Now with these major results about measures, it is time to specialize to the outer measure of Theorem
5.7.1. The next theorem describes some fundamental properties of Lebesgue measure on ℝ. The
conditions 5.16 and 5.17 given below are known respectively as inner and outer regularity.
Theorem 5.10.1Let ℱ denote the σ algebra of Theorem 5.8.4, associated with the outer measure μ inTheorem 5.7.1, on which μ is a measure. Then every open interval is in ℱ. So are all open and closed sets.Furthermore, if E is anyset in ℱ
μ(E) = sup {μ(K ) : K compact, K ⊆ E} (5.16)
(5.16)
μ (E ) = inf{μ(V) : V is an open set V ⊇ E } (5.17)
(5.17)
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 (5.18)
(5.18)
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))
≤ b − a+ a− c+ d − b
= d − 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 5.18 for S =
(c,d)
is
μ ((a,d))+ μ((c,a]) = d − a+ a− c = μ((c,d))
The last case is entirely similar. Thus 5.18 holds whenever S is an open interval. Now it is clear 5.18 also
holds if μ
Since ε is arbitrary, this shows 5.18 holds for any S and so any open interval is in ℱ.
It follows any open set is in ℱ. This follows from Theorem 2.7.10 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 > ∑ bi − 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 5.16. 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
( )
μ I ∩ EC + 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
VC ∩ I ⊆ (IC ∪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,
μ(V C ∩ I)+ ε ≥ μ(VC ∩ I)+ μ (E ∖ (VC ∩ 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 5.2.4 that for n large enough, μ
(E )
n
> l + ε. Now from what was
just shown, there exists K ⊆ E_{n} such that μ
= b−a. Also m is translationinvariantin the sense that if E is any Lebesgue measurable set, then m
(x+ E )
= m
(E)
.
Proof: The formula for the measure of an interval comes right away from Theorem 5.7.1. 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 5.10.1, 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 ).■
Definition 5.10.3There is an important idea which is often seen in the context of measures.Something happens a.e. (almost everywhere) means that it happens off a set of measure zero.