M}(X,\mathcal{A})$. Clearly
$u\in<f>+<g>$. Also $f\in<u>$, because $f=u.f_{1}$, where

$f_{1}(x)=\begin{cases}\frac{f(x)}{f^{2}(x)+g^{2}(x)}&\text{ if }x\notin
Z(f)\cap Z(g)\\\ 0&\text{ if }x\in Z(f)\cap Z(g)\end{cases}$

and similarly, $g\in<u>$. Hence, $<f>+<g>=<u>$ . i.e., $f,g$ are adjacent in
$\Gamma^{\prime}_{2}$. ∎

The following theorem provides a necessary and sufficient condition under
which a pair of vertices of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
admits of a third vertex adjacent to both of them.

######  Theorem 3.3.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then there is a vertex in
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ adjacent to both $f,g$ in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ if and only if
$\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$.

###### Proof.

Let $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$. Consider $h=1_{Z(f)\cup
Z(g)}\in\mathcal{M}(X,\mathcal{A})$. Clearly $Z(h)=X\setminus Z(f)\cap
X\setminus Z(g)\implies h\in\mathscr{D}$. Also $Z(f)\cap
Z(h)=\emptyset=Z(g)\cap Z(h)$ i.e., $h$ is adjacent to both $f,g$ in
$\Gamma^{\prime}_{2}$. Conversely, let $h\in\mathscr{D}$ be adjacent to both
$f,g$ in $\Gamma^{\prime}_{2}$. So $\mu(Z(f)\cap Z(h))=0=\mu(Z(g)\cap Z(h))$
and hence, $\mu(Z(h)\cap[Z(f)\cup Z(g)])=0$. Now $Z(h)=(Z(h)\cap[Z(f)\cup
Z(g)])\sqcup(Z(h)\cap X\setminus[Z(f)\cup Z(g)])\implies\mu(Z(h)\cap
X\setminus[Z(f)\cup Z(g)])=\mu(Z(h))>0$, as $h\in\mathscr{D}$. Consequently
$\mu(X\setminus[Z(f)\cup Z(g)])>0$; i.e., $\mu(X\setminus Z(f)\cap X\setminus
Z(g))>0$. ∎

######  Corollary 3.4.

Every edge $f-g$ in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
either an edge of a triangle or an edge of a square according as
$\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$ or $\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0$.

###### Proof.

If $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$, then by Theorem 3.3, there
exists a vertex adjacent to both $f,g$. If $\mu(X\setminus Z(f)\cap X\setminus
Z(g))=0$, then $f-g-2f-2g-f$ forms a square in $\Gamma^{\prime}_{2}$. ∎

######  Corollary 3.5.

$gr(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))\leq 4$.

The distance between two vertices in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is given below.

######  Theorem 3.6.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then

$d(f,g)=\begin{cases}1&\text{ if }\mu(Z(f)\cap Z(g))=0\\\ 2&\text{ if
}\mu(Z(f)\cap Z(g))>0\text{ and }\mu(X\setminus Z(f)\cap X\setminus Z(g))>0\\\
3&\text{ if }\mu(Z(f)\cap Z(g))>0\text{ and }\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0\end{cases}$

###### Proof.

The conditions for $d(f,g)=1$ or $2$ are straightforward consequences of the
adjacency relation and Theorem 3.3. From these two results,
$d(f,g)=3\implies\mu(Z(f)\cap Z(g))>0$ and $\mu(X\setminus Z(f)\cap X\setminus
Z(g))=0$. Conversely let $\mu(Z(f)\cap Z(g))>0$ and $\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0$. Then certainly, $d(f,g)>2$. Let $h_{1}=1_{Z(f)}$ and
$h_{2}=1_{Z(g)}$. Then $h_{1},h_{2}\in\mathscr{D}$ and $Z(h_{1})=X\setminus
Z(f),Z(h_{2})=X\setminus Z(g)\implies Z(f)\cap Z(h_{1})=\emptyset=Z(h_{2})\cap
Z(g)$ and also $\mu(Z(h_{1})\cap Z(h_{2}))=0$. Therefore $f-h_{1}-h_{2}-g$ is
a path in $\Gamma^{\prime}_{2}$. Consequently, $d(f,g)=3$. ∎

######  Corollary 3.7.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a connected graph.

######  Lemma 3.8.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.

  1. (1)

If $\mu(Z(f)\triangle Z(g))=0$, then $f,g$ are not adjacent in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$.

  2. (2)

If $\mu(Z(f)\setminus A)=0$ and $\mu(Z(g)\cap A)=0$ for some
$A\in\mathcal{A}$, then $f,g$ are adjacent in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$.

  3. (3)

If $f,g$ are adjacent in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$,
then for any $f_{1}\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ with
$\mu(Z(f_{1})\triangle Z(f))=0$ and for any
$g_{1}\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ with $\mu(Z(g_{1})\triangle
Z(g))=0$, $f_{1},g_{1}$ are adjacent in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$.

###### Proof.

  1. (1)

Let $\mu(Z(f)\triangle Z(g))=0$. If $f,g$ are adjacent in
$\Gamma^{\prime}_{2}$, then $\mu(Z(f)\cap Z(g))=0$. Now $Z(f)\subset(Z(f)\cap
Z(g))\cup(Z(f)\triangle Z(g))\implies\mu(Z(f))=0$, a contradiction to
$f\in\mathscr{D}$. Therefore, $f,g$ are not adjacent in $\Gamma^{\prime}_{2}$.

  2. (2)

Let $\mu(Z(f)\setminus A)=0$ and $\mu(Z(g)\cap A)=0$ for some
$A\in\mathcal{A}$. Now $Z(f)\cap Z(g)\subset[A\cup(Z(f)\setminus
A)]\cap[(X\setminus A)\cup(Z(g)\cap A)]\implies\mu(Z(f)\cap Z(g))=0$.
Therefore, $f,g$ are adjacent in $\Gamma^{\prime}_{2}$.

  3. (3)

Let $f,g$ be adjacent in $\Gamma^{\prime}_{2}$ and $\mu(Z(f_{1})\triangle
Z(f))=0=\mu(Z(g_{1})\triangle Z(g))$ for some $f_{1},g_{1}\in\mathscr{D}$.
Since $f,g$ are adjacent, $\mu(Z(f)\cap Z(g))=0$. Now $Z(f_{1})\cap
Z(g_{1})\subset[Z(f)\cup(Z(f_{1})\triangle
Z(f))]\cap[Z(g)\cup(Z(g_{1})\triangle Z(g))]$. Therefore, $f_{1},g_{1}$ are
adjacent in $\Gamma^{\prime}_{2}$.

∎

######  Lemma 3.9.

A partition of $X$ by two atoms $A,B$ yields a partition of
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ by two sets
$\\{f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A})):\mu(Z(f)\triangle A)=0\\}$ and
$\\{f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A})):\mu(Z(f)\triangle B)=0\\}$.

###### Proof.

To prove this result, it is enough to show that for each $f\in\mathscr{D}$,
either $\mu(Z(f)\triangle A)=0$ or $\mu(Z(f)\triangle B)=0$. Let
$f\in\mathscr{D}$. Since $A$ is an atom, either $\mu(A\cap Z(f))=0$ or
$\mu(A\cap X\setminus Z(f))=0$. At first, let $\mu(A\cap Z(f))=0$. By
hypothesis, $A=X\setminus B$ and so, $\mu(X\setminus B\cap Z(f))=0$. If
$\mu(B\cap Z(f))=0$, then $\mu(Z(f))=0$ which contradicts $f\in\mathscr{D}$.
Hence, $\mu(B\cap Z(f))>0$ which implies $\mu(B\cap X\setminus Z(f))=0$, as
$B$ is also an atom. Consequently, $\mu(B\triangle Z(f))=0$. Analogously, the
hypothesis $\mu(A\cap X\setminus Z(f))=0$ would lead to the conclusion
$\mu(Z(f)\triangle A)=0$.∎

######  Theorem 3.10.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite
graph if and only if $X$ is partitioned into two atoms.

###### Proof.

If $\Gamma^{\prime}_{2}$ is a complete bipartite graph, then it has a
bipartion, say $V_{1},V_{2}$. Since both $V_{1},V_{2}$ are non-empty, we
choose and fix $f\in V_{1}$ and $g\in V_{2}$. $\Gamma^{\prime}_{2}$ being
complete bipartite, no $h\in\mathscr{D}$ is adjacent to both $f,g$ and hence
by Lemma 3.3, $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$. Also $f,g$ are
adjacent in $\Gamma^{\prime}_{2}\implies\mu(Z(f)\cap Z(g))=0$. We claim that
$Z(f)$ is an atom. To prove this, let $A\in\mathcal{A}$ with $\mu(A)>0$. If
$\mu(X\setminus A)=0$, then $\mu(X\setminus A\cap Z(f))=0$. Now let
$\mu(X\setminus A)>0$. Then $1_{A},1_{X\setminus A}\in\mathscr{D}$ and
$1_{A},1_{X\setminus A}$ are adjacent in $\Gamma^{\prime}_{2}$. If $1_{A}\in
V_{1}$, then $1_{X\setminus A}\in V_{2}\implies 1_{X\setminus A},f$ are
adjacent i.e., $\mu(A\cap Z(f))=0$. If $1_{A}\in V_{2}$, then $1_{A},f$ are
adjacent and so $\mu(X\setminus A\cap Z(f))=0$. In any case either $\mu(A\cap
Z(f))=0$ or $\mu(X\setminus A\cap Z(f))=0$. Consequently, $Z(f)$ is an atom.
Similarly we can show that $Z(g)$ is an atom. Let $A=Z(f)\setminus
Z(g)=Z(f)\setminus[Z(f)\cap Z(g)]$ and $B=Z(g)\cup X\setminus
Z(f)=Z(g)\cup[X\setminus Z(f)\cap X\setminus Z(g)]$. Then $X=A\sqcup B$ and
$A,B$ are atoms. Conversely let $A,B$ be two atoms in $X$ such that $X=A\sqcup
B$. Let $V_{1}=\\{f\in\mathscr{D}:\mu(Z(f)\triangle A)=0\\}$ and
$V_{2}=\\{f\in\mathscr{D}:\mu(Z(f)\triangle B)=0\\}$. By Lemma 3.9,
$\mathscr{D}=V_{1}\sqcup V_{2}$. Again by Lemma 3.8(1), $V_{1},V_{2}$ are
stable sets in $\Gamma^{\prime}_{2}$. Let $f\in V_{1}$ and $g\in V_{2}$. Then
$\mu(Z(f)\triangle A)=0=\mu(Z(g)\triangle B)\implies\mu(Z(f)\setminus
A)=0=\mu(Z(g)\setminus B)=\mu(Z(g)\cap A)$, as $B=X\setminus A$. By Lemma
3.8(2), $f,g$ are adjacent in $\Gamma^{\prime}_{2}$. Consequently,
$\Gamma^{\prime}_{2}$ is a complete bipartite graph. ∎

######  Corollary 3.11.

$X$ is partitioned into two atoms if and only if for each
$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, both $Z(f)$ and $X\setminus
Z(f)$ are atoms.

###### Proof.

Let $X=A\sqcup B$, where $A,B$ are two atoms. Then by Theorem 3.10,
$\mu(Z(f)\triangle A)=0$ or $\mu(Z(f)\triangle B)=0$. Without loss of
generality let $\mu(Z(f)\triangle A)=0\implies\mu(Z(f)\setminus A)=0$. Now
$Z(f)=A\setminus(A\triangle Z(f))\cup Z(f)\setminus A\implies Z(f)$ is an
atom. Let $g=1_{Z(f)}$. Then $g\in\mathscr{D}$ and $Z(g)=X\setminus Z(f)$.
Similarly we can show that $Z(g)$ i.e., $X\setminus Z(f)$ is an atom.
Conversely let $X$ can not be partitioned into two atoms and
$f\in\mathscr{D}$. Since $X=Z(f)\sqcup X\setminus Z(f)$, either $Z(f)$ or
$X\setminus Z(f)$ is not an atom. ∎

Replacing $Z(f)$ by $X\setminus Z(f)$ in the proof of Theorem 3.10 and making
some suitable modification we can prove that
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph if and only
if $X$ is partitioned into two atoms. This observation is recorded in the
following theorem.

######  Theorem 3.12.

The following statements are equivalent:

  1. (1)

$\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph.

  2. (2)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite
graph.

  3. (3)

$X$ is partitioned into two atoms.

In the next few theorems we find the eccentricity, diameter and the lengths of
possible cycles in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
determine how far the measure $\mu$ is responsible to make
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ triangulated,
hypertriangulated and complemented.

######  Theorem 3.13.

Let $f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then

$ecc(f)=\begin{cases}2&\text{ if }Z(f)\text{ is an atom}\\\ 3&\text{
otherwise}\end{cases}$

###### Proof.

Let $Z(f)$ be an atom and $g\in\mathscr{D}$. Then either $\mu(Z(f)\cap
Z(g))=0$ or $\mu(Z(f)\cap X\setminus Z(g))=0$. If $\mu(Z(f)\cap Z(g))=0$, then
$f,g$ are adjacent. If $\mu(Z(f)\cap X\setminus Z(g))=0$, then $\mu(X\setminus
Z(f)\cap X\setminus Z(g))>0$, for otherwise $\mu(X\setminus Z(g))=0$. By
Theorem 3.6, $d(f,g)=2$. Thus $ecc(f)=2$. Conversely, let $Z(f)$ be not an
atom i.e., there exist $A,B\in\mathcal{A}$ with $\mu(A),\mu(B)>0$ such that
$Z(f)=A\sqcup B$. Clearly, $1_{A}\in\mathscr{D}$, $Z(f)\cap Z(1_{A})=B$ and
$X\setminus Z(f)\cap X\setminus Z(1_{A})=\emptyset$ i.e., $\mu(Z(f)\cap
Z(1_{A}))>0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(1_{A})=0$. By Theorem
3.6, $d(f,1_{A})=3\implies ecc(f)=3$. ∎

######  Corollary 3.14.

The diameter of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is

$diam(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))=\begin{cases}2&\text{
if }X\text{ is partitioned into two atoms}\\\ 3&\text{ otherwise}\end{cases}$

###### Proof.

If $X$ is partitioned into two atoms, then by Theorem 3.10,
$\Gamma^{\prime}_{2}$ is a complete bipartite graph and hence
$diam(\Gamma^{\prime}_{2})=2$. If $X$ can not be partitioned into two atoms,
then for any $f\in\mathscr{D}$, either $Z(f)$ or $X\setminus Z(f)$ is not an
atom. By Theorem 3.13, $ecc(f)=3$, if $Z(f)$ is not an atom and
$ecc(1_{Z(f)})=3$, if $X\setminus Z(f)$ is not an atom. In any case, the
diameter of $\Gamma^{\prime}_{2}$ is $3$. ∎

######  Theorem 3.15.

$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ is a vertex of a triangle in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ if and only if $X\setminus
Z(f)$ is not an atom.

###### Proof.

If $X\setminus Z(f)$ is not an atom, then there exist $A,B\in\mathcal{A}$ with
$\mu(A),\mu(B)>0$ and $X\setminus Z(f)=A\sqcup B$. Let $g=1_{X\setminus A}$
and $h=1_{X\setminus B}$. Then $g,h\in\mathscr{D}$ and $Z(g)=A,Z(h)=B\implies
Z(f)\cap Z(g)=\emptyset=Z(g)\cap Z(h)=Z(h)\cap Z(f)$. Thus $f-g-h-f$ is a
triangle in $\Gamma^{\prime}_{2}$. Conversely if $f$ is a vertex of a triangle
in $\Gamma^{\prime}_{2}$, then there exists $g,h\in\mathscr{D}$ such that
$f-g-h-f$ is a triangle in $\Gamma^{\prime}_{2}$. Since $f,g$ are adjacent,
$\mu(Z(f)\cap Z(g))=0\implies\mu(X\setminus Z(f)\cap Z(g))>0$, for otherwise
$\mu(Z(g))=0$. Similarly the adjacency of $f,h$ implies $\mu(X\setminus
Z(f)\cap Z(h))>0$. Let $A=X\setminus Z(f)\cap Z(g)$ and $B=X\setminus Z(f)\cap
X\setminus Z(g)$. Then $\mu(A)>0$ and $X\setminus Z(f)=A\sqcup B$. Since $g,h$
are adjacent, $\mu(Z(g)\cap Z(h))=0\implies\mu(A\cap Z(h))=0$. Now $X\setminus
Z(f)\cap Z(h)=(A\cap Z(h))\sqcup(B\cap Z(h))\implies\mu(B\cap
Z(h))=\mu(X\setminus Z(f)\cap Z(h))>0\implies\mu(B)>0$. Thus $X\setminus
Z(f)=A\sqcup B$ and $\mu(A),\mu(B)>0$, which prove that $X\setminus Z(f)$ is
not an atom. ∎

######  Corollary 3.16.

The girth of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is

$gr(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))=\begin{cases}4&\text{ if
}X\text{ is partitioned into two atoms}\\\ 3&\text{ otherwise}\end{cases}$

.

###### Proof.

If $X$ is partitioned into two atoms, then by Theorem 3.10,
$\Gamma^{\prime}_{2}$ is a complete bipartite graph and hence
$gr(\Gamma^{\prime}_{2})=4$. If $X$ can not be partitioned into two atoms,
then for each $f\in\mathscr{D}$, either $Z(f)$ or $X\setminus Z(f)$ is not an
atom. Consequently, by Theorem 3.15, $1_{Z(f)}$ or $f$ is a vertex of a
triangle in $\Gamma^{\prime}_{2}$. So, $gr(\Gamma^{\prime}_{2})=3$. ∎

######  Theorem 3.17.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is triangulated if and only
if $\mu$ is non-atomic.

###### Proof.

If $\mu$ is non-atomic, then $X\setminus Z(f)$ is not an atom for all
$f\in\mathscr{D}$. By Theorem 3.15, every $f\in\mathscr{D}$ is a vertex of a
triangle. In other words, $\Gamma^{\prime}_{2}$ is triangulated. Conversely
let $X$ contain an atom, say $A$. Then $\mu(X\setminus A)>0$, for otherwise,
$X$ will be an atom. Thus $1_{A}\in\mathscr{D}$ and $X\setminus Z(1_{A})=A$ is
an atom. Hence by Theorem 3.15, $1_{A}$ is not a vertex of a triangle; i.e.,
$\Gamma^{\prime}_{2}$ is not triangulated. ∎

######  Theorem 3.18.

Given $f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, there exists an edge
containing $f$ which is not an edge of any triangle in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$.

###### Proof.

For each $f\in\mathscr{D}$, consider $g=1_{Z(f)}\in\mathscr{D}$. Then $f,g$
are adjacent and $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$. By Theorem 3.3,
there does not exist any vertex in $\mathscr{D}$ adjacent to both $f$ and $g$.
Therefore, the edge $f-g$ is not an edge of any triangle in
$\Gamma^{\prime}_{2}$. ∎

######  Corollary 3.19.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is not hypertriangulated.

If $X$ is partitioned into two atoms, then by Theorem 3.10,
$\Gamma^{\prime}_{2}$ is a complete bipartite graph and hence $c(f,g)=4$ for
all $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. We now calculate $c(f,g)$
for $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, when $X$ is not
partitioned into two atoms.

######  Theorem 3.20.

Suppose $X$ can not be partitioned into two atoms and
$f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then

$c(f,g)=\begin{cases}3&\text{ if }\mu(Z(f)\cap Z(g))=0\text{ and
}\mu(X\setminus Z(f)\cap X\setminus Z(g))>0\\\ 4&\text{ if }\mu(Z(f)\cap
Z(g))=0\text{ and }\mu(X\setminus Z(f)\cap X\setminus Z(g))=0\\\ &\text{ or if
}\mu(Z(f)\cap Z(g))>0\text{ and }\mu(X\setminus Z(f)\cap X\setminus Z(g))>0\\\
6&\text{ if }\mu(Z(f)\cap Z(g))>0\text{ and }\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0\end{cases}$

###### Proof.

It is clear that $c(f,g)=3$ if and only if $f-g$ is an edge of a triangle. So,
$c(f,g)=3$ if and only if $f,g$ are adjacent in $\Gamma^{\prime}_{2}$ and
there exists a vertex in $\Gamma^{\prime}_{2}$ adjacent to both $f$ and $g$;
i.e., by Theorem 3.2 and Theorem 3.3, $\mu(Z(f)\cap Z(g))=0$ and
$\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$.
Let $c(f,g)=4$. If $f,g$ are adjacent in $\Gamma^{\prime}_{2}$, then there
exists no vertex adjacent to both $f$ and $g$, for otherwise $c(f,g)=3$.
Therefore, $\mu(Z(f)\cap Z(g))=0$ and $\mu(X\setminus Z(f)\cap X\setminus
Z(g))=0$. If $f,g$ are not adjacent, then there exists vertex adjacent to both
$f,g$ in $\Gamma^{\prime}_{2}$, because $c(f,g)=4$. Therefore, $\mu(Z(f)\cap
Z(g))>0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$. Conversely let
$\mu(Z(f)\cap Z(g))=0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$. Then
$f,g$ are adjacent and there does not exist any vertex adjacent to both
$f,g\implies c(f,g)>3$. Clearly $f-g-2.f-2.g-f$ is a $4$-cycle in
$\Gamma^{\prime}_{2}$ which implies $c(f,g)=4$. Again let $\mu(Z(f)\cap
Z(g))>0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$. Then $f,g$ are non-
adjacent and there exists $h\in\mathscr{D}$ adjacent to both $f,g$ in
$\Gamma^{\prime}_{2}$. Clearly, $c(f,g)>3$ and $f-h-g-2.h-f$ is a $4$-cycle in
$\Gamma^{\prime}_{2}\implies c(f,g)=4$.
If $c(f,g)=6$, then it follows from the previous two cases that $\mu(Z(f)\cap
Z(g))>0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$. Conversely let
$\mu(Z(f)\cap Z(g))>0$ and $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$. So,
$f,g$ are not adjacent and there does not exist any vertex adjacent to both
$f,g$. Hence there does not exist any $5$-cycle in $\Gamma^{\prime}_{2}$
containing $f,g$ as vertices. In other words, $c(f,g)>5$. By Theorem 3.6,
$d(f,g)=3$; i.e., there exist $h_{1},h_{2}\in\mathscr{D}$ such that
$f-h_{1}-h_{2}-g$ is a path in $\Gamma^{\prime}_{2}$. Consequently,
$f-h_{1}-h_{2}-g-2.h_{2}-2.h_{1}-f$ is a $6$-cycle in
$\Gamma^{\prime}_{2}\implies c(f,g)=6$. ∎

The visual representation of the above Theorem is exhibited as follows:

$f$$g$$f$$g$$\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$$\text{ and }f,g\text{
are adjacent}$$\mu(X\setminus Z(f)\cap X\setminus Z(g))=0$$\text{ and
}f,g\text{ are adjacent}$$f$$g$$f$$g$$\mu(X\setminus Z(f)\cap X\setminus
Z(g))>0$$\text{ and }f,g\text{ are non-adjacent}$$\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0$$\text{ and }f,g\text{ are non-adjacent}$

From the definition of orthogonality and from Theorem 3.3, it naturally
follows that for $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, $f\perp g$
in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ if and only if
$\mu(Z(f)\cap Z(g))=0=\mu(X\setminus Z(f)\cap X\setminus Z(g))$. As a result,
for each $f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, $f\perp 1_{Z(f)}$,
where $1_{Z(f)}\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. This observation
leads to the following fact.

######  Theorem 3.21.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a complemented graph.

######  Lemma 3.22.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then $f,g$ are adjacent
to the same set of vertices in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ if and only if
$\mu(Z(f)\triangle Z(g))=0$.

###### Proof.

Let $\mu(Z(f)\triangle Z(g))=0$. Then $\mu(Z(f)\cap X\setminus
Z(g))=0=\mu(X\setminus Z(f)\cap Z(g))$. Let $h\in\mathscr{D}$ be adjacent to
$f$, i.e., $\mu(Z(f)\cap Z(h))=0$. Now $Z(g)=(Z(g)\cap Z(f))\sqcup(Z(g)\cap
X\setminus Z(f))\implies Z(g)\cap Z(h)\subset[Z(g)\cap X\setminus
Z(f)]\cup[Z(f)\cap Z(h)]\implies\mu(Z(g)\cap Z(h))=0$; i.e., $g,h$ are
adjacent. Similarly, if $h\in\mathscr{D}$ is adjacent to $g$, then it can also
be adjacent to $f$. Therefore, $f,g$ are adjacent to the same set of vertices
in $\Gamma^{\prime}_{2}$. Conversely let $\mu(Z(f)\triangle Z(g))>0$. Then
either $\mu(Z(f)\cap X\setminus Z(g))>0$ or $\mu(X\setminus Z(f)\cap Z(g))>0$.
Without loss of generality let $\mu(Z(f)\cap X\setminus Z(g))>0$. Choose
$h=1_{X\setminus Z(f)\cup Z(g)}\in\mathscr{D}$. Then $Z(h)=Z(f)\cap X\setminus
Z(g)\implies Z(h)\cap Z(g)=\emptyset$ and $Z(h)\subset
Z(f)\implies\mu(Z(h)\cap Z(g))=0$ and $\mu(Z(h)\cap Z(f))=\mu(Z(h))>0$. Hence
$h$ is adjacent to $g$ but $h$ is not adjacent to $f$. ∎

######  Theorem 3.23.

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a uniquely complemented
graph.

###### Proof.

Let $f,g,h\in\mathscr{D}$ be such that $f\perp g$ and $f\perp h$ in
$\Gamma^{\prime}_{2}$. We claim that $\mu(Z(g)\triangle Z(h))=0$. If possible
let $\mu(Z(g)\cap X\setminus Z(h))>0$, then $k=1_{X\setminus Z(g)\cup
Z(h)}\in\mathscr{D}$. Now $Z(k)=Z(g)\cap X\setminus Z(h)$ and so $Z(k)\cap
Z(h)=\emptyset\implies\mu(Z(k)\cap Z(h))=0$ and also $Z(k)\cap Z(f)\subset
Z(g)\cap Z(f)\implies\mu(Z(k)\cap Z(f))=0$, as $f\perp g$. Therefore $k$ is
adjacent to both $f,h$, which contradicts that $f\perp h$. Therefore,
$\mu(Z(g)\cap X\setminus Z(h))=0$ and similarly $\mu(X\setminus Z(g)\cap
Z(h))=0$. Hence $\mu(Z(g)\triangle Z(h))=0$. By Lemma 3.22, $g,h$ are adjacent
to the same set of vertices in $\Gamma^{\prime}_{2}$. Consequently,
$\Gamma^{\prime}_{2}$ is uniquely complemented. ∎

We now introduce special subgraphs of $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ which enable us to find
conditions under which $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ become isomorphic. On
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, we define a relation “ $f\sim g$ if
and only if $\mu(Z(f)\triangle Z(g))=0$”. It is easy to see that $\sim$ is an
equivalence relation on $\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ and hence it
partitions $\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ into disjoint equivalence
classes, denoted by$[f]$, for each
$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Certainly, for each
$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, $f\sim 1_{X\setminus Z(f)}$. We
choose a subset $V$ of $\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ obeying the
following conditions:

  1. (1)

every element of $V$ is of the form $1_{A}$ where $\mu(A)>0$ and
$\mu(X\setminus A)>0$.

  2. (2)

$1_{A},1_{B}$ are distinct elements in $V$ if and only if $\mu(A\triangle
B)>0$.

Then $V$ is a collection of distinct class representatives corresponding to
the equivalence relation $\sim$ on $\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.
Let $G_{2}$ be the induced subgraph of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ whose set of vertices is
$V$. We first observe the following:

######  Observation 3.24.

  1. (1)

$[1_{A}]$ is a stable set in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, for each $1_{A}\in V$.
(Follows from Lemma 3.8(1)).

  2. (2)

For any two distinct stable sets $[1_{A}],[1_{B}]$ in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, either
$[1_{A}]\sqcup[1_{B}]$ is a stable set or $[1_{A}]\sqcup[1_{B}]$ forms a
complete bipartite subgraph of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ (Follows from Lemma 3.8(2)).

Thus $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a $|V|$-partite
graph such that for any two stable sets $V_{1},V_{2}$, either $V_{1}\sqcup
V_{2}$ is a stable set or $V_{1}\sqcup V_{2}$ is a complete bipartite subgraph
of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$. Therefore, $G_{2}$ is a
subgraph of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ where two
vertices $1_{A}$, $1_{B}$ are adjacent if and only if the corresponding two
stable sets $[1_{A}]$ and $[1_{B}]$ make a complete bipartite subgraph. We jot
down in the next theorem some of the features highlighting the behaviour of
$G_{2}$ as a graph and observe the analogy with its parent graph:

######  Theorem 3.25.

  1. (1)

Let $1_{A},1_{B}\in V$. There is a vertex in $V$ adjacent to both
$1_{A},1_{B}$ in $G_{2}$ if and only if $\mu(A\cap B)>0$.

  2. (2)

The distance between two vertices $1_{A},1_{B}$ in $G_{2}$ is given by

$d_{G_{2}}(1_{A},1_{B})=\begin{cases}1&\text{ if }\mu(X\setminus A\cap
X\setminus B)=0\\\ 2&\text{ if }\mu(X\setminus A\cap X\setminus B)>0\text{ and
}\mu(A\cap B)>0\\\ 3&\text{ if }\mu(X\setminus A\cap X\setminus B)>0\text{ and
}\mu(A\cap B)=0\end{cases}$

  3. (3)

$G_{2}$ is a connected graph.

  4. (4)

$|V|=2$ i.e., $G_{2}=K_{2}$ if and only if $X$ can be partitioned into two
atoms.

  5. (5)

The diameter of $G_{2}$ is $diam(G_{2})=\begin{cases}2&\text{ if }|V|=2\\\
3&\text{ otherwise}\end{cases}$

  6. (6)

The eccentricity of $1_{A}\in V$ in $G_{2}$ is
$ecc_{G_{2}}(1_{A})=\begin{cases}2&\text{ if }X\setminus A\text{ is an
atom}\\\ 3&\text{ otherwise}\end{cases}$

  7. (7)

A vertex $1_{A}\in V$ is a vertex of a triangle in $G_{2}$ if and only $A$ is
not an atom.

  8. (8)

The girth of $G_{2}$ is $=\begin{cases}\infty&\text{ if }|V|=2\\\ 3&\text{
otherwise}\end{cases}$

  9. (9)

$G_{2}$ is triangulated if and only if $\mu$ is a non-atomic measure.

  10. (10)

$G_{2}$ is not hypertriangulated.

  11. (11)

$1_{A},1_{B}\in V$ are orthogonal in $G_{2}$ if and only if $\mu(A\cap
B)=0=\mu(X\setminus A\cap X\setminus B)$.

  12. (12)

For every $1_{A}\in V$ there exists unique $1_{B}\in V$ such that
$1_{A}\perp_{G_{2}}1_{B}$ [uniqueness follows from Lemma 3.22].

  13. (13)

$G_{2}$ is uniquely complemented.

To compare the clique number of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ with that of its subgraph
$G_{2}$, we need the following lemma.

######  Lemma 3.26.

If $M$ is a complete subgraph of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, then there exists a
complete subgraph $M^{\prime}$ of $G_{2}$ such that $|M|=|M^{\prime}|$.

###### Proof.

Let $M$ be a complete subgraph of $\Gamma^{\prime}_{2}$. For each vertex $f$
in $M$, there is a vertex $1_{A}\in V$ such that $f\sim 1_{A}$. Let $f,g$ be
distinct vertices in $M$ and $1_{A},1_{B}\in V$ are such that $1_{A}\sim f$
and $1_{B}\sim g$. Since $M$ is a complete subgraph of $\Gamma^{\prime}_{2}$,
$f,g$ are adjacent in $\Gamma^{\prime}_{2}$. By Lemma 3.8(3), $1_{A},1_{B}$
are adjacent. Consequently, $1_{A},1_{B}$ are distinct vertices in $G_{2}$ and
they are adjacent in $G_{2}$. Let $M^{\prime}$ be the subgraph of $G_{2}$
whose vertex set is $\\{1_{A}\in V:1_{A}\sim f\text{ for some vertex }f\text{
in }M\\}$. Clearly, $M^{\prime}$ is a complete subgraph of $G_{2}$ and
$|M|=|M^{\prime}|$. ∎

Since $G_{2}$ is a subgraph of $\Gamma^{\prime}_{2}$, $cl(G_{2})\leq
cl(\Gamma^{\prime}_{2})$. So, using Lemma 3.26, we get the following:

######  Theorem 3.27.

The clique number of $G_{2}$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are the same.

As $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a $|V|$-partite graph,
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is $|V|$-colorable and
consequently
$\chi(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))$$\leq|V|=|G_{2}|$. We
conclude that the chromatic number of both
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and $G_{2}$ are equal from
the following lemma.

######  Lemma 3.28.

Let $\alpha$ be a cardinal number. If $G_{2}$ is $\alpha$-colorable, then we
can color $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ by using
$\alpha$-many colors.

###### Proof.

For each $f\in\mathscr{D}$, there exists $1_{A}\in V$ such that $f\sim 1_{A}$.
Since $G_{2}$ is $\alpha$-colorable, $1_{A}$ already gets a color. We color
each $f\in[1_{A}]$ by the color of $1_{A}\in V$. As a consequence, each
$f\in\mathscr{D}$ gets a color from the available $\alpha$ many colors . It
only remains to show that this coloring on $\Gamma^{\prime}_{2}$ is
consistent. Let $f,g\in\mathscr{D}$ be colored by the same color, say the
color of $1_{A}$. By our method of coloring of $\Gamma^{\prime}_{2}$, $f\sim
1_{A}$ and $g\sim 1_{A}$. Since $[1_{A}]$ is a stable set in
$\Gamma^{\prime}_{2}$, $f,g$ are non-adjacent in $\Gamma^{\prime}_{2}$.
Consequently, $\Gamma^{\prime}_{2}$ is $\alpha$-colorable. ∎

Since $G_{2}$ is a subgraph of $\Gamma^{\prime}_{2}$,
$\chi(G_{2})\leq\chi(\Gamma^{\prime}_{2})$. From Lemma 3.28, we get the
following result.

######  Theorem 3.29.

The chromatic number of $G_{2}$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are equal.

######  Theorem 3.30.

$dt(G_{2})\leq dt(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))$.

###### Proof.

Let $D\subset\mathscr{D}$ be a dominating set in $\Gamma^{\prime}_{2}$. Let
$V^{\prime}=\\{1_{A}\in V:1_{A}\sim f\text{ for some }f\in D\\}$. Clearly,
$|V^{\prime}|\leq|D|$ and $D\cap V\subset V^{\prime}$. To show that
$V^{\prime}$ is a dominating set in $G_{2}$, let $1_{B}\in V\setminus
V^{\prime}$. If $1_{B}\in D$, then $1_{B}\in V^{\prime}$, which is not. So,
$1_{B}\in\mathscr{D}\setminus D$. Since $D$ is a dominating set in
$\Gamma^{\prime}_{2}$, there exists $f\in D$ such that $f,1_{B}$ are adjacent
in $\Gamma^{\prime}_{2}$. Let $1_{A}\in V^{\prime}$ such that $1_{A}\sim f$.
By Observation 3.24, $1_{A},1_{B}$ are adjacent in $\Gamma^{\prime}_{2}$ and
hence they are adjacent in $G_{2}$. Therefore, $V^{\prime}$ is a dominating
set in $G_{2}$. Now $dt(G_{2})\leq|V^{\prime}|\leq|D|$ and this holds for
every dominating set $D$ in $\Gamma^{\prime}_{2}$. Consequently,
$dt(G_{2})\leq dt(\Gamma^{\prime}_{2})$. ∎

######  Lemma 3.31.

Let $D\subset\mathscr{D}$.

  1. (1)

Every total dominating set in $G_{2}$ is also a total dominating set in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$.

  2. (2)

If $D$ is a total dominating set in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, then there exists a total
dominating set $D^{\prime}$ in $G_{2}$ such that $|D|\geq|D^{\prime}|$.

###### Proof.

  1. (1)

Let $V^{\prime}$ be a total dominating set in $G_{2}$ and $f\in\mathscr{D}$.
Let $1_{A}\in V$ such that $f\sim 1_{A}$. Since $V^{\prime}$ is a total
dominating set in $G_{2}$, there exists $1_{B}\in V^{\prime}$ such that
$1_{A},1_{B}$ are adjacent in $G_{2}$. By Lemma 3.8(3), $f,1_{B}$ are adjacent
in $\Gamma^{\prime}_{2}$. Thus $V^{\prime}$ be a total dominating set in
$\Gamma^{\prime}_{2}$.

  2. (2)

Let $D$ be a total dominating set in $\Gamma^{\prime}_{2}$ and
$D^{\prime}=\\{1_{A}\in V:1_{A}\sim f\text{ for some }f\in D\\}$. Clearly,
$|D^{\prime}|\leq|D|$ and $D\cap V\subset D^{\prime}$. Let $1_{A}\in V$. Then
$1_{A}\in\mathscr{D}$ and hence there exists $f\in D$ such that $f,1_{A}$ are
adjacent in $\Gamma^{\prime}_{2}$. Consequently, $1_{A}$ is adjacent to
$1_{B}$ in $G_{2}$ where $f\sim 1_{B}\in D^{\prime}$. Hence, $D^{\prime}$ is a
a total dominating set in $G_{2}$.

∎

From Lemma 3.31 the following result is immediate.

######  Theorem 3.32.

The total dominating number of $G_{2}$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are equal.

For any two subsets $A,B$ of $X$, as $A\triangle B=(X\setminus
A)\triangle(X\setminus B)$ holds, we get a similar induced subgraph $G$ of the
zero-divisor graph $\Gamma(\mathcal{M}(X,\mathcal{A}))$, arising from the same
equivalence relation $\sim$ on $\mathscr{D}$ and considering the same set of
vertices $V$ as in the case of comaximal graph. In this case too, each
equivalence class $[1_{A}]$ is a stable set in
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ and for any two distinct equivalence
classes $[1_{A}]$ and $[1_{B}]$, either $[1_{A}]\sqcup[1_{B}]$ is a stable set
in $\Gamma(\mathcal{M}(X,\mathcal{A}))$ or they make a complete bipartite
subgraph of $\Gamma(\mathcal{M}(X,\mathcal{A}))$. Hence,
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ is also a $|V|$-partite graph. This
inherent similarity between the induced subgraphs $G$ and $G_{2}$ leads to a
more stronger conclusion, as seen in the next proposition.

######  Theorem 3.33.

$G$ and $G_{2}$ are graph isomorphic.

###### Proof.

Let us define a function $\psi:V\to V$ as follows $\psi(1_{A})=1_{\psi(A)}$
where $1_{\psi(A)}\sim 1_{X\setminus A}$. Since for $1_{A}\in
V\implies\mu(A),\mu(X\setminus A)>0$, there exists $1_{B}\in V$ such that
$1_{B}\sim 1_{X\setminus A}$. Also if $1_{A}$ and $1_{B}$ are two distinct
elements in $V$, then $\mu(A\triangle B)>0\implies\mu(X\setminus A\triangle
X\setminus B)>0$. As $1_{\psi(A)}\sim 1_{X\setminus A}$ and $1_{\psi(B)}\sim
1_{X\setminus B}$, $\mu(\psi(A)\triangle\psi(B))>0$ i.e., $1_{\psi(A)}$ and
$1_{\psi(B)}$ are distinct in $V$. Consequently, $\psi$ is well-defined and
injective. It is also clear that, for each $1_{B}\in V$, $\psi(1_{A})=1_{B}$,
where $1_{A}\sim 1_{X\setminus B}$. Thus $\psi$ is surjective and hence $\psi$
is a bijection on $V$. We now claim that, $\psi$ preserves the adjacency
relation between $G$ and $G_{2}$. Let $1_{A},1_{B}\in V$ be adjacent in $G$;
i.e., $\mu(A\cap B)=0$. It suffices to show that $1_{\psi(A)},1_{\psi(B)}$ are
adjacent in $G_{2}$. We have $1_{\psi(A)}\sim 1_{X\setminus A}$ and
$1_{\psi(B)}\sim 1_{X\setminus B}$. Since $\mu(A\cap B)=0$, $1_{X\setminus
A},1_{X\setminus B}$ are adjacent in $G_{2}$. By Lemma 3.8(3),
$\psi(1_{A}),\psi(1_{B})$ are adjacent in $G_{2}$. Similarly if $1_{A},1_{B}$
are adjacent in $G_{2}$, then $\psi^{-1}(1_{A}),\psi^{-1}(1_{B})$ are adjacent
in $G$. Hence $\psi$ is an isomorphism between the two graphs $G$ and $G_{2}$.
∎

Theorem 3.33 prompted us to suspect that $\Gamma(\mathcal{M}(X,\mathcal{A}))$
and $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are isomorphic. But, in
the last section of this paper we could find an example of a measure space on
which these two graphs are not isomorphic. However, we establish a sufficient
condition for these two graphs to be isomorphic in general, as given below.
If $|[1_{A}]|=|[1_{X\setminus A}]|$ for each $1_{A}\in V$, then there exists a
bijection between $[1_{A}]$ and $[1_{\psi(A)}]$, because $[1_{X\setminus
A}]=[1_{\psi(A)}]$. For each $1_{A}\in V$, let
$\phi_{A}:[1_{A}]\to[1_{\psi(A)}]$ be a bijection. Define
$\phi:\mathscr{D}(\mathcal{M}(X,\mathcal{A}))\to\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$
by $\phi(f)=\phi_{A}(f)$ whenever $f\sim 1_{A}$. Since each $\phi_{A}$ is a
bijection and $V$ is the collection of the class representatives under the
equivalence relation $\sim$, $\phi$ is a bijective map. Let
$f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ be adjacent in
$\Gamma(\mathcal{M}(X,\mathcal{A}))$. Let $1_{A},1_{B}\in V$ be such that
$f\sim 1_{A}$ and $g\sim 1_{B}$. Since $f,g$ are adjacent, it can be proved
easily that $1_{A},1_{B}$ are adjacent in $G$. Therefore
$\psi(1_{A}),\psi(1_{B})$ are adjacent in $G_{2}$. By the definition of
$\phi$, $\phi(f)\sim\psi(1_{A})$ and $\phi(g)\sim\psi(1_{B})$. Therefore
$\phi(f),\phi(g)$ are adjacent in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$. Similarly if $f,g$ are
adjacent in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$, then
$\phi^{-1}(f),\phi^{-1}(g)$ are adjacent in
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ (follows from Lemma 3.8(3)). Hence,
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are graph isomorphic.
We record this fact in the form of a theorem.

######  Theorem 3.34.

If $|[1_{A}]|=|[1_{X\setminus A}]|$ for each $1_{A}\in V$, then
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are graph isomorphic.

##  4\. The annihilator graph $AG(\mathcal{M}(X,\mathcal{A}))$ of
$\mathcal{M}(X,\mathcal{A})$

We redefine the annihilator graph of the ring $\mathcal{M}(X,\mathcal{A})$ as
follows:

######  Definition 4.1.

The annihilator graph $AG(\mathcal{M}(X,\mathcal{A}))$ (in short $AG$) of
$\mathcal{M}(X,\mathcal{A})$ is a simple graph whose set of vertices is
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ and the adjacency relation is given
by the following rule : $f,g$ are adjacent if and only if $ann(f)\cup
ann(g)\subsetneqq ann(f.g)$.

It is quite clear that for any two $f,g\in\mathcal{M}(X,\mathcal{A})$,
$ann(f)\cup ann(g)\subseteq ann(f.g)$. So, two vertices $f,g$ are non-adjacent
when and only when $ann(f)\cup ann(g)=ann(f.g)$. We first observe that the
adjacency of two vertices of $AG(\mathcal{M}(X,\mathcal{A}))$ can be
determined by measuring the differences between their corresponding zero sets.

######  Theorem 4.2.

$f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ are adjacent in
$AG(\mathcal{M}(X,\mathcal{A}))$ if and only if $\mu(Z(f)\setminus Z(g))>0$
and $\mu(Z(g)\setminus Z(f))>0$.

###### Proof.

Let $f,g$ be not adjacent in $AG$. Then $ann(f)\cup ann(g)=ann(f.g)$. Since
union of two ideals is an ideal, either $ann(f)\subset ann(g)$ or
$ann(g)\subset ann(f)$. By Theorem 2.5, either $\mu(Z(f)\setminus Z(g))=0$ or
$\mu(Z(g)\setminus Z(f))=0$. Conversely, for definiteness sake, let
$\mu(Z(f)\setminus Z(g))=0$. Then by Theorem 2.5, $ann(f)\subset ann(g)$. Let
$h\in ann(f.g)$. Then $\mu(X\setminus Z(h)\cap X\setminus Z(f)\cap X\setminus
Z(g))=0$. Now, $X\setminus Z(h)\cap X\setminus Z(g)=(X\setminus Z(h)\cap
X\setminus Z(g)\cap Z(f))\cup(X\setminus Z(h)\cap X\setminus
Z(g)\cap(X\setminus Z(f)))\subset(Z(f)\setminus Z(g))\cup(X\setminus Z(h)\cap
X\setminus Z(g)\cap X\setminus Z(f))$ implies $\mu(X\setminus Z(h)\cap
X\setminus Z(g))=0$. Hence, $h\in ann(g)$. i.e., $ann(g)=ann(f)\cup
ann(g)=ann(f.g)$ which proves $f,g$ are non-adjacent in AG. ∎

The next theorem sets conditions for the existence of a third vertex, adjacent
to a given pair of vertices in $AG(\mathcal{M}(X,\mathcal{A}))$.

######  Theorem 4.3.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.

  1. (1)

If $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$ or $\mu(Z(f)\cap Z(g))>0$,
then there is a vertex adjacent to both $f$ and $g$ in
$AG(\mathcal{M}(X,\mathcal{A}))$.

  2. (2)

Let $\mu(X\setminus Z(f)\cap X\setminus Z(g))=0=\mu(Z(f)\cap Z(g))$. Then
there exists a vertex adjacent to both $f$ and $g$ if and only if $Z(f)$ and
$Z(g)$ are not atoms.

###### Proof.

It is easy to check that $\mu(X\setminus Z(f)\cap X\setminus Z(g))>0$ or
$\mu(Z(f)\cap Z(g))>0$ imply that $1_{Z(f)\cup Z(g)}$ or $1_{Z(f)\cap Z(g)}$
is adjacent to both $f$, $g$ in $AG$ respectively. Let $\mu(X\setminus
Z(f)\cap X\setminus Z(g))=0=\mu(Z(f)\cap Z(g))$. Without loss of generality
let, $Z(f)=X\setminus Z(g)$. If both $Z(f)$ and $Z(g)$ are not atoms, then
there exist $A_{1},A_{2},B_{1},B_{2}\in\mathcal{A}$ each with positive measure
such that $Z(f)=A_{1}\sqcup A_{2}$ and $Z(g)=B_{1}\sqcup B_{2}$. Consider
$h=1_{A_{1}\cup B_{1}}$. Clearly $h\in\mathscr{D}$ and $A_{2}\cup B_{2}\subset
Z(h)$. Therefore, $\mu(Z(f)\setminus Z(h))>0$ and $\mu(Z(h)\setminus Z(f))>0$
i.e., $f,h$ are adjacent in $AG$. Similarly $g,h$ are adjacent in $AG$.
Conversely let $h$ be adjacent to both $f,g$ in $AG$. Then $\mu(Z(f)\setminus
Z(h))>0$ and $\mu(Z(h)\setminus Z(g))>0$. Now $Z(f)=(Z(f)\cap
Z(h))\sqcup(Z(f)\cap X\setminus Z(h))=(Z(h)\setminus Z(g))\sqcup(Z(f)\setminus
Z(h))$, as $Z(f)=X\setminus Z(g)$. Therefore, $Z(f)$ is not an atom and
similarly, $Z(g)$ is also not an atom. ∎

######  Corollary 4.4.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then

$d(f,g)=\begin{cases}1&\text{ if }f,g\text{ are adjacent}\\\ 2&\text{
otherwise}\end{cases}$

######  Corollary 4.5.

$ecc(f)=2$ for all $f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.

######  Corollary 4.6.

The diameter of $AG(\mathcal{M}(X,\mathcal{A}))$ is $2$.

We have already seen that the vertex sets of the comaximal graph and the zero-
divisor graph of $\mathcal{M}(X,\mathcal{A})$ are also
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$; i.e., the same as the vertex set of
AG. If $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ are adjacent in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ then $\mu(Z(f)\cap Z(g))=0$.
So, $(Z(f)\setminus Z(g))\sqcup(Z(f)\cap Z(g))=Z(f)\implies\mu(Z(f)\setminus
Z(g))=\mu(Z(f))>0$. Similarly, $\mu(Z(g)\setminus Z(f))>0$. This indicates
that $f,g$ are adjacent in $AG(\mathcal{M}(X,\mathcal{A}))$. By an analogous
argument we also get that the vertices which are adjacent in
$\Gamma(\mathcal{M}(X,\mathcal{A}))$, are adjacent in
$AG(\mathcal{M}(X,\mathcal{A}))$. We record these observations in the form of
a theorem :

######  Theorem 4.7.

Both $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are subgraphs of
$AG(\mathcal{M}(X,\mathcal{A}))$.

######  Theorem 4.8.

$AG(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph if and only if
$X$ is partitioned into two atoms.

###### Proof.

Let $AG$ be complete bipartite by bipartion $V_{1},V_{2}$. Since
$V_{1}\neq\emptyset$ and $V_{2}\neq\emptyset$, fix $f\in V_{1}$ and $g\in
V_{2}$. Since $AG$ is a complete bipartite graph, there does not exit any
vertex adjacent to both $f,g$ and hence by Theorem 4.3, $\mu(X\setminus
Z(f)\cap X\setminus Z(g))=0=\mu(Z(f)\cap Z(g))$ and either $Z(f)$ or $Z(g)$ is
an atom. Suppose $Z(f)$ is an atom. We claim that $Z(g)$ is also an atom. If
possible let $Z(g)=A_{1}\sqcup A_{2}$ where $A_{1},A_{2}\in\mathcal{A}$ with
$\mu(A_{1}),\mu(A_{2})>0$. Consider $h=1_{X\setminus(Z(f)\cup A_{1})}$. Then
$h\in\mathscr{D}$ and $Z(h)=Z(f)\cup A_{1}$. Clearly, $h$ is adjacent to both
$f,g$, which is a contradiction. Thus $Z(g)$ is an atom. Let $A=Z(f)\setminus
Z(g)=Z(f)\setminus(Z(f)\cap Z(g))$ and $B=Z(g)\cup(X\setminus
Z(f))=Z(g)\cup((X\setminus Z(f))\cap(X\setminus Z(g)))$. Thus $A,B$ are atoms
and $X=A\sqcup B$. Conversely let $A,B$ be two atoms of $X$ such that
$X=A\sqcup B$. Then by Lemma 3.9, $\mathscr{D}$ is partitioned into two sets,
say $V_{1},V_{2}$, where $V_{1}=\\{f\in\mathscr{D}:\mu(A\triangle Z(f))=0\\}$
and $V_{2}=\\{f\in\mathscr{D}:\mu(B\triangle Z(f))=0\\}$. Let $f_{1},f_{2}\in
V_{1}$. Then $\mu(Z(f_{1})\setminus A)=0=\mu(A\setminus Z(f_{2}))$. Now
$Z(f_{1})\subset A\sqcup Z(f_{1})\setminus A\implies Z(f_{1})\setminus
Z(f_{2})\subset A\setminus Z(f_{2})\sqcup Z(f_{1})\setminus
A\implies\mu(Z(f_{1})\setminus Z(f_{2}))=0$ i.e., $f_{1},f_{2}$ are non-
adjacent in $AG$. Consequently $V_{1}$ is a stable set in $AG$ and similarly
$V_{2}$ is a stable set. Now let $f\in V_{1}$ and $g\in V_{2}$. Then
$\mu(Z(f)\setminus A)=0=\mu(Z(g)\setminus B)$. Now, $Z(f)\subset A\sqcup
Z(f)\setminus A$ and $Z(g)\subset B\sqcup Z(g)\setminus B$. Therefore,
$Z(f)\cap Z(g)\subset(A\sqcup Z(f)\setminus A)\cap(B\sqcup Z(g)\setminus
B)\subset(A\cap B)\cup(Z(f)\setminus A)\cup(Z(g)\setminus
B)\implies\mu(Z(f)\cap Z(g))=0$. i.e., $f,g$ are adjacent in
$\Gamma^{\prime}_{2}$. By Theorem 4.7, $f,g$ are adjacent in $AG$ and hence
$AG$ is a complete bipartite graph. ∎

Combining Theorem 3.12 and Theorem 4.8, we get the following result.

######  Corollary 4.9.

The following statements are equivalent:

  1. (1)

$\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph.

  2. (2)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite
graph.

  3. (3)

$AG(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph.

  4. (4)

$X$ is partitioned into two atoms.

######  Theorem 4.10.

The following statements are equivalent:

  1. (1)

$\Gamma(\mathcal{M}(X,\mathcal{A}))=AG(\mathcal{M}(X,\mathcal{A}))$.

  2. (2)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))=AG(\mathcal{M}(X,\mathcal{A}))$.

  3. (3)

$X$ is partitioned into two atoms.

###### Proof.

If $X$ is partitioned into two atoms, then all the three graphs are complete
bipartite with same bipartion and hence they are equal. Suppose $X$ can not be
partitioned into two atoms. Consider an $A\in\mathcal{A}$ such that
$\mu(A),\mu(X\setminus A)>0$. Then either $A$ or $X\setminus A$ is not an
atom. Suppose $X\setminus A$ is not an atom. Then there exist
$A_{1},A_{2}\in\mathcal{A}$ with $\mu(A_{1}),\mu(A_{2})>0$ such that
$X\setminus A=A_{1}\sqcup A_{2}$. Let $f_{1}=1_{A\sqcup A_{1}}\in\mathscr{D}$
and $f_{2}=1_{A\sqcup A_{2}}\in\mathscr{D}$. Then $X\setminus Z(f_{1})\cap
X\setminus Z(f_{2})=A$ and $Z(f_{1})\cap Z(f_{2})=\emptyset$. Therefore,
$f_{1},f_{2}$ are not adjacent in $\Gamma$, but they are adjacent in
$\Gamma^{\prime}_{2}$ and hence they are adjacent in $AG$. Therefore,
$\Gamma\neq AG$. Again let $g_{1}=1_{A_{1}}\in\mathscr{D}$ and
$g_{2}=1_{A_{2}}\in\mathscr{D}$. Then $Z(g_{1})\cap Z(g_{2})=A$ and
$X\setminus Z(g_{1})\cap X\setminus Z(g_{2})=\emptyset$. Therefore,
$g_{1},g_{2}$ are not adjacent in $\Gamma^{\prime}_{2}$, but are adjacent in
$\Gamma$ and hence they are adjacent in $AG$. Therefore,
$\Gamma^{\prime}_{2}\neq AG$. ∎

######  Theorem 4.11.

A vertex $f$ in $AG(\mathcal{M}(X,\mathcal{A}))$ is a vertex of a triangle if
and only if either $Z(f)$ or $X\setminus Z(f)$ is not an atom.

###### Proof.

If $X\setminus Z(f)$ is not an atom, then by Theorem 3.15, $f$ is a vertex of
triangle in $\Gamma^{\prime}_{2}$. By Theorem 4.7, $f$ is a vertex of a
triangle in $AG$. Again if $Z(f)$ is not an atom, then there exists
$A_{1},A_{2}\in\mathcal{A}$ with $\mu(A_{1}),\mu(A_{2})>0$ such that
$Z(f)=A_{1}\sqcup A_{2}$. Clearly $1_{A_{1}},1_{A_{2}}\in\mathscr{D}$ and
$f.1_{A_{1}}\equiv 0\equiv 1_{A_{1}}.1_{A_{2}}\equiv 1_{A_{2}}.f$ on $X$.
i.e., $f-1_{A_{1}}-1_{A_{2}}-f$ is a triangle in $\Gamma$. By Theorem 4.7,
$f-1_{A_{1}}-1_{A_{2}}-f$ is a triangle in $AG$. Let both $Z(f)$ and
$X\setminus Z(f)$ be atoms. Then by Theorem 4.8, $AG$ is complete bipartite.
Hence $f$ can not be a vertex of a triangle. ∎

######  Corollary 4.12.

The girth of $AG(\mathcal{M}(X,\mathcal{A}))$ is given by

$gr(AG(\mathcal{M}(X,\mathcal{A})))=\begin{cases}4&\text{ if }X\text{ is
partitioned into two atoms}\\\ 3&\text{ otherwise}\end{cases}$

.

######  Corollary 4.13.

$AG(\mathcal{M}(X,\mathcal{A}))$ is triangulated if and only if $X$ is not
partitioned into two atoms.

The next result directly follows from Theorem 4.3.

######  Theorem 4.14.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. Then $f\perp g$ in
$AG(\mathcal{M}(X,\mathcal{A}))$ if and only if $\mu(X\setminus Z(f)\cap
X\setminus Z(g))=0=\mu(Z(f)\cap Z(g))$ and either $Z(f)$ or $Z(g)$ is an atom.

######  Corollary 4.15.

An edge $f-g$ in $AG(\mathcal{M}(X,\mathcal{A}))$ is an edge of a triangle if
and only if $f\not\perp g$.

If $X$ can be partitioned into two atoms, then
$AG(\mathcal{M}(X,\mathcal{A}))$ is never hypertriangulated. On the contrary,
if $X$ can not be partitioned into two atoms, then for any $A\in\mathcal{A}$
with $\mu(A)>0$ and $\mu(X\setminus A)>0$, either $A$ or $X\setminus A$ is not
an atom. This condition helps in determining exactly when the graph
$AG(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated.

######  Theorem 4.16.

$AG(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated if and only if $\mu$ is
non-atomic.

###### Proof.

If $\mu$ is non-atomic, then for any pair of vertices $f,g$ in $AG$,
$f\not\perp g$ (by Theorem 4.14), because $Z(f)$ and $Z(g)$ are not atoms.
Consequently, by Corollary 4.15, $AG$ is hypertriangulated. Conversely, let
$A\in\mathcal{A}$ be an atom. Then, $\mu(X\setminus A)>0$, as $X$ is not an
atom. Thus, $1_{A},1_{X\setminus A}\in\mathscr{D}$. By Theorem 4.14,
$1_{A}\perp 1_{X\setminus A}$. Therefore, $1_{A}-1_{X\setminus A}$ is not an
edge of a triangle in $AG$. ∎

######  Corollary 4.17.

$AG(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated if and only if
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is triangulated.

######  Theorem 4.18.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. In
$AG(\mathcal{M}(X,\mathcal{A}))$,

$c(f,g)=\begin{cases}3&f,g\text{ are adjacent and }f\not\perp g\\\ 4&\text{
either }f,g\text{ are not adjacent or }f\perp g\end{cases}$

###### Proof.

Let $f,g\in\mathscr{D}$. If $f,g$ are adjacent and $f\not\perp g$, then there
exists $h\in\mathscr{D}$ adjacent to both $f,g$; i.e., $f-g-h-f$ is a triangle
in $AG$. Thus $c(f,g)=3$. If $f,g$ are not adjacent in $AG$, then they are not
adjacent in $\Gamma^{\prime}_{2}$ also, because $\Gamma^{\prime}_{2}$ is a
subgraph of $AG$. So, $\mu(Z(f)\cap Z(g))>0$. By Theorem 4.3, there exists
$h\in\mathscr{D}$ adjacent to both $f,g$. Thus $f-h-g-2.h-f$ is a square in
$AG\implies c(f,g)\leq 4$. Since $f,g$ are not adjacent in $AG$, $c(f,g)=4$.
If $f\perp g$, then $f-g-2.f-2.g-f$ is a square in $AG$. Hence, $c(f,g)\leq
4$. Since $f\perp g$, no vertex is adjacent to both $f,g$; i.e., $c(f,g)>3$.
Therefore, $c(f,g)=4$. ∎

The visual representation of the above Theorem is exhibited as follows:

$f$$g$$f$$g$$f$$g$$f,g\text{ are adjacent}$$\text{ and }f\not\perp
g$$f,g\text{ are adjacent}$$\text{ and }f\perp g$$f,g\text{ are non-adjacent}$

######  Theorem 4.19.

$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ has an orthogonal complement if
and only if either $Z(f)$ or $X\setminus Z(f)$ is an atom.

###### Proof.

If one of $Z(f)$ and $X\setminus Z(f)$ is an atom, then by Theorem 4.14,
$f\perp 1_{Z(f)}$ in AG. Conversely let $f\perp g$ for some $g\in\mathscr{D}$.
Then $\mu(Z(f)\cap Z(g))=0=\mu(X\setminus Z(f)\cap X\setminus Z(g))$ and
either $Z(f)$ or $Z(g)$ is an atom. If $Z(g)$ is an atom, then $X\setminus
Z(f)$ is an atom, because $X\setminus Z(f)=[X\setminus Z(f)\cap X\setminus
Z(g)]\cup Z(g)\setminus[Z(f)\cap Z(g)]$. Therefore either $Z(f)$ or
$X\setminus Z(f)$ is an atom. ∎

######  Lemma 4.20.

If $X$ is partitioned into three atoms, then for each
$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, either $Z(f)$ or $X\setminus
Z(f)$ is an atom.

###### Proof.

Let $X=A\sqcup B\sqcup C$, where $A,B,C$ are atoms. If possible let there be
some $f\in\mathscr{D}$ such that both $Z(f),X\setminus Z(f)$ are not atoms. So
there are $A_{1},A_{2},B_{1},B_{2}\in\mathcal{A}$, such that $Z(f)=A_{1}\sqcup
A_{2}$, $X\setminus Z(f)=B_{1}\sqcup B_{2}$ and
$\mu(A_{1}),\mu(A_{2}),\mu(B_{1}),\mu(B_{2})>0$. Then
$\mathscr{B}=\\{A,B,C\\}$ and
$\mathscr{B}^{\prime}=\\{A_{1},A_{2},B_{1},B_{2}\\}$ constitute partitions of
$X$ by sets with positive measure. Hence there exist $E\in\mathscr{B}$ and
$F_{1},F_{2}\in\mathscr{B}^{\prime}$ such that $\mu(E\cap F_{1}),\mu(E\cap
F_{2})>0$. Now $E=(E\cap F_{1})\sqcup(E\setminus F_{1})$. Since, $F_{1},F_{2}$
are disjoint, $E\cap F_{2}\subset E\setminus F_{1}$ and hence $\mu(E\setminus
F_{1})>0$. This contradicts that $E$ is an atom. ∎

######  Theorem 4.21.

$AG(\mathcal{M}(X,\mathcal{A}))$ is complemented if and only if $X$ is
partitioned into either two or three atoms.

###### Proof.

We prove this in three cases.
Case I: $X=A\sqcup B$, where $A,B$ are atoms.
By Theorem 4.8, $AG$ is a complete bipartite graph and hence $AG$ is
complemented.
Case II: $X=A\sqcup B\sqcup C$, where $A,B,C$ are atoms.
Then by Lemma 4.20, either $Z(f)$ or $X\setminus Z(f)$ is an atom for each
$f\in\mathscr{D}$. By Theorem 4.19, each $f\in\mathscr{D}$ has an orthogonal
complement. Consequently, $AG$ is complemented.
Case III: Neither Case I nor Case II.
Subcase I: $X$ contains no atom. Then for any pair $f,g\in\mathscr{D}$,
$f\not\perp g$. Hence, $AG$ is not complemented.
Subcase II: $X$ contains an atom, say $A\in\mathcal{A}$. Then $\mu(X\setminus
A)>0$, for otherwise $X$ will be an atom. By our hypothesis, $X\setminus A$ is
not an atom. Then $X\setminus A=B_{1}\sqcup B_{2}$, for some
$B_{1},B_{2}\in\mathcal{A}$ with $\mu(B_{1}),\mu(B_{2})>0$. Then again, one of
$B_{1},B_{2}$ is not an atom. Without loss of generality, let $B_{1}$ be not
an atom. Then $B_{1}=C_{1}\sqcup C_{2}$, for some $C_{1},C_{2}\in\mathcal{A}$
with $\mu(C_{1}),\mu(C_{2})>0$. Consider $f\in\mathscr{D}$ such that
$Z(f)=C_{1}\sqcup B_{2}$. Then $X\setminus Z(f)=C_{2}\sqcup A$. Therefore,
both of $Z(f)$ and $X\setminus Z(f)$ are not atoms. By Theorem 4.19,
$f\in\mathscr{D}$ has no orthogonal complement. Consequently, $AG$ is not
complemented. ∎

######  Lemma 4.22.

Let $f,g,h\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ be such that $f\perp g$
and $f\perp h$ in $AG$. Then $g,h$ are adjacent to the same set of vertices.

###### Proof.

By Theorem 4.14, $\mu(Z(f)\cap Z(g))=0=\mu(X\setminus Z(f)\cap X\setminus
Z(g))$ and $\mu(Z(f)\cap Z(h))=0=\mu(X\setminus Z(f)\cap X\setminus Z(h))$.
i.e., $\mu(Z(g)\triangle(X\setminus Z(f)))=0=\mu(Z(h)\triangle(X\setminus
Z(f)))$. This implies $\mu(Z(g)\triangle Z(h))=0$. Let $k\in\mathscr{D}$ be
adjacent to $g$; i.e., $\mu(Z(k)\setminus Z(g))>0$ and $\mu(Z(g)\setminus
Z(k))>0$. Now $Z(g)\setminus Z(k)=(Z(g)\cap Z(h)\cap X\setminus
Z(k))\sqcup(Z(g)\setminus Z(h)\cap X\setminus Z(k))$. Since $\mu(Z(g)\setminus
Z(h))=0$, $\mu(Z(g)\cap Z(h)\cap X\setminus Z(k))=\mu(Z(g)\setminus
Z(k))>0\implies\mu(Z(h)\setminus Z(k))\geq\mu(Z(g)\cap Z(h)\cap X\setminus
Z(k))>0$ i.e., $\mu(Z(h)\setminus Z(k))>0$. Similarly, $\mu(Z(k)\setminus
Z(h))>0$. Thus $k$ is adjacent to $h$. Likewise if $k$ is adjacent to $h$,
then it is adjacent to $g$ also. Hence $g,h$ are adjacent to the same set of
vertices. ∎

As a consequence of this lemma, we get the following theorem.

######  Theorem 4.23.

$AG(\mathcal{M}(X,\mathcal{A}))$ is complemented if and only if
$AG(\mathcal{M}(X,\mathcal{A}))$ is uniquely complemented.

The fact that $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma(\mathcal{M}(X,\mathcal{A}))$ are subgraphs of
$AG(\mathcal{M}(X,\mathcal{A}))$ helps us to determine the dominating sets in
$AG(\mathcal{M}(X,\mathcal{A}))$, as seen in the next result.

######  Theorem 4.24.

$dt(AG(\mathcal{M}(X,\mathcal{A})))=2$.

###### Proof.

Fix an $A\in\mathcal{A}$ with $\mu(A),\mu(X\setminus A)>0$. We claim that
$\\{1_{A},1_{X\setminus A}\\}$ is a dominating set in $AG$. Suppose
$f\in\mathscr{D}$ is not adjacent to $1_{X\setminus A}$ in $AG$; i.e., either
$\mu(Z(f)\setminus A)=0$ or $\mu(A\setminus Z(f))=0$. If $\mu(Z(f)\setminus
A)=0$, then $\mu(Z(f)\cap X\setminus A)=0$. So $f,1_{A}$ are adjacent in
$\Gamma^{\prime}_{2}$ and hence by Theorem 4.7, $f,1_{A}$ are adjacent in
$AG$. If $\mu(A\setminus Z(f))=0$, then $\mu(X\setminus Z(f)\cap A)=0$. Thus
$f,1_{A}$ are adjacent in $\Gamma$. Again by Theorem 4.7, $f,1_{A}$ are
adjacent in $AG$. Hence, in any case, $f$ is adjacent to $1_{A}$ in $AG$.
Consequently, $\\{1_{A},1_{X\setminus A}\\}$ is a dominating set in $AG$. Also
for each $f\in\mathscr{D}$, $f,2.f$ are not adjacent in $AG$ which proves
singleton sets are not dominating sets in $AG$. Hence $dt(AG)=2$. ∎

In the above theorem, we note that the dominating set $\\{1_{A},1_{X\setminus
A}\\}$ is a total dominating set in $AG(\mathcal{M}(X,\mathcal{A}))$. We
record this observation in the following result.

######  Corollary 4.25.

$dt_{t}(AG(\mathcal{M}(X,\mathcal{A})))=2$.

In Theorem 4.10, we have seen that
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))=AG(\mathcal{M}(X,\mathcal{A}))$
if and only if $X$ is partition into two atoms. We now show that these two
graphs are identical if and only if they are isomorphic.

######  Theorem 4.26.

The graphs $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$AG(\mathcal{M}(X,\mathcal{A}))$ are isomorphic if and only if $X$ is
partitioned into two atoms.

###### Proof.

If $X$ is partitioned into two atoms, then by Theorem 4.10, these two graph
are equal and so, they are isomorphic.
If $X$ is not partitioned into two atoms then two cases may arise; either $X$
is partitioned into three atoms or $X$ is $X$ is not partitioned into three
atoms. First we assume that $X$ is partitioned into three atoms, say
$X=A\sqcup B\sqcup C$ where $A,B,C$ are atoms. If possible let $\psi$ be a
graph isomorphism between the graphs
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
$AG(\mathcal{M}(X,\mathcal{A}))$. Now $1_{A}$ is a vertex in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and $ecc(1_{A})=3$ in
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ (by Theorem 3.13), because
$Z(1_{A})=B\sqcup C$, is not an atom. Since $\psi$ is a graph isomorphism,
$ecc(\psi(1_{A}))=3$ in $AG(\mathcal{M}(X,\mathcal{A}))$, which contradicts
the fact that $ecc(f)=2$ for every vertex $f$ in
$AG(\mathcal{M}(X,\mathcal{A}))$ (Corollary 4.5). Therefore, there does not
exist any graph isomorphism between these two graphs.
Finally, let $X$ be neither partitioned into two atoms nor into three atoms.
Then $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a complemented graph
(by Theorem 3.21), though $AG(\mathcal{M}(X,\mathcal{A}))$ is not
complemented. Thus, they are not isomorphic as graphs. ∎

######  Corollary 4.27.

$\Gamma(\mathcal{M}(X,\mathcal{A}))$ and $AG(\mathcal{M}(X,\mathcal{A}))$ are
isomorphic if and only if $X$ is partitioned into two atoms.

##  5\. The weakly zero-divisor graph $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ of
$\mathcal{M}(X,\mathcal{A})$

The weakly zero-divisor graph $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ (in short
$W\Gamma$) of $\mathcal{M}(X,\mathcal{A})$, is defined as a graph with
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ as the set of vertices and two
vertices $f,g$ are adjacent if there exist $h_{1}\in
ann(f)\cap\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ and $h_{2}\in
ann(g)\cap\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ such that
$h_{1}.h_{2}\equiv 0$ a.e. on $X$.

The adjacency relation in $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is interpreted
via the measure $\mu$ as follows:

######  Theorem 5.1.

Let $f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. In
$W\Gamma(\mathcal{M}(X,\mathcal{A}))$,

  1. (1)

if $\mu(Z(f)\triangle Z(g))>0$, then $f,g$ are adjacent.

  2. (2)

if $\mu(Z(f)\triangle Z(g))=0$ and $Z(f)$ is not an atom, then $f,g$ are
adjacent.

  3. (3)

if $\mu(Z(f)\triangle Z(g))=0$ and $Z(f)$ is an atom, then $f,g$ are not
adjacent.

###### Proof.

  1. (1)

Since $\mu(Z(f)\triangle Z(g))>0$, either $\mu(Z(f)\setminus Z(g))>0$ or
$\mu(Z(g)\setminus Z(f))>0$. Without loss of generality let $\mu(Z(f)\setminus
Z(g))>0$. Consider $h_{1}=1_{Z(f)\setminus Z(g)}$ and $h_{2}=1_{Z(g)}$. Then
$h_{1},h_{2}\in\mathscr{D}$ and $h_{1}.f=0=h_{2}.g$ on $X$. So, $h_{1}\in
ann(f)\cap\mathscr{D}$ and $h_{2}\in ann(g)\cap\mathscr{D}$. Also
$h_{1}.h_{2}=0$ on $X$. Hence, $f,g$ are adjacent in $W\Gamma$.

  2. (2)

Since $Z(f)$ is not an atom, $Z(f)=A\sqcup B$ for some $A,B\in\mathcal{A}$
with $\mu(A),\mu(B)>0$. Clearly $1_{A},1_{B}\in\mathscr{D}$ and
$1_{A}.f=0=1_{B}.f$ on $X$ ; i.e., $1_{A},1_{B}\in ann(f)$. Since
$\mu(Z(f)\triangle Z(g))=0$, by Corollary 2.6, $ann(f)=ann(g)\implies 1_{B}\in
ann(g)$. Also, $1_{A}.1_{B}=0$ on $X\implies f,g$ are adjacent in $W\Gamma$.

  3. (3)

If possible let $f$ and $g$ be adjacent. Then there exist $h_{1}\in
ann(f)\cap\mathscr{D}$ and $h_{2}\in ann(g)\cap\mathscr{D}$ such that
$\mu(X\setminus Z(h_{1})\cap X\setminus Z(h_{2}))=0$. Since $\mu(Z(f)\triangle
Z(g))=0$, by Lemma 2.6, $ann(f)=ann(g)$. Therefore $h_{1},h_{2}\in ann(f)$;
i.e., there exists a zero measurable set $E\in\mathcal{A}$ such that
$X\setminus Z(h_{1})\cap X\setminus E\subset Z(f)$ and $X\setminus
Z(h_{2})\cap X\setminus E\subset Z(f)$. Let $A=X\setminus Z(h_{1})\cap
X\setminus E$ and $B=Z(f)\setminus A$. Clearly, $A,B\in\mathcal{A}$ and
$Z(f)=A\sqcup B$. Since $h_{1}\in\mathscr{D}$ and $\mu(E)=0$, $\mu(A)>0$.
Again, $(X\setminus E\cap X\setminus Z(h_{2}))\setminus(X\setminus
Z(h_{1})\cap X\setminus Z(h_{2}))\subset B$. Since $h_{2}\in\mathscr{D}$ and
$\mu(X\setminus Z(h_{1})\cap X\setminus Z(h_{2}))=0=\mu(E)$, $\mu(B)>0$. This
contradicts the hypothesis that $Z(f)$ is an atom.

∎

From Theorem 5.1, it follows that $f,f$ are adjacent if and only if $Z(f)$ is
not an atom. In other words,

######  Corollary 5.2.

$f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$ is self-adjacent in
$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ if and only if $Z(f)$ is not an atom.

To make $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ a simple graph, we redefine the
vertex set of the graph as
$\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))=\\{f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A})):Z(f)\text{
is an atom}\\}$. If $\mu$ is non-atomic, then $\mathcal{A}$ does not contain
any atom and hence $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is an empty graph.
For the non-emptiness of $W\Gamma(\mathcal{M}(X,\mathcal{A}))$, we assume that
$\mathcal{A}$ contains atleast one atom $A$ with $\mu(X\setminus A)>0$. Then
the adjacency relation between two vertices in
$\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$ takes the form as described
in the next theorem.

######  Theorem 5.3.

Let $f,g\in\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$. Then $f,g$ are
adjacent in $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ if and only if
$\mu(Z(f)\triangle Z(g))>0$.

Consider the equivalence relation $\sim$ on
$\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$ given by: “$f\sim g$ if and
only if $\mu(Z(f)\triangle Z(g))=0$”. For each
$f\in\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$, let $[f]$ denote the
equivalence class of $f$ under the relation $\sim$ on
$\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$. From Theorem 5.3, it
follows that

  1. (1)

for each $f\in\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$, $[f]$ is a
stable set in $W\Gamma(\mathcal{M}(X,\mathcal{A}))$.

  2. (2)

if $[f],[g]$ are distinct classes, then for all $f_{1}\in[f]$ and for all
$g_{1}\in[g]$, $f_{1},g_{1}$ are adjacent in
$W\Gamma(\mathcal{M}(X,\mathcal{A}))$.

If $W$ is the collection of distinct class representatives under $\sim$ on
$\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$, then we get the following
theorem.

######  Theorem 5.4.

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete $|W|$-partite graph.

It is known that if $A\in\mathcal{A}$ is an atom, then $\mu(X\setminus A)>0$.
Moreover, two atoms $A,B\in\mathcal{A}$ are said to be distinct if
$\mu(A\triangle B)>0$. So, $|W|=$ number of distinct atoms in $\mathcal{A}$.
This observation leads to the following theorem.

######  Theorem 5.5.

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete bipartite graph if and
only if $\mathcal{A}$ contains exactly two distinct atoms.

######  Theorem 5.6.

If $\mathcal{A}$ contains atleast three distinct atoms, then the following
hold:

  1. (1)

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is triangulated.

  2. (2)

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is hypertriangulated.

  3. (3)

The girth of $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is $3$.

  4. (4)

No pair of vertices in $\mathscr{D}^{\prime}(\mathcal{M}(X,\mathcal{A}))$ are
orthogonal in $W\Gamma(\mathcal{M}(X,\mathcal{A}))$.

  5. (5)

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is not a complemented graph.

We list the values of certain important graph parameters of
$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ in the following theorem:

######  Theorem 5.7.

  1. (1)

The dominating number of $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is $2$.

  2. (2)

The chromatic number of $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is $|W|$.

  3. (3)

The clique number of $W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is $|W|$.

##  6\. Illustration via two familiar Measure Spaces

###  6.1. On the Counting Measure Space

Let $X$ be a non-empty set, $\mathcal{A}=\mathscr{P}(X)$, the power set of $X$
and $\mu$, the counting measure on $X$, be defined as follows:

$\mu(E)=\begin{cases}|E|\text{ if }E\text{ is finite}\\\ \infty\text{
otherwise}\end{cases}$

Then $\mathcal{M}(X,\mathcal{A})=\mathbb{R}^{X}$ and
$\mathscr{D}(\mathcal{M}(X,\mathcal{A}))=\\{f\in\mathcal{M}(X,\mathcal{A}):Z(f)\neq\emptyset,X\\}$.
For the non-emptiness of $\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$, we always
consider $|X|>1$. Clearly, each singleton set is an atom in
$(X,\mathcal{A},\mu)$.

Choosing $(X,\mathcal{A},\mu)$ as the counting measure space, the following
two theorems list the important features of the comaximal graph and the
annihilator graph of $\mathcal{M}(X,\mathcal{A})$.

######  Theorem 6.1.1.

Consider the comaximal graph $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
of $\mathcal{M}(X,\mathcal{A})$ and
$f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.

  1. (1)

There exists a vertex adjacent to both $f$ and $g$ if and only if $Z(f)\cup
Z(g)\neq X$.

  2. (2)

$d(f,g)=\begin{cases}1&Z(f)\cap Z(g)=\emptyset\\\ 2&Z(f)\cap
Z(g)\neq\emptyset\text{ and }Z(f)\cup Z(g)\neq X\\\ 3&Z(f)\cap
Z(g)\neq\emptyset\text{ and }Z(f)\cup Z(g)=X\end{cases}$

  3. (3)

$ecc(f)=\begin{cases}2&\text{ if }|Z(f)|=1\\\ 3&\text{ otherwise}\end{cases}$

  4. (4)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is complete bipartite if and
only if $|X|=2$.

  5. (5)

The diameter of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
$\begin{cases}2&|X|=2\\\ 3&|X|\geq 3\end{cases}$

  6. (6)

The girth of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
$\begin{cases}4&|X|=2\\\ 3&|X|\geq 3\end{cases}$

  7. (7)

$f$ is a vertex of a triangle if and only if $|X\setminus Z(f)|>1$.

  8. (8)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is never triangulated and
never hypertriangulated.

  9. (9)

$f\perp g$ if and only if $Z(f)=X\setminus Z(g)$.

######  Theorem 6.1.2.

Consider the annihilator graph $AG(\mathcal{M}(X,\mathcal{A}))$ of
$\mathcal{M}(X,\mathcal{A})$ and
$f,g\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$.

  1. (1)

There exists a vertex adjacent to both $f,g$ if and only if either $Z(f)\cup
Z(g)\neq X$ or $Z(f)\cap Z(g)\neq\emptyset$ or $|Z(f)|,|Z(g)|\geq 2$.

  2. (2)

$d(f,g)=\begin{cases}1&Z(f)\setminus Z(g)\neq\emptyset\text{ and
}Z(g)\setminus Z(f)\neq\emptyset\\\ 2&\text{ otherwise}\end{cases}$

  3. (3)

$ecc(f)=2$

  4. (4)

$AG(\mathcal{M}(X,\mathcal{A}))$ is complete bipartite if and only if $|X|=2$.

  5. (5)

$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))=AG(\mathcal{M}(X,\mathcal{A}))$
if and only if $|X|=2$ if and only if
$\Gamma(\mathcal{M}(X,\mathcal{A}))=AG(\mathcal{M}(X,\mathcal{A}))$.

  6. (6)

The girth of $AG(\mathcal{M}(X,\mathcal{A}))$ is $\begin{cases}4&|X|=2\\\
3&|X|\geq 3\end{cases}$

  7. (7)

$AG(\mathcal{M}(X,\mathcal{A}))$ is not hypertriangulated.

  8. (8)

$f$ has an orthogonal complement in $AG(\mathcal{M}(X,\mathcal{A}))$ if and
only if either $|Z(f)|=1$ or $|X\setminus Z(f)|=1$.

  9. (9)

$AG(\mathcal{M}(X,\mathcal{A}))$ is uniquely complemented if and only if
$|X|\leq 3$.

For this measure space, the Theorem 5.4 takes the following form:

######  Theorem 6.1.3.

$W\Gamma(\mathcal{M}(X,\mathcal{A}))$ is a complete $|X|$-partite graph where
the stable sets are $W_{x}=\\{f\in\mathcal{M}(X,\mathcal{A}):Z(f)=\\{x\\}\\}$,
$x\in X$.

In the counting measure space, the equivalence relation described before,
reads as “$f\sim g$ if and only if $Z(f)=Z(g)$” and therefore
$[f]=\\{g\in\mathcal{M}(X,\mathcal{A}):Z(f)=Z(g)\\}$. Hence, the vertex set of
the induced graph $G_{2}$ of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
becomes $V=\\{1_{A}:A\neq\emptyset,X\\}$. In $G_{2}$, two distinct vertices
$1_{A},1_{B}$ are adjacent if and only if $X\setminus A\cap X\setminus
B=\emptyset$.

In the next few results we observe that in the counting measure space, the
chromatic number and clique number of
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ solely depend on the
cardinality of the underlying space $X$. We prove these results using the
induced graph $G_{2}$ of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ and
Theorem 3.29 and Theorem 3.27 of Section 3.

######  Theorem 6.1.4.

The chromatic number of $G_{2}$ is $|X|$.

###### Proof.

Clearly, $\\{1_{X\setminus\\{x\\}}:x\in X\\}$ is a complete subgraph of
$G_{2}$. Therefore, $\chi(G_{2})\geq|X|$. We begin with coloring each
$1_{X\setminus\\{x\\}}\in V$ by using distinct colors. Let $1_{A}\in
V\setminus\\{1_{X\setminus\\{x\\}}:x\in X\\}$. Then $|X\setminus A|\geq 2$. We
color $1_{A}$ by one of the colors of $1_{X\setminus\\{x\\}}$, where $x\in
X\setminus A$. Suppose $1_{A},1_{B}\in V$ are colored by the same color, say
by the color of $1_{X\setminus\\{x\\}}$. Then by hypothesis, $x\in X\setminus
A\cap X\setminus B$. Consequently, $1_{A},1_{B}$ are not adjacent in $G_{2}$.
It gives a consistent coloring of $G_{2}$ by $|X|$-many colors. Therefore,
$\chi(G_{2})\leq|X|$. ∎

######  Corollary 6.1.5.

The chromatic number of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
$|X|$.

######  Theorem 6.1.6.

The clique number of $G_{2}$ is $|X|$.

###### Proof.

Since $\\{1_{X\setminus\\{x\\}}:x\in X\\}$ is a complete subgraph of $G$,
$cl(G_{2})\geq|X|$. For any graph $H$, we know that $cl(H)\leq\chi(H)$. Hence
by Theorem 6.1.4, $cl(G_{2})\leq|X|$. ∎

######  Corollary 6.1.7.

The clique number of $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is
$|X|$.

######  Remark 6.1.8.

A graph is said to be weakly perfect if the clique number and the chromatic
number are equal. Since
$cl(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))=|X|=\chi(\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A})))$,
in this measure space, $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ is a
weakly perfect graph.

We now compare the comaximal graph and the zero-divisor graph of
$\mathcal{M}(X,\mathcal{A})$ in the counting measure space. For this, we need
the following lemmas.

######  Lemma 6.1.9.

For each $f\in\mathcal{M}(X,\mathcal{A})$, $|[f]|=|\mathbb{R}^{|X\setminus
Z(f)|}|$.

###### Proof.

Let $X\setminus Z(f)=\\{x_{\alpha}:\alpha\in\Lambda\\}$. Define
$\eta:[f]\to(\mathbb{R}\setminus\\{0\\})^{|\Lambda|}$ by
$\eta(g)=\Pi_{\alpha\in\Lambda}(g(x_{\alpha}))$. $\eta$ is well defined, since
for all $g\in[f]$, $X\setminus Z(g)=X\setminus Z(f)$. Let $g_{1},g_{2}\in[f]$
be such that $\eta(g_{1})=\eta(g_{2})$; i.e.,
$g_{1}(x_{\alpha})=g_{2}(x_{\alpha})$ for all $\alpha\in\Lambda$. Also for all
$x\notin X\setminus Z(f)$, $g_{1}(x)=0=g_{2}(x)\implies g_{1}=g_{2}$ on $X$.
Therefore $\eta$ is injective. Let
$y\in(\mathbb{R}\setminus\\{0\\})^{|\Lambda|}$. Then $y$ can be written as
$y=(y_{\alpha})_{\alpha\in\Lambda}$, where
$y_{\alpha}\in\mathbb{R}\setminus\\{0\\}$ for all $\alpha\in\Lambda$. Let
$g(x)=\begin{cases}y_{\alpha}&\text{ if }x=x_{\alpha}\\\ 0&\text{
otherwise}\end{cases}$. Since $y_{\alpha}\neq 0$ for all $\alpha\in\Lambda$,
$X\setminus Z(g)=\\{x_{\alpha}:\alpha\in\Lambda\\}=X\setminus Z(f)\implies
g\in[f]$. Clearly $\eta(g)=y$. Consequently $\eta$ is a bijection i.e.,
$|[f]|=|(\mathbb{R}\setminus\\{0\\})^{|\Lambda|}|=|\mathbb{R}^{|X\setminus
Z(f)|}|$. ∎

Let $f\in\mathscr{D}(\mathcal{M}(X,\mathcal{A}))$. By Theorem 6.1.1(3) it
follows that, $ecc(f)=2$ in $\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$
if and only if $Z(f)$ is a singleton set. Similarly, it can be proved that,
$ecc(f)=2$ in $\Gamma(\mathcal{M}(X,\mathcal{A}))$ if and only if $X\setminus
Z(f)$ is a singleton set. These observations lead to the following important
lemma.

######  Lemma 6.1.10.

If $\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are graph isomorphic, then
for each $x\in X$ there exists $y\in X$ such that
$|[1_{x}]|=|[1_{X\setminus\\{y\\}}]|$.

###### Proof.

Let $\psi:\Gamma\to\Gamma^{\prime}_{2}$ be a graph isomorphism. Fix an $x\in
X$. Let $f\in[1_{x}]$. Then $X\setminus Z(f)=\\{x\\}\implies ecc(f)=2$ in
$\Gamma$. Since $\psi$ is a graph isomorphism, $ecc(\psi(f))=2$ in
$\Gamma^{\prime}_{2}\implies Z(\psi(f))$ is a singleton set i.e.,
$Z(\psi(f))=\\{y\\}$ for some $y\in X$. Therefore,
$\psi(f)\in[1_{X\setminus\\{y\\}}]$. Conversely let
$g\in[1_{X\setminus\\{y\\}}]$. Then $Z(g)=\\{y\\}\implies ecc(g)=2$ in
$\Gamma^{\prime}_{2}\implies ecc(\psi^{-1}(g))=2$ in $\Gamma\implies
X\setminus Z(\psi^{-1}(g))$ is a singleton set, i.e., $X\setminus
Z(\psi^{-1}(g))=\\{x^{\prime}\\}$ for some $x^{\prime}\in X$. If $x\neq
x^{\prime}$, then $1_{x},\psi^{-1}(g)$ are adjacent in $\Gamma$. Therefore
their images under $\psi$ are adjacent in $\Gamma^{\prime}_{2}$, i.e.,
$\psi(1_{x}),g$ are adjacent in $\Gamma^{\prime}_{2}$. But
$1_{x}\in[1_{x}]\implies\psi(1_{x})\in[1_{X\setminus\\{y\\}}]$, by our
definition. Therefore, $Z(\psi(1_{x}))=\\{y\\}=Z(g)\implies Z(\psi(1_{x}))\cap
Z(g)=\\{y\\}$. So, $\psi(1_{x}),g$ are not adjacent in $\Gamma^{\prime}_{2}$,
which is a contradiction. Therefore, $X\setminus
Z(\psi^{-1}(g))=\\{x\\}\implies\psi^{-1}(g)\in[1_{x}]$. Hence, the restriction
map $\psi:[1_{x}]\to[1_{X\setminus\\{y\\}}]$ is a bijection, proving
$|[1_{x}]|=|[1_{X\setminus\\{y\\}}]|$. ∎

######  Theorem 6.1.11.

$\Gamma(\mathcal{M}(X,\mathcal{A}))$ and
$\Gamma^{\prime}_{2}(\mathcal{M}(X,\mathcal{A}))$ are graph isomorphic if and
only if $X$ is atmost countable.

###### Proof.

If $X$ is atmost countable, then for any $f\in\mathscr{D}$, $|X\setminus
Z(f)|$ is countable and hence by Lemma 6.1.9, $|[f]|=|\mathbb{R}^{|X\setminus
Z(f)|}|=|\mathbb{R}|=\mathfrak{c}$. So, $|[f]|=|[g]|$ for all
$f,g\in\mathscr{D}$. Therefore by Theorem 3.34, $\Gamma$ and
$\Gamma^{\prime}_{2}$ are isomorphic. Conversely let $X$ be uncountable. Then
for any $x,y\in X$, $X\setminus\\{x\\}$ is uncountable
$\implies|\mathbb{R}^{|X\setminus\\{x\\}}||>|\mathbb{R}^{|\\{y\\}|}|$ ; i.e.,
$|[1_{X\setminus\\{x\\}}]|>|[1_{y}]|$. By Lemma 6.1.10, $\Gamma$ and
$\Gamma^{\prime}_{2}$ are not graph isomorphic. ∎

######  Remark 6.1.12.

Let $(X,\mathcal{A},\mu)$ be the counting measure space, where $X$ is an
uncountable set. Then by Theorem 6.1.11, the zero-divisor graph and the
comaximal graph of $\mathcal{M}(X,\mathcal{A})$ are never isomorphic as
graphs. But $G$ and $G_{2}$ are graph isomorphic.

###  6.2. On the Lebesgue Measure Space

Let $(\mathbb{R},\mathcal{A},\ell)$ be the Lebesgue measure space on
$\mathbb{R}$ and $\mathcal{A}$, the collection of all Lebesgue measurable sets
on $\mathbb{R}$. We know that, the Lebesgue measure on $\mathbb{R}$ is a non-
atomic measure. So we have the following results:

######  Theorem 6.2.1.

  1. (1)

The diameter of the comaximal graph
$\Gamma^{\prime}_{2}(\mathcal{M}(\mathbb{R},\mathcal{A}))$ of
$\mathcal{M}(\mathbb{R},\mathcal{A})$ is $3$.

  2. (2)

The eccentricity of every vertex in
$\Gamma^{\prime}_{2}(\mathcal{M}(\mathbb{R},\mathcal{A}))$ is $3$.

  3. (3)

The girth of $\Gamma^{\prime}_{2}(\mathcal{M}(\mathbb{R},\mathcal{A}))$ is
$3$.

  4. (4)

$\Gamma^{\prime}_{2}(\mathcal{M}(\mathbb{R},\mathcal{A}))$ is always a
triangulated graph.

  5. (5)

The annihilator graph $AG(\mathcal{M}(\mathbb{R},\mathcal{A}))$ of
$\mathcal{M}(\mathbb{R},\mathcal{A})$ is always triangulated as well as
hypertriangulated.

  6. (6)

$AG(\mathcal{M}(\mathbb{R},\mathcal{A}))$ can never be a complemented graph.

  7. (7)

The zero-divisor graph and the comaximal graph of
$\mathcal{M}(\mathbb{R},\mathcal{A})$ are not isomorphic to the annihilator
graph of $\mathcal{M}(\mathbb{R},\mathcal{A})$.

  8. (8)

The weakly zero-divisor graph of $\mathcal{M}(\mathbb{R},\mathcal{A})$ is an
empty graph.

As $\mathcal{A}$ contains $2^{\mathfrak{c}}$-many distinct Lebesgue measurable
sets, $|\mathcal{M}(\mathbb{R},\mathcal{A})|=2^{\mathfrak{c}}$. We now show
that the comaximal graph and the zero-divisor graph of
$\mathcal{M}(\mathbb{R},\mathcal{A})$ are isomorphic.

######  Lemma 6.2.2.

Let $A\in\mathcal{A}$ be such that $A$ is compact in $\mathbb{R}$ (with
respect to the usual topology). Then there exist $a,b\in A$ such that
$\ell(A\cap[a,b])=\ell(A)$.

###### Proof.

Since $A$ is compact in $\mathbb{R}$, $A$ is closed and $A\subset[x,y]$ for
some $x,y\in\mathbb{R}$. Clearly, $\ell(A)\leq y-x$, finite. Let
$B=\\{z\in[a,b]:\ell(A\cap[z,y])=\ell(A)\\}$ and $a=\sup B$. Since $x\in B$,
$B\neq\emptyset$ and hence such $a$ exists and $x\leq a<y$. If possible let
$a\notin A$, then $a$ is not a limit point of $A$. There exists $\delta>0$
such that $(a-\delta,a+\delta)\cap A=\emptyset$. Therefore
$\ell((a-\delta,a+\delta)\cap A)=0$. Since $a$ is the supremum of the set $B$,
there exists $a^{\prime}\in B$ such that $a-\delta<a^{\prime}<a$. Now
$a^{\prime}\in B\implies\ell(A\cap[a^{\prime},y])=\ell(A)$. Therefore
$\ell(A\cap[x,a^{\prime}))=0$. This along with the fact
$\ell((a-\delta,a+\delta)\cap A)=0$ imply
$\ell(A\cap[x,a+\delta))=0\implies\ell(A\cap[a+\delta,y])=\ell(A)$ i.e.,
$a+\delta\in B$, which contradicts that $a=\sup B$. Therefore $a\in A$. Also
$[x,a)=\bigcup\limits_{n=1}^{\infty}[x,a-\frac{1}{n}]\implies\ell(A\cap[x,a))\leq\sum\limits_{n=1}^{\infty}\ell(A\cap[x,a-\frac{1}{n}])$.
Since $a=\sup B$, $\ell(A\cap[x,a-\frac{1}{n}])=0$ for all $n$ and hence
$\ell(A\cap[x,a))=0\implies\ell(A\cap[a,y])=\ell(A)$. Similarly if
$b=\inf\\{z\in[a,y]:\ell(A\cap[a,z])=\ell(A)\\}$, then $b\in A$ and
$\ell(A\cap[a,b])=\ell(A)$. ∎

######  Lemma 6.2.3.

Let $A\in\mathcal{A}$ and $\ell(A)>0$. Then for every $r\in[0,\ell(A)]$, there
exists $A_{r}\in\mathcal{A}$ such that $A_{r}\subset A$ and $\ell(A_{r})=r$.

###### Proof.

We consider the function $\psi:\mathbb{R}\to[0,\ell(A)]$ given by
$\psi(x)=\ell(A\cap(-\infty,x])$. Since $|\psi(x)-\psi(y)|\leq|x-y|$ for all
$x,y\in\mathbb{R}$, $\psi$ is continuous on $\mathbb{R}$. Also
$\lim\limits_{x\to-\infty}\psi(x)=0$ and
$\lim\limits_{x\to\infty}\psi(x)=\ell(A)$. Therefore $\psi$ takes every value
on $[0,\ell(A)]$. i.e., For each $r\in[0,\ell(A)]$, there exists
$x_{r}\in\mathbb{R}$ such that $\psi(x_{r})=r$. So,
$\ell(A\cap(-\infty,x_{r}])=r$. ∎

In the Lemma 6.2.3, if $A$ is a compact set, then for some $a,b\in A$,
$\psi:[a,b]\rightarrow[0,\ell(A)]$ may be taken as $\psi(x)=\ell(A\cap[a,x])$.
Also for each $r\in[0,\ell(A)]$, there exists $y_{r}\in A\cap[a,b]$ such that
$\ell(A\cap[a,y_{r}])=r$.

######  Theorem 6.2.4.

If $\ell(A)>0$ for some $A\in\mathcal{A}$, then $A$ contains an uncountable
set of measure zero.

###### Proof.

Without loss of generality let $\ell(A\cap[0,1])>0$. By the regularity of
$\ell$, there exists a compact set $K$ in $\mathbb{R}$ such that $\ell(K)>0$
and $K\subset A\cap[0,1]$. By Lemma 6.2.2, there exists $a,b\in K\cap[0,1]$
such that $\ell(K\cap[a,b])=\ell(K)$. By the observation after Lemma 6.2.3,
there exist $x_{11},x_{12}\in K\cap[a,b]$ such that
$\ell(K\cap[a,x_{11}])=\frac{1}{3}\ell(K)$ and
$\ell(K\cap[a,x_{12}])=\frac{2}{3}\ell(K)$. Let $I_{11}=K\cap[a,x_{11}]$,
$I_{12}=K\cap[x_{11},x_{12}]$ and $I_{13}=K\cap[x_{12},b]$. Then
$\ell(I_{11})=\frac{1}{3}\ell(K)=\ell(I_{12})=\ell(I_{13})$; i.e., we trisect
$K\cap[a,b]$ into three parts $I_{11},I_{12},I_{13}$ with equal measure. Let
$E_{1}=I_{11}\sqcup I_{13}$. Then $E_{1}\in\mathcal{A}$ and
$\ell(E_{1})=\frac{2}{3}\ell(K)$. Similarly we trisect $I_{11}$ into
$I_{21}=K\cap[a,x_{21}],I_{22}=K\cap[x_{21},x_{22}]$ and
$I_{23}=K\cap[x_{22},x_{11}]$ with equal measure, where $x_{21},x_{22}\in
K\cap[a,b]$. We also trisect $I_{13}$ into
$I_{24}=K\cap[x_{12},x_{23}],I_{25}=K\cap[x_{23},x_{24}]$ and
$I_{26}=K\cap[x_{24},b]$ with equal measure, where $x_{23},x_{24}\in
K\cap[a,b]$. Let $E_{2}=(I_{21}\sqcup I_{23})\sqcup(I_{24}\sqcup I_{26})$.
Then $E_{2}\in\mathcal{A}$ and $\ell(E_{2})=(\frac{2}{3})^{4}\ell(K)$.
Continuing this process, at the $n$-th stage, we get $E_{n}\in\mathcal{A}$
which is the union of $2^{n}$-many disjoint measurable sets
$I_{n1},\ldots,I_{n2^{n}}$ each of which has measure $\frac{1}{3^{n}}\ell(K)$.
Thus $\ell(E_{n})=(\frac{2}{3})^{2^{n}}\ell(K)$. Let
$E=\bigcap\limits_{n=1}^{\infty}E_{n}$. Then $E\in\mathcal{A}$ and
$\ell(E)=\lim\limits_{n\to\infty}\ell(E_{n})=\ell(K).\lim\limits_{n\to\infty}(\frac{2}{3})^{2^{n}}=0$.
For every chain of Lebesgue measurable sets of the form $I_{mn}$,
$x_{mn}\in\bigcap I_{mn}$ and $\bigcap
I_{mn}\subset\bigcap\limits_{n=1}^{\infty}E_{n}\subset E\implies
E\neq\emptyset$. We claim that $E$ is uncountable. If possible let $E$ be
countable, say $E=\\{x_{n}:n\in\mathbb{N}\\}$. Since $x_{1}\in E$, $x_{1}\in
E_{1}\implies$ there exists $k\in\\{1,3\\}$ such that $x_{1}\notin
I_{1k_{1}}$. Similarly $x_{2}\in E_{2}\implies x_{2}\notin I_{2k_{2}}$ where
$I_{2k_{2}}$ is one of the measurable sets in $E_{2}$ such that
$I_{2k_{2}}\subset I_{1k_{1}}$. Thus for each $n\in\mathbb{N}$, there exists
some $k_{n}\in\\{1,2,...,2^{n}\\}$ such that $x_{n}\notin I_{nk_{n}}$ and
$I_{nk_{n}}\subset I_{(n-1)k_{n-1}}$. Since $I_{nk_{n}}\subset
I_{(n-1)k_{n-1}}$ for all $n\in\mathbb{N}$,
$\bigcap\limits_{n=1}^{\infty}I_{nk_{n}}\neq\emptyset$, say
$x\in\bigcap\limits_{n=1}^{\infty}I_{nk_{n}}$. Then $x\in E$ and
$x\notin\\{x_{1},x_{2},...,x_{n},....\\}=E$, a contradiction. Hence, $E$ is an
uncountable set. ∎

######  Corollary 6.2.5.

If $\ell(A)>0$ for some $A\in\mathcal{A}$, then $A$ contains exactly
$2^{\mathfrak{c}}$-many Lebesgue measurable sets.

###### Proof.

Clearly, $A$ cannot contain more than $2^{\mathfrak{c}}$-many sets, as
$A\subset\mathbb{R}$. By Theorem 6.2.4, $A$ contains an uncountable set of
measure zero, say $E$. Since $\ell$ is a complete measure, and $E$ is
uncountable, $E$ contains $2^{\mathfrak{c}}$-many Lebesgue measurable sets.
Consequently, $A$ contains $2^{\mathfrak{c}}$-many Lebesgue measurable sets. ∎

We already made a partition of
$\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$ by the equivalence relation
“$\sim$” given by: “$f\sim g$ if and only if $\ell(Z(f)\triangle Z(g))=0$”.
For each $f\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$, we consider
$[f]$ as the equivalence class of $f$ under the equivalence relation “$\sim$”
which is precisely the set
$\\{g\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A})):\ell(Z(f)\triangle
Z(g))=0\\}$. Clearly, for each
$f\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$,
$|[f]|\leq|\mathcal{M}(\mathbb{R},\mathcal{A})|=2^{\mathfrak{c}}$. $V$ is the
collection of all class representatives of the form $1_{A}$ under $\sim$, as
in all the previous cases.

######  Lemma 6.2.6.

For each $f\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$,
$|[f]|=2^{\mathfrak{c}}$.

###### Proof.

Let $f\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$. Then
$\ell(X\setminus Z(f))>0$. By Corollary 6.2.5, $X\setminus Z(f)$ contains
exactly $2^{\mathfrak{c}}$-many Lebesgue measurable sets. Let
$\\{A_{\alpha}:\alpha\in\Lambda\\}$ be the collection of all Lebesgue
measurable subsets of $X\setminus Z(f)$, where $|\Lambda|=2^{\mathfrak{c}}$.
For each $\alpha\in\Lambda$, let $f_{\alpha}:\mathbb{R}\to\mathbb{R}$ be
defined by

$f_{\alpha}(x)=\begin{cases}0&\text{ if }x\in Z(f)\\\ 1&\text{ if }x\in
A_{\alpha}\\\ 2&\text{ otherwise}\end{cases}$

Clearly $f_{\alpha}\in[f]$ for all $\alpha\in\Lambda$. Also for distinct
$\alpha,\beta\in\Lambda$, $f_{\alpha}\neq f_{\beta}$. Therefore $|[f]|\geq
2^{\mathfrak{c}}$. Hence $|[f]|=2^{\mathfrak{c}}$. ∎

######  Theorem 6.2.7.

In the Lebesgue measure space $(\mathbb{R},\mathcal{A},\ell)$, the zero-
divisor graph and the comaximal graph on $\mathcal{M}(\mathbb{R},\mathcal{A})$
are isomorphic.

###### Proof.

By Lemma 6.2.6, $|[f]|=2^{\mathfrak{c}}$ for all
$f\in\mathscr{D}(\mathcal{M}(\mathbb{R},\mathcal{A}))$. In particular
$|[1_{A}]|=|[1_{X\setminus A}]|$ for each $1_{A}\in V$. Hence by Theorem 3.34,
the zero-divisor graph and the comaximal graph on
$\mathcal{M}(\mathbb{R},\mathcal{A})$ are isomorphic. ∎

Since the Lebesgue measure on $\mathbb{R}$ is a non-atomic measure, we fell
tempted to ask the following question.

######  Question 6.2.8.

Suppose $(X,\mathcal{A},\mu)$ is a non-atomic measure space. Are the zero-
divisor graph and the comaximal graph of $\mathcal{M}(X,\mathcal{A})$
isomorphic?

## References

  * [1] Acharyya, S. K., Bag, S. and Sack, J., _Ideals of rings and intermediate rings of measurable functions_ , J. of Alg. and Its App., Vol. 19, No. 2 (2020), 2050038.
  * [2] Acharyya, S., Acharyya, S. K., Bag, S. and Sack, J., _Recent progress in rings and subrings of real valued measurable functions_ , Quaes. Math., Vol. 43(7) (2020), 959–973.
  * [3] Acharyya, S., Acharyya, S. K., Bharati, R. and Deb Ray, A. _Some algebraic and topological properties of rings of measurable functions_ , Houst. J. Math., Vol. 47, No. 3 (2021), 633-657.
  * [4] Anderson, D. F. and Livingston, P. S., _The zero-divisor graph of a commutative ring_ , J. of Alg., Vol. 217, Issue 2 (1999), 434–447.
  * [5] Badawi, A., _On the Annihilator Graph of a Commutative Ring_ , Comm. in Alg., Vol. 42, Issue 1(2014), 108–121.
  * [6] Badie, M., _Comaximal graph of $C(X)$_, Comm. Math. Univ. Carolin., Vol. 57, Issue 3 (2016), 353–364.
  * [7] Beck, I., _Coloring of commutative rings_ , J. of Alg., vol. 116, Issue 1 (1988), 208–226.
  * [8] Diestel, R., _Graph Theory_ , Springer, 2017.
  * [9] Hejazipour, H. and Naghipour, A. R., _Zero-divisor graph of the rings of real measurable functions with the measures_ , J. of Alg. Sys., Vol. 9, No. 2 (2022), 175–192.
  * [10] Johnson, R. A., _Atomic and nonatomic measures_ , Proc. Amer. Math. Soc., Vol 25 (1970), 650–655.
  * [11] Nikmehr, M. J., Azadi, A. and Nikandish, R., _The weakly zero-divisor graph of a commutative ring_ , Revi. de la Uni. Mat. Arg., Vol. 62, no. 1 (2021), 105–116.
  * [12] Sharma, P. K. and Bhatwadekar, S. M., _A Note on Graphical Representation of Rings_ , J. of Alg., Vol. 176, Issue 1 (1995), 124–127.