March 27, 2026
By using a similar pattern of arguments, we show that in three categories the collection of isomorphisms forms a residual subset of the space of morphisms. We first consider surjective continuous mappings on Cantor spaces. Next, we look at measure preserving maps on Polish measure spaces. Finally, we examine continuous, measure preserving maps on Cantor spaces equipped with so-called good measures.
Consider the tent map \(f : [0,1] \to [0,1]\), defined by \[\label{eq14601} f(t) = \begin{cases} 2t \;\;\;\text{for} \;0 \le t \le \frac{1}{2}, \\ 2(1 - t) \;\text{for} \; \frac{1}{2} \le t \le 1. \end{cases}\tag{1}\]
This is a surjective, continuous map. It is clear that it cannot be uniformly approximated by a homeomorphism. If \(J\) is a non-trivial closed interval contained in \((0,1)\), then \(f^{-1}(J)\) is union of two disjoint intervals \(J_1 \subset (0,\frac{1}{2})\) and \(J_2 \subset (\frac{1}{2},1)\). If \(g : [0,1] \to [0,1]\) is a continuous map close to \(f\), then \(g(J_1)\) and \(g(J_2)\) are each intervals close to \(J\). They may not be equal but they will still overlap and so \(g\) will not be injective.
One can perform this folding procedure for the Cantor set as well. In addition, the tent map is measure preserving with respect to the Lebesgue measure \(\lambda\) on \([0,1]\), i.e. \(\lambda(f^{-1}(E)) = \lambda(E)\) for every Borel subset of \([0,1]\). As we will see, in each of these cases, \(f\) can be approximated arbitrarily closely by a suitable invertible map.
We will consider three different categories.
The objects are Cantor spaces and the morphisms are surjective continuous maps, with the topology of uniform convergence on the space of maps.
The objects are measure spaces obtained from complete separable metric spaces and the morphisms are measure preserving measurable maps. The topology is that of pointwise convergence on the Borel sets.
The objects are again Cantor spaces but equipped with certain special “good” measures. The morphisms are continuous, measure preserving maps and the topology is that of uniform convergence.
In each case, we show that the invertible maps, i.e. the isomorphisms in the category, comprise a dense \(G_{\delta}\) subset of the space of maps. In each case, the space of maps is a complete metric space and so the Baire Category Theorem applies.
The argument in each case consists of three steps. We illustrate them with the Cantor set case.
Any two Cantor spaces are homeomorphic and so, in particular, any two nonempty clopen (=simultaneously closed and open) subsets are homeomorphic.
In the space of surjective continuous maps between Cantor spaces, the invertible maps comprise a \(G_{\delta}\) subset.
In the space of surjective continuous maps between Cantor spaces, the invertible maps form a dense subset.
In the other two categories, these same three steps of Uniqueness, \(G_{\delta}\), and density occur in the appropriate, albeit more complicated, ways.
Notes: The importance of those dense sets which are \(G_{\delta}\) comes from the Baire Category Theorem which says that in a completely metrizable space, e.g. a compact metric space, the countable intersection of dense open sets is dense and so the countable intersection of dense \(G_{\delta}\) sets is dense. By contrast, the two sets which are merely dense may have an empty intersection, e.g. the rationals and the irrationals in the real line. For a subset \(A\) of a completely metrizable space \(X\) we say that \(A\) is residual when it contains a \(G_{\delta}\) set dense in \(X\). We then say that the members of \(A\) are generic in \(X\). The complement of a residual set is called meagre. Thus, the generic real number is irrational.
In a measure space there is an alternate notion of genericity. A subset is said to have full measure when its complement has measure zero. The countable intersection of subsets of full measure again has full measure. If the measure space has a topology such that every nonempty open subset is measurable with positive measure, then the measure is called full and a subset with full measure is dense.
Unfortunately, these two notions conflict. For example, on the unit interval \(I\) Lebesgue measure is full. While the classical Cantor set has measure zero, there exists for every positive integer \(n\), a Cantor subset \(C_n\) of \(I\) with measure \(1 - \frac{1}{n}\), obtained by removing “middle third divided by \(n\)” intervals. Since no \(C_n\) contains an interval of positive length, the complement of each is dense. The union of the \(C_n\)’s is then a meagre subset with full measure.
A classical, and less artificial, example is the set of normal numbers in \(I\). A number is normal when every sequence of \(n\) digits occurs with frequency \(1/10^n\) in its decimal expansion. That the set of normal numbers has full measure was proved by Emile Borel in 1909 [1]. However, it is a meagre subset of \(I\), see, e.g. [2].
In a recent paper, [3], Bobok et al. consider the space of continuous maps on the unit interval which preserve Lebesgue measure. These maps are far from invertible. In fact, the only homeomorphisms of the interval which preserve Lebesgue measure are the identity \(id_I\) and \(1 - id_I\). Among other generic properties the authors show that for such a generic map \(f\), the point-inverse \(f^{-1}(t)\) contains a perfect subset (and so is uncountable) for every \(t \in (0,1)\), i.e. for all points except the end-points.
When you first meet the classical Cantor set obtained by removing successively the “middle thirds”, the points you see are the endpoints of the open intervals which were removed. There are countably many such points. One then observes that the points of the Cantor set are those points of the unit interval who have a ternary expansion consisting only of \(0'\)s and \(2\)’s, no \(1\)’s, and note that this expansion is unique. This shows that the set is uncountable.
Thus, in addition to the obvious endpoints, there are uncountably many hidden points. The distinction between these two sorts of points makes it hard to credit that the Cantor set is homogeneous, i.e. for any two points of the Cantor set there exists a homeomorphism of the set taking one to the other. However, the distinction between the two sorts of points is an artifact of the order structure inherited from the interval. When you consider the ternary expansions and replace the \(2\)’s by \(1\)’s you map the Cantor set to the countable product \(\{0, 1 \}^{\mathbb{N}}\). With the product topology this a compact metrizable space and the map is a homeomorphism. On the other hand \(\{0, 1 \}\), regarded as the group of integers mod \(2\), is an additive group and with the product topology \(\{0, 1 \}^{\mathbb{N}}\) becomes a compact topological group via coordinate-wise addition. Such a group is always homogeneous. The map \(x \mapsto x + (b - a)\) is a homeomorphism taking \(a\) to \(b\).
****Definition** 1**. A Cantor space* \(X\) is a topological space satisfying the following properties.*
\(X\) is compact and metrizable.
\(X\) is zero-dimensional, i.e. the clopen sets form a basis for the topology.
\(X\) is perfect, i.e. there are no isolated points, or, equivalently, every non-empty open subset is infinite.
Recall that any admissible metric on a compact, metrizable space is complete and all such metrics are uniformly equivalent. A compact metric space admits a finite cover by open sets of diameter less than \(1/n\) for any positive integer \(n\). Combining a sequence of such covers we obtain a basis for the topology and so a compact, metrizable space is second countable, i.e. it admits a countable basis. It then follows that in a compact, metrizable space the collection of clopen subsets is countable since each is a finite union of elements of the basis. A compact, metrizable space is zero-dimensional if and only if it is totally disconnected, i.e. the only connected subsets are singletons. Notice that with the relative topology any nonempty clopen subset of a Cantor space satisfies the above properties and so is itself a Cantor space.
We will use \(C = \{ 0. 1 \}^{\mathbb{N}}\) as our standard Cantor space. It clearly satisfies the above conditions as does the classical Cantor set. An element \(z \in C\) is an infinite string of \(0\)’s and \(1\)’s. With \([n] = \{ 1,2, \dots, n \}\), the set \(\{ 0, 1 \}^{[n]}\) consists of the \(2^n\) \(n\)-strings, i.e. strings of length \(n\). Given a string \(z\) which is infinite or of length at least \(m\), we call the restriction \(z|[m]\) the initial \(m\)-string of \(z\). Given an \(m\)-string \(w\) the set \(C_w = \{ z \in C : z|[m] = w \}\) is a clopen set and the collection of \(C_w\)’s as \(w\) varies over all finite strings provides \(C\) with a countable basis of clopen sets.
A partition \(\mathcal{P}\) of a set \(X\) is a finite, pairwise disjoint collection of nonempty subsets which covers \(X\). For partitions \(\mathcal{P}_1\) and \(\mathcal{P}_2\) of \(X\), \(\mathcal{P}_1\) refines \(\mathcal{P}_2\), written \(\mathcal{P}_1 \le \mathcal{P}_2\), when every element of \(\mathcal{P}_1\) is contained in a, necessarily unique, element of \(\mathcal{P}_2\), or, equivalently, when every element of \(\mathcal{P}_2\) is a union of elements of \(\mathcal{P}_1\). A sequence of partitions \(\{ \mathcal{P}_n \}\) is monotone when \(m \ge n\) implies \(\mathcal{P}_m \le \mathcal{P}_n\). By transitivity of the refinement relation it suffices \(\mathcal{P}_{n+1}\) refines \(\mathcal{P}_n\) for all \(n\). For partitions \(\mathcal{P}_1\) and \(\mathcal{P}_2\), the common refinement is given by \[\label{eq24601a} \mathcal{P}_1 \wedge \mathcal{P}_2 = \{ U_1 \cap U_2 : U_1 \in \mathcal{P}_1, U_2 \in \mathcal{P}_2 \} \setminus \{ \emptyset \}.\tag{2}\]
If \(X\) is a compact metric space, then \(\mathcal{P}\) is a clopen partition when the elements are clopen subsets and the mesh of the partition \(\mathcal{P}\), written \(|\mathcal{P}|\), is the maximum of the diameters of the elements of \(\mathcal{P}\).
****Proposition** 2**. (Uniqueness) Any two Cantor spaces are homeomorphic.
Proof. Given a Cantor space \(X\) it suffices to obtain a homeomorphism from \(X\) onto our standard space \(C\). Choose a metric \(d\) on \(X\).
Because \(X\) is compact and zero-dimensional there exist clopen partitions of \(X\) with arbitrarily small mesh. We will construct a monotone sequence \(\{\mathcal{P}_n \}\) of clopen partitions, an increasing sequence \(\{ m_n \}\) of positive integers and a labelling bijection \(w : \mathcal{P}_n \to \{ 0, 1\}^{[m_n]}\) so that if \(V \in \mathcal{P}_{n+1}\) is contained in \(U \in \mathcal{P}_n\) then the label of \(U\) is an initial string of the label of \(V\). Thus, since \(\mathcal{P}_{n+1}\) is a refinement of \(\mathcal{P}_n\), it follows that \[\label{eq24601} V \subset U \qquad \Longleftrightarrow \qquad w(U) = w(V)|[m_n].\tag{3}\] In addition, \[\label{eq24602} |\mathcal{P}_n| \le 1/n.\tag{4}\]
Let \(\mathcal{P}_1\) be an arbitrary clopen partition of mesh at most \(1\). Now assume that \(\mathcal{P}_n\) has been constructed. First choose for each \(U \in \mathcal{P}_n\) a clopen partition of \(U\) with mesh at most \(1/(n+1)\). Because the Cantor space \(U\) is perfect we can further subdivide to obtain a partition of any number of elements larger than what we started with. Thus, we can choose \(k_n\) large enough that any \(U\) has been subdivided into exactly \(2^{k_n}\) pieces. Putting these together we obtain our clopen partition \(\mathcal{P}_{n+1}\). Let \(m_{n+1} = m_n + k_n\). Label the elements of \(\mathcal{P}_{n+1}\) which are contained in \(U\) using those \(m_{n+1}\)-strings whose initial \(m_n\)-strings are equal to \(w(U)\). Thus, (3 ) holds.
For each point \(x \in X\), there is a unique sequence \(\{ U_n \in \mathcal{P}_n \}\) with \(x \in U_n\). Because the sequence \(\{ \mathcal{P}_n \}\) is monotone, \(U_{n+1} \subset U_n\). Thus, there exists a unique \(z(x) \in C\) such that \(z(x)|[m_n] = w(U_n)\). On the other hand, if \(z \in C\), there is a sequence \(\{ U_n \in \mathcal{P}_n \}\) with \(z|[m_n] = w(U_n)\). From (3 ) it follows that \(U_{n+1} \subset U_n\). The decreasing sequence \(\{ U_n \}\) of nonempty compacta has a nonempty intersection and (4 ) implies that the intersection is a singleton \(x\). Clearly, \(z(x) = z\). Thus, the map \(x \mapsto z(x)\) is a bijection from \(X\) to \(C\). From \[\label{eq24603} x \in U_n \qquad \Longleftrightarrow \qquad z(x)|[m_n] = w(U_n),\tag{5}\] it follows that the map is a homeomorphism. ◻
For our next result, assume that \(f : X \to Y\) is a set map, if \(A \subset X, B \subset Y\) with \(U = X \setminus A, V = Y \setminus B\), then, \[\label{eq24604}\begin{align} f(A) \cap B = \emptyset \qquad \Longleftrightarrow \qquad f(A) \subset V \qquad \Longleftrightarrow \qquad A \subset f^{-1}(V)\\ \Longleftrightarrow \qquad A \cap f^{-1}(B) = \emptyset \qquad \Longleftrightarrow \qquad f^{-1}(B) \subset U. \end{align}\tag{6}\]
For compact metrizable spaces \(X\) and \(Y\) we let \(\mathcal{C}(X,Y)\) be the set of continuous maps from \(X\) to \(Y\). We denote by \(\mathcal{C}_{sur}(X,Y)\) and by \(\mathcal{C}_{iso}(X,Y)\) the subsets of surjective maps and bijective maps, respectively. By compactness, a bijective continuous map is a homeomorphism. Given a metric \(d\) on \(Y\), we define the metric \(d\) on \(\mathcal{C}(X,Y)\) by \[\label{eq24605} d(f,g) = \sup \{ d(f(x),g(x)) : x \in X \}.\tag{7}\] The associated topology is that of uniform convergence and with this metric \(\mathcal{C}(X,Y)\) is a complete metric space. It is not hard to show that that \(\{ f : f(\overline{U}) \subset V \}\) with \(U\) and \(V\) varying over countable bases in \(X\) and \(Y\), respectively, provides a countable subbase for \(\mathcal{C}(X,Y)\). Thus, \(\mathcal{C}(X,Y)\) is separable as well.
****Proposition** 3**. (\(\mathbf{G_{\delta}}\)) For compact metrizable spaces \(X\) and \(Y,\)
\(\mathcal{C}_{sur}(X,Y)\) is a closed subset of \(\mathcal{C}(X,Y)\) and \(\mathcal{C}_{iso}(X,Y)\) is a \(G_{\delta}\)
subset of \(\mathcal{C}(X,Y)\).
Proof. If \(A, B\) are closed subsets of \(X\) and \(Y\) respectively with open complements \(U\) and \(V\) then the compact sets \(f(A)\) and \(B\) are a positive distance apart if they are disjoint. Hence, \(\{ f \in \mathcal{C}(X,Y) : f(A) \cap B = \emptyset \}\) is an open subset of \(\mathcal{C}(X,Y)\).
For any proper open subset \(V\) of Y, it follows from (6 )that \(\{ f \in \mathcal{C}: f(X) \subset V \}\) is an open set and taking the union over all such proper open sets \(V\), we obtain the open set \(\mathcal{C}\setminus \mathcal{C}_{sur}\). Hence, \(\mathcal{C}_{sur}\) is closed.
Clearly, \(f \mapsto f \times f\) is a continuous map from \(\mathcal{C}(X,Y)\) to \(\mathcal{C}(X \times X,\\ Y \times Y)\). With the metric \(d\) on \(X\), the set \(V_{1/n} = \{ (x_1,x_2) : d(x_1,x_2) < 1/n \}\) is open in \(X \times X\). With \(\Delta_Y = \{ (y,y) \}\), the closed diagonal in \(Y,\) it follows from (6 )that \(\{ f \in \mathcal{C}:(f \times f)^{-1}(\Delta_Y) \subset V_{1/n} \}\) is open. Intersecting over \(n\) we see that the set of injective maps is \(G_{\delta}\). Since a closed set in a metric space is always \(G_{\delta}\), it follows that \(\mathcal{C}_{iso}\) is \(G_{\delta}\). ◻
****Proposition** 4**. (Density) Let \(f : X \to Y\) be a surjective, continuous map with \(X\) and \(Y\) Cantor spaces. Let \(d\) be an admissible metric on \(Y\). Given \(\epsilon> 0\), there exists a homeomorphism \(g\) of \(X\) onto \(Y\) such that \[\label{eq24606} x \in X \qquad \Longrightarrow \qquad d(f(x),g(x)) \le \epsilon.\qquad{(1)}\] .
Proof. Let \(\mathcal{P}= \{U_1, \dots, U_n \}\) be a clopen partition of \(Y\) with mesh at most \(\epsilon\). Since \(f\) is surjective and each \(U_i\) is nonempty, it follows that each \(f^{-1}(U_i)\) is a nonempty clopen subset of \(X\) and so \(\{ f^{-1}(U_1), \dots, f^{-1}(U_n) \}\) is a clopen partition of \(X\). Each \(U_i\) and each \(f^{-1}(U_i)\) is a Cantor space and so Proposition 2 implies we can choose a homeomorphism \(g_i\) from \(f^{-1}(U_i)\) onto \(U_i\). Concatenating we obtain the homeomorphism \(g\) from \(X\) onto \(Y\). Clearly, for all \(x \in X\) \[\label{eq24607} f(x) \in U_i \qquad \Longleftrightarrow \qquad g(x) \in U_i.\tag{8}\] Since the partition \(\mathcal{P}\) has mesh at most \(\epsilon\), (?? ) follows. ◻
From Propositions 3 and 4 we immediately obtain:
****Theorem** 5**. If \(X\) and \(Y\) are Cantor spaces, then the set of homeomorphisms \(\mathcal{C}_{iso}(X,Y)\) is a \(G_{\delta}\) subset of the completely metrizable space \(\mathcal{C}(X,Y)\) and it is dense in the closed subset \(\mathcal{C}_{sur}(X,Y)\) of surjective maps. Thus, the generic surjective map is invertible.
Notes: The uniqueness of Cantor space is a classical result, first proved by L. E. J. Brouwer [4] in 1909. Metrizability is not redundant. There exist many distinct examples of compact, separable, first countable, zero-dimensional, perfect spaces. The Sorgenfrey Double Arrow construction yields a linearly ordered topological space with these properties. In it every point is an endpoint and there are uncountably many clopen subsets. For a discussion of such spaces and the connected linearly ordered spaces in which they live, see [5] and its sequel [6].
For the density results it suffices to consider the case with \(Y = X\). As part of a study of the dynamics of surjective self-maps of the Cantor space in [7] it is proved that the group of invertible self-maps is a dense \(G_{\delta}\) subset. In fact, it then follows from the work of Kechris and Rosenthal [8] that there is a single conjugacy class which is a dense \(G_{\delta}\) subset. For an explicit description of a member of this class, see [9].
A measure space is a triple \((X, \mathcal{B}, \mu)\) with \(\mathcal{B}\) a \(\sigma\)-algebra of subsets of \(X\) and \(\mu\) a finite measure on the elements of \(\mathcal{B}\) with \(\mu(X) > 0\). It is normalized when \(\mu(X) = 1\). An element \(A \in \mathcal{B}\) has full measure when \(\mu(A) = \mu(X)\) or, equivalently, when \(X \setminus A\) has measure zero.
An atom \(A\) is an element of \(\mathcal{B}\) with \(\mu(A) > 0\) such that \[\label{eq34601} B \in \mathcal{B}\; \text{with} \;B \subset A \qquad \Longrightarrow \qquad \mu(B) = \mu(A) \;\text{or} \;\mu(B) = 0.\tag{9}\] The measure is nonatomic when it has no atoms.
On \(\mathcal{B}\), the measure induces a pseudo-metric \(d\) given by \[\label{eq34602} d(A,B) = \mu(A + B)\tag{10}\] where \(A + B\) is the Boolean sum \(= (A \cup B) \setminus (A \cap B)\). The set operations: union, intersection and complementation, are continuous with respect this metric. If \(\{ A_n \}\) is a monotonically increasing (or decreasing) sequence of elements of \(\mathcal{B}\) then the sequence converges to \(\bigcup_n A_n\) ( or \(\bigcap_n A_n\), respectively).
The associated measure algebra \(\mathcal{M}_X\) is the associated metric space. That is, the elements are the equivalence classes of sets in \(\mathcal{B}\) which differ by a set of measure zero. The set operations and the measure are well-defined using representatives from the classes. We denote by \(0\) the class of sets of measure zero. In general, we will write \(A\) both for an element of \(\mathcal{M}_X\) and for any element of \(\mathcal{B}_X\) which represents it.
The measure space is separable when there is a countable subset \(\mathcal{B}_0 \subset \mathcal{B}\), which we may take to be a subalgebra, such that every element of \(\mathcal{B}\) can be arbitrarily closely approximated by an element of \(\mathcal{B}_0\). That is, given \(A \in \mathcal{B}\) and \(\epsilon> 0\), there exists \(B \in \mathcal{B}_0\) such that \(d(A,B) < \epsilon\). This exactly says that the classes of \(\mathcal{B}_0\) form a dense subset of the metric space \(\mathcal{M}_X\). Thus, the measure space is separable exactly when the measure algebra is a separable metric space.
****Proposition** 6**. For a measure space, the associated measure algebra is a complete metric space.
Proof. Let \(\{ A_n \in \mathcal{B}\}\) represent a Cauchy sequence in \(\mathcal{M}\). The sequence of characteristic functions \(\{ 1_{A_n} \}\) is clearly fundamental in measure in the sense of [10] Section 22 of Chapter IV. By Theorem E of that section the sequence \(\{ 1_{A_n} \}\) converges in measure to a measurable function \(k\) and a subsequence converges almost uniformly, and so pointwise a.e., to the function \(k\). It then follows that \(k\) is the characteristic function of some set \(A\) and that \(\{ d(A_n,A) \}\) has limit zero. ◻
For \((X, \mathcal{B}, \mu)\) a measure space partition for \(\mathcal{P}\) is a finite collection \(\{ A_1, \dots , A_k \in \mathcal{B}\}\) such that \(\mu(A_i \cap A_j) = 0\) when \(i \not= j\) and \(\bigcup_i A_i\) is a set of full measure. These represent a measure algebra partition \(\{ A_1, \dots , A_k \in \mathcal{M}\}\) such that \(A_i \cap A_j = 0\) when \(i \not= j\) and \(\bigcup_i A_i = X\) in \(\mathcal{M}\). In either case, the mesh \(|\mathcal{P}|\) is the maximum of the measures \(\mu(A_i)\).
For measure spaces \((X, \mathcal{B}_X, \mu)\) and \((Y, \mathcal{B}_Y, \nu)\) a mapping \(T : X \to Y\) is a measure space map when it is a measurable function which preserves measure, i.e. \[\label{eq34603} B \in \mathcal{B}_Y \qquad \Longrightarrow \qquad T^{-1}(B) \in \mathcal{B}_X \;\text{with} \;\mu(T^{-1}(B)) = \nu(B).\tag{11}\]
The map \(T\) is a measure space isomorphism when there exists a measure space map \(\hat{T} : (Y, \mathcal{B}_Y, \nu) \to (X, \mathcal{B}_X, \mu)\) such that \(\hat{T} \circ T = id_X\) ae and \(T \circ \hat{T} = id_Y\) ae.
****Proposition** 7**. Let \(T : (X, \mathcal{B}_X, \mu) \to (Y, \mathcal{B}_Y, \nu)\) be a measure space map.
(a) If \(T : (X, \mathcal{B}_X, \mu) \to (Y, \mathcal{B}_Y, \nu)\) is a measure space isomorphism with inverse \(\hat{T}\) then there exist subsets \(X_1 \subset X\) and \(Y_1 \subset Y\) of full measure such that the restriction \(T|X_1\) is a bijection onto \(Y_1\) with inverse the restriction \(\hat{T}|Y_1\).
(b) Conversely, if there exist subsets \(X_1 \subset X\) and \(Y_1 \subset Y\) of full measure such that the restriction \(T|X_1\) is a bijection onto \(Y_1\) with a measurable inverse, then \(T\) is a measure space isomorphism.
Proof. (a) Let \(X_0 \subset X\) and \(Y_0 \subset Y\) be subsets of full measure such that \(\hat{T} \circ T\) is the identity on \(X_0\) and \(T \circ \hat{T}\) is the identity on \(Y_0\). Let \(X_1 = X_0 \cap T^{-1}(Y_0)\) and \(Y_1 = Y_0 \cap \hat{T}^{-1}(X_0)\). If \(x \in X_1\) then \(T(x) \in Y_0\) and since \(x \in X_0\), \(\hat{T}(T(x)) = x\). Hence, \(T(x) \in \hat{T}^{-1}(X_0)\). That is, \(T(X_1) \subset Y_1\) and similarly, \(\hat{T}(Y_1) \subset X_1\). The maps \(T : X_1 \to Y_1\) and \(\hat{T} : Y_1 \to X_1\) are clearly inverses.
(b) Define \(\hat{T}\) to be \((T|X_1)^{-1}\) on \(Y_1\) and map \(Y \setminus Y_1\) to a point \(x\) of \(X_1\). Since \(\hat{T}\) maps into \(X_1\), we have for \(A \in \mathcal{B}_X\), \(\hat{T}^{-1}(A) = T(A \cap X_1)\) if \(x \not\in A\) or \(T(A \cap X_1) \cup (Y \setminus Y_0)\) if \(x \in A\). These are measurable because \((T|X_1)^{-1}\) is assumed measurable (\(T(A \cap X_1) = ((T|X_1)^{-1})^{-1}(A \cap X_1)\)). Since \(A \cap X_1 = T^{-1}(T(A \cap X_1)) \cap X_1\) we have \[\label{eq34603a} \mu(A) = \mu(A \cap X_1) = \mu(T^{-1}(T(A \cap X_1)) = \nu(T(A \cap X_1)) = \nu(\hat{T}^{-1}(A)).\tag{12}\] Hence, \(\hat{T} : (Y, \mathcal{B}_Y, \nu) \to (X, \mathcal{B}_X, \mu)\) is a measure space map. Since \(\hat{T} \circ T\) is the identity on \(X_1\) and \(T \circ \hat{T}\) is the identity on \(Y_1\), \(\hat{T}\) is the inverse of \(T\). ◻
In particular, if \(A_1 \subset A \in \mathcal{B}\) with \(\mu(A_1) = \mu(A) > 0\), then the inclusion map from \(A_1\) to \(A\) is a measure space isomorphism.
Note that a measurable bijection need not have a measurable inverse. For example, if \(\mathcal{B}_0\) is a \(\sigma\)-algebra of subsets of \(X\) which is properly contained in the \(\sigma\)-algebra \(\mathcal{B}\), then the identity map on \(X\) is measurable from \(\mathcal{B}\) to \(\mathcal{B}_0\) but the inverse is not. However, if a bijection has a measurable inverse and preserves measures, then the inverse clearly preserves measures as well.
If \(T : (X, \mathcal{B}_X, \mu) \to (Y, \mathcal{B}_Y, \nu)\) is a measure space map, then the map from \(\mathcal{B}_Y\) to \(\mathcal{B}_X\) given by \(B \mapsto T^{-1}(B)\) induces a map \(T^* : \mathcal{M}_Y \to \mathcal{M}_X\), which is an example of a measure algebra map.
****Definition** 8**. For measure algebras \(\mathcal{M}_X\) and \(\mathcal{M}_Y\) a mapping \(S : \mathcal{M}_Y \to \mathcal{M}_X\) is a measure algebra map* when it satisfies:*
\(S(Y) = X\) and \(S(0) = 0\).
\(S(B_1 \cup B_2) = S(B_1) \cup S(B_2)\) and \(S(Y \setminus B) = X \setminus S(B)\) for all \(B_1, B_2, B \in \mathcal{M}_Y\).
\(\mu(S(B)) = \nu(B)\) for all \(B \in \mathcal{M}_Y\).
It follows that \(S\) preserves intersection and Boolean sums. Hence, \(S\) preserves the metrics as well. That is, \(S : \mathcal{M}_Y \to \mathcal{M}_X\) is a metric space isometry. In particular, a measure algebra map is always injective. When it is surjective as well, then the inverse mapping is clearly a measure algebra map from \(\mathcal{M}_X\) to \(\mathcal{M}_Y\) and so \(S\) is then a measure algebra isomorphism.
The arguments that follow will all be at the measure algebra level. At the end of the section, we will describe the results needed to lift the arguments to measure space conclusions.
Our standard example of a normalized, separable measure space will be the unit interval \(I = [0,1]\) with Lebesgue measure \(\lambda\) on \(\mathcal{B}_I\) the Borel sets in \(I\) and measure algebra \(\mathcal{M}_I\). The measure space \((I,\mathcal{B}_I,\lambda)\) is separable and nonatomic.
****Proposition** 9**. (Uniqueness) Any two normalized, separable,
nonatomic measure algebras are isomorphic.
Proof. It suffices to show that a normalized, separable, nonatomic measure algebra \(\mathcal{M}\) is isomorphic to \(\mathcal{M}_I\). Let \(\{E_1, E_2, \dots \}\) be a dense sequence in \(\mathcal{M}\). We construct a monotone sequence of measure algebra partitions \(\{ \mathcal{P}_n \}\) and maps \(S : \mathcal{P}_n \to \mathcal{B}_I\) such that
The mesh \(|\mathcal{P}_n|\) is at most \(1/n\) and \(E_n\) is a union of elements of \(\mathcal{P}_n\).
The set \(S(\mathcal{P}_n) = \{ S(A) : A \in \mathcal{P}_n \}\) is a partition of \(I\) by closed intervals with \(\lambda(S(A)) = \mu(A)\).
If \(A \in \mathcal{P}_n, B \in \mathcal{P}_{n+1}\) with \(B \subset A\), then \(S(B) \subset S(A)\).
Thus, \(\{ S(\mathcal{P}_n) \}\) is a monotone sequence of partitions of \(\mathcal{M}_I\) with \(|S(\mathcal{P}_n)| = |\mathcal{P}_n|\).
We begin the inductive construction with \(\mathcal{P}_1 = \{ E_1, X \setminus E_1 \}\). Let \(S(E_1)\) be the interval \([0,\mu(E_1)]\) and let \(S(X \setminus E_1)\) be the interval \([\mu(E_1),1]\). Since \(\mathcal{M}_X\) is normalized, \(\mu(X \setminus E_1) = 1 - \mu(E_1)\).
Now assume that \(\mathcal{P}_k\) and \(S : \mathcal{P}_k \to \mathcal{B}_I\) have been constructed for \(k \le n\). Because \(\mathcal{M}\) is nonatomic, we can choose a refinement \(\mathcal{P}_{n+1}\) of the common refinement \(\mathcal{P}_n \wedge \{ E_{n+1}, X \setminus E_{n+1} \}\) such that \(|\mathcal{P}_{n+1}| \le 1/(n+1)\). Given \(A \in \mathcal{P}_n\), \(\mathcal{P}_{n+1}\) contains a partition \(\{ B_1, \dots,B_k \}\) of \(A\). Now \(S(A)\) is an interval \([a,b]\) of length \(\lambda(S(A)) = \mu(A)\). Choose \(S(B_1) = [a,a+\mu(B_1)], S(B_2) = [a+\mu(B_1),a+\mu(B_1)+\mu(B_2)],\) etc to obtain a partition of \([a,b]\) by closed intervals. Letting \(A\) vary over \(\mathcal{P}_n\) we obtain the partition \(\mathcal{P}_{n+1}\). This completes the inductive construction.
Let \(\mathcal{M}_0\) be the subalgebra of \(\mathcal{M}\) generated by \(\bigcup_n \{ \mathcal{P}_n \}\). If \(A \in \mathcal{B}_0\) then for some \(n\) \(A\) is a union of elements of \(\mathcal{P}_n\) and we let \(S(A)\) be the union of the corresponding intervals. Since the sequences of partitions are monotone, the definition of \(S(A)\) is independent of the choice of large enough \(n\). Thus, we obtain a measure algebra map \(S : \mathcal{M}_0 \to \mathcal{M}_I\). As \(S\) is an isometry and \(\mathcal{M}_0\) is dense in \(\mathcal{M}\), \(S\) extends to an isometry from \(\mathcal{M}\) to the complete metric space \(\mathcal{M}_I\). From continuity of the algebraic operations it follows that \(S\) is a measure algebra map on all \(\mathcal{M}\). Finally, since the mesh of the \(S(\mathcal{P}_n)\) tend to zero, every interval in \(I\) can be arbitrary closely approximated by a union of intervals in \(\bigcup_n \{ S(\mathcal{P}_n) \}\). Thus, \(S(\mathcal{M}_0)\) is dense in \(\mathcal{M}_I\) and so \(S : \mathcal{M}\to \mathcal{M}_I\) is surjective and so is an isomorphism. ◻
For measure algebras \(\mathcal{M}_Y\) and \(\mathcal{M}_X\) let \(\mathcal{S}(\mathcal{M}_Y,\mathcal{M}_X)\) be the set of measure algebra maps from \(\mathcal{M}_Y\) to \(\mathcal{M}_X\) and let \(\mathcal{S}_{iso}(\mathcal{M}_Y,\mathcal{M}_X)\) be the subset of isomorphisms. On \(\mathcal{S}\) we use the topology of pointwise convergence. That is, a net \(\{ S_i \}\) converges to \(S\) when \(\{ S_i(A) \}\) converges to \(S(A)\) for all \(A \in \mathcal{M}_Y\). Since the elements of \(\mathcal{S}\) are isometries, it suffices to check this on a dense subset of \(\mathcal{M}_Y\). Thus, in the separable case with \(\{ E_1, E_2, \dots \}\) a dense sequence in \(\mathcal{M}_Y\), the topology is given by the metric \[\label{eq34604} d(S_1, S_2) = \max_{n=1}^{\infty} \;d(S_1(E_n),S_2(E_n))/n.\tag{13}\] Notice that if \(\{ S_n \}\) is a Cauchy sequence in \(\mathcal{S}\), then for each \(A \in \mathcal{M}_Y\), \(\{ S_n(A) \}\) is a Cauchy sequence in the complete space \(\mathcal{M}_X\) and so it converges to an element which we label \(S(A)\). It follows from continuity of the measure and of the algebraic operations that \(S\) is a measure algebra map and so is the limit in \(\mathcal{S}\) of the sequence. Thus, if \(\mathcal{M}_X\) and \(\mathcal{M}_Y\) are separable, then \(\mathcal{S}(\mathcal{M}_X,\mathcal{M}_Y)\) is a completely metrizable space.
Furthermore, the countable product of separable metric spaces is a separable metrizable space with the product topology. From the metric 13 we see that the map \(S \mapsto \{ S(E_i) \}\) is an embedding of \(\mathcal{S}(\mathcal{M}_Y, \mathcal{M}_X)\) into the product space \(\mathcal{M}_X^{\{ E_i \}}\). Thus, \(\mathcal{S}(\mathcal{M}_Y, \mathcal{M}_X)\) is a completely metrizable, separable space.
****Proposition** 10**. (\(\mathbf{G_{\delta}}\)) If \(\mathcal{M}_Y\) and \(\mathcal{M}_X\) are separable, nonatomic measure algebras, then the set of isomorphisms \(\mathcal{S}_{iso}(\mathcal{M}_Y.M_X)\) is a \(G_{\delta}\) subset of \(\mathcal{S}(\mathcal{M}_Y.M_X)\).
Proof. Let \(\{E_i \}\) be a dense sequence in \(\mathcal{M}_X\). If \(S \in \mathcal{S}\), then because it is an isometry, the range \(S(\mathcal{M}_Y)\) is a closed subset of \(\mathcal{M}_X\). Since \(S\) is not an isomorphism exactly when it is not surjective, this occurs when some \(E_i\) is at a positive distance from the range of \(S\). Define for positive integers \(i, k\) \[\label{eq34605} \mathcal{S}_{ik} = \{ S : d(E_i, S(A)) \ge 1/k \;\text{for all} \;A \in \mathcal{M}_Y \}.\tag{14}\] This is a closed set and \(\mathcal{S}\setminus \mathcal{S}_{iso}\) is the union of the countable collection obtained by varying \(i\) and \(k\). Since \(\mathcal{S}\setminus \mathcal{S}_{iso}\) is an \(F_{\sigma}\) set, the complement \(\mathcal{S}_{iso}\) is a \(G_{\delta}\). ◻
****Proposition** 11**. (Density) Let \(S : \mathcal{M}_Y \to \mathcal{M}_X\) be a measure algebra map with \(\mathcal{M}_X\) and \(\mathcal{M}_Y\) normalized, separable, nonatomic measure spaces. If \(\{ E_1, E_2, \dots \}\) is a dense sequence in \(\mathcal{M}_Y\), then for any positive integer \(n\), there exists a measure algebra isomorphism \(T_n : \mathcal{M}_Y \to \mathcal{M}_X\) such that \(S(E_i) = T_n(E_i)\) for \(i = 1, \dots, n\). The sequence \(\{ T_n \}\) converges to \(S\) in \(\mathcal{S}(\mathcal{M}_Y,\mathcal{M}_X)\).
Proof. Let \(\mathcal{P}= \{ A_1, \dots, A_m \}\) be the common refinement partition \(\bigwedge_{i = 1}^n \;\{E_i, Y \setminus E_i \}\) and let \(S(\mathcal{P}) = \{ S(A_1), \dots, S(A_m) \}\) be the image partition on \(\mathcal{M}_X\). For each \(j\), \(\mathcal{M}_{A_j} = \{ A \in \mathcal{M}_Y : A \subset A_j \}\) and \(\mathcal{M}_{S(A_j)} = \{ B \in \mathcal{M}_X : B \subset S(A_j) \}\) are nonatomic measure algebras. Each is separable as any subspace of a separable metric space is separable. We can normalize by dividing the measures by \(\nu(A_j) = \mu(S(A_j))\) and then apply Proposition 9 to get an isomorphism \(\hat{T}_j : \mathcal{M}_{A_j} \to \mathcal{M}_{S(A_j)}\). We obtain the isomorphism \(T_n\) by defining for \(A \in \mathcal{M}_Y\) and \(B \in \mathcal{M}_X\): \[\label{eq34606}\begin{align} T_n(A) = \bigcup_j \;\hat{T}_j(A \cap A_j), \\ T_n^{-1}(B) = \bigcup_j \;(\hat{T}_j)^{-1}(B \cap S(A_j)). \end{align}\tag{15}\] Since \(T_n(A_j) = S(A_j)\) for all \(j\) and each \(E_i\) with \(i \le n\) is a union of the \(A_j\)’s it follows that \(T_n(E_i) = S(E_i)\).
For each \(i\), the sequence \(\{ T_n(E_i) \}\) is constant at \(S(E_i)\) once \(n \ge i\). Since \(\{ E_i \}\) is dense in \(\mathcal{M}_Y\), the sequence \(\{ T_n \}\) converges to \(S\). ◻
From Propositions 10 and 11 we obtain:
****Theorem** 12**. If \(\mathcal{M}_Y\) and \(\mathcal{M}_X\) are separable, nonatomic measure algebras, then the set \(\mathcal{S}_{iso}(\mathcal{M}_Y,\mathcal{M}_X)\) of measure algebra isomorphisms is a dense, \(G_{\delta}\) subset of the completely metrizable space \(\mathcal{S}(\mathcal{M}_Y,\mathcal{M}_X)\) of measure algebra mappings. Thus, the generic measure algebra map is invertible.
Our lifting results will concern measure space maps between nonatomic measure spaces of Borel measures on Polish spaces. A topological space is a Polish space when it is separable and admits a complete metric. A subset \(A\) of a Polish space \(X\) has a Polish relative topology if and only if \(A\) is a \(G_{\delta}\) subset of \(X\). When \((X,\mathcal{B}_X,\mu)\) is a measure space with \(\mathcal{B}_X\) the Borel sets and \(X\) a Polish space we will call \((X,\mathcal{B}_X,\mu)\) a Polish measure space.
Our results are taken from Royden [11] Chapter 15, Mappings of Measure Spaces.
For a Borel measure \(\mu\) on a separable metrizable space \(X\), the support of \(\mu\), written \(supp(\mu)\), is the complement of the union of all open sets of measure zero. Since we may take the union just over the sets in a countable base, we see that the union has measure zero and so the closed set \(supp(\mu)\) has full measure and every open set which meets the support has positive measure. The measure on \(X\) is called full when \(supp(\mu) = X\) and so when every nonempty open subset has positive measure.
Notice that if \(A\) is an atom for \(\mu\), then the support of the restriction \(\mu|A\) must be a singleton. Thus, \(\mu\) is nonatomic exactly when each singleton has measure zero, or, equivalently, all countable subsets have measure zero. So if \(\mu\) is a nonatomic measure on a separable, metrizable space \(X\), then \(X\) is uncountable. If \(\mu\) is full, then there are no isolated points and so \(X\) is perfect.
We first observe that the functor from measure space maps to measure algebra maps is injective.
****Proposition** 13**. Let \(T_1, T_2 : (X,\mathcal{B}_X,\mu) \to (Y, \mathcal{B}_Y, \nu)\) be measure space maps with \(\nu\) a Borel measure on the separable metric space \(Y\). If \(T_1^* = T_2^* : \mathcal{M}_Y \to \mathcal{M}_X\), then \(T_1 = T_2\) ae.
Proof. The product map \(T_1 \times T_2 : X \to Y \times Y\) is measurable because the Borel subsets of \(Y \times Y\) are generated by the sets \(B_1 \times B_2\) with \(B_1, B_2 \in \mathcal{B}_Y\). Hence, \(T_1 = T_2\) ae if and only if the set \(\{ x : T_1(x) \not= T_2(x) \}\) has measure zero. If this is not true, then for some positive integer \(k\), the set \(A = \{ x : d(T_1(x),T_2(x)) > 1/k \}\) has positive measure.
Let \(\mathcal{P}\) be a partition of \(Y\) by measurable sets each of metric diameter at most \(1/k\). For some \(B \in \mathcal{P}\), the set \(A \cap T_1^{-1}(B)\) has positive measure. If \(x \in A \cap T_1^{-1}(B)\) then \(T_1(x) \in B\) and \(d(T_2(x),T_1(x)) > 1/k\) implies \(T_2(x) \not\in B\). That is, \(T_2^{-1}(B)\) is disjoint from \(A \cap T_1^{-1}(B)\) and so \(T_1^*(B) \not= T_2^*(B)\). ◻
For the following deep result we refer to [11] Proposition 15.19 (Theorem 15.11 of the second edition).
****Theorem** 14**. Let \((X,\mathcal{B}_X,\mu)\) and \((Y,\mathcal{B}_Y,\nu)\) be separable measure spaces with \(X\) and \(Y\) uncountable and \((Y,\mathcal{B}_Y,\nu)\) Polish. If \(S : \mathcal{M}_Y \to \mathcal{M}_X\) is a measure algebra map, then there exists \(T : (X,\mathcal{B}_X,\mu) \to(Y,\mathcal{B}_Y,\nu)\) a measure space map with \(T^* = S\).
****Corollary** 15**. (a) Let \((X,\mathcal{B}_X,\mu)\) and \((Y,\mathcal{B}_Y,\nu)\) be Polish measure spaces with \(X\) and \(Y\) uncountable. If \(S : \mathcal{M}_Y \to \mathcal{M}_X\) is a measure algebra isomorphism, then there exists \(T : (X,\mathcal{B}_X,\mu) \to(Y,\mathcal{B}_Y,\nu)\) a measure space isomorphism with \(T^* = S\).
(b) Let \(T : (X,\mathcal{B}_X,\mu) \to(Y,\mathcal{B}_Y,\nu)\) be a measure space map between Polish measure spaces with \(X\) and \(Y\) uncountable. If \(T^*\) is a measure algebra isomorphism, then \(T\) is a measure space isomorphism.
Proof. (a) Applying Theorem 14 to \(S\) and \(S^{-1}\) we obtain measure space maps \(T : (X,\mathcal{B}_X,\mu) \to(Y,\mathcal{B}_Y,\nu)\) and \(\hat{T} : (Y,\mathcal{B}_Y,\nu) \to(X,\mathcal{B}_X,\mu)\) such that \(T^* = S\) and \(\hat{T}^* = S^{-1}\). Hence, \((\hat{T} \circ T)^* = T^* \circ \hat{T}^* = id_X^*\) and so by Proposition 13 \(\hat{T} \circ T = id_X\) ae. Similarly, \(T \circ \hat{T} = id_Y\) ae. Thus, \(T\) and \(\hat{T}\) are inverse isomorphisms.
(b) By (a) there is a measure space isomorphism \(\hat{T}\) with \(\hat{T}^* = T^*\). By Proposition 13 \(T = \hat{T}\) ae. ◻
By [11] Proposition 15.11 a Borel measure \(\mu\) on a metrizable space \(X\) is regular, ie. for any Borel set \(A\), \(\mu(A) = \inf \{ \mu(U): U \;\) is open and \(\;A \subset U \}\). It follows that \(A\) is contained in a \(G_{\delta}\) set \(A_1\) with \(\mu(A_1 \setminus A) = 0\). The inclusion of \(A\) into \(A_1\) thus induces a measure space isomorphism. If \(X\) is Polish, then \(A_1\) is Polish as well. It then follows from Proposition 9 and Theorem 14that if \((X,\mathcal{B}_X,\mu)\) are \((Y,\mathcal{B}_Y,\nu)\) are nonatomic Polish measure spaces and \(A \in \mathcal{B}_X, B \in \mathcal{B}_Y\) with \(\mu(A) = \nu(B) > 0\), then the measures spaces restricted to \(A\) and \(B\) are isomorphic. Notice that since the measures are nonatomic, sets of positive measure like \(A\) and \(B\) are uncountable.
For Polish measure spaces the map \(T \mapsto T^*\) from measure space maps to measure algebra maps is a bijection, we use the topology from the space of measure algebra maps and so a sequence \(\{ T_n \}\) of measure space maps converges to \(T\) when for every Borel subset \(E\) of the range we have \(\mu(T^{-1}_n(E) + T^{-1}(E)) \to 0\). Thus, we have:
****Theorem** 16**. If \((X,\mathcal{B}_X,\mu)\) and \((Y,\mathcal{B}_Y,\nu)\) are nonatomic Polish measure spaces (and so with \(X\) and \(Y\) uncountable) the generic measure space map between them is invertible.
Notes: The measure algebra approach is here adapted from Halmos [10] and [12]. The isomorphism result Proposition 9 is described there and is attributed by Royden to Caratheodory ([11] Theorem 15.4). Again we may assume that \(Y = X\). In [13] Eisner proves that the group of invertible measure space self-maps of a standard measure space is a dense \(G_{\delta}\) subset of the space of measure space sef-maps. In fact, the proof of Propostion 10 is essentially hers. Instead of the measure algebra isometry \(T^*\) on \(\mathcal{M}_X\), Eisner uses the Koopman operator on \(\mathcal{L}^2(X,\mu)\) given by \(T^*(u) = u \circ T,\) for \(u \in \mathcal{L}^2\) as a proxy for the measure space map. Note that \(d'(A,B) = \sqrt{\mu(A + B)}\) is a metric on \(\mathcal{M}_X\) uniformly equivalent to \(d\). The map associating to \(A \in \mathcal{M}_X\) the characteristic function \(1_A \in \mathcal{L}^2(X,\mu)\) is then an isometry from \(\mathcal{M}_X\) into \(\mathcal{L}^2(X,\mu)\). It maps \(T^*\) to the restriction of the Koopman operator.
The results leading to the lifting theorem Theorem 14, i.e. to Proposition 15.19 in [11], is the work of many hands. Royden attributes various results to Sikorski, Kuratowski, von Neumann and Halmos.
In [3] the authors show that the generic continuous map of the interval which preserves Lebesgue measure has measure-theoretic entropy zero. A measure space map with entropy zero is invertible a.e., see [14] Corollary 4.14.3. They have thus shown that the generic continuous measure-preserving map of the interval, while wildly non-invertible in the topological sense, is nonetheless invertible a.e. They are using the topology of uniform convergence. In addition, within the space of measure-preserving maps of the interval, with the topology we are using, the continuous maps are dense and so every measure-preserving map of the interval can be approximated by a continuous measure preserving map which is invertible a.e. This provides an alternate approach to the density portion of Theorem 16
We will call a pair \((X,\mu)\) a Cantor measure space when \(\mu\) is a finite, full, nonatomic Borel measure on a Cantor space \(X\). Thus, every nonempty open subset has positive measure and every countable subset has measure zero. The clopen subsets form a countable algebra which is a basis for the topology and which generates the Borel \(\sigma\)-algebra. Whenever we need one, we will fix a metric \(d\) on \(X\). For a clopen partition we will call the maximum diameter the \(d\)-mesh and the maximum measure the \(\mu\)-mesh.
****Lemma** 17**. For every \(\epsilon> 0\) there exists \(\delta> 0\) such that any Borel set with diameter less than \(\delta\) has measure less than \(\epsilon\). In particular, if a clopen partition has \(d\)-mesh less than \(\delta\), then it has \(\mu\)-mesh less than \(\epsilon\).
Proof. If \(\{ A_i \}\) were a sequence of Borel sets with diameter tending to zero but with measures at least \(\epsilon\), then for any limit point \(x\), every neighborhood would have measure at least \(\epsilon\) and so \(x\) would be an atom. ◻
A Cantor measure space map \(f : (X,\mu) \to (Y,\nu)\) is a continuous map from \(X\) to \(Y\) which preserves measure. To see that measure is preserved, it suffices to check: \[\label{eq44601} \mu(f^{-1}(A)) = \nu(A) \;\text{for all clopen subsets} \;A \;\text{of} \;Y.\tag{16}\] If \(f\) is a measure space map with \(f\) a homeomorphism, then \(f^{-1}\) satisfies (16 ) and so \(f\) is a measure space isomorphism. Notice that if continuous maps \(f_1, f_2 : X \to Y\) are equal ae, then they are equal everywhere because \(\{ x : f_1(x) \not= f_2(x) \}\) is an open set and so has positive measure if it is nonempty. Thus, conversely, if \(f : (X,\mu) \to (Y,\nu)\) is a Cantor measure space isomorphism, then \(f\) is a homeomorphism from \(X\) to \(Y\).
Since \(f^{-1}(Y) = X\), the existence of such a map requires \(\mu(X) = \nu(Y)\). Furthermore, \(f(X) = Y\) because \(\nu\) is full and so \(f(X)\) meets every nonempty open subset. Thus, \(f\) is necessarily surjective.
****Definition** 18**. For a Cantor measure space \((X,\mu)\) the clopen values set* \(S(\mu)\) is defined by \[\label{eq44602} S(\mu) = \{ \mu(A) : \;\text{for} \;A \;\text{clopen in} \;X \}.\tag{17}\] *
****Proposition** 19**. (a) If \((X,\mu)\) is a Cantor measure space, then \(S(\mu)\) is a countable, dense subset of \([0,\mu(X)]\) which contains the endpoints.
(b) If \(f : (X,\mu) \to (Y,\nu)\) is a Cantor measure space map, then \(S(\nu) \subset S(\mu)\) with equality if \(f\) is an isomorphism.
Proof. (a) Given \(\epsilon> 0\) and \(\delta> 0\) from Lemma 17 we choose \(\{A_1, \dots, A_n \}\) is a clopen partition with \(d\)-mesh less than \(\delta\) then \[\{ a_0 = 0, a_1 = \mu(A_1), a_2 = \mu(A_1 \cup A_2), \dots, a_n = \mu(A_1 \cup \dots \cup A_n) = \mu(X) \}\] is an increasing sequence in \([0,\mu(X)]\) with \(a_n - a_{n-1} < \epsilon\) for \(n = 1, \dots, n\). As \(\epsilon\) was arbitrary, it follows that \(S(\mu)\) is dense. It is clearly countable.
(b) This is obvious from (16 ). ◻
Our results will be applied to the Cantor measure spaces which satisfy a homogeneity condition, the Subset Condition.
****Definition** 20**. For a Cantor measure space \((X,\mu)\), the measure \(\mu\) is called good* and \((X,\mu)\) is called a good Cantor measure space when for every pair of clopen subsets \(A, B\) of \(X\) \[\label{eq44603}\begin{align} \mu(A) \le \mu(B) \quad \Longrightarrow\\ \text{there exists a clopen} \;A_1 \subset B \;\text{with} \;\mu(A_1) = \mu(A). \end{align}\tag{18}\] *
It is clear that if \((X,\mu)\) is a good Cantor measure space and \(A\) is a nonempty clopen subset of \(X\), then \((A,\mu|A)\) is a good Cantor measure space with \[\label{eq44604a} S(\mu|A) = S(\mu) \cap [0,\mu(A)].\tag{19}\]
A subset \(S\) of an interval \([0,m]\) is called group-like when \(m \in S\) and \(S\) is the intersection of the interval with an additive subgroup of \(\mathbb{R}\).
****Lemma** 21**. If \(S\) is a subset of the unit interval \([0,1]\) with \(0, 1 \in S\), then \((S + \mathbb{Z}) \cap [0,1] = S\). Furthermore, \(S + \mathbb{Z}\) is a subgroup of \(\mathbb{R}\), and so \(S\) is group-like, if and only if for all \(s, t \in S\) \[\label{eq44604} s \le t \qquad \Longrightarrow \qquad t - s \in S.\qquad{(2)}\]
Proof. If \(0 < t < 1\) and \(n \in \mathbb{Z}\), then \(n \ge 1\) implies \(n + t > 1\) and \(n \le -1\) implies \(n +t < 0\). Hence, \(n+t \in [0,1]\) implies \(n = 0\). Thus, \((S + \mathbb{Z}) \cap [0,1] = S\).
If \(S\) is group-like then (?? ) obviously holds. Assume, conversely, that (?? ) is true. Since \(1 \in S\), \(1 - t \in S\) when \(t \in S\). If \(s + t \le 1\), then \(s \le 1 - t\) and so \(s + t = 1 - [(1 - t) - s] \in S\). If \(s + t > 1\), then \((1 - s) + (1 - t) = 2 - (s + t) \le 1\). Hence, \(s + t - 1 = 1 - [(1 - s) + (1 - t)] \in S\). It follows that \(S + \mathbb{Z}\) is closed under addition. Furthermore, \(-(n + t) = -(n+1) + (1 - t)\) and so \(S + \mathbb{Z}\) is closed under negation. ◻
We now describe the crucial properties of good measures.
****Theorem** 22**. (a) If \((X,\mu)\) is a good Cantor measure space, then \(S(\mu)\) is group-like.
(b) If \(S\) is a dense, group-like subset of \([0,1]\) with \(0, 1 \in S\), then there exists a normalized good Cantor measure space with \(S(\mu) = S\).
(c) Assume that \((X,\mu)\) and \((Y,\nu)\) are good Cantor measure spaces with \(\mu(X) = \nu(Y)\).
There exists a Cantor measure space isomorphism \(f : (X,\mu) \to (Y,\nu)\) if and only if \(S(\nu) = S(\mu)\).
There exists a Cantor measure mapping \(f : (X,\mu) \to (Y,\nu)\) if and only if \(S(\nu) \subset S(\mu)\).
Proof. (a) A subset \(S\) is group-like in \([0.m]\) if and only if \((1/m)S\) is group-like in \([0,1]\). So we may assume \((X,\mu)\) is normalized. Let \(s = \mu(A)\) and \(t = \mu(B)\) with clopens \(A, B\). If \(s \le t\) then there exists \(A_1 \subset B\) clopen with \(\mu(A_1) = \mu(A)\). Then \(\mu(B \setminus A_1) = t - s\) and so (?? ) holds for \(S(\mu)\) and \(S(\mu)\) is group-like by Lemma 21.
(b) This result is Theorem 2.6 of [15] taken together with Corollary 3.3 of [16].
(c) The conditions on the clopen values sets are necessary by Proposition 19. By normalizing, i.e. dividing the measures by \(\mu(X) = \nu(Y)\) we may assume they are normalized. The results are then in Theorem 2.9 of [15]. ◻
Thus, when we restrict ourselves to good Cantor measure spaces, the clopen values set is a complete invariant. Furthermore, if \((X,\mu)\) and \((Y,\nu)\) are good with \(S(\mu) = S(\nu)\) and if \(A \subset X\) and \(A_1 \subset Y\) are clopens with \(\mu(A) = \nu(A_1)\), then there exists an isomorphism from \((A,\mu|A)\) to \((A_1,\nu|A_1)\). The maximum values in \(S(\mu) = S(\nu)\) are \(\mu(X) = \nu(Y)\) and so we can combine the isomorphism from \(A\) to \(A_1\) with an isomorphism from \(X \setminus A\) to \(Y \setminus A_1\) to obtain an isomorphism from \((X,\mu)\) to \((Y,\nu)\) which moves \(A\) to \(A_1\). In particular, with \(X = Y\) we see that the homogeneity condition defining a good measure is not as mild as it first appears.
Theorem 22 provides the uniqueness results that we need in this context.
****Proposition** 23**. (\(\mathbf{G_{\delta}}\)) Let \((X,\mu)\) and \((Y,\nu)\) be Cantor measure spaces. The set of Cantor measure space maps is a closed subset of \(\mathcal{C}_{sur}(X,Y)\), the set of continuous surjections. The set of Cantor measure space isomorphisms is a \(G_{\delta}\) subset.
Proof. We saw above that any Cantor measure space map is surjective. A continuous function \(f : X \to Y\) preserves measure exactly when for every continuous function \(u : Y \to \mathbb{R}, \;\int (u \circ f) \; d\mu = \int u \;d\nu\). This is clearly a closed condition and so \(\mathcal{S}\), the set of Cantor measure space maps is a closed subset of \(\mathcal{C}_{sur}(X,Y)\).
By Proposition 3 \(\mathcal{C}_{iso}(X,Y)\) is a \(G_{\delta}\) subset and so the intersection \(\mathcal{S}\cap \mathcal{C}_{iso}(X,Y)\), is a \(G_{\delta}\) subset. The intersection clearly contains the set of Cantor measure space isomorphisms. On the other hand, if \(f : (X,\mu) \to (Y,\nu)\) is a Cantor measure space map with \(f : X \to Y\) a homeomorphism then, as remarked above, \(f\) is an isomorphism. That is, the intersection equals the set of isomorphisms. ◻
Now suppose that \((X,\mu)\) is a normalized, good Cantor measure space such that \(t \in S(\mu)\) implies \(t/2 \in S(\mu)\). For example, this is true if \(S(\mu) = \mathbb{Q}\cap [0,1]\). In that case, \(2 \times ([0,\frac{1}{2}] \cap S(\mu)) = S(\mu)\). If \(A_1\) is a clopen set with \(\mu(A_1) = \frac{1}{2}\) and \(A_2 = X \setminus A_1\), then there exist isomorphisms \(f_1 : (A_1,2\mu|A_1) \to (X,\mu)\) and \(f_2 : (A_2,2\mu|A_2) \to (X,\mu)\). Combining these we obtain a measure space map \(f : (X,\mu) \to (X,\mu)\) which is the analogue in this context of the tent map.
****Proposition** 24**. (Density) Let \(f : (X,\mu) \to (Y,\nu)\) be a Cantor measure space map with \(S(X,\mu) = S(Y,\nu)\). Let \(d\) be a metric on \(Y\). Given \(\epsilon> 0\), there exists a Cantor measure space isomorphism \(g :(X,\mu) \to (Y,\nu)\) such that \[\label{eq44608} x \in X \qquad \Longrightarrow \qquad d(f(x),g(x)) \le \epsilon.\qquad{(3)}\] .
Proof. The proof is almost the same as that of Proposition 4. Let \(\mathcal{P}= \{U_1, \dots, U_n \}\) be a clopen partition of \(Y\) with \(d\)-mesh at most \(\epsilon\). As before \(\{ f^{-1}(U_1), \dots, f^{-1}(U_n) \}\) is a clopen partition of \(X\). For each \(i\), \((U_i,\nu|U_i)\) and \((f^{-1}(U_i),\mu|f^{-1}(U_i)),\) are good Cantor measure spaces with the same clopen values set and so Theorem 22 implies we can choose an isomorphism \(g_i\) from \((f^{-1}(U_i),\mu|f^{-1}(U_i))\) onto \((U_i,\nu|U_i)\). Concatenating we obtain the isomorphism \(g : (X,\mu) \to (Y,\nu)\). Again for all \(x \in X\) \[\label{eq44609} f(x) \in U_i \qquad \Longleftrightarrow \qquad g(x) \in U_i.\tag{20}\] and so (?? ) follows. ◻
Thus, we obtain
****Theorem** 25**. If \((X,\mu)\) and \((Y,\nu)\) are good Cantor measure spaces with the same clopen values set, then the set of Cantor measure isomorphisms from \((X,\mu)\) to \((Y,\nu)\) is a dense \(G_{\delta}\) subset of the completely metrizable space of Cantor measure space maps from \((X,\mu)\) to \((Y,\nu)\). Thus, the generic Cantor measure space map between them is invertible.
Notes: The theory of normalized good Cantor measure spaces is laid out in [16] and [15]. In Theorem 2.18 of [15] an example is given of two Bernoulli measures with the same group-like clopen values set one of which is good and the other is not and so they are not isomorphic. In [17] those Bernoulli measures which are good are characterized.
2010 Mathematical Subject Classification 54C05, 28C15, 28D05↩︎