Kalmbach measurability in D-lattices

Abstract

We extend to D-lattices the definition of Kalmbach measurable elements with respect to an outer measure $\mu$. We prove, in case $\mu$ is faithful, that Kalmbach measurable elements are central, thus generalizing a result known for orthomodular lattices.

1 Introduction

In Kalmbach (1990), G. Kalmbach defines a measurability condition for elements of a dimension lattice, with respect to a $\sigma$-subadditive outer measure, and shows that elements satisfying such condition form a Boolean algebra.

Later on, in d’Andrea et al. (1994), the authors consider the same condition—which they call Kalmbach measurability—in any orthomodular lattice L, and with respect to any outer measure $\mu$. They show that, in case $\mu$ is faithful, the Kalmbach measurable elements form a (Boolean) subalgebra of the centre of L.

The present paper is concerned with D-lattices (or, equivalently, lattice-ordered effect algebras). We introduce a suitable generalization of Kalmbach measurability for elements of a D-lattice L, with respect to an outer measure $\mu$. Assuming that $\mu$ is faithful, we prove that Kalmbach measurable elements are precisely those central elements which are measurable in a quite natural sense; we also show that the Kalmbach measurable elements still form a (Boolean) subalgebra of the centre of L, thus extending the result of d’Andrea et al. (1994).

We recall that D-lattices have been introduced by Chovanec and Kôpka (1994) and effect algebras by Bennett and Foulis (1994). The equivalence of the two structures, D-lattices and lattice-ordered effect algebras, is shown in Dvurečenskij and Pulmannová (2000), 1.3.4.

D-lattices are a common generalization of orthomodular lattices and of MV algebras, hence of Boolean algebras. Then the investigation of modular measures on D-lattices (see for instance Avallone et al. 2003, 2009, 2006; Avallone and Vitolo 2003, 2005, 2009) gave results which may be applied in both fuzzy measure theory and noncommutative measure theory.

For the significance of D-lattices and the related bibliography, we refer to Dvurečenskij and Pulmannová (2000).

2 Preliminaries

Throughout, let L be a D-lattice, i.e. a lattice, with a greatest element 1 and a smallest element 0, endowed with a partial binary operation $\ominus$ such that for all $a, b, c \in L$

• (D1) $b\ominus a$ is defined if and only if $a\le b$,

• (D2) If $a\le b$, then $b\ominus a \le b$ and $b\ominus (b\ominus a) = a$,

• (D3) If $a\le b\le c$, then $c\ominus b\le c\ominus a$ and $(c\ominus a)\ominus (c\ominus b) = b\ominus a$.

For any $a,\,b\in L$, we set $a^\perp = 1\ominus a$ and call two elements $a, b \in L$ orthogonal if $a\le b^\perp$. When a and b are orthogonal, we write $a\perp b$. We have $\left( a^\perp \right) ^\perp = a$, and $a\le b$ implies $a^\perp \ge b^\perp$.

One defines in L a partial operation $\oplus$ by $a\oplus b = (a^\perp \ominus b)^\perp$ if $a\perp b$. The operation $\oplus$ is commutative and associative, and the structure $(L,\oplus ,0,1)$ is an effect algebra (see Bennett and Foulis 1994). As it is well known, a D-lattice is equivalent to a lattice-ordered effect algebra.

The following result summarizes some properties of D-lattices. For the proofs, we refer to Dvurečenskij and Pulmannová (2000).

Proposition 2.1

Let $a,b,c\in L$. Then

• (i) If $a\perp b$, then $a\le a\oplus b$ and $(a\oplus b)\ominus a=b$;

• (ii) If $a\le b$ and $b\perp c$, then $a\perp c$ and $a\oplus c\le b\oplus c$;

• (iii) If $a\le b$ and $b\perp c$, then $(b\oplus c)\ominus a=(b\ominus a)\oplus c$;

• (iv) If $a\le b\le c$, then $b\ominus a\le c\ominus a$ and $(c\ominus a)\ominus (b\ominus a)=c\ominus b$;

• (v) If $a\perp b$, then $a\vee b\perp a\wedge b$, and $a\oplus b=(a\vee b)\oplus (a\wedge b)$ (in particular, if $a\wedge b=0$, then $a\vee b=a\oplus b$);

• (vi) If $a\le c$ and $b\le c$, then $(c\ominus a)\wedge (c\ominus b)=c\ominus (a\vee b)$ and $(c\ominus a)\vee (c\ominus b)=c\ominus (a\wedge b)$.

Observe that letting $c=1$ in Proposition 2.1 (vi) gives the de Morgan laws:

$$\begin{aligned} a^\perp \wedge b^\perp =(a\vee b)^\perp \text { and }a^\perp \vee b^\perp =(a\wedge b)^\perp . \end{aligned}$$

We say that $p\in L$ is sharp if $p\wedge p^\perp =0$ (equivalently: $p\vee p^\perp =1$). The following fact is well known.

Theorem 2.2

An orthomodular lattice can be equivalently viewed as a D-lattice L in which all elements are sharp. In this case, we have $a\ominus b=a\wedge b^\perp$ for every $a,b\in L$ with $a\ge b$.

For $a,b\in L$, we say that a is compatible with b, and write $a \leftrightarrow b$ if there exist $a_*,b_*, c\in L$ such that

• (C1) $a_*\perp b_*$ and $a_*\oplus b_*\perp c$;

• (C2) $a=a_*\oplus c$ and $b=b_*\oplus c$.

It follows immediately from the definition that compatibility is a (reflexive and) symmetric relation. Compatibility in D-lattices con be characterized as follows (see for instance Chovanec and Kôpka 1997 or Dvurečenskij and Pulmannová 2000, Theor. 1.10.6).00

Proposition 2.3

For any $a\leftrightarrow b\in L$, one has $a, b$ if and only if

$$\begin{aligned} (a\vee b)\ominus b=a\ominus (a\wedge b). \end{aligned}$$

(2.1)

An element $s\in L$ is said to be central if for every $b\in L$ we have $b=(b\wedge s)\vee \left( b\wedge s^\perp \right)$. The set C(L) of all central elements is said to be the centre of L. By Dvurečenskij and Pulmannová (2000), Theor. 1.9.14, C(L) is a Boolean algebra and a subalgebra of L.

The following result has been proved by Z. Riečanová in Riečanová (1999), Theor. 2.5.

Theorem 2.4

An element $p\in L$ is central if and only the following conditions hold:

• (a) p is sharp;

• (b) for every $a\in L$ one has $a\leftrightarrow p$.

If $\mathbb G$ is a topological group, a measure on a D-lattice L is a map $\mu :A\rightarrow \mathbb G$ such that $\mu (a\ominus b)=\mu (a)-\mu (b)$ whenever a, b are elements of L with $b\le a$.

It is immediate to verify that $\mu (0)=0$ for any measure $\mu$. Also, using the effect algebra structure of L, it is clear that a measure can be characterized by the following property:

$$\begin{aligned} \mu (a\oplus b)=\mu (a)+\mu (b), \text { whenever } a\perp b. \end{aligned}$$

(2.2)

Moreover, any nonnegative real-valued measure $\mu$ on A is monotone.

Following (d’Andrea et al. 1994; Kalmbach 1990), we define an outer measure as a monotone (nonnegative) real-valued function $\mu$ on L, with $\mu (0)=0$, which is also subadditive, i.e.

$$\begin{aligned} \mu (a\oplus b)\le \mu (a)+\mu (b), \text { whenever } a\perp b. \end{aligned}$$

(2.3)

In view of Proposition 2.1 (v) and Theorem 2.2, when L is an orthomodular lattice, $\mu$ is subadditive if and only if

$$\begin{aligned} \mu (a\vee b)\le \mu (a)+\mu (b), \text { whenever } a\wedge b=0. \end{aligned}$$

We say that $\mu$ is faithful if $\mu (a)=0$ implies $a=0$.

3 Basic facts

We collect here some results which will be useful in the sequel. Most of them can be found in the literature—our major reference is (Dvurečenskij and Pulmannová 2000). Anyway, in some cases we prefer to give a proof, in order to make the present paper more self-contained.

Proposition 3.1

If an element $c\in L$ is central then, for every $a\in L$,

$$\begin{aligned} a=(a\wedge c)\oplus \left( a\wedge c^\bot \right) , \text { that is }a\wedge c=a\ominus \left( a\wedge c^\bot \right) . \end{aligned}$$

Proof

See (Dvurečenskij and Pulmannová 2000, Lemma 1.9.12). $\square$

Proposition 3.2

For every $a,b\in L$, the following hold:

• (1) $\left( \bigl ((a\vee b)\ominus b\bigr )\vee \bigl ((b\vee a)\ominus a\bigr )\right) \ominus \bigl ((b\vee a)\ominus a\bigr )=a\ominus (a\wedge b)$;

• (2) $\bigl ((a\vee b)\ominus b\bigr )\ominus \left( \bigl ((a\vee b)\ominus b\bigr ) \wedge \bigl ((b\vee a)\ominus a\bigr )\right) =(a\vee b)\ominus b$.

Proof

Indeed using Proposition 2.1 (vi) and (D3), we have

$$\begin{aligned}&\left( \bigl ((a\vee b)\ominus b\bigr )\vee \bigl ((b\vee a)\ominus a\bigr )\right) \ominus \bigl ((b\vee a)\ominus a\bigr )\\&\quad =\bigl ((a\vee b)\ominus (a\wedge b)\bigr )\ominus \bigl ((a\vee b)\ominus a\bigr )\\&\quad =a\ominus (a\wedge b), \end{aligned}$$

which proves (1). To prove (2) it suffices to observe that, by Proposition 2.1 (vi), we have $\bigl ((a\vee b)\ominus b\bigr )\wedge \bigl ((b\vee a)\ominus a\bigr )=0$. $\square$

An alternative proof of the foregoing proposition may be obtained by applying the de Morgan laws to the interval $[0,a\vee b]$ (which is a D-lattice on its own).

Lemma 3.3

For every $c,d\in L$, we have

$$\begin{aligned} \left( \,c\vee \bigl ((c\vee d)\ominus d\bigr )\right) \ominus \bigl ((c\vee d)\ominus d\bigr ) \le d. \end{aligned}$$

Proof

Indeed, applying (D3) and (D2)

$$\begin{aligned}&\left( \,c\vee \bigl ((c\vee d)\ominus d\bigr )\right) \ominus \bigl ((c\vee d)\ominus d\bigr )\\&=\left( (c\vee d)\ominus \bigl ((c\vee d)\ominus d\bigr )\right) \\&\quad \ominus \Bigl ( (c\vee d)\ominus \left( \,c\vee \bigl ((c\vee d)\ominus d\bigr )\right) \Bigr )\\ {}&=d\ominus \Bigl ((c\vee d)\ominus \left( \,c\vee \bigl ((c\vee d)\ominus d\bigr )\right) \Bigr )\\ {}&\le d. \end{aligned}$$

$\square$

The previous lemma also may be proved arguing from a different viewpoint: simply do the calculations in the interval $[0,c\vee d]$.

Lemma 3.4

For every $c,d\in L$, if $(c\vee d)\ominus d\le c$ then $c\ominus (c\wedge d)\le (c\vee d)\ominus d$.

Proof

Let $s=(c\vee d)\ominus d$. Since $s\le c$ by assumption, (D2) gives $c\ominus s\le c$, and

$$\begin{aligned} c\ominus (c\ominus s)=s. \end{aligned}$$

(3.1)

From the previous lemma, we get $c\ominus s=(c\vee s)\ominus s\le d$, so that $c\ominus s\le c\wedge d$. Therefore, by (3.1) and Proposition 2.1 (iv), we obtain

$$\begin{aligned} c\ominus (c\wedge d)\le c\ominus (c\ominus s)=s. \end{aligned}$$

$\square$

Proposition 3.5

Let $a\leftrightarrow b\in L$. Then $a\leftrightarrow b$ if and only if $(a\vee b)\ominus b\le a$.

Proof

Suppose first $a\leftrightarrow b$. From Proposition 2.3, it follows that $(a\vee b)\ominus b=a\ominus (a\wedge b)$. Thus, $(a\vee b)\ominus b\le a$ by virtue of (D2).

Conversely, assume $(a\vee b)\ominus b\le a$. The previous lemma (with a in place of c and b in place of d) yields

$$\begin{aligned} a\ominus (a\wedge b)\le (a\vee b)\ominus b \end{aligned}$$

(3.2)

Now let $c=(a\vee b)\ominus b$ and $d=(a\vee b)\ominus a$. By Proposition 3.2 (1) and (3.2), we have

$$\begin{aligned} (c\vee d)\ominus d=a\ominus (a\wedge b)\le (a\vee b)\ominus b=c; \end{aligned}$$

therefore, applying the previous lemma and Proposition 3.2, we obtain

$$\begin{aligned} (a\vee b)\ominus b=c\ominus (c\wedge d)\le (c\vee d)\ominus d=a\ominus (a\wedge b) \end{aligned}$$

which, together with (3.2), gives $(a\vee b)\ominus b=a\ominus (a\wedge b)$, and hence, $a \leftrightarrow b$ in view of Proposition 2.3. $\square$

Corollary 3.6

For $a,b\in L$, let

$$\begin{aligned} F(a,b)\!=\! \left( a^\bot \ominus \left( a^\bot \wedge b^\bot \right) \right) \!\ominus \! \left( \left( a^\bot \ominus \left( a^\bot \wedge b^\bot \right) \right) \!\wedge \! b\right) .\nonumber \\ \end{aligned}$$

(3.3)

Then $a\leftrightarrow b$ if and only if $F(a,b)=0$.

Proof

Set $G(a,b)=a^\bot \ominus \left( a^\bot \wedge b^\bot \right)$; from (D3) and Proposition 2.1 (vi), it follows that $G(a,b)=(a\vee b)\ominus a$. Now, since $F(a,b)= G(a,b)\ominus \bigl (G(a,b)\wedge b\bigr )$, we have $F(a,b)=0$ if and only if $G(a,b)=G(a,b)\wedge b$, that is

$G(a,b)\le b$. By the previous proposition, this is in turn equivalent to $b\leftrightarrow a$, i.e. $a\leftrightarrow b$. $\square$

Remark 3.7

For the sake of brevity, let $u(a,b)=a\ominus (a\wedge b)$ and $v(a,b)=(a\vee b)\ominus b$. In the foregoing proof, we have seen that $G(a,b)=u \left( a^\bot ,b^\bot \right) =v(b,a)$.

In Jenča and Pulmannová (2001), the following notion of e-commutator is introduced:

$$\begin{aligned} {{\,\mathrm{com}\,}}_e(a,b)=u\bigl (u(a,b),v(a,b)\bigr )\vee u\bigl (v(a,b),u(a,b)\bigr ). \end{aligned}$$

Using the notation of the above corollary, we now show that ${{\,\mathrm{com}\,}}_e(a,b)\ge F(b,a)$.

Indeed, observing that $u(a,b)\le a$, we have

$$\begin{aligned}&{{\,\mathrm{com}\,}}_e(a,b)\ge u\bigl (v(a,b),u(a,b)\bigr )\\&\quad =v(a,b)\ominus \bigl (v(a,b)\wedge u(a,b)\bigr )\\&\quad \ge v(a,b)\ominus \bigl (v(a,b)\wedge a\bigr ) =G(b,a)\ominus \bigl (G(b,a)\wedge a\bigr )\\&\quad =F(b,a). \end{aligned}$$

4 Kalmbach measurable elements

In what follows, $\mu :L\rightarrow \mathbb {R}$ will always be a fixed outer measure.

• We say that an element $p\in L$ is Kalmbach measurable (with respect to $\mu$) if for every $a\in L$

  $$\begin{aligned} \mu (a)=\mu \left( a\ominus \left( a\wedge p^\bot \right) \right) +\mu \left( a\ominus (a\wedge p)\right) . \end{aligned}$$

• We say that $p\in L$ is measurable (with respect to $\mu$) if for every $a\in L$

  $$\begin{aligned} \mu (a)=\mu (a\wedge p)+\mu \bigl (a\ominus (a\wedge p)\bigr ). \end{aligned}$$

  (4.1)

For the sake of brevity, we denote by $K_{\mu }(L)$ the set of Kalmbach measurable elements and by $M_{\mu }(L)$ the set of measurable ones.

Our definition of measurable elements is motivated by the following fact.

Proposition 4.1

An outer measure $\mu$ is a measure if and only if every element is measurable.

Proof

Suppose $\mu$ is a measure, and fix $a\in L$. Since for every $p\in L$ we has $a=(a\wedge p)\oplus \bigl (a\ominus (a\wedge p)\bigr )$, applying (2.2) we get immediately (4.1).

Conversely, we assume that every element in L is measurable and prove that (2.2) holds. Consider any $a,b\in L$, with $a\perp b$, and let $s=a\oplus b$; one readily sees that $s\wedge a=a$ and $s\ominus (s\wedge a)=s\ominus a=b$. Hence, measurability of a yields

$$\begin{aligned}&\mu (a\oplus b)=\mu (s)=\mu (s\wedge a)+\mu \bigl (s\ominus (s\wedge a)\bigr )\\&\quad =\mu (a)+\mu (b). \end{aligned}$$

$\square$

In view of Proposition 2.1 (v) and Theorem 2.2, when L is an orthomodular lattice, p is Kalmbach measurable if and only if

$$\begin{aligned} \forall a\in L\quad \mu (a)=\mu \left( a\wedge \left( a^\bot \vee p \right) \right) +\mu \left( a\wedge \left( a^\bot \vee p^\bot \right) \right) .\nonumber \\ \end{aligned}$$

(4.2)

Actually this is how Kalmbach measurable elements are defined in Kalmbach (1990) and in d’Andrea et al. (1994), where orthomodular lattices are dealt with.

In the general case, our definition of Kalmbach measurable elements is to be preferred to property (4.2), which is not very effective, as we are going to see.

Proposition 4.2

Suppose that $\mu$ is faithful and (4.2) is satisfied for $p=0$. Then L is an orthomodular lattice.

Proof

Indeed, letting $p=0$ in (4.2) gives

$$\begin{aligned} \forall a\in L\quad \mu (a)=\mu (a\wedge a^\bot )+\mu (a) \end{aligned}$$

whence $\mu (a\wedge a^\bot )=0$, so that $a\wedge a^\bot =0$ because $\mu$ is faithful. Now Theorem 2.2 yields the conclusion. $\square$

Now in order to approach our main result, some preliminary facts are needed.

Lemma 4.3

If $p\in L$ is Kalmbach measurable and $s\le p\wedge p^\bot$, then

• (1) $\mu (s)=0$;

• (2) for every $r\perp s$ we have $\mu (r\oplus s)=\mu (r)$;

• (3) for every $t\ge s$ we have $\mu (t\ominus s)=\mu (t)$.

Proof

First of all, since $s\wedge p=s$ and $s\wedge p^\bot =s$, we have

$$\begin{aligned}&\mu (s)=\mu \left( s\ominus \left( s\wedge p^\bot \right) \right) +\mu \bigl (s\ominus (s\wedge p)\bigr )\\&\quad =\mu (s\ominus s)+\mu (s\ominus s)=\mu (0)+\mu (0)=0 \end{aligned}$$

and (1) is proved. Now note that (2) and (3) are equivalent, thus we just prove (2).

Let $t=r\oplus s$, so that $t\ge r$. Using (1), we get

$$\begin{aligned} \mu (r)\le \mu (t) =\mu (r\oplus s) \le \mu (r)+\mu (s)=\mu (r). \end{aligned}$$

$\square$

Lemma 4.4

For every $a,b\in L$, the following hold:

$$\begin{aligned} a\wedge b\le \left( a\ominus \left( a\wedge b^\bot \right) \right) \oplus \left( a\wedge b\wedge b^\bot \right) . \end{aligned}$$

Proof

Indeed, applying Items (v), (iii) and (ii) of Proposition 2.1, we obtain:

$$\begin{aligned}&a\wedge b\\ {}&\quad =\left( (a\wedge b)\oplus \left( a\wedge b^\bot \right) \right) \ominus \left( a\wedge b^\bot \right) \\ {}&\quad =\left( \left( (a\wedge b)\vee \left( a\wedge b^\bot \right) \right) \oplus \left( (a\wedge b)\wedge \left( a\wedge b^\bot \right) \right) \right) \\&\qquad \ominus \left( a\wedge b^\bot \right) \\ {}&\quad =\left( \left( (a\wedge b)\vee \left( a\wedge b^\bot \right) \right) \oplus \left( a\wedge b\wedge b^\bot \right) \right) \ominus \left( a\wedge b^\bot \right) \\ {}&\quad =\left( \left( (a\wedge b)\vee \left( a\wedge b^\bot \right) \right) \ominus \left( a\wedge b^\bot \right) \right) \oplus \left( a\wedge b\wedge b^\bot \right) \\ {}&\quad \le \left( a\ominus \left( a\wedge b^\bot \right) \right) \oplus \left( a\wedge b\wedge b^\bot \right) . \end{aligned}$$

$\square$

Proposition 4.5

If $p\in L$ is Kalmbach measurable, then for every $x\in L$ we have

• (1) $\mu (x\wedge p)=\mu \left( x\ominus \left( x\wedge p^\bot \right) \right)$;

• (2) $\mu \left( x\wedge p^\bot \right) =\mu \left( x\ominus \left( x\wedge p\right) \right)$.

Proof

Clearly p is Kalmbach measurable if and only if $p^\bot$ is: therefore, it suffices to prove (1). Suppose $p\in K_{\mu }(L)$, and consider any $x\in L$. First, observe that

$$\begin{aligned} x=\left( x\ominus \left( x\wedge p \right) \right) \oplus (x\wedge p) \end{aligned}$$

and hence

$$\begin{aligned}&\mu \left( x\ominus \left( x\wedge p^\bot \right) \right) +\mu \left( x\ominus (x\wedge p)\right) \\&\quad =\mu (x)\le \mu \left( x\ominus \left( x\wedge p \right) \right) +\mu (x\wedge p). \end{aligned}$$

It follows that

$$\begin{aligned} \mu (x\wedge p)\ge \mu \left( x\ominus \left( x\wedge p^\bot \right) \right) . \end{aligned}$$

(4.3)

Now, by Lemma 4.4 and Lemma 4.3, we have

$$\begin{aligned} \begin{aligned} \mu (x\wedge p)&\le \mu \left( \left( x\ominus \left( x\wedge p^\bot \right) \right) \oplus \left( x\wedge p\wedge p^\bot \right) \right) \\ {}&=\mu \bigl (x\ominus (x\wedge p^\bot )\bigr ). \end{aligned} \end{aligned}$$

(4.4)

Putting together (4.3) and (4.4) gives the assertion. $\square$

A consequence of the foregoing proposition is that $K_{\mu }(L)\subseteq M_{\mu }(L)$.

Corollary 4.6

If $p\in L$ is Kalmbach measurable, then for every $x\in L$ we have

• (1) $\mu (x)=\mu (x\wedge p)+\mu \bigl (x\ominus (x\wedge p)\bigr )$, i.e. p is measurable;

• (2) $\mu (x)=\mu (x\wedge p)+\mu \left( x\wedge p^\bot \right)$.

Proof

It follows immediately from Proposition 4.5. $\square$

The following is the final step towards our main result.

Proposition 4.7

Let $p\in L$ be Kalmbach measurable. For every $e\in L$, we have

$$\begin{aligned} \mu \left( \left( e\ominus \left( e\wedge p^\bot \right) \right) \ominus \left( \left( e\ominus \left( e\wedge p^\bot \right) \right) \wedge p\right) \right) =0. \end{aligned}$$

(4.5)

Proof

Fix $e\in L$, and define $e_*=e\ominus \left( e\wedge p^\bot \right)$, so that (4.5) becomes

$$\begin{aligned} \mu \bigl (e_*\ominus (e_*\wedge p)\bigr )=0. \end{aligned}$$

(4.6)

First note that from Lemma 4.4, using Proposition 2.1 (iv), we get

$$\begin{aligned} (e\wedge p)\ominus \left( e\wedge p\wedge p^\bot \right) \le e\ominus \left( e\wedge p^\bot \right) =e_* \end{aligned}$$

hence

$$\begin{aligned} (e\wedge p)\ominus \left( e\wedge p\wedge p^\bot \right)&=\left( (e\wedge p)\ominus \left( e\wedge p\wedge p^\bot \right) \right) \wedge p\\ {}&\le e_*\wedge p\le e\wedge p, \end{aligned}$$

and applying Lemma 4.3, we have

$$\begin{aligned} \mu (e\wedge p)&=\mu \left( (e\wedge p)\ominus \left( e\wedge p\wedge p^\bot \right) \right) \\ {}&\le \mu (e_*\wedge p)\le \mu (e\wedge p) \end{aligned}$$

that is $\mu (e_*\wedge p)=\mu (e\wedge p)$.

Therefore using Proposition 4.5 and Corollary 4.6(1) , we obtain

$$\begin{aligned} \mu (e\wedge p)&=\mu \left( e\ominus \left( e\wedge p^\bot \right) \right) =\mu (e_*)\\ {}&=\mu (e_*\wedge p)+\mu \left( e_*\ominus (e_*\wedge p)\right) \\ {}&=\mu (e\wedge p)+\mu \left( e_*\ominus (e_*\wedge p)\right) \end{aligned}$$

which gives (4.6). $\square$

Now we are ready to prove that $K_{\mu }(L)=M_{\mu }(L)\cap C(L)$, assuming that $\mu$ is faithful.

Theorem 4.8

Let $p\in L$ be measurable and central. Then p is Kalmbach measurable. If in addition $\mu$ is faithful, the converse also holds.

Proof

Suppose p is measurable and central. By Proposition 3.1, we have, for every $a\in L$,

$$\begin{aligned}&\quad \mu (a)=\mu (a\wedge p)+\mu \bigl (a\ominus (a\wedge p)\bigr )\\&\quad =\mu \left( a\ominus \left( a\wedge p^\bot \right) \right) +\mu \left( a\ominus (a\wedge p)\right) . \end{aligned}$$

Conversely, if p is Kalmbach measurable then it is measurable by Corollary 4.6 (1) (without extra assumptions on $\mu$). Now we assume that $\mu$ is faithful. From Lemma 4.3 (1), we get $\mu \left( p\wedge p^\bot \right) =0$. Hence, $p\wedge p^\bot =0$, i.e. p is sharp. Taking into account Theorem 2.4, it remains to prove that every element of L is compatible with p.

Let a be any element of L. In view of Corollary 3.6, we have to show that $F(a,p)=0$, where F is defined in (3.3) above. Setting $e=a^\bot$, one immediately sees that Proposition 4.7 gives $\mu \left( F(a,p)\right) =\mu \left( F \left( e^\bot ,p \right) \right) =0$: hence, $F(a,p)=0$ because $\mu$ is faithful. This completes the proof. $\square$

Corollary 4.9

A central element is Kalmbach measurable if and only if it is measurable.

Proof

We have $K_{\mu }(L)\subseteq M_{\mu }(L)$ by virtue of Corollary 4.6 (1), and the previous theorem says that $M_{\mu }(L)\cap C(L)\subseteq K_{\mu }(L)$. Hence, $K_{\mu }(L)\cap C(L)=M_{\mu }(L)\cap C(L)$. (Notice that there is no need to assume $\mu$ faithful.) $\square$

Another characterization of Kalmbach measurable elements among central elements is the following.

Proposition 4.10

A central element p is Kalmbach measurable if and only if, for each $x\in L$

$$\begin{aligned} \mu (x)=\mu (x\wedge p)+\mu (x\wedge p^\bot ). \end{aligned}$$

(4.7)

Proof

Necessity follows from Corollary 4.6(2). Conversely, suppose that (4.7) hold. Since $p^\bot$ also is central, Proposition 3.1 gives, for all $x\in L$,

$$\begin{aligned} x\wedge p=x\ominus \left( x\wedge p^\bot \right) \text { and } x\wedge p^\bot =x\ominus (x\wedge p); \end{aligned}$$

hence, $\mu (x)=\mu \left( x\ominus \left( x\wedge p^\bot \right) \right) +\mu \left( x\ominus \left( x\wedge p \right) \right)$. $\square$

Consequently, we can see that $K_{\mu }(L)\cap C(L)$ is a subalgebra of the centre of L.

Proposition 4.11

The Kalmbach measurable central elements of L form a subalgebra of C(L).

Proof

Let p and q be Kalmbach measurable central elements of L; clearly, $p^\bot$ and $q^\bot$ are Kalmbach measurable and central, too; since C(L) is a Boolean subalgebra of L, it suffices to show that $p\wedge q$ is Kalmbach measurable.

Using the fact that C(L) is a Boolean algebra, we may write

$$\begin{aligned}&\left( p^\bot \vee q^\bot \right) \wedge p = \left( p^\bot \wedge p \right) \vee \left( q^\bot \wedge p \right) \nonumber \\&\quad =0\vee \left( q^\bot \wedge p \right) =p\wedge q^\bot . \end{aligned}$$

(4.8)

Now, for every $x\in L$, since both p and q are Kalmbach measurable, applying Proposition 3.1, (4.8) and Corollary 4.6(2), we have

$$\begin{aligned}&\mu (x\wedge p\wedge q)+\mu \left( x\wedge \left( p\wedge q \right) ^\bot \right) \\&\quad =\mu (x\wedge p\wedge q)+\mu \left( x\wedge \left( p^\bot \vee q^\bot \right) \right) \\&\quad =\mu (x\wedge p\wedge q)+\mu \left( \left( x\wedge \left( p^\bot \vee q^\bot \right) \wedge p \right) \right) \\&\qquad +\mu \left( \left( x\wedge \left( p^\bot \vee q^\bot \right) \wedge p^\bot \right) \right) \\&\quad =\mu (x\wedge p\wedge q)+\mu \left( x\wedge p\wedge q^\bot \right) +\mu \left( x\wedge p^\bot \right) \\&\quad =\mu (x\wedge p)+\mu \left( x\wedge p^\bot \right) =\mu (x). \end{aligned}$$

In view of Proposition 4.10, we conclude that $p\wedge q$ is Kalmbach measurable, as desired. $\square$

Corollary 4.12

If $\mu$ is faithful, then the Kalmbach measurable elements of L form a subalgebra of C(L).

Proof

Indeed, the previous theorem says that $K_{\mu }(L)\cap C(L)$ is a subalgebra of C(L); but, if $\mu$ is faithful, we have seen in Theorem 4.8 that $K_{\mu }(L)\cap C(L)=K_{\mu }(L)$. $\square$

As a conclusion, it might be interesting to ask whether the results of the present paper can be extended to more general structures than D-lattices, in particular effect algebras. We plan to investigate this in a subsequent paper.