METRIC SPACE VALUED RANDOM VARIABLES
CLINTON T. CONLEY
Given a bounded metric space (V, d) and a probability space (X, µ), we denote by
MALG(X, µ) the σ-algebra of µ-measurable subsets of X (taken modulo null sets),
and we denote by L(X, µ, V ) the space of V -valued random variables, i.e., the set
of µ-measurable functions from X to V . After making the standard identification
between functions agreeing µ-almost everywhere, L(X, µ, V ) may be equipped with
R
e g) = d(f (x), g(x)) dµ(x). For v ∈ V , let cv ∈ L(X, µ, V ) denote
the metric d(f,
the constant v function (so cv (x) = v for all x ∈ X), and let CV = {cv : v ∈ V }.
Henceforth, we fix a standard probability space (X, µ).
Let add(null) denote the smallest cardinal
κ for which there exists a sequence
S
(Aα )α<κ of µ-null subsets of X such that α<κ Aα is not µ-null. If A ⊆ MALG(X, µ)
is a sub-σ-algebra, we denote by µA the restriction of µ to A. Observe that the
measure algebra of (X × X, µ × µA ) is separable and non-atomic, and is therefore
isomorphic to MALG(X, µ) (see, e.g., Exercise 17.44 in [1]). If π is an isomorphism
witnessing this, and if (W, d) is a bounded metric space with |W | < add(null), we
have an isometry f 7→ f˜ between L(X × X, µ × µA , W ) and L(X, µ, W ) defined by
f˜−1 (w) = π(f −1 (w)).
We immediately see the following.
Theorem 1. Suppose that (V, dV ) and (W, dW ) are bounded metric spaces with
|W | < add(null), and suppose there exists some sub-σ-algebra A ⊆ MALG(X, µ)
such that V ∼
= L(X, µA , W ). Then L(X, µ, V ) ∼
= L(X, µ, W ).
Proof. Fix a sub-σ-algebra A ⊆ MALG(X, µ) such that V is isometric to L(X, µA , W ).
We see by Fubini that
L(X, µ, V ) ∼
=
∼
=
∼
=
L(X, µ, L(X, µA , W ))
L(X × X, µ × µA , W )
L(X, µ, W ),
completing the proof.
2
The main focus of this paper is to examine when the converse of Theorem 1
holds. Given any metric space (W, d), we may associate with each w ∈ W a partial
order ≤w of W defined by
w1 ≤w w2 ⇔ d(w1 , w) + d(w1 , w2 ) − d(w2 , w) = 0.
1
2
Let us confirm that ≤w is transitive. Suppose that w1 , w2 , w3 ∈ W are such that
w1 ≤w w2 and w2 ≤w w3 . Twice using the triangle inequality, we see
d(w1 , w) + d(w1 , w3 ) ≥
=
=
≥
d(w3 , w)
d(w2 , w) + d(w2 , w3 )
d(w1 , w) + d(w1 , w2 ) + d(w2 , w3 )
d(w1 , w) + d(w1 , w3 ),
so d(w1 , w) + d(w1 , w3 ) = d(w3 , w) and hence w1 ≤w w3 .
We may view w1 ≤w w2 as expressing that w1 “lies on a geodesic” between w
and w2 . For convenience of notation, we fix some element 0 ∈ W and suppress
the subscript in ≤0 . If (W, d) is a bounded metric space, we use the symbol 4w
to denote the partial order ≤cw of L(X, µ, W ). Again, when there is no danger of
confusion we write 4 instead of 40 . It is immediate that w1 ≤w w2 if and only if
cw1 4w cw2 . The following proposition generalizes this fact.
Proposition 2. Suppose that (W, d) is a bounded metric space, w ∈ W , and f, g ∈
L(X, µ, W ). Then f 4w g if and only if f (x) ≤w g(x) on a set of full measure.
Proof. Note thatR q : x 7→ d(f (x), w) + d(f (x), g(x)) − d(g(x), w) is a non-negative
function. Thus, q(x) dµ(x) = 0 if and only if ∀µ x (q(x) = 0). Then
f 4w g
e cw ) + d(f,
e g) − d(g,
e cw ) = 0
d(f,
Z
⇔
q(x) dµ(x) = 0
⇔
⇔ ∀µ x (q(x) = 0)
⇔ ∀µ x (f (x) ≤w g(x)).
2
Given a poset (P, ≤P ), we say that a set A ⊆ P is closedVunder meets in (P, ≤P )
if for all sequences (ai )i∈I in A with |I| < add(null), if i∈I ai exists in P then
it is in A. Analogously, we say that A is closed
W under joins in (P, ≤P ) if for all
sequences (ai )i∈I in A with |I| < add(null), if i∈I ai exists in P then it is in A.
Proposition 3. Suppose that (W, d) is a bounded metric space. Then for all w ∈
W , the set of constant functions is closed under meets and joins in (L(X, µ, W ), 4w ).
Proof. Without loss of generality, we may assume w = 0 and work with the partial
order 4. We show that the set CW = {cw : w ∈ W } is closed under meets in
(L(X, µ, W ), 4). The argument that C is closed under joins is similar.
Fix
V a sequence (wi )i∈I , |I| < add(null), of elements of W , and suppose that
f = i∈I cwi exists in L(X, µ, W ). By Proposition 2, we know that for all i ∈ I,
∀µ x (f (x) ≤ wi ). Since |I| < add(null), this implies that ∀µ x ∀i (f (x) ≤ wi ).
Equivalently, ∀µ x ∀i (cf (x) 4 cwi ). Since f is the greatest lower bound of (cwi )i∈I ,
we have that ∀µ x (cf (x) 4 f ). Upon appealing once more to Proposition 2, we see
that ∀µ x ∀µ y (f (x) ≤ f (y)), implying that f is constant on a set of full measure.2
Given a metric space (W, d) and w0 , w1 ∈ W , we let w0 w1 denote the set {w ∈
W : w ≤w0 w1 }, that is, the set of points between w0 and w1 . We say that w2 ∈ W
discerns the endpoints of w0 w1 if w2 6∈ w0 w1 and, for i ∈ {0, 1},
∀w ∈ w0 w1 \ {wi } (w ∧w2 wi = w2 ).
3
Intuitively, w2 discerns the endpoints of w0 w1 if for all w strictly between w0 and w1 ,
the only point simultaneously between w2 and w and between w2 and an endpoint
of w0 w1 is w2 . Note that if w2 discerns the endpoints of w0 w1 then {w0 , w1 , w2 } is a
triangle in the following sense: for any σ ∈ S3 , wσ(0) ∧wσ(1) wσ(2) = wσ(1) , provided
that the meets exist. We say that W is self discerning if for all w ∈ W , binary
meets always exist in (W, ≤w ) and for all w0 , w1 ∈ W there exists some w2 ∈ W
which discerns the endpoints of w0 w1 . For example, all nonlinear subspaces of a
Euclidean metric space are self discerning, as is any set of cardinality at least 3
equipped with the discrete metric.
The following technical lemma is the heart of the main theorem.
Lemma 4. Suppose that (W, d) is a bounded, self-discerning metric space, |W | <
add(null), and F ⊆ L(X, µ, W ). Suppose further that for all w ∈ W , cw ∈ F and
that F is closed under meets and joins in (L(X, µ, W ), 4w ). Then F = L(X, µA , W )
for some sub-σ-algebra A ⊆ MALG(X, µ).
Proof. If |W | = 1 the lemma is trivial, so we assume that |W | > 1.
For each w 6= 0 in W , we fix w⊥ discerning the endpoints of 0w. Let Aw =
−1
{f (w) : f ∈ F and f 4 cw } and let Bw = {f −1 (w) : f ∈ F and µ(f −1 ({0, w})) =
1}. Obviously, Bw ⊆ Aw ; we now check Aw ⊆ Bw . Fix A ∈ Aw and f ∈ F with
f 4 cw and A = f −1 (w). Let g = (f ∧w⊥ cw ) ∧ cw . Then f −1 (w) ⊆ g −1 (w) and
f −1 (W \ {w}) ⊆ g −1 (0), so f −1 (w) = g −1 (w) and f −1 (W \ {w}) = g −1 (0) and thus
g witnesses that A ∈ Bw .
For all w0 , w1 6= 0, we in fact have that Aw0 = Aw1 . Obviously, this is true when
w0 = w1 , so suppose they are distinct. Fix A ∈ Aw0 and f ∈ F with A = f −1 (w0 ).
In light of the above paragraph, we may assume f takes only values in {0, w0 }. We
consider two cases contingent upon the value of w0 ∧w1 0. If w0 ∧w1 0 = 0 (i.e., if 0 ∈
w0 w1 ), then fix w2 discerning the endpoints of w0 w1 . Then let g = (f ∧w2 c0 )∧w1 c0 ,
noting that f −1 (w0 ) = g −1 (w1 ) and f −1 (0) = g −1 (0), showing that A ∈ Aw1 . On
the other hand, if w0 ∧w1 0 6= 0, we let g = ((f ∧w1 c0 ) ∧w1⊥ c0 ) ∧w1 c0 , again noting
that f −1 (w0 ) = g −1 (w1 ) and f −1 (0) = g −1 (0) as required.
Since Aw is independent of the choice of w 6= 0, we henceforth denote it by A.
We now check that A is a σ-algebra. Suppose (An )n∈N is a sequence of elements
of A. Fix w 6= 0 and a sequence (fn )n∈N of elements of F such that
V each fn takes
values only in {0, w} and for each n, An = fn−1 (w). Letting
g
=
n∈N fn , we see
T
T
that g takes values only in {0, w} and that g −1 (w) = n∈N An , so n∈N An ∈ A.
To check closure under complementation, we let h = (f0 ∧w⊥ cw ) ∧ cw⊥ . Then
f0−1 (0) = h−1 (w⊥ ) and f0−1 (w) = h−1 (0), so X \ A0 ∈ Aw⊥ = A.
Finally, we check that F = L(X, µA , W ). First, suppose that f ∈ F . For each
w 6= 0, we see that
[
f −1 (w) = (f ∧ cw )−1 (w) \
(f ∧ cw0 )−1 (w0 ) ∈ A,
w<w0
where w < w0 means w ≤ w0 and w 6= w0 . Since |W | < add(null), this ensures that
f ∈ L(X, µA , W ). For the reverse inclusion, suppose that f ∈ L(X, µA , W ). For
−1
−1
each w 6= 0,
W fix fw ∈ F taking values only in {0, w} such that fw (w) = f (w).
Then f = w∈W fw ∈ F .
2
4
We say that a bounded metric space W is L-minimal if whenever L(X, µ, W ) ∼
=
L(X, µ, V ), there exists an isometry ϕ : L(X, µ, W ) → L(X, µ, V ) such that ϕ[CW ] ⊆
CV . For example, all sets of cardinality less than add(null), equipped with the discrete metric, are L-minimal.
Theorem 5. Suppose that (V, dV ) is a bounded metric space, and suppose that
(W, dW ) is a bounded, self-discerning, L-minimal metric space with |W | < add(null).
Then the following are equivalent:
1. V is isometric to L(X, µA , W ) for some sub-σ-algebra A ⊆ MALG(X, µ);
2. L(X, µ, V ) is isometric to L(X, µ, W ).
Proof. (1) ⇒ (2) is simply an instance of Theorem 1.
(2) ⇒ (1): Fix an isometry ϕ : L(X, µ, W ) → L(X, µ, V ) such that ϕ[CW ] ⊆ CV .
Let F = ϕ−1 (CV ), so F ∼
= V . We have that F contains CW and, by Proposition 3, F
is closed under meets and joins. Thus, by Proposition 4, there exists a sub-σ-algebra
A ⊆ MALG(X, µ) such that F ∼
2
= L(X, µA , W ) as desired.
Corollary 6. Suppose that V and W satisfy the hypotheses of Theorem 5. Suppose
in addition that W is a metric group. If L(X, µ, V ) ∼
= L(X, µ, W ), then V can be
given a group structure compatible with its metric.
Proof. By Theorem 5, we have that V ∼
= L(X, µA , W ) for some sub-σ-algebra
A ⊆ MALG(X, µ). V then forms a metric group under the operations of pointwise
multiplication and inversion.
2
If we equip {0, 1} with the discrete metric, it is easily seen that L(X, µ,{0,1}) ∼
=
(MALG(X, µ), dµ ), where dµ (A, B) = µ(A4B). As mentioned, {0, 1} is L-minimal,
but it fails to be self discerning. Nevertheless, the analogue of Theorem 5 remains
true.
Theorem 7. Suppose that (V, d) is a bounded metric space. The following are
equivalent:
1. V is isometric to the restriction of MALG(X, µ) to a sub-σ-algebra;
2. L(X, µ, V ) is isometric to MALG(X, µ).
Proof. (1) ⇒ (2) is yet another instance of Theorem 1.
(2) ⇒ (1): Fix an isometry ϕ : L(X, µ, V ) → MALG(X, µ). We may assume
ϕ(c0 ) = ∅: if this is not the case, we replace ϕ by ϕ0 defined by ϕ0 (f ) = ϕ(f )4ϕ(c0 ).
We will show that A = {ϕ(cv ) : v ∈ V } forms a sub-σ-algebra of MALG(X, µ).
Since v 7→ ϕ(cv ) is an isometry between V and A, this will complete the proof of
the theorem.
We first notice that if we have such a ϕ, then for each v ∈ V there is a unique
v ∈ V with d(v, v) = 1. To see existence, we consider f = ϕ−1 (X \ ϕ(cv )). Since
e v , f ) = dµ (ϕ(cv ), ϕ(f )) = 1, we may find a v ∈ V with d(v, v) ≥ 1. Of course,
d(c
we must have that the diameter of V is at most 1, so d(v, v) = 1.
On the other hand, if v 1 and v 2 are such that d(v, v 1 ) = d(v, v 2 ) = 1, then
dµ (ϕ(cv ), ϕ(cv1 )) = dµ (ϕ(cv ), ϕ(cv1 )) = 1. Thus, ϕ(cv1 ) and ϕ(cv2 ) are both complements of ϕ(cv ), and hence equal, so v 1 = v 2 . Since dµ (ϕ(cv ), ϕ(cv )) = 1, we
conclude that A is closed under complements.
5
To see A is closed under countable intersections, we simply observe that ϕ
induces an isomorphism between the posets (L(X, µ, V ), 4) and (MALG(X, µ), ⊆).
Then, since MALG(X, µ) is closed under countable intersections, Proposition 3
implies that A is as well.
2
Corollary 8. Suppose that (W, d) is a bounded metric space such that |W | < c. If
L(X, µ, W ) is isometric to MALG(X, µ), then W is finite.
Proof. The smallest infinite sub-σ-algebra of MALG(X, µ) has size c.
2
Regrettably, there is not yet a simple characterization of metric spaces which are
both L-minimal and self discerning. Additionally, in light of Theorem 7, it appears
that the definition of a self discerning space is a bit too restrictive, as it excludes
the smallest nontrivial space for which the theorem is true.
References
[1] A.S. Kechris. Classical descriptive set theory, volume 156 of Graduate Texts in
Mathematics. Springer-Verlag, New York, 1995.