21.5 Measures On Hausdorff Spaces
In the case of a Hausdorff topological space, the following lemma gives conditions under which
the σ algebra of μ measurable sets for an outer measure μ contains the Borel sets. In words, it
assumes the outer measure is inner regular on open sets and outer regular on all sets. Also
it assumes you can approximate the measure of an open set with a compact set
and the measure of a compact set with an open set. Recall that the Borel sets are
those sets in the smallest σ algebra that contains the open sets. The big result is
the Riesz representation theorem for positive linear functionals and the following
lemma is the technical part of the proof of this big theorem in addition to being
interesting for its own sake. It holds in a Hausdorff space, not just one which is locally
Lemma 21.5.1 Let Ω be a Hausdorff space and suppose μ is an outer measure
satisfying μ is finite on compact sets and the following conditions,
- μ = inf
for all E. (Outer regularity.)
- For every open set V,μ = sup
on open sets.)
- If A,B are compact disjoint sets, then μ =
Then the following hold.
- If ε > 0 and if K is compact, there exists V open such that V ⊇ K and
- If ε > 0 and if V is open with μ
< ∞, there exists a compact subset K of V such
- The μ measurable sets S contains the Borel sets and also μ is inner regular on every
open set and for every E ∈S with μ(E) < ∞. Here S consists of those subsets of Ω, E,
with the property that for any subset S of Ω,
Proof: First we establish 1 and 2 and use them to establish the last assertion. Consider 2.
Suppose it is not true. Then there exists an open set V having μ
but for all
K ⊆ V,μ
for some ε >
0. By inner regularity on open sets, there exists
K1 ⊆ V,K1
compact, such that μ
Now by assumption, μ
and so by
inner regularity on open sets again, there exists compact K2 ⊆ V ∖ K1
Continuing this way, there is a sequence of disjoint compact sets
contained in V
2. Now this is an obvious contradiction
because by 3
for each n, contradicting μ
Next consider 1. By outer regularity, there exists an open set W ⊇ K such that
, there exists compact K1 ⊆ W ∖K
such that μ
Then consider V ≡ W ∖ K1.
This is an open set containing K
and from what was just
Now consider the last assertion.
for all compact K.
First it will be shown the compact sets are in S. From this it will follow the closed sets are
in S1. Then you show S1 = S. Thus S1 = S is a σ algebra and so it contains the Borel sets.
Finally you show the inner regularity assertion.
Claim 1: Compact sets are in S.
Proof of claim: Let V be an open set with μ
. I will show that for C
The various sets are illustrated in the following diagram. By 2, there
exists a compact set K ⊆ V ∖ C
and a compact set H ⊆ V such that
Since ε is arbitrary, this shows that
Of course 21.5 is exactly what needs to be shown for arbitrary S in place of V . It suffices
to consider only S having μ
. If S ⊆
Ω, with μ
) < ∞
, let V ⊇ S
) + ε > μ
Then from what was just shown, if C
is arbitrary, this shows the compact sets are in S
. This proves the claim.
As discussed above, this verifies the closed sets are in S1 because if H is closed and C is
compact, then H ∩C ∈S. If S1 is a σ algebra, this will show that S1 contains the Borel sets.
Thus I first show S1 is a σ algebra.
To see that S1 is closed with respect to taking complements, let E ∈S1 and K a compact
Then from the fact, just established, that the compact sets are in S,
S1 is closed under countable unions because if K is a compact set and En ∈S1,
because it is a countable union of sets of S. Thus S1 is a σ algebra.
Therefore, if E ∈S and K is a compact set, just shown to be in S, it follows K ∩ E ∈S
because S is a σ algebra which contains the compact sets and so S1 ⊇S. It remains to verify
S1 ⊆S. Recall that
Let E ∈S1 and let V be an open set with μ(V ) < ∞ and choose K ⊆ V such that
μ(V ∖ K) < ε. Then since E ∈S1, it follows E ∩ K,EC ∩ K ∈S and so
is arbitrary, this shows
which would show E ∈S if V were an arbitrary set.
Now let S ⊆ Ω be such an arbitrary set. If μ(S) = ∞, then
If μ(S) < ∞, let
Since ε is arbitrary, this shows that E ∈S and so S1 = S. Thus S⊇ Borel sets as
From 2 μ is inner regular on all open sets. It remains to show that
for all F ∈S with μ(F) < ∞. It might help to refer to the following crude picture to keep
things straight. It also might not help. I am not sure. In the picture, the green marks the
boundary of V while the shaded red marks U ∖ F and black marks F and V C ∩ K. This
last set is as shown because K is a compact subset of U such that μ(U ∖ K) < ε.
and let U
be an open set, U ⊇ F,μ
) < ∞
. Let V
be open, V ⊇ U ∖ F
(This can be obtained as follows, because μ is a measure on S.
Thus from the outer regularity of μ, 1 above, there exists V such that it contains U ∖ F
Since V ⊇ U ∩ FC,V C ⊆ UC ∪ F so U ∩ V C ⊆ U ∩ F = F. Hence U ∩ V C is a subset of
F. Now let K ⊆ U,μ(U ∖ K) < ε. Thus K ∩ V C is a compact subset of F and
is arbitrary, this proves the second part of the lemma. ■