I was once wondering about definitions of Radon measures. They are Borel measures on a topological space X that interact nicely with the topology on X. However, several authors have different definitions, and they are only guaranteed to coincide on locally compact Hausdorff spaces.
According to Wikipedia, a Radon measure on X is a measure μ on X such that:
- μ is finite on compact sets, namely such that μ(K)<∞ for every compact subset K of X;
- μ is outer regular (on Borel sets), namely, for every Borel subset E of X, μ(E) = inf {μ(U) | U is an open neighborhood of E};
- and μ is inner regular on open sets, namely for every open subset U of X, μ(U) = sup {μ(K) | K is compact and included in U}.
But you can also find the following definition, for example (see https://math.stackexchange.com/questions/1858615/what-is-actually-the-standard-definition-for-radon-measure) a measure μ on X is a Radon measure if and only if:
- μ is inner regular (on all Borel sets, not just the open sets): for every Borel subset E of X, μ(E) = sup {μ(K) | K is compact and included in E};
- μ is locally finite: every point x of X has an open neighborhood U such that μ(U)<∞.
Those definitions are fine, except that they do not really make sense outside of Hausdorff spaces. If X is Hausdorff, then every compact set K is closed, hence Borel, and therefore μ(K) makes sense. But there is no reason why compact sets should be Borel in more general, non-Hausdorff spaces. A standard fix is to consider closed compact sets, or compact Gδ sets. A more sophisticated one is due to Klaus Keimel and Jimmie Lawson [1, Definition 7.2].
Let me not go into those extra complications, which we will avoid by focusing on Hausdorff spaces, so that every compact subset is closed… and despite the title of this site, “Non-Hausdorff topology and domain theory”.
There are some relations between the various notions involved. For example,
- Inner regularity implies inner regularity on the open sets; this is immediate.
- If μ is locally finite, then μ is finite on compact sets.
Indeed, let K be a compact subset of X. Every point x has an open neighborhood Ux such that μ(Ux)<∞. Finitely many of them cover K, since K is compact, and therefore μ(K) is less than or equal to the sum of those finitely many values μ(Ux), hence is finite.
But does the converse hold? Is it true that if μ is finite on compact sets, then μ is locally finite?
That certainly holds if X is locally compact: for every point x of X, find an open neighborhood U of x (e.g., X itself), a compact neighborhood K of x included in U; then int(K) is the desired open neighborhood of x, since μ(int(K)) ≤ μ(K) < ∞.
Well, no, this converse does not hold. There is a Hausdorff space X, and a measure μ on X that is finite on compact sets but not locally finite.
The key is to find the right space X, and X must not be locally compact, to start with.
We will use Antoine Appert’s “espace (u)” [2], a classic counterexample in topology [3, Counterexample 98]. You would be right in thinking that what I said about measures above is a convenient excuse to talk about this particular space.
This space is X ≝ N, the set of natural numbers, with a topology that we define right away.
For every subset E of N, for every n ∈ N, let N(n,E) be the cardinality of E ∩ {0,1, ··· ,n}. Then N(n,E)/(n+1) is the probability of finding an element that falls in E, over all numbers in {0,1, ··· ,n}, drawn at random uniformly.
We define the density d(E) of E as liminfn∈N N(n,E)/(n+1), in other words as the directed supremum supm∈N infn≥m N(n,E)/(n+1). Steen and Seebach [3] define it as the limit of the probabilities N(n,E)/(n+1) when n tends to infinity, but there is no reason why this limit should exist; the liminf will work fine instead.
Definition (Appert space). Appert space is N, with the topology whose open subsets are:
- the subsets of N–{0} (all of them);
- and the subsets of N that contains 0 and whose density is equal to 1.
The second condition informally says “every open subset that contains 0 must contain almost all natural numbers”.
Fact. This is a topology.
In order to see this, we first consider unions. Every union of open sets of the first kind, namely that are included in N–{0}, is also included in N–{0}. Given any family of open sets such that one of them is of the second kind, namely contains 0 and has density 1, the union of the family also contain 0 and has density 1.
Second, we turn to finite intersections. The intersection of the empty family is the whole of N, which has density 1. Let U and V be two open sets. If U or V is of the first kind, namely is included in N–{0}, then so is U ∩ V. Otherwise, U and V contain 0 and have density 1, and we need to show that d(U ∩ V)=1. This is the core of the argument.
For every ε > 0, since d(U)=1, there is a number m such that N(n, U) > (1-ε/2) (n+1) for every n≥m; similarly, there is a number m’ (and we can take m=m’, by replacing both m and m’ by their maximum) such that N(n, V) > (1-ε/2) (n+1) for every n≥m. In other words, for every n≥m, there are less than (n+1) ε/2 numbers below n outside of U, and less than (n+1) ε/2 numbers below n outside of V. Hence there are at most (n+1) ε numbers below n outside of U or outside of V, which means that N(n, U ∩ V) ≥ (1-ε) (n+1). As ε>0 is arbitrary, d (U ∩ V)=1, so U ∩ V is open.
Fact. Appert space is Hausdorff.
Proof. Given any two points m, n ∈ N, we wish to find an open neighborhood U of m and an open neighborhood V of n such that U ⋂ V = ∅. If m, n ≠ 0, we simply take {m} and {n}.
Otherwise, and exchanging m and n if necessary, we may assume that m=0 and n≠0. We take U ≝ N–{n} and V ≝ {n}. It is easy to see that V is open. In order to see that U is also open, we must show that d(U)=1. In {0, 1, …, k}, there are k+1 elements in U if k<n, and k if k≥n. Hence N(k, U) is equal to 1 for k<n and to k/(k+1) = 1 – 1/(k+1) for k≥n, and this tends to 1 as k tends to ∞. ☐
Proposition A. The compact subsets of Appert space are its finite subsets.
Proof. Every finite subset of a topological space is compact, and the only thing we have to show is that every compact subset is finite, or equivalently that no infinite set can be compact.
Let E be an infinite subset of Appert space. If E does not contain 0, then the sets {m} with m ∈ E form an open cover of E without any proper subcover, so E is not compact. Hence we will assume that 0 is in E in the following.
The core of the argument now consists in finding a subset U of N containing 0, of density 1, and which misses infinitely many points from E. I will explain why later. To this end, we define a sequence of points from E as follows. We start by defining m0 as any point of E larger than or equal to 1. Having defined mk, we pick mk+1 as any number in E larger than or equal to max (mk+1, 2k). This is possible: since E is infinite, there are infinitely many points in E larger than any given number. Now we let U ≝ N–{mk | k ∈ N}.
Since m0 < m1 < … < mk < … and mk ≥ 2k for every k ∈ N, the only numbers mk smaller than or equal to a given n must be such that 2k ≤ n, and there are about log2 n of them, plus or minus 1. Hence N(n, U) is of the order of n+1–log2 n, and then N(n, U)/(n+1) = 1–log2 n/(n+1) tends to 1 as n tends to ∞. Therefore d(U)=1. Since m0 ≥ 1, U does not contain 0, and is therefore open.
By definition, U misses the infinitely many points mk from E. Hence U plus the open sets {mk}, k ∈ N, form a cover of E (in fact of the whole of N) which has no proper subcover, in particular no finite subcover. Therefore E is not compact. ☐
Corollary B. Appert space is not locally compact: 0 has no compact neighborhood.
Proof. If K is a compact neighborhood of 0, then it contains an open neighborhood U of 0, by definition. By Proposition A, K is finite, hence so is U. But the only open neighborhoods of 0 have density 1, and are in particular infinite. ☐
Lemma C. The Borel σ-algebra of Appert space consists of all the subsets of N.
Proof. All the subsets of N–{0} are open hence Borel. {0} is Borel since it is the complement of N–{0} (hence closed). Finally, all the subsets of N that contain 0 can be obtained as the union of {0} with a set of the first kind. In summary, we have shown much more than what Lemma C claims: every subset of Appert space is UCO (a notion I have already talked about several times, for example in the August 2019 post), namely the union of an open and a closed set. ☐
Hence we don’t need to be careful about which subsets are Borel or not: they all are.
We now consider an extremely simple measure on Appert space: the counting measure c. This maps every (necessarily Borel) subset E of N to its cardinality: its number of elements if E is finite, and ∞ otherwise. It is easy to see that c is a measure, namely that c(∅)=0 and that for every countable family (An)n ∈ N of subsets of N, c(∪n ∈ N An) = ∑n ∈ N c(An).
Proposition D. The counting measure c on Appert space is inner regular.
Proof. This is trivial. For every subset E of N, the cardinality of E is equal to the supremum of the cardinalities of its finite (=compact by Proposition A) subsets, because any set E is the union of the chain of sets E ⋂ {0, 1, …, n}, n ∈ N. ☐
Proposition E. For every compact subset K of Appert space, c(K)<∞, but c is not locally finite.
Proof. By Proposition A, every compact set K is finite, hence c(K)<∞. But 0 has no open neighborhood U such that c(U)<∞, since every such U has density 1, hence is infinite. ☐
And that is it: as promised, we have found a measure, and even an inner regular measure, such that the measure of every compact subset is finite but that is not locally finite.
By the way, c is not outer regular: with E ≝ {0}, c(E)=1, but for every open neighborhood U of E, U has density 1, hence is infinite; therefore inf {μ(U) | U is an open neighborhood of E} = ∞.
Some topological properties of Appert space
I have already mentioned that Appert space is Hausdorff, and not locally compact. Let me list a few other properties and non-properties of Appert space, as listed by Steen and Seebach [3], although I will omit some.
A Hausdorff space is normal or T4 if and only if for every pair of disjoint closed subsets A and B, there are disjoint open neighborhoods U and V of A and B respectively. It is completely normal or T5 if and only if all of its subspaces are normal. Here is an alternative (classical) characterization. We say that two subsets A and B of a space X are separated if and only if each one is disjoint from the closure of the other; we have already seen the notion in the February 2025 post, where we wrote A|B for the notioni=n.
Lemma F. For a topological space X, the following are equivalent:
- X is completely normal, namely every subspace of X is normal;
- every open subspace of X is normal;
- for every pair of separated subsets A and B, there are disjoint open neighborhoods U and V of A and B respectively.
Proof. 1⇒2 is trivial.
2⇒3. Let A and B be separated subsets of X. We write cl() for closure in X, and we build the subspace Y ≝ X–(cl(A) ⋂ cl(B)). By item 2, which we assume, Y is normal.
We check that A ⊆ Y. Because A and B are separated, A is disjoint from cl(B), so A ⊆ X–cl(B) ⊆ Y. Similarly, B ⊆ Y.
Let A’ ≝ cl(A) ⋂ Y and B’ ≝ cl(B) ⋂ Y. Then A’ and B’ are closed in Y, and they are disjoint, because A’ ⋂ B’ = (cl(A) ⋂ cl(B)) ⋂ Y = ∅. Since Y is normal, there are two disjoint open neighborhoods of A’ and of B’ in Y; we write them as U ⋂ Y and as V ⋂ Y respectively, where U and V are open in X. But Y is open in X, so U ⋂ Y and V ⋂ Y are disjoint open sets in X. Then A is included in cl(A) and in Y, hence in A’, and A’ is included in U ⋂ Y, so U ⋂ Y is an open neighborhood of A in X; symmetrically, V ⋂ Y is an open neighborhood of B in X, and we have seen that they are disjoint.
3⇒1. We show that the negation of 1 implies the negation of 3. In other words, we assume that X is not completely normal, and we exhibit a pair of separated subsets A and B with no disjoint open neighborhoods. Since X is not completely normal, it has a subspace Y that contains two closed (in Y) subsets A and B with no disjoint open neighborhoods. While A and B are closed in Y, there is no reason why they should be closed in X.
However, they are separated in X. We argue as follows. We have A ⊆ cl(A) ⋂ Y. Since A is closed in Y, we can write A as C ⋂ Y for some closed subset C of X. But A ⊆ C, so cl(A) ⊆ C, since C is closed. Therefore cl(A) ⋂ Y ⊆ C ⋂ Y = A. Therefore A = cl(A) ⋂ Y. If cl(A) intersected B, then it would intersect it at some point of Y, so cl(A) ⋂ Y would intersect B; which is impossible since cl(A) ⋂ Y = A and A is disjoint from B. Hence cl(A) is disjoint from B, and symmetrically, cl(B) is disjoint from A. In other words, A and B are separated.
It remains to show that A and B do not have pairwise disjoint open neighborhoods. If they had, namely if we could find an open neighborhood U of A and an open neighborhood V of B such that U ⋂ V = ∅, then U ⋂ Y and V ⋂ Y would be disjoint open neighborhoods of A and B in Y, respectively; and that was assumed to be nonexistent. ☐
We turn to Appert space proper.
Proposition G. Appert space is completely normal, hence normal in particular.
Proof. Let A and B be separated subsets of Appert space. We wish to find disjoint open neighborhoods U and V of A and B respectively. Since A and B are separated, they are disjoint, so at most one of them contains 0. In other words, one of them does not contain 0. By symmetry, let us assume that B does not contain 0.
Then B is open. Hence we can take V≝B. Since A and B are separated, cl(B) is disjoint from A, so A is included in U≝N–cl(B). And clearly U and V are disjoint. ☐
At this point, we may wonder what interior and closure look like in Appert space, and they are amazingly simple operations.
Lemma H. In Appert space, the interior of a subset A is:
- A itself if d(A)=1 or if 0 ∉ A,
- and A–{0} if 0 ∈ A and d(A)<1.
Proof. If d(A)=1 or if 0 ∉ A, then A is already open, hence is equal to its own interior. Otherwise, A–{0} is open, and is the largest open subset of A, since the only strictly larger subset of A is A, which is not open. ☐
Taking complements, we obtain the following corollary.
Corollary I. In Appert space, the closure of a subset A is:
- A itself if d(N–A)=1 or if 0 ∈ A,
- and is A ∪ {0} if 0 ∉ A and d(N–A)<1.
Proposition J. Appert space is not first-countable (hence not second-countable): the point 0 does not have any countable base of open neighborhoods.
Proof. Let us imagine that 0 had a countable base of open neighborhoods U0, U1, …, Un, … We now follow an argument that is similar to an argument we have seen in the proof of Proposition A.
Since every Un is open and contains 0, its density is equal to 1, and in particular Un is infinite. We pick mn from Un so that mn≥2n, for each n ∈ N. Let U be the complement of {mn | n ∈ N}. The elements of U that are in {0, 1, …, k} are elements mn which must satisfy 2n≤k, since mn≥2n. Hence there are k–log2 k points in U ⋂ {0, 1, …, k}, plus or minus 1. The limit of (k–log2 k)/(k+1) when k tends to ∞ is 1, so d(U)=1, and therefore U is open.
But U contains none of the sets Un, because mn is in Un, but not in U. ☐
For the next proposition, we need to recall what a scattered space is. We have already seen the notion in the May 2022 post.
An isolated point of a subset D of a space X is any point x of D that has an open neighborhood U in X such that U ∩ D = {x}; if you prefer, such that {x} is open in D seen as a subspace of X. The points of D that are not isolated are called the limit points of D, and those are exactly the points of D that are limits of nets of points of D–{x}.
A subset D of X is dense-in-itself if and only if every point of D is a limit point of D, namely if and only if it has no isolated point. A scattered space is a space whose only dense-in-itself subset is the empty set.
Proposition K. Appert space is scattered.
Proof. Let D be a subset of N. Every point n ∈ D–{0} is isolated in D, since {n} is open in Appert space. Hence if D is dense-in-itself, then it must be empty or equal to {0}. But {0} is not dense-in-itself, since 0 is isolated in {0}. Therefore the only dense-in-itself subset of Appert space is the empty set, and therefore Appert space is scattered. ☐
A 0-dimensional space is one whose topology has a base consisting of clopen subsets, namely of subsets that are both open and closed. A totally disconnected space is a space whose only connected subspaces have cardinality at most 1. Every 0-dimensional Hausdorff space X is totally disconnected: given any subspace Y with two distinct elements a and b, we can find two disjoint open neighborhoods U and V of a and b in X respectively; since X is totally disconnected, U and V contain further clopen neighborhoods of a and b, so we can assume that U and V are themselves clopen without loss of generality; but then U ⋂ Y and Y–U (or Y–V and V ⋂ Y) are disjoint open neighborhoods of a and b in Y whose union is the whole of Y, showing that Y is not connected.
Proposition L. Appert space is 0-dimensional, hence totally disconnected.
Proof. As a base, we take the one-element sets {n} where n≠0, plus all the open sets that contain 0. This is a base of the topology: given any point n ∈ N, and any open neighborhood U of n, either n≠0 and n is in the basic open set {n}, which is included in U; or n=0, and U itself is a basic open neighborhood of n.
It remains to see that all those sets are clopen. For every n≠0, the complement of {n} is a set that contains 0, and whose density is equal to 1, hence an open set; therefore {n} is closed, and hence clopen. For every open neighborhood U of 0, the complement of U is included in N–{0}, hence is open; therefore U is closed, and hence clopen. ☐
A space is extremally disconnected if and only if the closure of every open set is open. In order to compare this notion to totally disconnectedness, in a totally disconnected space, the closure of every open from any chosen base of clopen sets is closed, vacuously; but total disconnectedness does not say anything about closures of open sets that are not in the base.
Proposition M. Appert space is not extremally disconnected.
Proof. We consider a subset U of N–{0}. This is open, and if d(N–U)≠1, then the closure of U will be U ∪ {0} by Corollary I. We wish to find such a U, in such a way that the closure U ∪ {0} is not open, namely such that d(U ∪ {0})≠1. One possibility is to take the set of odd natural numbers for U: then d(N–U) = 1/2 ≠ 1, and d(U ∪ {0}) = 1/2 ≠ 1. ☐
- Klaus Keimel and Jimmie Lawson. (2005). Measure extension theorems for T0-spaces. Topology and its Applications, 149(1–3), 57–83.
- Antoine Appert. (1934). Propriétés des espaces abstraits les plus généraux. PhD thesis, Faculté des sciences de l’université de Paris. Série A, numéro 1465, numéro d’ordre 2331. Available on NumDam (http://www.numdam.org/item?id=THESE_1934__156__1_0), in the “thèses de l’entre-deux-guerres” series.
- Lynn Arthur Steen and J. Arthur Seebach, Jr. (1978). Counterexamples in Topology (Second ed.) Springer-Verlag.
— Jean Goubault-Larrecq (March 20th, 2026)