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Consider the following generalization of the concept of a measure:

Let $L = (X, \lor, \land, \bot)$ be a semi-bounded lattice. Let $M = (Y, \bullet, e)$ be a commutative monoid. An $(L, M)$-measure is a function $\mu : X \to Y$ that satisfies the following properties.

$\mu$ is modular: \begin{align} \mu(a \lor b) \bullet \mu(a \land b) = \mu(a) \bullet \mu(b) \end{align}

$\mu$ maps bottom to identity: \begin{align} \mu(\bot) = e \end{align}

$\mu$ is continuous: Assume $X$ and $Y$ are convergence spaces. Let $I$ be a directed set and $f : I \to X$ be a net. Then \begin{align} f \to a \quad \Longrightarrow \quad \mu \circ f \to \mu(a) \end{align}

Examples: Let $L = (\Sigma, \cup, \cap, \varnothing)$ where $\Sigma$ is a sigma algebra and $M = (Y, +, 0)$. Then:

$Y$ name
$\mathbb{R}_{\geq 0}$ plain measure
$\mathbb{R}$ signed measure
$\mathbb{C}$ complex measure
a vector space vector measure

Does this concept have a name? Has it been discussed in the literature?

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I do not have reference for your type of measure, but I have a version of the Riesz representation theorem for all such measures when the lattice is distributive. In this post, we shall not assume that the mapping $\mu$ maps the least element of the lattice to the identity of the monoid.

From each lattice $(X,\wedge,\vee)$, we construct a commutative monoid $\mathbf{M}(X)$ with $X\subseteq\mathbf{M}(X)$ such that each modular map $\mu:X\rightarrow M$ extends to a monoid homomorphism $\overline{\mu}:\mathbf{M}(X)\rightarrow M$ in the usual way. When $X$ is distributive, $\mathbf{M}(X)$ is the structure monoid for a set theoretic solution to the Yang-Baxter equation.

Suppose that $T$ is a set and $T:X^2\rightarrow X^2$. Let $X^\ast$ denote the collection of all possibly empty sequences $(x_1,\dots,x_n)$ of elements in $X$. Let $\simeq$ be the smallest congruence on the monoid $X^\ast$ where $(x,y)\simeq T(x,y)$ whenever $x,y\in X$, and let $\mathbf{M}(X,T)=X^\ast/\simeq$. We shall write $\mathbf{M}(X,\ast_1,\ast_2)$ whenever $\ast_1,\ast_2$ are the binary operations where $T(x,y)=(x\ast_1y,x\ast_2y)$, and we shall write $\simeq_T$ to specify the function $T$.

Suppose that $T(x,y)=(x\vee y,x\wedge y)$. We observe that $[x,y]=[x\vee y,x\wedge y]=[y,x]$, so the monoid $\mathbf{M}(X,T)=\mathbf{M}(X,\vee,\wedge)$ is commutative. If $\mu:X\rightarrow M$, then we extend $\mu$ to a monoid homomorphism $\overline{\mu}:\mathbf{M}(X,T)\rightarrow M$ by setting $\overline{\mu}([x_1,\dots,x_r])=\mu(x_1)+\dots+\mu(x_r)$.

Suppose that $X$ is a set and $T:X^2\rightarrow X^2$ is a function. Then we say that $T$ is a set theoretic solution to the Yang-Baxter equation if $$(1_X\times T)\circ(T\times 1_X)\circ(1_X\times T)=(T\times 1_X)\circ(1_X\times T)\circ(T\times 1_X).$$

The motivation behind the Yang-Baxter equation is that a function $T:X^2\rightarrow X^2$ is a set theoretic solution to the Yang-Baxter equation precisely when for all $n$, we have an action of the positive braid monoid $B_n^+$ on $X^n$ defined by setting $$(x_1,\dots,x_n)\bullet\sigma_i=(x_1,\dots,x_{i-1},T(x_i,x_{i+1}),x_{i+2}\dots,x_n).$$

Whenever $b,c\in B_n^+$, there are $r,s\in B_n^+$ where $bc=rs$. As a corollary, if $T$ is a set theoretic solution to the Yang-Baxter equation, then $(x_1,\dots,x_n)\simeq_T(y_1,\dots,y_n)$ precisely when there are positive braids $b,c\in B_n^+$ where $(x_1,\dots,x_n)\bullet b=(y_1,\dots,y_n)\bullet c$.

Lemma: (median identity) A lattice is distributive if and only if it satisfies the following identity known as the median identity: $$(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)\wedge(y\vee z).$$

Proposition: In a lattice $(X,\wedge,\vee)$, the operation $(x,y)\mapsto(x\vee y,x\wedge y)$ satisfies the Yang-Baxter equation if and only if the lattice is distributive.

Proof: We have $(a,b,c)\bullet\sigma_1\sigma_2\sigma_1=(a\vee b,a\wedge b,c)\bullet\sigma_2\sigma_1=(a\vee b,(a\wedge b)\vee c,a\wedge b\wedge c)\bullet\sigma_1=(a\vee b\vee c,(a\vee b)\wedge((a\wedge b)\vee c),a\wedge b\wedge c)$.

$(a,b,c)\bullet\sigma_2\sigma_1\sigma_2=(a,b\vee c,b\wedge c)\bullet\sigma_1\sigma_2 =(a\vee b\vee c,a\wedge(b\vee c),b\wedge c)\bullet\sigma_2=(a\vee b\vee c,(b\wedge c)\vee((a\wedge(b\vee c)),a\wedge b\wedge c).$

Therefore, $(X,\wedge,\vee)$ satisfies the Yang-Baxter equation if and only if the lattice $X$ satisfies the identity $(a\vee b)\wedge((a\wedge b)\vee c)=(b\wedge c)\vee((a\wedge(b\vee c))$ which we shall call the YB-distributivity identity.

In a distributive lattice, using the median identity, we have $$(a\vee b)\wedge((a\wedge b)\vee c) =(a\vee b)\wedge(a\vee c)\wedge(b\vee c) =(b\wedge c)\vee(a\wedge b)\vee(a\wedge c) =(b\wedge c)\vee((a\wedge(b\vee c)),$$ so every distributive lattice satisfies the YB-distributivity identity.

Now, suppose that we are in a lattice that satisfies the YB-distributivity identity. Then $(a\vee b)\wedge((a\wedge b)\vee c)=(b\wedge c)\vee(a\wedge(b\vee c))=(a\vee c)\wedge((a\wedge c)\vee b)$. More generally, $$(a_{\sigma(1)}\vee a_{\sigma(2)})\wedge((a_{\sigma(1)}\wedge a_{\sigma(2)})\vee a_{\sigma(3)})=(a_{\tau(1)}\wedge a_{\tau(2)})\vee((a_{\tau(1)}\vee a_{\tau(2)})\wedge a_{\tau(3)})$$ whenever $\sigma,\tau\in S_3$. Thus, $$(a_1\vee a_2)\wedge(a_1\vee a_3)\wedge(a_2\vee a_3)=\bigwedge_{\sigma\in S_3}((a_{\sigma(1)}\vee a_{\sigma(2)})\wedge((a_{\sigma(1)}\wedge a_{\sigma(2)})\vee a_{\sigma(3)}))$$ $$=\bigvee_{\tau\in S_3}((a_{\tau(1)}\wedge a_{\tau(2)})\vee((a_{\tau(1)}\vee a_{\tau(2)})\wedge a_{\tau(3)}))=(a_1\wedge a_2)\vee(a_1\wedge a_3)\vee(a_2\wedge a_3).$$

Since the lattice satisfies the median identity, the lattice must also be distributive.

Q.E.D.

In the case when $X$ is a linearly ordered set, the operation $(x,y)\mapsto(x\vee y,x\wedge y)$ is just a sorting operation, and the action of the positive braid monoid on $X$ is simply the sorting operation. Let $\delta_n=\sigma_n\dots\sigma_1$ and let $\Delta_n=\delta_1\dots\delta_{n-1}$. Then the list $(x_1,\dots,x_n)\bullet\Delta_n$ is the list obtained from $(x_1,\dots,x_n)$ by sorting $(x_1,\dots,x_n)$ from the greatest element to the least element. In particular, if $b$ is a positive braid, then $(x_1,\dots,x_n)\bullet\Delta_n\cdot b=(x_1,\dots,x_n)\bullet\Delta_n$. Since the variety of distributive lattices is generated by the 2 element linearly ordered distributive lattice, we conclude that $(x_1,\dots,x_n)\bullet\Delta_n\cdot b=(x_1,\dots,x_n)\bullet\Delta_n$ in the variety of all distributive lattices. As a consequence, for distributive lattices, we conclude that $(x_1,\dots,x_n)\simeq(y_1,\dots,y_n)$ if and only if $(x_1,\dots,x_n)\bullet\Delta_n=(y_1,\dots,y_n)\bullet\Delta_n$. Furthermore, in distributive lattices, for each $(x_1,\dots,x_n)\in X^*$, the element $(x_1,\dots,x_n)\bullet\Delta_n$ is the unique member $(y_1,\dots,y_n)$ of the equivalence class $[x_1,\dots,x_n]$ with $y_1\geq\dots\geq y_n$.

If $X$ is a distributive lattice, then let $\mathbf{M}^\sharp(X)$ be the monoid consisting of all (possibly empty) sequences $(x_1,\dots,x_n)$ where $x_1\geq\dots\geq x_n$ and with an operation $\bullet$ defined by letting $$(x_1,\dots,x_m)\bullet(y_1,\dots,y_n)=(x_1,\dots,x_m,y_1,\dots,y_n)\bullet\Delta_{m+n-1}.$$ Then the monoid $\mathbf{M}^\sharp(X)$ is canonically isomorphic to the monoid $\mathbf{M}(X,\vee,\wedge)$.

Suppose now that $X$ is a set and $\mathcal{A}$ is a collection of subsets of $X$ closed under taking the union of two sets and the intersection of two sets.

Let $\mathbf{M}^!(X,\mathcal{A})$ be the collection of pairs $(f,n)$ where $n\geq 0$ and $f:X\rightarrow\\{0,\dots,n\\}$ is a function such that $f^{-1}[\\{i,\dots,n\\}]\in\mathcal{A}$ for $1\leq i\leq n$.

Give $\mathbf{M}^!(X,\mathcal{A})$ the operation $+$ defined by letting $(f,m)+(g,n)=(f+g,m+n)$. The operation $+$ is clearly associative and commutative.

Define a $\Gamma:\mathbf{M}^!(X,\mathcal{A})\rightarrow \mathbf{M}(\mathcal{A})$ by setting $$\Gamma((f,n))=([f^{-1}[\\{1,\dots,n\\}],\dots,f^{-1}[\\{n\\}]],n).$$

Define a function $\Delta:\mathbf{M}(\mathcal{A})\rightarrow\mathbf{M}^!(X,\mathcal{A})$ by setting $$\Delta([A_1,\dots,A_n])=(\chi_{A_1}+\dots+\chi_{A_n},n).$$ We observe that this definition does not depend on the equivalence class that we choose.

Proposition: The functions $\Gamma,\Delta$ are inverse monoid homomorphisms.

Proof: Observe that $\chi_{f^{-1}[\\{1,\dots,n\\}]}+\dots+\chi_{f^{-1}[\\{n\\}]}=f.$

Therefore, $$\Delta\Gamma(f,n)=\Delta([f^{-1}[\{1,\dots,n\}],\dots,f^{-1}[\{n\}]],n) =(\chi_{f^{-1}[\{1,\dots,n\}]}+\dots+\chi_{f^{-1}[\{n\}]},n)=(f,n).$$

Suppose that $\mathbf{x}\in\mathbf{M}(\mathcal{A})$. Then there are $A_1\supseteq A_2\supseteq\dots\supseteq A_n$ with $\mathbf{x}=[A_1,\dots,A_n]$. In this case, $(\chi_{A_1}+\dots\chi_{A_n})^{-1}[\{i,\dots,n\}]=A_i$, so $$\Gamma\Delta(\mathbf{x})=\Gamma\Delta([A_1,\dots,A_n])=\Gamma(\chi_{A_1}+\dots+\chi_{A_n},n)=[(\chi_{A_1}+\dots+\chi_{A_n})^{-1}[\\{1,\dots,n\\}],\dots,(\chi_{A_1}+\dots+\chi_{A_n})^{-1}[\\{n\\}]]=[A_1,\dots,A_n].$$

Now observe that $$\Delta([A_1,\dots,A_m]+[B_1,\dots,B_n])=\Delta([A_1,\dots,A_m,B_1,\dots,B_n])$$

$$=(\chi_{A_1}+\dots+\chi_{A_m}+\chi_{B_1}+\dots+\chi_{B_n},m+n) =(\chi_{A_1}+\dots+\chi_{A_m},m)+(\chi_{B_1}+\dots+\chi_{B_n},n).$$

$$=\Delta([A_1,\dots,A_m])+\Delta([B_1,\dots,B_n]).$$

Q.E.D.

If we define integration by setting $$\int_n(\chi_{A_1}+\dots+\chi_{A_n})d\mu=\mu(A_1)+\dots+\mu(A_n),$$ then $$\overline{\mu}(\Gamma(f,n))=\int_n f d\mu$$ for all functions $f$.

Topological monoids and lattices

Our construction respects the topological structure on distributive lattices and monoids.

Suppose that $X$ is a topological distributive lattice and $M$ is a topological commutative monoid. This simply means that the operations $\vee,\wedge$ on $X$ and the addition operation on $M$ are continuous. We can give $X^n$ the product topology, and we can give $\bigcup_{n=0}^\infty X^n$ the disjoint union topology where $U\subseteq\bigcup_{n=0}^\infty X^n$ is open precisely when $U\cap X^n$ is open for each $n\geq 0$. Now associate $\mathbf{M}(X)$ with $\mathbf{M}^\sharp(X)$ and give $\mathbf{M}^\sharp(X)$ the topology induced by the topology on $\bigcup_{n=0}^\infty X^n$. Then one can easily verify that the mapping $\mu:X\rightarrow M$ is continuous if and only if the extension $\overline{\mu}:\mathbf{M}^\sharp(X)\rightarrow M$ is a continuous monoid homomorphism.

Median algebras

The least upper bound and greatest lower bounds are in symmetric positions in the modularity identity; for distributive lattices, we can even formulate the modularity identity in terms of the median operation where there is no notion of which way is up.

A median algebra is an algebraic structure $(X,m)$ that satisfies the identities $m(x,x,y)=x,m(x,y,z)=m(y,z,x),m(x,y,z)=m(y,x,z)$ and the distributivity identity $x\wedge_a(y\wedge_bz)=(x\wedge_ay)\wedge_b(x\wedge_az)$ where we define $m(x,y,z)=x\wedge_yz$.

In a distributive lattice $(X,\wedge,\vee)$ define the ternary median operation $m$ by setting $m(x,y,z)=(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)$; in this case $(X,m)$ is a median algebra.

If $(X,m)$ is a median algebra, $G$ is a semigroup, and $\mu:X\rightarrow G$ is a function, then we say that $\mu$ is modular with respect to the median operation $m$ when $\mu(m(a,x,y))\bullet\mu(m(b,x,y))=\mu(x)\bullet\mu(y)$ whenever $x=m(a,b,x),y=m(a,b,y)$. If $\mu$ is modular with respect to $m$, then the subsemigroup $\langle\mu[G]\rangle$ is commutative, so from now on, we shall assume that all semigroups are commutative, and we shall use the $+$ symbol for the semigroup operation.

Lemma: In a median algebra, if $x=m(a,b,x),y=m(a,b,y)$, then $m(a,x,y)\vee m(b,x,y)=x\vee y$ and $m(a,x,y)\wedge m(b,x,y)=x\wedge y$.

Proof: We have $$m(a,x,y)\vee m(b,x,y)=(a\wedge x)\vee(a\wedge y)\vee(b\wedge x)\vee(b\wedge y)\vee(a\wedge b)$$

$$=((a\vee b)\wedge(x\vee y))\vee(a\wedge b)$$

$$=[(a\vee b)\wedge x]\vee[(a\vee b)\wedge y]\vee(a\wedge b)$$

$$=(a\wedge x)\vee(b\wedge x)\vee(a\wedge y)\vee(b\wedge y)\vee(a\wedge b)$$ $$=x\vee y.$$

The identity, $m(a,x,y)\wedge m(b,x,y)=x\wedge y$, can be obtained by switching the places of the operation $\wedge,\vee$. Q.E.D

Proposition: Let $(X,\wedge,\vee)$ be a distributive lattice with median operation $m$, and let $G$ be a commutative semigroup. Let $\mu:X\rightarrow G$ be a function. Then $\mu$ is modular with respect to the median operation $m$ if and only if $\mu$ is modular with respect to the lattice operations $\wedge,\vee$.

Proof: $\rightarrow.$ Let $a=x\wedge y,b=x\vee y$. Then $m(a,b,x)=(a\wedge x)\vee(b\wedge x)\vee(a\wedge b)=(x\wedge y)\vee x\vee(x\wedge y)=x$ and $m(a,b,y)=y$ as well. Therefore, $\mu(x)+\mu(y)=\mu(m(a,x,y))+\mu(m(b,x,y))=\mu(x\wedge y)+\mu(x\vee y)$.

$\leftarrow$. Suppose that $x=m(a,x,y),y=m(b,x,y)$. Then $$\mu(m(a,x,y))+\mu(m(b,x,y))=\mu(m(a,x,y)\vee m(b,x,y))+\mu(m(a,x,y)\wedge m(b,x,y))=\mu(x\vee y)+\mu(x\wedge y)=\mu(x)+\mu(y).$$

Q.E.D.

While the construction of $\mathbf{M}(X)$ is valid for median algebras, for a general median algebra $X$, I have not been able to put the elements of $\mathbf{M}(X)$ in a normal form.

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The OP asked "has it been discussed in the literature". The two mentioned below are, indeed, known in the literature. But they are perhaps not precisely the same as the OP.

Sion, Maurice, A theory of semigroup valued measures, Lecture Notes in Mathematics. 355. Berlin-Heidelberg-New York: Springer-Verlag. V, 140 p. DM 16.00; $ 6.60 (1973). ZBL0312.28016.

and

Carathéodory, Constantin, Algebraic theory of measure and integration, New York: Chelsea Publishing Company, 378 p. (1963). ZBL0106.26403.

This is the English translation of

Carathéodory, Constantin, Mass und Integral und ihre Algebraisierung, Lehrbücher und Monographien aus dem Gebiete der exakten Wissenschaften, Math. Reihe, Band 10. Basel-Stuttgart: Verlag Birkhäuser (1956). ZBL0074.04003.

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  • $\begingroup$ Did you mean to start your post with a line reading just "comments", followed by "see also"? $\endgroup$
    – LSpice
    Sep 21, 2023 at 20:47

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