The Yankov von Neumann Aumann theorem deals with the projection of a product
measurable set. It is a very difficult but interesting theorem. The material of this chapter
is taken from [?], [?], [?], and [?]. We use the standard notation that for S and ℱ σ
algebras, S×ℱ is the σ algebra generated by the measurable rectangles, the product
measure σ algebra. The next result is fairly easy and the proof is left for the
reader.
Lemma 46.0.1Let
(X, d)
be a metric space. Then if d_{1}
(x,y)
=
-d(x,y)-
1+d(x,y)
, it followsthat d_{1}is a metric on X and the basis of open balls taken with respect to d_{1}yieldsthe same topology as the basis of open balls taken with respect to d.
Theorem 46.0.2Let
(Xi,di)
denote a complete metric space and let X ≡∏_{i=1}^{∞}X_{i}.Then X is also a complete metric space with the metric
∞∑ d (x ,y)
ρ(x,y) ≡ 2−i--i--i--i--.
i=1 1+ di(xi,yi)
Also, if X_{i}is separable for each i then so is X.
Proof:It is clear from the above lemma that ρ is a metric on X. We need to
verify X is complete with this metric. Let
{xn }
be a Cauchy sequence in X.
Then it is clear from the definition that
{xn}
i
is a Cauchy sequence for each i
and converges to x_{i}∈ X_{i}. Therefore, letting ε > 0 be given, we choose N such
that
We need to verify that X is separable. Let D_{i} denote a countable dense set in
X_{i},D_{i}≡
{ }
rik
_{k=1}^{∞}. Then let
D ≡ D × ⋅⋅⋅× D × {rk+1} ×{rk+2} × ⋅⋅⋅
k 1 k 1 1
Thus D_{k} is a countable subset of X. Let D≡∪_{k=1}^{∞}D_{k}. Then D is countable and we can
see D is dense in X as follows. The projection of D_{k} onto the first k entries is dense in
∏_{i=1}^{k}X_{i} and for k large enough the remaining component’s contribution to the metric,
ρ is very small. Therefore, obtaining d ∈D close to x ∈ X may be accomplished by
finding d ∈D such that d is close to x in the first k components for k large enough. Note
that we do not use ∏_{k=1}^{∞}D_{k}!
Definition 46.0.3A complete separable metric space is called a polish space.
Theorem 46.0.4Let X be a polish space. Then there exists f : ℕ^{ℕ}→ X which is ontoand continuous. Here ℕ^{ℕ}≡∏_{i=1}^{∞}ℕ and a metric is given according to the abovetheorem. Thus forn,m∈ ℕ^{ℕ},
∑∞
ρ(n,m ) ≡ 2−i--|ni-− mi-|-.
i=1 1 +|ni − mi|
Proof:Since X is polish, there exists a countable covering of X by closed sets having
diameters no larger than 2^{−1},
{B(i)}
_{i=1}^{∞}. Each of these closed sets is also a polish
space and so there exists a countable covering of B
(i)
by a countable collection of closed
sets,
{B (i,j)}
_{j=1}^{∞} each having diameter no larger than 2^{−2} where B
(i,j)
⊆ B
(i)
≠∅ for
all j. Continue this way. Thus
B (n ,n ,⋅⋅⋅,n ) = ∪∞ B (n ,n ,⋅⋅⋅,n ,i)
1 2 m i=1 1 2 m
and each of B
(n1,n2,⋅⋅⋅,nm,i)
is a closed set contained in B
(n1,n2,⋅⋅⋅,nm )
whose
diameter is at most half of the diameter of B
(n1,n2,⋅⋅⋅,nm )
. Now we define our
mapping from ℕ^{ℕ} to X. If n =
{nk}
_{k=1}^{∞}∈ ℕ^{ℕ}, we let f
(n )
≡∩_{m=1}^{∞}B
(n1,n2,⋅⋅⋅,nm)
.
Since the diameters of these sets converge to 0, there exists a unique point in this
countable intersection and this is f
(n)
.
We need to verify f is continuous. Let n ∈ ℕ^{ℕ} be given and suppose m is very close to
n. The only way this can occur is for n_{k} to coincide with m_{k} for many k. Therefore, both
f
(n )
and f
(m )
must be contained in B
(n1,n2,⋅⋅⋅,nm)
for some fairly large m. This
implies, from the above construction that f
(m )
is as close to f
(n)
as 2^{−m},
proving f is continuous. To see that f is onto, note that from the construction,
if x ∈ X, then x ∈ B
(n1,n2,⋅⋅⋅,nm )
for some choice of n_{1},
⋅⋅⋅
,n_{m} for each
m. Note nothing is said about f being one to one. It probably is not one to
one.
Definition 46.0.5We call a topological space X a Suslin space if X is a Hausdorffspaceand there exists a polish space, Z and a continuous function f which maps Z ontoX.
Z f o→nto X
continuous
These Suslin spaces are also calledanalytic sets in some contexts but we will use the termSuslin space in referring to them.
Corollary 46.0.6X is a Suslin space, if and only if there exists a continuousmapping from ℕ^{ℕ}onto X.
Proof:We know there exists a polish space Z and a continuous function,
h : Z → X which is onto. By the above theorem there exists a continuous map,
g : ℕ^{ℕ}→ Z which is onto. Then h ∘ g is a continuous map from ℕ^{ℕ} onto X.
The “if” part of this theorem is accomplished by noting that ℕ^{ℕ} is a polish
space.
Lemma 46.0.7Let X be a Suslin space and suppose X_{i}is a subspace of X whichis also a Suslin space. Then ∪_{i=1}^{∞}X_{i} and ∩_{i=1}^{∞}X_{i}are also Suslin spaces. Alsoevery Borel set in X is a Suslin space.
Proof:Let f_{i} : Z_{i}→ X_{i} where Z_{i} is a polish space and f_{i} is continuous and onto.
Without loss of generality we may assume the spaces Z_{i} are disjoint because if not, we
could replace Z_{i} with Z_{i}×
{i}
. Now we define a metric, ρ, for Z ≡∪_{i=1}^{∞}Z_{i} as
follows.
ρ(x,y) ≡ 1 if x ∈ Zi,y ∈ Zk,i ⁄= k
di(x,y)
ρ(x,y) ≡ 1-+-d-(x,y) if x,y ∈ Zi.
i
Here d_{i} is the metric on Z_{i}. It is easy to verify ρ is a metric and that
(Z, ρ)
is a polish
space. Now we define f : Z →∪_{i=1}^{∞}X_{i} as follows. For x ∈ Z_{i}, f
(x)
≡ f_{i}
(x)
. This is
well defined because the Z_{i} are disjoint. If y is very close to x it must be that x and y are
in the same Z_{i} otherwise this could not happen. Therefore, continuity of f follows from
continuity of f_{i}. This shows countable unions of Suslin subspaces of a Suslin space are
Suslin spaces.
If H ⊆ X is a closed subset, then, letting f : Z → X be onto and continuous, it
follows f : f^{−1}
(H )
→ H is onto and continuous. Since f^{−1}
(H )
is closed, it follows
f^{−1}
(H )
is a polish space. Therefore, H is a Suslin space.
Now we show countable intersections of Suslin spaces are Suslin. It is clear that
θ : ∏_{i=1}^{∞}Z_{i}→∏_{i=1}^{∞}X_{i} given by θ
(z)
≡ x =
{xi}
where x_{i} = f_{i}
(zi)
is continuous
and onto, this with respect to the usual product topology. Note that ∏_{i=1}^{∞}Z_{i} is a polish
space because of the assumption that each Z_{i} is and the above considerations. Therefore,
∏_{i=1}^{∞}X_{i} is a Suslin space. Now let P ≡
∏ ∞
{y ∈ i=1 fi(Zi) : yi = yj for all i,j}
(This is
how you get it on the intersection. I guess this must be the case where each X_{i}⊆ X).
Then P is a closed subspace of a Suslin space and so it is Suslin. Then we define
h : P →∩_{i=1}^{∞}X_{i} by h
(y)
≡ f_{i}
(yi)
. This shows ∩_{i=1}^{∞}X_{i} is Suslin because h is
continuous and onto. (h ∘ θ : θ^{−1}
(P )
→∩_{i=1}^{∞}X_{i} is continuous and θ^{−1}
(P )
being a
closed subset of a polish space is polish.)
Next let U be an open subset of X. Then f^{−1}
(U )
, being an open subset of a polish
space, can be obtained as an increasing limit of closed sets, K_{n}. Therefore,
U = ∪_{n=1}^{∞}f
(Kn )
. Each f
(Kn )
is a Suslin space because it is the continuous image of a
polish space, K_{n}. Therefore, by the first part of the lemma, U is a Suslin space. Now
let
ℱ ≡ {E ⊆ X : both EC and E are Suslin}.
We see that ℱ is closed with respect to taking complements. The first part of
this lemma shows ℱ is closed with respect to countable unions. Therefore, ℱ
is a σ algebra and so, since it contains the open sets, must contain the Borel
sets.
It turns out that Suslin spaces tend to be measurable sets. In order to develop this
idea, we need a technical lemma.
Lemma 46.0.8Let
(Ω, ℱ,μ)
be a measure space and denote by μ^{∗}the outer measuregenerated by μ. Thus
μ∗(S) ≡ inf{μ(E ) : E ⊇ S, E ∈ ℱ}.
Then μ^{∗}is regular, meaning that for every S, there exists E ∈ℱ such that E ⊇ S andμ
(E )
= μ^{∗}
(S )
. If S_{n}↑ S, it follows that μ^{∗}
(Sn)
↑ μ^{∗}
(S)
. Also if μ
(Ω )
< ∞, then a set,E is measurable if and only if
∗ ∗ ∗
μ (Ω) ≥ μ (E)+ μ (Ω ∖ E).
Proof:First we verify that μ^{∗} is regular. If μ^{∗}
(S)
= ∞, let E = Ω. Then
μ^{∗}
(S )
= μ
(E )
and E ⊇ S. On the other hand, if μ^{∗}
(S )
< ∞, then we can obtain E_{n}∈ℱ
such that μ^{∗}
(S)
+
1
n
≥ μ
(E )
n
and E_{n}⊇ S. Now let F_{n} = ∩_{i=1}^{n}E_{
i}. Then F_{n}⊇ S and so
μ^{∗}
(S )
+
1
n
≥ μ
(F )
n
≥ μ^{∗}
(S)
. Therefore, letting F = ∩_{
k=1}^{∞}F_{
k}∈ℱ it follows
μ
(F )
= lim_{n→∞}μ
(F )
n
= μ^{∗}
(S)
.
Let E_{n}⊇ S_{n} be such that E_{n}∈ℱ and μ
(En)
= μ^{∗}
(Sn)
. Also let E_{∞}⊇ S
such that μ
(E∞ )
= μ^{∗}
(S)
and E_{∞}∈ℱ. Now consider B_{n}≡∪_{k=1}^{n}E_{k}. We
claim
μ (Bn ) = μ (Sn ). (46.0.1)
(46.0.1)
Here is why:
μ (E ∖ E ) = μ (E )− μ(E ∩ E ) = μ∗(S )− μ∗(S ) = 0.
1 2 1 1 2 1 1
Therefore,
μ (B ) = μ(E ∪ E ) = μ (E ∖E )+ μ(E ) = μ (E ) = μ∗ (S ).
2 1 2 1 2 2 2 2
Continuing in this way we see that 46.0.1 holds. Now let B_{n}∩ E_{∞}≡ C_{n}. Then
C_{n}↑ C ≡∪_{k=1}^{∞}C_{n}∈ℱ and μ
(Cn)
= μ^{∗}
(Sn )
. Since S_{n}↑ S and each C_{n}⊇ S_{n}, it
follows C ⊇ S and therefore,
Now we verify the second claim of the lemma. It is clear the formula holds whenever
E is measurable. Suppose now that the formula holds. Let S be an arbitrary set. We need
to verify that
μ∗ (S ) ≥ μ∗(S ∩E )+ μ∗(S ∖E ).
Let F ⊇ S, F ∈ℱ, and μ
(F )
= μ^{∗}
(S )
. Then since μ^{∗} is subadditive,
∗ ∗ ∗( C C )
μ (Ω ∖F ) ≤ μ (E ∖ F)+ μ Ω ∩ E ∩ F . (46.0.2)
(46.0.2)
Since F is measurable,
μ∗(E ) = μ∗(E ∩ F)+ μ∗ (E ∖ F) (46.0.3)
(46.0.3)
and
μ ∗(Ω ∖ E) = μ ∗(F ∖ E) +μ ∗(Ω∩ EC ∩ F C) (46.0.4)
showing that all the inequalities must be equal signs. Hence, referring to the top and
fourth lines above,
∗ ∗ ∗
μ(Ω ) = μ (Ω ∖F )+ μ (F ∖E )+ μ (E ∩F ).
Subtracting μ^{∗}
(Ω ∖F)
= μ
(Ω ∖F )
from both sides gives
∗ ∗ ∗ ∗ ∗
μ (S ) = μ(F ) = μ (F ∖E )+ μ (E ∩F ) ≥ μ (S ∖E )+ μ (E ∩ S),
This proves the lemma.
The next theorem is a major result. It states that the Suslin subsets are measurable
under appropriate conditions. This is sort of interesting because something being a Suslin
subset has to do with topology and this topological condition implies that the set is
measurable.
Theorem 46.0.9Let Ω be a metric space and let
(Ω,ℱ, μ)
be a complete Borelmeasure space with μ
(Ω )
< ∞. Denote by μ^{∗}the outer measure generated by μ.Then if A is a Suslin subset of Ω, it follows that A is μ^{∗}measurable. Since theoriginal measure space is complete, it follows that the completion produces nothingnew and so in fact A is in ℱ. See Proposition 10.1.5.
Proof: We need to verify that
μ∗(Ω) ≥ μ∗(A)+ μ∗ (Ω ∖A ).
We know from Corollary 46.0.6, there exists a continuous map, f : ℕ^{ℕ}→ A which is onto.
Let
Since ε is arbitrary, this will conclude the proof. Therefore, we only need to verify that
C ⊆ A.
What we know is that each f
(E(m1, m2,⋅⋅⋅,mk −1,mk))
is contained in A. We do
not know their closures are contained in A. We let m ≡
{mi}
_{i=1}^{∞} where the m_{i} are
defined above. Then letting
{ }
K ≡ n ∈ ℕℕ : ni ≤ mi for all i ,
we see that K is a closed, hence complete subset of ℕ^{ℕ} which is also totally bounded due
to the definition of the distance. Therefore, K is compact and so f
(K )
is also compact,
hence closed due to the assumption that Ω is a Hausdorff space and we know that
f