# Existence of a strange measure

The answer to this question must be known, but I do not know where to find it. It is related to the Ulam measures I believe.

Question. Is there a finitely additive measure defined on all subsets of positive integers $$\mathbb{N}$$, with values into $$\{0,1\}$$, (only two values) that is $$\mu:2^{\mathbb{N}}\to \{0,1\}$$ such that $$\mu(\{n\})=0$$ for every $$n\in\mathbb{N}$$ and $$\mu(\mathbb{N})=1$$ ?

I believe such a result is needed in a proof of the co-area inequality that is stated in Coarea inequality, Eilenberg inequality. I am working with my student on some generalizations of that result so this is a research related question.

• This is called non-principal ultrafilter (namely $\mu^{-1}(\{1\})$)
– YCor
Feb 18, 2020 at 20:52
• I don't know (it's extensively documented anyway), but here's a proof (of its existence on any infinite set $X$): the set of ultrafilters is closed in the pointwise convergence topology, hence is compact, and its subset of principal ultrafilters (=Dirac measures) is an infinite discrete subset, hence is not compact and hence is a proper subset. Verifications are straightforward.
– YCor
Feb 18, 2020 at 21:31
• An approach that functional analysts might like is to observe that $l^\infty/c_0$ is a C*-algebra. So let $\phi: l^\infty/c_0 \to \mathbb{C}$ be a unital *-homomorphism, and then define $\mu(S) = \phi(1_S + c_0)$. Feb 18, 2020 at 22:59
• @DavidHandelman: A non-principal ultrafilter isn’t any kind of “complicated notion” — it’s just spelling out the details of what “maximal ideal” means in this special case (or rather, the dual conditions). Feb 19, 2020 at 17:07
• @DavidHandelman: Sure, I wouldn’t quibble with describing ultrafilters as, say, less familiar. Feb 20, 2020 at 0:23

The answer is yes. I wrote a proof using YCor's comment.

Theorem. There a finitely additive measure defined on all subsets of positive integers $$\mathbb{N}$$, with values into $$\{0,1\}$$, (only two values) that is $$\mu:2^{\mathbb{N}}\to \{0,1\}$$ such that $$\mu(\{n\})=0$$ for every $$n\in\mathbb{N}$$ and $$\mu(\mathbb{N})=1$$.

Proof. Let $$\mathcal{F}_0=\{A\subset\mathbb{N}:\, \mathbb{N}\setminus A\ \ \text{is finite}.\}$$ $$\mathcal{F}_0$$ is a filter. By a filter we mean here a family $$\mathcal{F}\subset 2^\mathbb{N}$$ with the following properties

1. $$\mathbb{N}\in \mathcal{F}$$,
2. $$\emptyset\not\in\mathcal{F}$$,
3. $$A\in\mathcal{F}$$, $$A\subset B$$ $$\Rightarrow$$ $$B\in\mathcal{F}$$,
4. $$A,B\in\mathcal{F}$$ $$\Rightarrow$$ $$A\cap B\in\mathcal{F}$$.

Lemma. There is a family $$\mathcal{F}\subset 2^{\mathbb{N}}$$ such that $$\mathcal{F}_0\subset\mathcal{F}$$ and $$\mathcal{F}$$ has properties 1.-4. and also property

1. $$A\subset\mathbb{N}$$ $$\Rightarrow$$ $$A\in\mathcal{F}$$ or $$\mathbb{N}\setminus A\in\mathcal{F}$$.

Remark. A family $$\mathcal{F}\subset 2^\mathbb{N}$$ satisfying properties 1.-5. is called an ultrafilter.

Proof. Filters containing $$\mathcal{F}_0$$ are ordered by the inclusion. The union of a chain of filters is a filter (this is obvious). Therefore by the Kuratowski-Zorn lemma there is a maximal filter $$\mathcal{F}$$ that contains $$\mathcal{F}_0$$. It remains to show that $$\mathcal{F}$$ satisfies property 5. (it has properties 1.-4. since it is a filter).

Suppose to the contrary that there is $$A\subset\mathbb{N}$$ such that $$(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A\not\in\mathcal{F} \quad \text{and} \quad \mathbb{N}\setminus A\not\in \mathcal{F}.$$ Define $$\widetilde{\mathcal{F}}= \{ E\subset \mathbb{N}:\, \exists B\in\mathcal{F}\ \ A\cap B\subset E\}.$$ Note that $$\mathcal{F}\subset\widetilde{\mathcal F}$$, because for $$B\in\mathcal{F}$$, $$A\cap B\subset B$$ so $$E=B\in\widetilde{\mathcal{F}}$$. Also $$\mathcal{F}\subsetneq\widetilde{F}$$ is a propert subset, because $$\mathbb{N}\in\mathcal{F}$$, $$A\cap\mathbb{N}\subset A$$ so $$A\in\widetilde{\mathcal F}$$ by definition and hence $$A\in\widetilde{\mathcal F}\setminus\mathcal{F}$$.

It remains to prove that $$\widetilde{\mathcal F}$$ is a filter (has properties 1.-4.) to reach a contradiction with the maximality of $$\mathcal{F}$$.

Clearly $$\widetilde{\mathcal F}$$ has properties 1., 3., and 4. (by property 4. for $$\mathcal{F}$$).

It remains to prove property 2. Suppose to the contrary that $$\emptyset\in \widetilde{\mathcal F}$$. Then $$A\cap B=\emptyset$$ for some $$B\in\mathcal{F}$$. Then $$B\subset\mathbb{N}\setminus A$$ so $$\mathbb{N}\setminus A\in\mathcal{F}$$ by property 3. for $$\mathcal{F}$$ which contradicts $$(*)$$. The proof is complete. $$\Box$$

Now we define the finitely additive measure $$\mu:2^{\mathbb{N}}\to \{0,1\}$$ such that $$\mu(\{n\})=0$$ for every $$n\in\mathbb{N}$$ and $$\mu(\mathbb{N})=1$$ as follows $$\mu(A)= \begin{cases} 1 & \text{if A\in \mathcal{F}},\\ 0 & \text{if A\not\in\mathcal{F}}. \end{cases}$$ Finite-additivity of $$\mu$$ follows from the fact that no two subsets with measure $$1$$ can be disjoint.

Note: Countable additivity fails because $$\mathbb{N} = \bigcup_{n=1}^\infty \{n\} \quad \text{but} \quad \mu(\mathbb{N}) \neq \sum_{n=1}^\infty \mu(\{n\}) \ .$$

$$\Box$$

• This interpretation of ultrafilter was used by Kleiner and Leeb in “Rigidity of quasi-isometries ...” Feb 19, 2020 at 2:59

Just about any form of the axiom of choice can be used to prove this. I like topology, so here's a proof using Tychonoff's theorem.

Consider the space $$2^{2^\mathbb{N}}$$ of all functions $$2^{\mathbb{N}}\to\{0,1\}$$ under the product topology. By Tychonoff's theorem this is compact, and it is easy to check from the definition of the product topology that the set $$\mathcal{F}$$ of all finitely-additive $$\{0,1\}$$-valued measures on $$\mathbb{N}$$ is closed in $$2^{2^\mathbb{N}}$$. Then the sequence of measures $$\mu_k\in\mathcal{F}$$ defined by $$\mu_k(S) = \begin{cases}1 & \text{if }k\in S,\\ 0 & \text{if }k\notin S\end{cases}$$ must have a limit point $$\mu\in\mathcal{F}$$. Then $$\mu(\mathbb{N})=1$$ since $$\mu_k(\mathbb{N})=1$$ for all $$k$$, and $$\mu(\{n\}) = 0$$ for all $$n\in\mathbb{N}$$ since $$\mu_k(\{n\})=0$$ whenever $$k>n$$.

• I like this proof! Here's a fun exercise to go with it: even though the closure of your sequence is compact, no subsequence of it converges! (This seems wrong at first, especially if you're accustomed to working with metric spaces -- but there's no contradiction here.) Feb 20, 2020 at 14:50
• This is basically the detailed version of the 3-line proof I gave in a comment.
– YCor
Feb 21, 2020 at 13:28
• @YCor Sorry, I hadn't noticed your comment, but you're right that it's the same proof as the one you suggest. Feb 21, 2020 at 13:53

My comment seems to be buried so I'd like to repeat it here. There is a simple C*-algebra construction that answers the question. The quotient space $$l^\infty/c_0$$ is a unital commutative C*-algebra, so there is a $$*$$-isomorphism $$\Phi: l^\infty/c_0 \cong C(X)$$ for some compact Hausdorff space $$X$$. For any $$x \in X$$, define $$\mu_x(S) = \Phi(1_S)(x)$$.

Every such $$\mu_x$$ works. In fact, this establishes a 1-1 correspondence between the set of non-principal ultrafilters on $$\mathbb{N}$$ and the spectrum of $$l^\infty/c_0$$.

• And $X$ is in fact the Stone-Čech remainder $\beta \mathbb{N} \setminus \mathbb{N}$, right? So there's another connection. Feb 21, 2020 at 2:22
• Right, the elements of $\beta\mathbb{N}\setminus\mathbb{N}$ correspond to free ultrafilters. Feb 21, 2020 at 2:36
• @NikWeaver: What is $c_0$? Feb 21, 2020 at 16:03
• @Bumblebee: it is one of the standard sequence spaces. Feb 21, 2020 at 19:37

This can be proved without introducing ultrafilters by name, by doing "finitary measure theory" and using Zorn's lemma.

An algebra $$A$$ on a set $$X$$ is just a $$\sigma$$-algebra without the $$\sigma$$, i.e. $$\newcommand{\powerset}{\mathcal{P}}A \subseteq \powerset(X)$$ and is closed under finite unions and complements (and therefore all other Boolean operations).

Let $$\newcommand{\N}{\mathbb{N}}F \subseteq \powerset(\N)$$ be the set of finite sets and their complements (so-called cofinite sets). This is an algebra. Furthermore, we can define a 2-valued finitely-additive measure $$\mu : F \rightarrow \{0,1\}$$ to be $$0$$ on the finite sets and $$1$$ on the cofinite sets. The existence of the required 2-valued finitely-additive measure on $$\powerset(\N)$$ then follows from:

Proposition For any algebra $$A \subseteq \powerset(X)$$ and finitely-additive 2-valued measure $$\mu : A \rightarrow \{0,1\}$$, there exists a finitely-additive measure $$\overline{\mu} : \powerset(X) \rightarrow \{0,1\}$$ extending $$\mu$$.

Proof: Most of the difficulty is in believing that it's true. We use Zorn's lemma. The poset consists of pairs $$(B,\nu)$$ where $$B \supseteq A$$ is an algebra of sets, and $$\nu : B \rightarrow \{0,1\}$$ is a finitely-additive measure extending $$\mu$$. The order relation $$(B_1,\nu_1) \leq (B_2,\nu_2)$$ is defined to hold when $$B_1 \subseteq B_2$$ and $$\nu_2$$ extends $$\nu_1$$. Every chain in this poset has an upper bound - we just take the union of algebras (this is the step that fails for $$\sigma$$-algebras) and define the measure on the union in the obvious way.

Let $$(B,\nu)$$ be a maximal element in the poset. Suppose for a contradiction that $$B \neq \powerset(X)$$, so there is some $$U \in \powerset(X) \setminus B$$. We contradict the maximality of $$B$$ by extending $$\nu$$ to a larger algebra $$B'$$ including $$U$$. Define $$B' = \{ (U \cap S_1) \cup (\lnot U \cap S_2) \mid S_1, S_2 \in B \}$$. It is clear that $$B \subseteq B'$$ and $$U \in B'$$, and with a little Boolean reasoning we can prove that for all $$S_1,S_2,T_1,T_2 \in B$$: $$((U \cap S_1) \cup (\lnot U \cap S_2)) \cup ((U \cap T_1) \cup (\lnot U \cap T_2))\\ = (U \cap (S_1 \cup T_1)) \cup (\lnot U \cap (S_2 \cup T_2))$$ and $$\lnot ((U \cap S_1) \cup (\lnot U \cap S_2)) = (U \cap \lnot S_1) \cup (\lnot U \cap \lnot S_2)$$ This proves that $$B'$$ is an algebra.

Now, define $$d \in \{0,1\}$$ to be the "outer measure" of $$U$$, i.e. $$d = 0$$ if there exists $$S \in B$$ such that $$U \subseteq S$$ and $$\nu(S) = 0$$, otherwise $$d = 1$$. Without loss of generality we can take $$d = 1$$, because we can exchange the roles of $$U$$ and $$\lnot U$$. We define $$\nu'((U \cap S_1) \cup (\lnot U \cap S_2)) = \nu(S_1)$$. This is well-defined because if $$(U \cap S_1) \cup (\lnot U \cap S_2) = (U \cap T_1) \cup (\lnot U \cap S_2)$$, then $$U \cap S_1 = U \cap S_2$$, so $$U \subseteq \lnot (S_1 \triangle S_2)$$, so as $$d = 1$$, $$\nu(\lnot (S_1 \triangle S_2)) = 1$$, and therefore $$\nu(S_1) = \nu(S_2)$$. The identities we used to prove that $$B'$$ is an algebra can then be used to prove that $$\nu'$$ is finitely additive, and it follows directly from the definition that it extends $$\nu$$. So we successfully contradicted the maximality of $$B$$. $$\square$$

Of course, I actually think ultrafilters are a good thing to know about, both in topology and logic. There is also no metamathematical benefit in doing it this way - over ZF the existence of a non-principal ultrafilter on $$\N$$ and a finitely-additive 2-valued measure on $$\powerset(\N)$$ are equivalent. The above proof is based on something I came up with while reproving Stone duality in the case where the points of the Stone space are defined to be Boolean homomorphisms into $$2$$, rather than ultrafilters. The proposition above is a special case of the fact that complete Boolean algebras (such as $$2$$) are injective objects in the category of Boolean algebras.