Chapter 7 Infinite product spaces

4 CHAPTER 7. INFINITE PRODUCT SPACES
h is an identity, while in the first case we note that g h is the bijection g h : (g
f ) n (A \ C) ! (g f ) n+1 (A \ C). Hence g h maps X bijectively onto
⇣ [
(g f ) n ⌘ ✓ ⇣ [
(X \ A) [ (g f ) n ⌘
(A \ C) [ \
(g f ) n ◆
(X) = A. (7.7)
n 1
n 0
Since all of these maps are measurable, so is the above partition of A. It follows
that h is measurable too.
The crux of the above theorem is that we only need to demonstrate the existence
of Borel isomorphism(s) onto subsets of the relevant spaces. We will eventually use
open, continuous bijections, i.e., homeomorphisms. Here it is important to characterize
the image of a Polish space under a homeomorphism. Recall that a Gd is a class
of sets in a topological space that are countable intersections of open sets.
Lemma 7.4 Let X and Y be Polish and let g : X ! B ⇢ Y be a homeomorphism—i.e., a
bijection that is open and continuous. Then B is a Gd subset of Y.
Proof. Let f = g 1 : B ! X. Define the oscilation of f by
n 0
osc f (y) =inf diamX( f (U)): U 3 y open . (7.8)
(This is defined at all y 2 Y.) Suppose y 2 ∂B is such that osc f (y) = 0. Then
for any yn ! y, yn 2 B, xn = f (yn) converges to x = limn!• f (yn). Since X is
complete, we have x 2 X and so y = limn!• yn = limn!• g(xn) =g(x). It follows
that y 2 B, i.e.,
B = B \ y 2 Y : osc f (y) =0 . (7.9)
But closed sets are Gd in metric spaces and {osc f = 0} is Gd because
{osc f = 0} = \
n 1
{osc f < 1/ n} (7.10)
and because osc f is upper semicontinuous, non-negative and so continuous on {osc f =
0}. Hence B is also a Gd set.
The actual use of Theorem 7.3 is enabled by the following observation:
Lemma 7.5 The spaces [0, 1], C =[0, 1] N , N = N N and [0, 1] N , equipped with their
natural Polish topologies, are Borel isomorphic.
Proof. This results from the string of open, continuous (and thus Borel) injections:
[0, 1] ,! C ,! N ,! [0, 1] N ,! [0, 1] (7.11)
The first map is the injection defined by writing each number in [0, 1] in dyadic
form. The second map is a simple inclusion. The third map is induced by the
continued fraction expansion: if x 2 [0, 1] \ Q, there exists a unique sequence (an) 2
{0, 1, . . . } N such that
1
x =
(7.12)
1
a1 +
a2 + 1
a3+...