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Measurable dense selections for fields of nonempty compact sets

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $(X, A, μ)$ be a standard Borel space with a sigma-finite measure (Standard Borel spaces, Measure spaces, Finite, sigma-finite, and semifinite measures). Let $(K, d)$ be a nonempty compact metric space (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) with a fixed dense sequence $(k_{m})_{m \in N}$ , and give $K$ its Borel sigma-algebra (The Borel sigma-algebra of a topological space). Suppose $g : X \times K \to [0, \infty)$ has measurable sections $x \mapsto g (x, k)$ for each fixed $k$ , continuous sections $k \mapsto g (x, k)$ for each fixed $x$ , and nonempty zero sets $C_{x} : = {k \in K : g (x, k) = 0}$ for every $x$ . Then: (1) there is a sequence of measurable maps $s_{j} : X \to K$ such that $s_{j} (x) \in C_{x}$ for all $x, j$ and $(s_{j} (x))_{j \in N}$ is dense in $C_{x}$ for every $x$ ; (2) for every $k \in K$ , the function $x \mapsto d (k, C_{x}) : = \inf_{c \in C_{x}} d (k, c)$ is measurable; and (3) for every open $U \subseteq K$ , the hit set ${x : C_{x} \cap U \neq \emptyset}$ is measurable.

Facts & Assumptions

Given: The measurable space $(X, A)$ is standard Borel, $μ$ is sigma-finite, $(K, d)$ is compact with its metric topology, the dense sequence $(k_{m})$ is fixed, and $g$ has the stated measurable and continuous sections with nonempty zero sets.

[F1]

A measurable space has a sigma-algebra of measurable sets; it is closed under countable unions and intersections. A standard Borel space is in particular a measurable space. The Borel sigma-algebra of $K$ is generated by its open sets and is minimal among sigma-algebras containing them (Standard Borel spaces, Measure spaces, Measures on sigma-algebras, Finite, sigma-finite, and semifinite measures, Measurable spaces and measurable sets, A measurable function between measurable spaces, The Borel sigma-algebra of a topological space, Sigma-algebras, Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal).

[F2]

Closed balls in a metric space are closed. In a compact space every family of closed sets with the finite intersection property has nonempty intersection (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection).

[F3]

Metric distances are nonnegative; every nonempty subset of $N$ has a least element; and for each $ρ > 0$ there is $n \geq 1$ with $2 / n < ρ$ , by applying the reciprocal bound to $ρ / 2$ (Order on the reals, Complete ordered field (least-upper-bound property), Maximum and minimum of a set, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The well-ordering principle).

[F4]

Measurable maps compose with Borel maps; pointwise sums, products, absolute values, maxima and minima of real measurable functions are measurable; and real-valued measurability follows from measurability of all strict sublevel sets (A measurable function between measurable spaces, Composition with a Borel measurable outer map preserves measurability, Arithmetic and lattice operations preserve measurability whenever they are defined, Threshold characterisations of real-valued and extended-real-valued measurability).

[F5]

AC is the explicit hypothesis in the Statement (The Axiom of Choice). The proof uses no choice: $(k_{m})$ is given, recursive indices are least natural numbers, and every limit selector is unique. The sigma-finite measure is not used.

Proof

technique · direct

1.1F1F2F3

First prove a closed-target hit claim for any field $h : X \times K \to [0, \infty)$ with measurable $x$ -sections, continuous $K$ -sections, and nonempty zero fibers $Z_{x} : = {k : h (x, k) = 0}$ . For a closed $F \subseteq K$ and $n \geq 1$ , let $M_{F, n} : = {m \geq 0 : \exists y \in F, d (k_{m}, y) < 1 / n}$ . Then ${x : Z_{x} \cap F \neq \emptyset} = ⋂_{n \geq 1} ⋃_{m \geq 0 : m \in M_{F, n}} {x : h (x, k_{m}) < 1 / n},$ where each union is read as an $N$ -indexed union with empty terms for $m \notin M_{F, n}$ . If $F = \emptyset$ , both sides are empty. If $z \in Z_{x} \cap F$ , continuity at $z$ and density of $(k_{m})$ provide, for each $n$ , an $m$ with $d (k_{m}, z) < 1 / n$ and $h (x, k_{m}) < 1 / n$ , so the right side holds. Conversely, suppose the right side holds and put $A_{r} : = {k_{m} : m \in M_{F, r}, h (x, k_{m}) < 1 / r}$ ; each $A_{r}$ is nonempty. The closed sets $L_{n} : = \overline{⋃_{r \geq n} A_{r}}$ are nonempty and nested. They have the finite intersection property, so [F2] gives $z \in ⋂_{n} L_{n}$ . For any $ρ > 0$ , choose $n$ with $2 / n < ρ$ by [F3]; since $z \in L_{n}$ , some $k_{m} \in A_{r}$ for an $r \geq n$ satisfies $d (k_{m}, z) < 1 / n$ , and $m \in M_{F, r}$ gives a $y \in F$ with $d (k_{m}, y) < 1 / r \leq 1 / n$ . Thus $d (z, y) < 2 / n < ρ$ , so $z \in F$ . Given $ϵ > 0$ , continuity at $z$ gives $δ > 0$ such that $d (w, z) < δ$ implies $∣ h (x, w) - h (x, z) ∣ < ϵ / 2$ . Choose $n$ with $1 / n < \min (δ, ϵ / 2)$ by [F3]. Some $k_{m} \in A_{r}$ for $r \geq n$ then satisfies $d (k_{m}, z) < 1 / n$ and $h (x, k_{m}) < 1 / r \leq 1 / n < ϵ / 2$ , so $0 \leq h (x, z) < ϵ$ . Since this holds for every $ϵ > 0$ , $h (x, z) = 0$ and $z \in Z_{x} \cap F$ . The displayed set is measurable by [F1] and the measurable-section hypothesis.

2.1F1F3F4step 1.1

Step 1.1 shows that every closed-target hit set is measurable. If $U \subseteq K$ is open, it is the union of the countable family of closed balls $\overset{ˉ}{B} (k_{m}, 1 / n)$ that are contained in $U$ : for $y \in U$ , choose $ρ > 0$ with $B (y, ρ) \subseteq U$ , choose $n$ with $2 / n < ρ$ , then choose $m$ with $d (k_{m}, y) < 1 / (2 n)$ . This gives $y \in B (k_{m}, 1 / n)$ and $\overset{ˉ}{B} (k_{m}, 1 / n) \subseteq B (y, ρ) \subseteq U$ . Thus ${x : Z_{x} \cap U \neq \emptyset} = ⋃_{m \geq 0} ⋃_{n \geq 1 : \overset{ˉ}{B} (k_{m}, 1 / n) \subseteq U} {x : Z_{x} \cap \overset{ˉ}{B} (k_{m}, 1 / n) \neq \emptyset},$ an iterated countable union, so open-target hit sets are measurable by [F1] and step 1.1. For fixed $k \in K$ and $q > 0$ , $d (k, Z_{x}) < q$ exactly when $Z_{x}$ meets $B (k, q)$ , by the definition of infimum; for $q \leq 0$ the strict sublevel set is empty by nonnegativity. The infimum exists in $R$ because these distances form a nonempty set bounded below by $0$ and $R$ is complete; it is finite because any one point of $Z_{x}$ gives a finite upper bound. Hence every strict sublevel set of $x \mapsto d (k, Z_{x})$ is measurable, and [F4] proves that this distance function is measurable.

3.1F1F2F3F4step 2.1

Apply steps 1.1–2.1 to a field $h$ as above, and put $G^{1} = h$ , $Z_{x}^{1} = Z_{x}$ . Since each $h (x, \cdot)$ is continuous, each initial zero fiber is closed. For $n \geq 2$ , define $m_{n} (x)$ to be the least $m \geq 0$ such that $d (k_{m}, Z_{x}^{n - 1}) < 1 / n$ , and set $G^{n} (x, k) : = G^{n - 1} (x, k) + \max {d (k, k_{m_{n} (x)}) - 1 / n, 0}, Z_{x}^{n} : = {k : G^{n} (x, k) = 0} = Z_{x}^{n - 1} \cap \overset{ˉ}{B} (k_{m_{n} (x)}, 1 / n) .$ This zero-set identity uses that both summands are nonnegative. Such an $m$ exists by density of $(k_{m})$ and nonemptiness of $Z_{x}^{n - 1}$ . For $M \geq 0$ , ${x : m_{n} (x) \leq M} = ⋃_{m = 0}^{M} {x : Z_{x}^{n - 1} \cap B (k_{m}, 1 / n) \neq \emptyset},$ which is measurable by step 2.1 applied to $G^{n - 1}$ . Differences give measurable singleton fibers of $m_{n}$ ; for any $S \subseteq N$ , $m_{n}^{- 1} (S)$ is the countable union of those fibers, taking the empty set for indices outside $S$ . Equip $N$ with its power-set sigma-algebra; then $m_{n}$ is measurable, and so is $x \mapsto k_{m_{n} (x)}$ by composition, since every map from this discrete measurable space into $K$ is measurable. For fixed $k$ , the scalar function $m \mapsto \max {d (k, k_{m}) - 1 / n, 0}$ is measurable on the discrete space, so [F4] makes the added term and then $G^{n} (\cdot, k)$ measurable. For fixed $x$ , $k \mapsto d (k, k_{m_{n} (x)})$ is continuous by the triangle inequality, and $t \mapsto \max {t - 1 / n, 0}$ is continuous because its two affine formulas agree at $1 / n$ ; thus $G^{n} (x, \cdot)$ is continuous. Its zero fiber is nonempty because $d (k_{m_{n} (x)}, Z_{x}^{n - 1}) < 1 / n$ means that $Z_{x}^{n - 1}$ meets the open ball, and it is closed as the intersection of a closed zero fiber with a closed ball. Its diameter is at most $2 / n$ .

4.1F1F2F3step 2.1step 3.1

For each $x$ , the nested nonempty closed sets $Z_{x}^{n}$ have the finite intersection property, so [F2] gives a point $σ (x) \in ⋂_{n \geq 1} Z_{x}^{n}$ . The diameter bound makes it unique: for every $n \geq 2$ , any two points in the intersection are at distance at most $2 / n$ , and [F3] makes these bounds arbitrarily small. Since $σ (x) \in Z_{x}^{n} \subseteq \overset{ˉ}{B} (k_{m_{n} (x)}, 1 / n)$ , the zero-indexed sequence $a_{q} (x) : = k_{m_{q + 2} (x)}$ converges to $σ (x)$ ; also $σ (x) \in Z_{x}^{1} = Z_{x}$ . For open $U \subseteq K$ , $σ^{- 1} (U) = ⋃_{m \geq 0} ⋃_{n \geq 1 : \overset{ˉ}{B} (k_{m}, 1 / n) \subseteq U} ⋃_{N \geq 2} ⋂_{q \geq N} {x : k_{m_{q} (x)} \in \overset{ˉ}{B} (k_{m}, 1 / n)} .$ These are successive countable unions and intersections over natural indices. Eventual membership in one of these closed balls forces the limit into $U$ . Conversely, if $σ (x) \in U$ , step 2.1 supplies a closed ball contained in $U$ with $σ (x)$ in its open ball, and convergence makes the sequence eventually lie in that closed ball. Each set on the right is measurable because $x \mapsto k_{m_{q} (x)}$ is measurable and the ball is Borel; all unions and intersections are countable. Since open sets generate the Borel sigma-algebra, [F1] proves that $σ : X \to K$ is measurable. This constructs an everywhere selection for every admissible field $h$ .

5.1F1F2F3F4step 1.1step 4.1

For $m \in N$ and $j \geq 1$ , let $X_{m, j} : = {x : C_{x} \cap \overset{ˉ}{B} (k_{m}, 1 / j) \neq \emptyset}$ , which is measurable by step 1.1, and define $h_{m, j} (x, k) : = g (x, k) + 1_{X_{m, j}} (x) \max {d (k, k_{m}) - 1 / j, 0} .$ Its fixed- $k$ sections are measurable by [F4]; its fixed- $x$ sections are continuous. Its zero fiber is $C_{x} \cap \overset{ˉ}{B} (k_{m}, 1 / j)$ on $X_{m, j}$ and $C_{x}$ off that set, so it is nonempty for every $x$ . Applying step 4.1 to this field gives a measurable $t_{m, j} : X \to K$ with $t_{m, j} (x) \in C_{x}$ for every $x$ , and $t_{m, j} (x) \in \overset{ˉ}{B} (k_{m}, 1 / j)$ when $x \in X_{m, j}$ . Given $y \in C_{x}$ and $ϵ > 0$ , choose $j$ with $2 / j < ϵ$ and $m$ with $d (k_{m}, y) < 1 / (2 j)$ . Then $x \in X_{m, j}$ and $d (t_{m, j} (x), y) \leq d (t_{m, j} (x), k_{m}) + d (k_{m}, y) \leq 1 / j + d (k_{m}, y) < 3 / (2 j) < 2 / j < ϵ .$ Therefore the family $(t_{m, j})_{m \in N, j \geq 1}$ is dense in each fiber. Enumerating pairs $(m, j) \in N \times N_{> 0}$ by increasing sum, and within each finite diagonal by increasing first coordinate, gives a sequence $(s_{q})_{q \in N}$ of measurable selections dense in every $C_{x}$ .

6.1F5step 2.1step 4.1step 5.1∎

Steps 4.1–5.1 prove assertion (1), and step 2.1 proves (2) and (3). The proof spends no form of Choice: the given dense sequence is fixed input, each recursive index is the least admissible natural number, and every selected limit point is unique. AC remains an explicit but unused hypothesis, and the sigma-finite measure is likewise unused.

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Sources

Bachir Bekka and Pierre de la Harpe, Unitary Representations of Groups, Duals, and Characters (arXiv:1912.07262v1, 16 December 2019; author-hosted complete book draft) (standard reference, not scraped)