Now with these major results about measures, it is time to specialize to the outer measure of
Theorem 7.8.3. The next theorem gives Lebesgue Stieltjes measure on ℝ. The conditions
7.17 and 7.18 given below are known respectively as inner and outer regularity.
Theorem 7.10.1Let ℱ denote the σ algebra of Theorem 7.9.4, associated withthe outer measure μ in Theorem 7.8.3, on which μ is a measure. Then every openinterval is in ℱ. So are all open and closed sets. Furthermore, if E is anyset inℱ
μ(E) = sup {μ(K ) : K compact, K ⊆ E} (7.17)
(7.17)
μ(E ) = inf{μ(V) : V is an open set V ⊇ E } (7.18)
(7.18)
Proof: The first task is to show
(a,b)
∈ℱ. I need to show that for every S ⊆ ℝ,
( )
μ(S) ≥ μ (S ∩ (a,b))+ μ S ∩(a,b)C (7.19)
(7.19)
Suppose first S is an open interval,
(c,d)
. If
(c,d)
has empty intersection with
(a,b)
or is
contained in
(a,b)
there is nothing to prove. The above expression reduces to nothing more
than μ
(S)
= μ
(S)
. Suppose next that
(c,d)
⊇
(a,b)
. In this case, the right side of the above
reduces to
μ((a,b))+ μ((c,a]∪[b,d))
≤ F (b− ) − F (a+) + F (a+ )− F (c+ )+ F (d− )− F (b− )
= F (d− )− F (c+) ≡ μ((c,d))
The only other cases are c ≤ a < d ≤ b or a ≤ c < d ≤ b. Consider the first of these cases.
Then the right side of 7.19 for S =
(c,d)
is
μ((a,d)) + μ((c,a]) = F (d− )− F (a+ )+ F (a+ )− F (c+ )
= F (d− )− F (c+ ) = μ((c,d))
The last case is entirely similar. Thus 7.19 holds whenever S is an open interval.
Now it is clear 7.19 also holds if μ
(S)
= ∞. Suppose then that μ
(S )
< ∞ and
let
S ⊆ ∪ ∞k=1(ak,bk)
such that
∞ ∞
μ (S )+ ε > ∑ (F (b − )− F (a + )) = ∑ μ((a ,b)).
k=1 k k k=1 k k
Then since μ is an outer measure, and using what was just shown,
Since ε is arbitrary, this shows 7.19 holds for any S and so any open interval is in
ℱ.
It follows any open set is in ℱ. This follows from Theorem 2.11.8 which implies that if U is
open, it is the countable union of disjoint open intervals. Since each of these open
intervals is in ℱ and ℱ is a σ algebra, their union is also in ℱ. It follows every
closed set is in ℱ also. This is because ℱ is a σ algebra and if a set is in ℱ then
so is its complement. The closed sets are those which are complements of open
sets.
The assertion of outer regularity is not hard to get. Letting E be any set μ
(E )
< ∞, there
exist open intervals covering E denoted by
{(ai,bi)}
_{i=1}^{∞} such that
∑∞ ∞∑
μ(E )+ ε > F (bi− )− F (ai+) = μ(ai,bi) ≥ μ (V )
i=1 i=1
where V is the union of the open intervals just mentioned. Thus
μ(E ) ≤ μ(V ) ≤ μ(E )+ ε.
This shows outer regularity. If μ
(E )
= ∞, there is nothing to show.
Now consider the assertion of inner regularity 7.17. Suppose I is a closed and bounded
interval and E ⊆ I with E ∈ℱ. By outer regularity, there exists open V containing I ∩ E^{C}
such that
( C)
μ I ∩ E + ε > μ(V )
Then since μ is additive on ℱ, it follows that μ
( ( ))
V ∖ I ∩ EC
< ε. Then K ≡ V^{C}∩ I is a
compact subset of E. This is because V ⊇ I ∩ E^{C} so V^{C}⊆ I^{C}∪ E and so
C ( C )
V ∩I ⊆ I ∪ E ∩ I = E ∩ I = E.
Also,
( ) ( )
E ∖ V C ∩ I = E ∩ V = V ∖EC ⊆ V ∖ I ∩ EC ,
a set of measure less than ε. Therefore,
( C ) ( C ) ( ( C ))
μ V ∩ I + ε ≥ μ V ∩ I + μ E ∖ V ∩I = μ(E),
so the desired conclusion holds in the case where E is contained in a compact interval.
Now suppose E is arbitrary and let l < μ
(E)
. Then choosing ε small enough,
l + ε < μ
(E)
also. Letting E_{n}≡ E ∩
[− n,n]
, it follows from Lemma 7.2.4 that for n large
enough, μ
(En)
> l + ε. Now from what was just shown, there exists K ⊆ E_{n} such that
μ
Definition 7.10.2When the integrator function is F
(x )
= x, the LebesgueStieltjes measure just discussed is known as one dimensionalLebesgue measure and isdenoted as m.
Proposition 7.10.3For m Lebesgue measure, m
([a,b])
= m
((a,b))
= b−a. Alsom is translation invariantin the sense that if E is any Lebesgue measurable set, thenm
(x+ E )
= m
(E)
.
Proof: The formula for the measure of an interval comes right away from Theorem 7.8.3.
From this, it follows right away that whenever E is an interval, m
(x + E)
= m
(E )
. Every
open set is the countable disjoint union of open intervals, so if E is an open set, then
m
(x+ E )
= m
(E)
. What about closed sets? First suppose H is a closed and bounded set.
Then letting
It follows right away that if G is the countable intersection of open sets, (G_{δ} set, pronounced
g delta set ) then
m (G ∩ (− n,n)+ x) = m (G ∩ (− n,n))
Now taking n →∞,m
(G + x)
= m
(G)
.Similarly, if F is the countable union of compact sets,
(F_{σ} set, pronounced F sigma set) then m
(F + x)
= m
(F)
. Now using Theorem 7.10.1, if E
is an arbitrary measurable set, there exist an F_{σ} set F and a G_{δ} set G such that F ⊆ E ⊆ G
and m
(F)
= m
(G )
= m
(E)
. Then
m (F) = m (x + F) ≤ m (x +E ) ≤ m (x+ G ) = m (G) = m (E ) = m (F). ■
Here is a general result. If you have a measure μ, then by Proposition 7.6.2, μ defined
there is an outer measure which agrees with μ on the σ algebra of measurable sets ℱ. What of
the measure determined by μ? Denoted still by μ. Is μ = μ on ℱ? Is ℱ a subset of the μ
measurable sets, those which satisfy the criterion of being measurable? Suppose E ∈ℱ. Is it
the case that
¯μ (S ) = ¯μ(S ∖E )+ ¯μ(S ∩E )?
As usual, there is nothing to show if μ
(S )
= ∞ so assume this does not happen. Let
F ⊇ S,F ∈ℱ. Then by Proposition 7.6.2,
Thus, indeed ℱ is a subset of the μ measurable sets. By Proposition 7.6.2, μ = μ on ℱ.
This gives a way to complete a measure space which is described in the following
proposition.
Proposition 7.10.4Let
(Ω, ℱ, μ )
be a measure space. Letμbe the outer measuredetermined by μ as in Proposition 7.6.2. Also denote asℱ, the σ algebra ofμmeasurablesets. Thus
( )
Ω,ℱ¯,μ¯
is a complete measure space in whichℱ⊇ ℱ andμ = μ on ℱ.Also, in this situation, ifμ
(E)
= 0, then E ∈ℱ. No new sets are obtained if
(Ω,ℱ, μ)
is already complete.
Proof: All that remains to show is the last claim. But this is obvious because if S is a set,
is complete. Let F ∈ℱ. Then there exists E ⊇ F such that
μ
(E )
= μ
(F )
. This is obvious if μ
(F )
= ∞. Otherwise, let E_{n}⊇ F,μ
(F)
+
1
n
> μ
(E )
n
. Just
let E = ∩_{n}E_{n}. Now μ
(E ∖F)
= 0. Now also, there exists a set of ℱ called W such that
μ
(W )
= 0 and W ⊇ E ∖F. Thus E ∖F ⊆ W, a set of measure zero. Hence by completeness of
(Ω, ℱ,μ)
, it must be the case that E ∖ F = E ∩ F^{C} = G ∈ℱ. Then taking complements of
both sides, E^{C}∪ F = G^{C}∈ℱ. Now take intersections with E. F ∈ E ∩ G^{C}∈ℱ.
■