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A countably-complete semiring is basically a semiring with some additional structure making infinite sums possible. Formally, it is a tuple $R=(|R|,\Sigma,\cdot,1)$, where $|R|$ is a set, $(R,\cdot,1)$ is a monoid, and $\Sigma$ is an operation which assigns to every countable family $(a_i)_{i \in I}$ of elements in $|R|$ an element $\sum_{i \in I} a_i$ in $|R|$, such that the following equations hold:

  • $\sum_{i \in \{j\}} a_i = a_j$
  • If a countable set $I$ is the disjoint union of countably many sets $I_j$, $j \in J$, then $\sum_{i \in I} a_i = \sum_{j \in J} \sum_{i \in I_j} a_i$.
  • $(\sum_{i \in I} a_i ) \cdot b = \sum_{i \in I} (a_i \cdot b)$
  • $b \cdot \sum_{i \in I} a_i = \sum_{i \in I} (b \cdot a_i)$

It is clear how to define homomorphisms of countably-complete semirings. Every countably-complete semiring has an underlying semiring: We define $0 := \sum_{i \in \emptyset} a_i$ for the unique $\emptyset$-indexed family $a$ and $a_1 + a_2 := \sum_{i \in \{1,2\}} a_i$. (The usual definition of a complete semiring in the literature starts with a semiring, but this is redundant.)

My interest for countably-complete semirings comes from the observation that the Lebesgue-integral for non-negative measurable functions has a purely algebraic characterization within countably-complete commutative monoids: We need monotone convergence, which just means that the integral is countably-additive, and the extension property $\int \chi_A \, d\mu = \mu(A)$. This also motivates the following:

Question. What is a description of the countably-complete semiring $(\mathbb{R}_{\geq 0} \cup \{\infty\},\Sigma,\cdot,1)$ by generators and relations? Here, we define $\sum_{i \in I} a_i := \sup_{F \subseteq I \text{ finite}} \sum_{i \in F} a_i$, where $\sup$ refers to the usual order on $\mathbb{R}_{\geq 0} \cup \{\infty\}$.

It is easy to check that the initial countably-complete semiring is $(\mathbb{N} \cup \{\infty\},\dotsc)$ and that the initial (usual) semiring in which all $n \in \mathbb{N}_{>0}$ are invertible is $(\mathbb{Q}_{\geq 0},\dotsc)$. Since $(\mathbb{R},\leq)$ is the order-completion of $(\mathbb{Q},\leq)$, I suspect that $(\mathbb{R}_{\geq 0} \cup \{\infty\},\dotsc)$ is the initial countably-complete semiring in which all $n \in \mathbb{N}_{>0}$ are invertible. But I don't think that this is trivial at all.

So here is explicitly what has to be proven for my claim: Let $R$ be a countably-complete semiring and assume that all $n \in \mathbb{N}_{>0}$ become invertible in $R$. We obtain a map $f : \mathbb{Q}_{\geq 0} \cup \{\infty\} \to |R|$. We want to define an extension $\overline{f} : \mathbb{R}_{\geq 0} \cup \{\infty\} \to |R|$ as follows: If $r$ is a non-negative real number, or $\infty$, find some countable sum expression $r = \sum_{i \in I} q_i$ with rational numbers $q_i$, and define $\overline{f}(r) := \sum_{i \in I} f(q_i)$.

Question. Why is $\overline{f}$ well-defined?

If this was the case, the rest of the proof would be clear. Well-definedness seems to have a quite strong interpretation, namely that every equation of infinite series of rational numbers is "purely algebraic". For example, the equation $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$ holds, in fact, in every countably-complete semiring containing the non-negative rationals.

Edit: I'm not so sure anymore about the equation $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$. Let $s$ denote the sum. Then one observes $2 \cdot s = s + 1$. But since we are in a semiring, this does not imply $s=1$! Also, I've tried to show the equation $\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ algebraically, i.e. in a countably-complete semiring containing $\mathbb{Q}_{\geq 0}$. Using prime factor decompositions, one reduces this to geometric series. But $s := \sum_{n=1}^{\infty} \frac{1}{q^n}$ (for $q \in \mathbb{Q}_{>1}$) only satisfies $q s = s + 1$, and I don't know how to deduce $s = \frac{1}{q-1}$.

Edit: Ok, my claim is wrong, but my question remains, namely how to describe $(\mathbb{R}_{\geq 0} \cup \{\infty\},\Sigma,\cdot,1)$ by generators and relations.

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  • $\begingroup$ This sounds interesting. Do you know of any articles, books (etc.) that take this perspective on the Lebesgue integral? $\endgroup$ – goblin Feb 23 '16 at 15:00
  • $\begingroup$ On a slightly related topic, its probably worth pointing out the similarities between the definition of a measure (in measure theory) and the notion of a multiplicative function in number theory; it seems that "kind of homomorphisms" between $\kappa$-complete commutative monoids are pretty important, if you catch my drift. $\endgroup$ – goblin Feb 23 '16 at 15:02
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Part 2

I just noticed that there is no axiom linking sums of finitely many elements to sums of infinitely many elements.

This means that we can have the following countably-complete semi-ring $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$ such that $\Sigma'$ is the usual sum when the set of indexes is finite and $\infty$ when the set of indexes is infinite (more precisely, when infinitely many of the terms of the sequence are non-zero.), or when $\infty$ is one of the summands.

This $\Sigma'$ satisfies the axioms because whenever a set of indexes is infinite in one side of the axioms there is also an infinite set of indexes on the other side of the axiom, turning it into $\infty=\infty$. When all set of indexes are finite the axioms are the usual properties of finite sums.

We see that each $n\in\mathbb{N}$ is invertible in $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$ with inverse $1/n$.

We can't have a morphism from $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma,\cdot,1)$ to $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$.

Such a morphism $f$ would have to send $f(1)=1$, and would have to satisfy $$1=f(1)=f\left(\sum_{n>0}2^{-n}\right)=\sum_{n>0}f(2^{-n})=\infty.$$

Therefore $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma,\cdot,1)$ can't be the initial object.

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  • $\begingroup$ Thank you! This shows that my claim was wrong, and that we need more relations. It should be pretty easy to write down the relations if we add the order structure (thereby also linking the infinite to the finite case), but this is exactly what I would like to avoid. $\endgroup$ – Martin Brandenburg Mar 29 '15 at 9:56
  • $\begingroup$ @MartinBrandenburg I think that, for the rationals, if one adds the relations corresponding to one infinite sum of rationals that add up to a rational and its tails then that implies all other relations for infinite sums of rationals that return rationals. The proof uses the technique I used in my other answer. $\endgroup$ – Nathanson Mar 29 '15 at 10:32
  • $\begingroup$ @MartinBrandenburg The above comment implies that one only needs one (set of) relation(s) (of a sum and its tails) for each equivalence class of the reals mod rational translations. $\endgroup$ – Nathanson Mar 29 '15 at 10:34
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Part 1

Assume $f:\mathbb{Q}_{\geq0}\cup\{\infty\}\to|R|$ is a morphism commutes with infinite sums of rationals that add up to a rational. Let us show that the extension to $\mathbb{R}_{\geq0}\cup\{\infty\}$ you defined above is well defined.


Assume that $r=\sum_{i\in I} q_i=\sum_{i\in I'} q_i'$.

Then you can do the following:

If $q_1=q_1'$ we look at the next index.

If $q_1>q_1'$ (let us assume that it is), then we write $q_1=q_1'+a$, we pass to look at $a$ from the first series and $q_2'$ from the second, and repeat:

If $a_1>q_2'$ we decompose $a_1=q_2'+a_2$. If $a_1<q_2'$ then it is $q_2'$ the one we decompose as $q_2'=a_1+a_2$.

In this manner we produce decompositions $q_i=\sum_{k\in I_i}a_k$ and $q_i'=\sum_{k\in I_i'}a_i$ and moreover $\{a_i|\ i\in\bigcup I_i\}=\{a_i|\ i\in\bigcup I_i'\}$ and the terms appear in the same order.

Notice that each $I_i,I_i'$ is finite except for possibly the case in which one of $I$ or $I'$ (say $I$) is finite and the decomposition of $q_i$ has infinitely many terms only for the last element of $I$.

We use now the second property of the countably complete semi-ring, and that $f(q_i)=f(\sum_{i\in I_i}a_i)=\sum_{i\in I_i}f(a_i)$ (which we have because $q_i$ is rational) and the same thing for the numbers with primes.

Each sum $\sum_{i\in I} f(q_i)$ and $\sum_{i\in I'} f(q_i')$ is equal to the sum $\sum_{i\in\bigcup I_i=\bigcup I_i'}f(a_i)$.

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Part 3 (a form of answer to your first question)

Lemma: Assume $q_i$, $q$ are in $\mathbb{Q}_{\geq0}$ such that $\sum_{i\geq0}q_i=q$ and we know that $\sum_{i\geq0}f(q_i)=f(q)$, and that $\sum_{i>n}f(q_i)=f(q-\sum_{i=0}^{n}q_i)$. Then, for every sequence $q_i'$ in $\mathbb{Q}_{\geq0}$ such that $\sum_{i\geq0}q_i'=q\in\mathbb{Q}_{i\geq0}$ we must have $\sum_{i\geq0}f(q_i')=f(q)$.

Proof: We first apply the technique in the other answer to split the terms in both series $\sum q_i$ and $\sum q_i'$ into a common series $\sum_{i\in I}a_i$ with $q_i=\sum_{k\in I_i}a_k$ and $q_i'=\sum_{k\in I_i'}a_k$.

By construction either all $I_i,I_i'$ are finite or say $q_i'$ were almost all zero and the last term $q_n'=\sum_{n>N}q_i$.

In each case, since we know that $f$ commutes with finite sums of rationals and we are given also the relations on the tails we get that

$$\begin{align}f(q)&=f\left(\sum_{i}q_i\right)\\&=\sum_if(q_i)\\&=\sum_if\left(\sum_{k\in I_i}a_i\right)\\&=\sum_i\left(\sum_{k\in I_i}f(a_k)\right)\\&=\sum_{i}\sum_{k\in I_i'}f(a_k)\\&=\sum_{i}f\left(\sum_{k\in I_i'}a_k\right)\\&=\sum_if(q_i')\end{align}$$

Lemma: If $\sum_i q_i=q$, with $q_i,q$ non-negative rationals and we know that $\sum_if(q_i)=f(q)$ (and its tails), then we also know a relation (and its tails) for every other rational $q'>q$.

Proof: In fact, $q'=(q'-q)+\sum_i q_i$ and we can apply $f$ to both sides to get

$$f(q')=f((q'-q)+\sum_iq_i)=f(q'-q)+f(\sum_iq_i)=f(q'-q)+\sum_if(q_i).$$

And the tails of this series are just tails for the series of $q=\sum_i q_i$.


Due to lack of cancellation for the sum I can't see right now if it is possible to get from the relation for the series for $q$, relations for series giving rational number smaller that $q$.

In any case, what the results we have so far imply is that it is enough to give the following set of relations:

Partition the non-negative reals according the orbits of translations by rational numbers. In for each equivalence class consider one sequence $r_1,r_2,...$ that tends to zero. For each element $r_i$ on each sequence give a relation $f(r_i)=\sum_jf(q_{i,j})$ corresponding to one series of non-negative rationals $\sum_jq_{i,j}=r_i$.

Given such a set of relations in $|R|$, the two lemmas above imply that $(\mathbb{R}_{\geq0},\Sigma,\cdot,1)$ embeds in $|R|$.

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