#### Appendix A.1. Definitions

Let W consist of all possible world states w and let V consist of all possible payoff values v.

**Notation** **A1.** For any natural number n, set $\underline{n}:=\{1,2,\dots ,n\}$.

We will take both spaces to be finite, so for some natural numbers $n,m$, we can represent W as a copy of $\underline{n}$ and V as a copy of $\underline{m}$. We are interested in fitness functions $f:W\to V$ or, equivalently, functions $f:\underline{n}\to \underline{m}$. In the following we will choose either, as convenient.

**Definition** **A1.** (**Quasi-Definition**). A (“first-order ”) homomorphism of the same kind of structure in V and W is a function $f:W\to V$ that preserves this structure.

This can take two forms, depending on the structure. For example, group structure is preserved under a “forward” homomorphism: If $W,V$ are groups, a homomorphism $f:W\to V$ preserves the group multiplication, so that $f(w\xb7{w}^{\prime})=f\left(w\right)\xb7f\left({w}^{\prime}\right)$, for all $w,{w}^{\prime}\in W$. Other examples are linear homomorphisms preserving vector space structure, monotonic functions preserving order and open mappings preserving openness of sets.

On the other hand, a continuous function between topological spaces is a “backward” homomorphism: If a subset $B\subset V$ is open in V, then ${f}^{-1}\left(B\right)$ is open in W. Another example of backward homomorphism is that of measurable functions preserving measurable structure.

We will need to define how one set G can act on another set X and how it can do so repeatedly. Suppose, firstly, we have the structure, in the acting set G, of an associative binary operation $\langle \xb7,\xb7\rangle :G\times G\to G$. This means that, for any pair $g,h\in G$ we have $\langle g,h\rangle \in G$, which we call the product of g and h (in that order); we can write this simply as $gh$. Associativity means that for any $g,h,k\in G$, we have $g\left(hk\right)=\left(gh\right)k$. Such a pair $(G,\langle \xb7,\xb7\rangle )$ is called a semigroup. If the product is clear from context, we will simply write G for the pair $(G,\langle \xb7,\xb7\rangle )$.

A monoid is a semigroup possessing an identity: i.e., $\exists \iota \in G$ such that $g\xb7\iota =\iota \xb7g=g$.

A group is a monoid with all inverses: $\forall g\in G,\exists {g}^{-1}\in G:{g}^{-1}\xb7g=g\xb7{g}^{-1}=\iota $.

Now suppose, moreover, that each element of G behaves as an operation on another set X, i.e., each element of G is a function, moving every element of X to some other element of X. If the pair $\langle G,\xb7\rangle $ is a semigroup, this allows us to be able to repeatedly “multiply” within G. The fact that each element acts on X means that the multiplication within G allows repeated such actions on X. We can think of the action of G on X also as a binary operation $G\times X\to X$, written as $g\xb7x$ for any $g\in G$ and $x\in X$. We would like the two binary operations to be consistent. This leads us to the precise definition of an action of G on X:

**Definition** **A2.** Let G be a semigroup. We will say that G acts (as a semigroup) on a set X if there is a binary operation $G\times X\to X$, written as $g\xb7x$ for $g\in G$ and $x\in X$, such that for any $g,h\in G$, and for any x in $\in X$, If G is a monoid, then we say that G acts as a monoid if it acts as a semigroup and also the identity ι acts, on all x in X, as Finally, if G is a group which acts as a monoid, then it automatically acts as a group: $\forall x\in X,\forall g\in G$, A homomorphism ϕ from G to H, where $G,H$ are both semigroups, both monoids or both groups, is just a (forward) first-order homomorphism from G to H: $\varphi :G\to H$ is a homomorphism if $\varphi \left(gh\right)=\varphi \left(g\right)\varphi \left(h\right),\forall g,h\in G$ (the product on the left being in G and that on right being in H).

In the following, we will assume that any G acting on a space will be a semigroup, a monoid, or a group. In this context, we introduce another kind of “homomorphism ”; namely, a ”second-order” homomorphism. Suppose G acts on each of two sets $X,Y$. A second-order homomorphism is a pair, consisting of a homomorphism $\varphi $ of G to itself, together with a function $f:X\to Y$; these are mutually consistent in the following sense:

**Definition** **A3.** If G acts on sets X and Y, a second-order G-homomorphism from X to Y is a pair $\langle \varphi ,f\rangle $ consisting of a homomorphism $\varphi :G\to G$ and a function $f:X\to Y$, such that $\forall g\in G$ and $\forall x\in X$, $f(g\xb7x)=\varphi \left(g\right)\dot{f}\left(x\right)$.

If the context makes it clear, we will abbreviate this by saying that “f respects $\varphi $. ”

A homomorphism of a semigroup, monoid or group G to another group H is just a (forward) first-order homomorphism from G to H: $\varphi $ is a homomorphism if $\varphi \left(gh\right)=\varphi \left(g\right)\varphi \left(h\right),\forall g,h\in G$.

Pictorially, the following diagram commutes, for all

$g\in G$:

Finally:

**Definition** **A4.** A function $f:\underline{n}\to \underline{m}$ is admissible if it achieves its maximum value: there is some $k\in \underline{n}$ with $f\left(k\right)=m.$

So admissible functions achieve the highest possible fitness value for some world state. In what follows we will compare the number of admissible homomorphic functions (in one of the above senses) to the number of all admissible functions, in the limit as the size n of the world grows to infinity.

#### Appendix A.3. Permutation Groups Theorem: Counting Functions Preserving Symmetry under the Symmetric Group S n

We will take $n=m$ and so only consider functions $f:\underline{n}\to \underline{n}$. We count the number of second-order homomorphisms of the symmetric group, acting on $\underline{n}$. These consist of certain functions, together with homomorphisms of ${S}_{n}$ to itself. We first classify the homomorphisms into three classes: within each class we count the number of functions respecting such homomorphisms and then sum over the three classes to get the total number of second-order homomorphisms. Then we compare this number with the admissible functions.

If $\varphi :{S}_{n}\to {S}_{n}$ is a homomorphism of ${S}_{n}$, then by the first group isomorphism theorem, the image of $\varphi $ is a subgroup of ${S}_{n}$, isomorphic to ${S}_{n}/\mathrm{ker}\varphi $, where the kernel $\mathrm{ker}\varphi $ of $\varphi $, is the set of elements sent to the identity by $\varphi $, and is a normal subgroup of ${S}_{n}$. Conversely, for any normal subgroup K, there is a canonical homomorphism from ${S}_{n}$ to ${S}_{n}/K$. So the set of homomorphisms of ${S}_{n}$ is in one-to-one correspondence with the set of all automorphisms of the groups ${S}_{n}/K$, as K ranges over the normal subgroups of ${S}_{n}$.

The normal subgroups of

${S}_{n}$ are, for

$n\ge 5$, the trivial subgroup

$\left\{\iota \right\}$, the alternating group

${A}_{n}$ and the whole group

${S}_{n}$ (Corollary G.33 in [

54], p.125). The corresponding quotient groups are isomorphic to

${S}_{n},{Z}_{2}$ and

$\left\{\iota \right\}$ respectively. Finding all homomorphisms is then a matter of finding the automorphisms of each of

${S}_{n}$,

${Z}_{2}$ and

$\left\{\iota \right\}$.

The group of automorphisms of the quotient group

${S}_{n}$, for

$n\ne 6$, is just the group of its

inner automorphisms [

55]; i.e., those of the form

$h\mapsto {g}^{-1}hg$ for some fixed

$g\in {S}_{n}$. Since different

g yield distinct automorphisms, the size of this group is the same as that of

${S}_{n}$; i.e.,

$n!$ (when

$n=6$ the number of automorphisms is

$2n!$; Theorem 3.5 below remains true when

n is at least 5). There is only one automorphism of the quotient group

${Z}_{2}$, namely, the identity automorphism, and the same is true of the quotient group

$\left\{\iota \right\}$.

We will denote the operation of an element g of the symmetric group ${S}_{n}$ on the element $k\in \underline{n}$ by $g\xb7k$. Recalling Definition 1.4, a function f respectful of the homomorphism $\varphi $ gives a second-order homomorphism $\langle \varphi ,f\rangle $:

**Lemma** **A2.** There are n functions f that respect the trivial homomorphism.

**Proof.** Since $\varphi \left(g\right)=\iota $, we have that $f(g\xb71)=f\left(1\right)$, for all $g\in {S}_{n}$. For any $k\in \underline{n}$ there is a g such that $g\xb71=k$, so f is constant. There are n constant functions. □

**Lemma** **A3.** There are $n!$ functions $f:\underline{n}\to \underline{n}$ that respect some inner automorphism ϕ of ${S}_{n}$. Indeed, if ϕ is given by $\varphi \left(g\right)=hg{h}^{-1}$, then $f\left(k\right):=h\xb7k$ is the only function respectful of ϕ.

**Proof.** Given the automorphism $\varphi \left(g\right)=hg{h}^{-1}$ and integer j, put $f\left(j\right):=h\xb7j$. Then $f(g\xb7k)=h\xb7(g\xb7k)=\left(hg\right)\xb7k=\left(hg{h}^{-1}h\right)\xb7k=\left(hg{h}^{-1}\right)\xb7(h\xb7k)=\varphi \left(g\right)\xb7f\left(k\right)$. Conversely, suppose $f:\underline{n}\to \underline{n}$ respects the inner automorphism $\varphi \left(g\right)=hg{h}^{-1}$ for some fixed $h\in {S}_{n}$: i.e., $f(g\xb7k)=hg{h}^{-1}\xb7f\left(k\right)$. As g runs over the whole group, the right-hand side runs over the whole of $\underline{n}$, since the group action is transitive and $hg{h}^{-1}$ runs over the whole group. So f is onto, and being a function from a finite set to itself, is also 1:1. Thus f is itself a permutation, so of the form $f\left(k\right)={h}^{\prime}\xb7k$ for some ${h}^{\prime}\in {S}_{n}$. Then, for any $g\in {S}_{n}$, ${h}^{\prime}\xb7g\xb7k=f(g\xb7k)=hg{h}^{-1}\xb7f\left(k\right)=hg{h}^{-1}{h}^{\prime}\xb7k$, or ${h}^{-1}{h}^{\prime}g\xb7k=g{h}^{-1}{h}^{\prime}\xb7k$. This being true for all $k,{h}^{-1}{h}^{\prime}g=g{h}^{-1}{h}^{\prime},\forall g$. So ${h}^{-1}{h}^{\prime}$ commutes with all g, which means that ${h}^{-1}{h}^{\prime}=e$. In particular, there is exactly one function respectful of any inner automorphism. Thus there are $n!$ such functions. □

**Lemma** **A4.** The function $f:\underline{n}\to \underline{n}$ is respectful of a homomorphism onto an order-2 subgroup only if it is constant, so that there are n such functions.

**Proof.** Suppose $H<{S}_{n}$ is of order two and $\psi :{S}_{n}\to H$ is a homomorphism. A function f respectful of $\psi $ satisfies $f(g\xb7k)=\psi \left(g\right)\xb7f\left(k\right)$. Since the kernel of $\psi $ consists of ${A}_{n}$, the even permutations, we have that for $g\in {A}_{n},f(g\xb7k)=f\left(k\right)$: f is invariant under the action of ${A}_{n}$. Let $g=(1,k)\in {A}_{n}$. Then $f\left(k\right)=f\left(1\right),k=1,\cdots n$: i.e., f is constant. □

Putting these facts together, we see that, for $n\ge 5$, the number of respectful functions is $2n+n!$

**Theorem** **A2.** The ratio of respectful functions to admissible ones has limit 0 as $n\to \infty $.

**Proof.** The ratio of respectful functions to admissible ones is

$(2n+n!)/({n}^{n}-{(n-1)}^{n})$ By Stirling’s approximation,

Since the first factor goes to

$1/(1-{e}^{-1})$, the expression goes to zero as

$n\to \infty $. □

#### Appendix A.5. Measurable Structure Theorem: Counting Measurable Functions, that is, (Backward) Homomorphisms Preserving Algebra or Partition Structure

For finite sets, probabilities can be consistently defined on subsets called events, if the collection of such subsets forms an algebra. (Because our sets are finite, we do not need to deal with sigma-algebras in order to define probabilities. The results of this section do not hold for uncountable sigma-algebras. For example, the interval $(0,1)$ on the Borel real line is not a countable union of elementary sets, so Proposition 5.4 does not hold.)

**Definition** **A6.** An algebra $\mathcal{W}$ on the set W is a collection of subsets that includes the empty set and is closed under intersections and complements. ($\mathcal{W}$)-measurable sets are the members of the algebra and the pair $(W,\mathcal{W})$ is called a measurable space. Given an algebra $\mathcal{V}$ on another set V, a function $f:W\to V$ is $\mathcal{W}/\mathcal{V}$ measurable if, for every $\mathcal{V}$-measurable set T, ${f}^{-1}\left(T\right)$ is $\mathcal{W}$-measurable.

As a consequence of the definition, the whole set is measurable, as are unions and differences of measurable sets. Measurable functions are then (backward) homomorphisms of the measurable, or algebra, structure. In this section we will compare the number of measurable, admissible functions to the totality of admissible functions and find its limit as its size W goes to infinity.

In the following, we shall assume that $(W,\mathcal{W})$ and $(V,\mathcal{V})$ are finite measurable sets.

**Definition** **A7.** Given any $j\in \underline{n}$, let ${U}_{j}$ be the smallest measurable set containing j, i.e., the intersection of all such sets.

Because of finiteness, ${U}_{j}\ne \varnothing $, for any j.

**Lemma** **A5.** If $j\in \underline{n}$ and $k\notin {U}_{j}$, then ${U}_{j}\cap {U}_{k}=\varnothing $.

**Proof.** Were j to be also an element of ${U}_{k}$, ${U}_{k}\setminus {U}_{j}$ would be a measurable set containing k but not j which is strictly smaller than ${U}_{k}$, contradicting the latter’s minimality. So we have $j\notin {U}_{k}$. Suppose ${U}_{j}\cap {U}_{k}\ne \varnothing $. Since $j\notin {U}_{k},j\notin {U}_{j}\cap {U}_{k}$. Thus $j\in {U}_{j}\setminus {U}_{k}$, but this is a contradiction as, by hypothesis, ${U}_{j}\setminus {U}_{k}$ is a measurable set containing j yet strictly smaller than the minimal ${U}_{j}$. □

**Proposition** **A1.** Algebras on $\underline{n}$ are in 1:1 correspondence with partitions of $\underline{n}$, consisting of the minimal measurable sets of Definition 5.2. In particular, a general measurable set is a (disjoint) union of some of those in the partition.

**Proof.** Let $\mathcal{A}$ be an algebra on $\underline{n}$, with minimal sets ${U}_{j}$ and let $j\in \underline{n}$. If any ${U}_{j}=\underline{n}$ we are done. If not, there is a $k\notin {U}_{j}$ and by the lemma, ${U}_{j}\cap {U}_{k}=\varnothing $. Continuing in this way, every $j\in \underline{n}$ is represented, so there is a finite subset $\{{j}_{1},\dots ,{j}_{k}\}\subset \underline{n}$ such that ${\left\{{U}_{{j}_{i}}\right\}}_{i=1,\dots k}$ is a partition of $\underline{n}$.

Let U be a measurable set. If U is empty, it is the empty union of of the sets in this partition. If U is non-empty, pick any element of U, call it ${j}_{1}$. By minimality, ${U}_{{j}_{1}}\subset U,U\setminus {U}_{{j}_{i}}$ is measurable and is either empty or contains some element, say ${j}_{2}$. We have that ${j}_{2}\notin {U}_{{j}_{1}}$ so ${U}_{{j}_{2}}\cap {U}_{{j}_{1}}=\varnothing $. Continuing in this way through the finite set $\underline{n}$, we see that, for some $m\le n$, U is a union of the disjoint measurable sets in the partition: $U={\bigcup}_{i=1}^{m}{U}_{{j}_{i}}$. □

Note that if W is countable and $,\mathcal{W}$ is a $\sigma $-algebra, the conclusion of this theorem holds, by induction.

**Definition** **A8.** - 1.
The collection of subsets in the partition corresponding to the algebra $\mathcal{W}$ on W will be termed the base of the algebra and will be written as $\{{W}_{1},\dots ,{W}_{k}\}$.

- 2.
The order of an algebra is the number of sets constituting its base. For example, the order of the trivial algebra is 1, and the order of the discrete algebra is the size of the underlying set.

- 3.
The characteristic of an algebra is the multiset giving the sizes of each of the elements ${W}_{i}$ of the base: we say that the characteristic is ${\{{m}_{i};{l}_{i}\}}_{i}$ if there are ${m}_{i}$ subsets of size ${l}_{i}$. Thus ${\sum}_{i}{m}_{i}{l}_{i}=n$, where n is the size of W. (In other words, the characteristic of an algebra is a partition, in the usual sense, of the number n. Saying that two algebras have the same characteristic is an equivalence relation on the collection of algebras: either algebra can be obtained from the other by a simple renumbering, or permutation, of the set W.)

Let the base of the algebra $\mathcal{W}$ on W be the partition $\{{W}_{1},\dots ,{W}_{k}\}$ and the base of the algebra $\mathcal{V}$ on V be the partition $\{{V}_{1},\dots ,{V}_{l}\}$. Saying that $f:W\to V$ is $\mathcal{W}/\mathcal{V}$ measurable is equivalent to saying, by Proposition 5.4, that for each base set ${V}_{j}\in \mathcal{V}$, we have ${f}^{-1}\left({V}_{j}\right)={\u2a06}_{i=1}^{p}{W}_{{j}_{i}}$, for some p-tuple of positive integers $({j}_{1},\dots ,{j}_{k})$, where $1\le p\le k$ and ⨆ means “disjoint union.” So f is measurable if the inverse image of the partition of V (made up of the base of V) is a new, coarse-grained, partition of W (made up unions of elements of the base of $\mathcal{W}$).

Note that when $\mathcal{W}$ is discrete (i.e., the minimal measurable sets are all the singletons), all functions $f:W\to V$ are $\mathcal{W}/\mathcal{V}$ measurable and their total number is therefore ${m}^{n}$. And this is also true of all functions when $\mathcal{V}$ is trivial (i.e., the only measurable sets are ∅ and V): the number of $\mathcal{W}/\mathcal{V}$ measurable functions $f:W\to V$ is again ${m}^{n}$. More generally:

**Lemma** **A6.** For any given algebra $\mathcal{V}$, the number of $\mathcal{W}/\mathcal{V}$ measurable functions $f:W\to V$ is determined solely by the order, and not the characteristic, of the algebra $\mathcal{W}$. (The number of such measurable functions may, however, depend on the details of the algebra $\mathcal{V}$, including its order m, as we will see below. )

**Proof.** We have seen above that this is true when $\mathcal{V}$ is trivial, so suppose $\mathcal{V}$ is not trivial: i.e., its base consists of two or more subsets.

If now $\mathcal{W}$ is discrete, there is only one characteristic of $\mathcal{W}$ (of order n), so the number ${m}^{n}$ is determined by the order n. Again, if $\mathcal{W}$ is trivial, there is only one characteristic of $\mathcal{W}$ (of order 1); a little thought shows that the number of measurable functions is now determined, and solely by the characteristic of $\mathcal{V}$.

So suppose in the following that $\mathcal{W}$ is neither discrete nor trivial.

Thus let the base of $\mathcal{W}$ be $\{{W}_{1},\dots ,{W}_{k}\}$ with $n>k>1$. Since $\mathcal{W}$ is not discrete, we can suppose that there is a base set, by renumbering call it ${W}_{1}$, of size at least two. This set has the form ${W}_{1}=\{a,b,\xb7\}$. Since $\mathcal{W}$ is not trivial, there is a distinct nonempty base set, call it ${W}_{2}$. We may suppose ${W}_{2}=\{c,\xb7\}$ (where, in either set, · represents zero or more elements!). We make a new algebra ${\mathcal{W}}^{\prime}$, also of order k, by taking only the element b and moving it to ${W}_{2}$: the base of ${\mathcal{W}}^{\prime}$ is $\{{W}_{1}^{\prime},\dots ,{W}_{k}^{\prime}\}$ with ${W}_{1}^{\prime}=\{a,\xb7\},{W}_{2}^{\prime}=\{b,c,\xb7\}$ and ${W}_{i}^{\prime}={W}_{i},\phantom{\rule{4pt}{0ex}}i>2$. We will call such a move:

Basic Move: $\{a,b,\xb7\},\{c,\xb7\}\dots \mapsto \{a,\xb7\},\{b,c,\xb7\}\dots $

Notice that the order of the algebra has not changed, but its characteristic certainly has. □

We will show that

**Claim 1**: Any algebra of order k can be obtained from any given one by means of a finite sequence of steps of the above kind.

**Proof.** Let us say that elements a and b are companions in the algebra $\mathcal{A}$ if they belong to the same base set of $\mathcal{A}$.

Suppose we have two algebras ${\mathcal{A}}_{1}$ and ${\mathcal{A}}_{2}$, both non-discrete, with the same order but different characteristics. We can convert one to the other using the following two-step algorithm.

In the first step, suppose that $\left\{a\right\}$ is a singleton in ${\mathcal{A}}_{2}$, but not in ${\mathcal{A}}_{1}$. Then we can use basic moves to remove all of its ${\mathcal{A}}_{1}$-companions, to any other basic set there, to get the new algebra ${\mathcal{A}}_{1}^{\prime}$, thus producing the singleton $\left\{a\right\}$ as a new basic set. While repeating this procedure for subsequent singletons, remove their putative companions to basic sets other than the singletons already created. In this way, we can produce all the singleton sets that are basic in ${\mathcal{A}}_{2}$. Of course, there may be extra singletons left over from ${\mathcal{A}}_{1}$, but these will be dealt with in the next step.

In the second step, whenever a and b are companions in the algebra ${\mathcal{A}}_{2}$, but not companions in ${\mathcal{A}}_{1}$, we can bring them together into a single (new) basic set by means of one of two processes:

If at least one of a and b belong to multi-element sets of ${\mathcal{A}}_{1}$, we can perform a basic move, as above, to bring them into the same basic set of the new algebra ${\mathcal{A}}_{1}^{\prime}$.

If both of a and b belong to singleton sets of ${\mathcal{A}}_{1}$, then because of the non-discreteness, there is another base set C with 2 or more elements. Pick an element $c\in C$. Make these elements companions by performing three basic moves in sequence, as follows:

$\left\{a\right\},\left\{b\right\},C\dots \mapsto \left\{a\right\},\{b,c\},C\setminus \left\{c\right\}\dots \mapsto \{a,b\},\left\{c\right\},C\setminus \left\{c\right\}\dots $

Of course, $C\setminus \left\{c\right\}$ is not empty, so these moves preserve the order of the new algebra.

Having brought together ${\mathcal{A}}_{2}$-companions a and b, we continue to use basic moves to bring all their other ${\mathcal{A}}_{2}$-companions into the same basic set. Once this is done, we move to a different basic set containing an element, at least one of whose ${\mathcal{A}}_{2}$-companions is not in that set. Now iterate the procedure for that element and all of its companions not already in the same basic set. Each of the operations involved maintains the algebra order. Two distinct collections of ${\mathcal{A}}_{2}$-companions could not end up being in the same basic set, because that would reduce the order by 1. When all operations are completed, we will have arrived at ${\mathcal{A}}_{2}$. □

**Claim 2**: If $\mathcal{W}$ and ${\mathcal{W}}^{\prime}$ have the same order, then the number of $\mathcal{W}/\mathcal{V}$ measurable functions is the same as the number of ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable functions.

**Proof.** We will show that the number of $\mathcal{W}/\mathcal{V}$ measurable functions that are not ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable equals the number of ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable functions that are not $\mathcal{W}/\mathcal{V}$ measurable. This establishes that the number of measurable functions is the same for both algebras and the theorem will be proved once this claim is established.

Assume that $\mathcal{W}$ and ${\mathcal{W}}^{\prime}$ are the algebras related by a single basic move of the kind preceding claim 1. Recall that $b\in {W}_{1}$ and $c\in {W}_{2}$, where ${W}_{1}$ and ${W}_{2}$ are distinct base sets. For f a $\mathcal{W}/\mathcal{V}$ measurable function, let ${X}_{1}$ and ${X}_{2}$ be the base sets of $\mathcal{V}$ such that $f\left(b\right)\in {X}_{1}$ and $f\left(c\right)\in {X}_{2}$.

First, suppose ${X}_{1}$ and ${X}_{2}$ are the same set. Then ${f}^{-1}\left({X}_{1}\right)={W}_{1}\bigsqcup {W}_{2}\bigsqcup {W}^{0}$, where ${W}^{0}$ is union of base sets disjoint from both ${W}_{1}$ and ${W}_{2}$, so that ${W}^{0}\in \mathcal{W}\cap {\mathcal{W}}^{\prime}$. But ${W}_{1}\bigsqcup {W}_{2}={W}_{1}^{\prime}\bigsqcup {W}_{2}^{\prime}$ so ${f}^{-1}\left({X}_{1}\right)={W}_{1}^{\prime}\bigsqcup {W}_{2}^{\prime}\bigsqcup {W}^{0}\in {\mathcal{W}}^{\prime}$. All other ${V}_{i}$ in the base of $\mathcal{V}$ are disjoint from ${X}_{1}={X}_{2}$, so the sets ${f}^{-1}\left({V}_{i}\right)$ are unions (possibly empty) of base sets ${W}_{j}$ other than ${W}_{1}$ and ${W}_{2}$; these being both in W and ${W}^{\prime}$, f is also ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable. As we have seen in the proof of claim 1 this will establish claim 2 whenever $\mathcal{W}$ and ${\mathcal{W}}^{\prime}$ are oft the same order.

Next, assume ${X}_{1}$ and ${X}_{2}$ are not the same, and therefore disjoint. We will show that then ${f}^{-1}\left({X}_{1}\right)\notin {\mathcal{W}}^{\prime}$, so that f is not ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable. Now ${f}^{-1}\left({X}_{1}\right)={W}_{1}\bigsqcup {W}_{1}^{0}$, where ${W}_{1}^{0}$ is a union of base sets disjoint from both ${W}_{1}$ and ${W}_{2}$ and so ${W}_{1}^{0}\in \mathcal{W}\cap {\mathcal{W}}^{\prime}$. Similarly, ${f}^{-1}\left({X}_{2}\right)={W}_{2}\bigsqcup {W}_{2}^{0}$, where ${W}_{2}^{0}$ is disjoint from both ${W}_{1}$ and ${W}_{2}$ and so ${W}_{2}^{0}\in \mathcal{W}\cap {\mathcal{W}}^{\prime}$. Additionally, all other ${f}^{-1}\left({V}_{i}\right)$ are unions (possibly empty) of base sets ${W}_{j}$ other than ${W}_{1}$ and ${W}_{2}$.

Now ${f}^{-1}\left({X}_{1}\right)={W}_{1}\bigsqcup {W}_{1}^{0}={W}_{1}^{\prime}\bigsqcup \left\{b\right\}\bigsqcup {W}_{1}^{0}$. But note that ${W}_{2}^{\prime}$ being a minimal set of ${\mathcal{W}}^{\prime}$, $\left\{b\right\}\u228a{W}_{2}^{\prime}$ is not in ${\mathcal{W}}^{\prime}$. Assume ${f}^{-1}\left({X}_{1}\right)={W}_{1}^{\prime}\bigsqcup \left\{b\right\}\bigsqcup {W}_{1}^{0}$ is in ${\mathcal{W}}^{\prime}$. Since ${W}_{1}^{\prime}\bigsqcup {W}_{1}^{0}\in {\mathcal{W}}^{\prime}$ and $b\notin {W}_{1}^{\prime}\bigsqcup {W}_{1}^{0}$, we would then have $\left\{b\right\}={f}^{-1}\left({X}_{1}\right)-\left({W}_{1}^{\prime}\dot{\cup}{W}_{1}^{0}\right)\in {\mathcal{W}}^{\prime}$, a contradiction. So f is not $\mathcal{W}/\mathcal{V}$ measurable.

Put ${v}_{1}=f\left(a\right)\in {X}_{1},{v}_{2}=f\left(b\right)\in {X}_{1}$, and ${v}_{3}=f\left(c\right)\in {X}_{2}$. By disjointness, ${v}_{3}\ne {v}_{1},{v}_{3}\ne {v}_{2}$. Let ${f}^{\prime}$ satisfy ${f}^{\prime}\left(a\right)={v}_{3},{f}^{\prime}\left(b\right)={v}_{1},{f}^{\prime}\left(c\right)={v}_{2}$ and ${f}^{\prime}\left(x\right)=f\left(x\right),\phantom{\rule{4pt}{0ex}}x\notin \{a,b,c\}$. Then, by an argument, mutatis mutandis that above, ${f}^{\prime}$ is ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable but not $\mathcal{W}/\mathcal{V}$ measurable. Thus the number of ${\mathcal{W}}^{\prime}/\mathcal{V}$ measurable functions is at least that of the $\mathcal{W}/\mathcal{V}$ measurable ones. Reversing the roles, mutatis mutandis, of $\mathcal{W}$ and ${\mathcal{W}}^{\prime}$, we see that the number is the same and the claim is proved. □

As the order of $\mathcal{V}$ increases by refinement, the number of measurable functions decreases:

**Lemma** **A7.** Fix the algebra $\mathcal{W}$ on W and assume it is not discrete. Let $\mathcal{V},{\mathcal{V}}^{\prime}$ be algebras on V such that ${\mathcal{V}}^{\prime}$ is a refinement of $\mathcal{V}$: each base set of ${\mathcal{V}}^{\prime}$ is contained in a base set of $\mathcal{V}$. If the order of ${\mathcal{V}}^{\prime}$ is strictly greater than that of $\mathcal{V}$, then the number of $\mathcal{W}/{\mathcal{V}}^{\prime}$ measurable functions is strictly less than the number of $\mathcal{W}/\mathcal{V}$ measurable functions.

**Proof.** Let the order of $\mathcal{V}$ be $l<m$, with base $\{{A}_{1},\dots ,{A}_{l}\}$. An algebra ${\mathcal{V}}^{\prime}$ of order $l+1$ can be made from $\mathcal{V}$ by extracting a nonempty proper subset from a base set of $\mathcal{V}$ that has at least two elements, thus creating two new sets. By an appropriate renumbering, let us call the original set ${A}_{1}$ and the piece that is removed ${A}_{l+1}^{\prime}$. The two new sets thus created are ${A}_{1}^{\prime}$ := ${A}_{1}\setminus {A}_{l+1}^{\prime}$ and ${A}_{l+1}^{\prime}$. If we take ${A}_{i}^{\prime}={A}_{i},\phantom{\rule{4pt}{0ex}}i=2,\dots ,l$, then the base of the new algebra ${\mathcal{V}}^{\prime}$ is $\{{A}_{1}^{\prime},\dots ,{A}_{l+1}\}.$

Now if f is a $\mathcal{W}/{\mathcal{V}}^{\prime}$ measurable function, it is automatically $\mathcal{W}/\mathcal{V}$ measurable: for $2\le i\le l$, ${f}^{-1}\left({A}_{i}\right)={f}^{-1}\left({A}_{i}^{\prime}\right)\in \mathcal{W}$, while, ${f}^{-1}\left({A}_{1}\right)={f}^{-1}({A}_{1}^{\prime}\bigsqcup {A}_{l+1}^{\prime})={f}^{-1}\left({A}_{1}^{\prime}\right)\bigsqcup {f}^{-1}\left({A}_{l+1}^{\prime}\right)\in \mathcal{W}$. Thus the number of $\mathcal{W}/{\mathcal{V}}^{\prime}$ measurable functions is no greater than the number of $\mathcal{W}/\mathcal{V}$ measurable functions.

Let ${W}_{0}$ be a base set of $\mathcal{W}$ which has two or more elements. Single out one element $x\in {W}_{0}$ and consider any function that takes all of ${W}_{0}\setminus \left\{x\right\}$ into ${A}_{1}^{\prime}$, takes x to ${A}_{l+1}^{\prime}$ and takes each remaining ${W}_{i}$ into ${A}_{1}^{\prime}$. Such a function is $\mathcal{W}/\mathcal{V}$ measurable but not $\mathcal{W}/{\mathcal{V}}^{\prime}$ measurable, so the decrease in number is strict. □

Observe that the collection of algebras on V is a lattice, ordered by $\mathcal{V}\ge {\mathcal{V}}^{\prime}$ if ${\mathcal{V}}^{\prime}$ is a refinement of $\mathcal{V}$. The proof above shows that the number of measurable functions is monotonically increasing with this partial order. Across algebras on V with the same order l, the number of measurable functions could be widely different. But as l increases to m, the size of V, that number will decrease, in each maximal linearly ordered sublattice, to a lowest number: that for the discrete algebra on V.

We seek an upper bound on the number of measurable functions. By Lemma 5.7, we need to maximize, given any fixed characteristic of a non-discrete algebra on W, this number over all possible characteristics of algebras on V of smallest order, i.e., order two.

**Theorem** **A4.** Suppose the measurable structure on W has order k and is neither trivial nor discrete. Additionally, suppose that the measurable structure on V is not trivial. Then the number of measurable functions $f:W\to V$ is bounded by **Proof.** Let $W=\underline{n}=\{1,\dots ,n\}$, and suppose $\mathcal{W}$ has base $\{{W}_{1},\dots ,{W}_{k}\}$, for $2\le k\le n-1$. Lemma 5.6 allows us to choose any characteristic of order k. Let us choose one with as many singletons as possible: ${W}_{1}=\{1,\dots ,n-k+1\}$ consists of the first $n-k+1$ elements, and for $2\le j\le k$, ${W}_{j}$ is just the singleton $\{n-k+j\}$.

Suppose, for the highest possible count, that the base of

$\mathcal{V}$ is

$\{{V}_{1},{V}_{2}\}$, such that

$|{V}_{1}|={m}_{1},\left|{V}_{2}\right|={m}_{2}=m-{m}_{1}$. For

$f:W\to V$ to be measurable, we need, for some subset

$A\subset \{1,\dots ,k\},$
where

${A}^{\prime}$ is the complement of

A. Distinguish two instances:

- (i)
$1\in A$. Then ${f}^{-1}\left({V}_{1}\right)={W}_{1}\cup C$ for $C\subset \{n-k+2,\dots ,n\}$ and ${f}^{-1}\left({V}_{2}\right)={C}^{\prime}$, where ${C}^{\prime}$ consists of the remaining elements of W: i.e., ${C}^{\prime}:=\{n-k+2,\dots ,n\}\setminus C$.

- (ii)
$1\notin A$. Then ${f}^{-1}\left({V}_{1}\right)={C}^{\prime}$ for $C\subset \{n-k+2,\dots ,n\}$ and ${f}^{-1}\left({V}_{2}\right)={W}_{1}\cup C$.

In either instance, the number of such functions is a product of two counts, summed over all possible C of size l, $l\in \{0,\dots ,k-1\}$. Once we have computed the count for instance (i), that for instance (ii) is that same count with ${m}_{1}$ and ${m}_{2}$ interchanged.

Consider instance (i). First, the count for

${f}^{-1}\left({V}_{1}\right)$: count the number of all functions from

${W}_{1}$ to

${V}_{1}$; times the number of functions from any fixed

C of size

l,

$l\in \{0,\dots ,k-1\}$, to

${V}_{1}$. This is

Second, the count for

${f}^{-1}\left({V}_{2}\right)$ is the number of all functions from

${C}^{\prime}$ to

${V}_{2}$. This is

Finally, we multiply by the number of subsets

C of size

l (of a set of size

$k-1$) and sum. The total count for instance (i) is then

Setting

$i=k-l-1$, this is the same as

For instance (ii), we interchange the roles of

${m}_{1}$ and

${m}_{2}$, so that the total number we seek is

By the binomial theorem this sums to

A little calculus shows that this has an extremum only at

${m}_{1}=m/2$, at which its value is a minimum. Thus the maximum occurs at the end-points

${m}_{1}\in \{1,m-1\}$, at both of which this count is

□

**Corollary** **A1.** For fixed $m\ge 2$, the ratio of the number of measurable functions to admissible functions goes to zero as n goes to infinity.

**Proof.** Since ${m}^{k-1}/({m}^{n}-{(m-1)}^{n})\to 0$ and ${(m/(m-1))}^{k-1}$ stays constant, we just need to check the limit of ${(m-1)}^{n}/({m}^{n}-{(m-1)}^{n})={\left({(m/(m-1))}^{n}-1\right)}^{-1}$ which, for $m\ge 2$, goes to zero. □

**Remark** **A1.** For m large and fixed, this goes to zero slowly as n goes to infinity. If we allow m to grow fast enough, as $O\left(n\right)$, say $m=n$, then the limit is $1/(e-1)\approx 0.58$.