Now with these major results about measures, it is time to specialize to the outer measure of Theorem
19.8.3. The next theorem gives Lebesgue Stieltjes measure on ℝ. The conditions 19.18 and 19.19 given
below are known respectively as inner and outer regularity.
Theorem 19.10.1Let ℱ denote the σ algebra of Theorem 19.9.4, associated with the outer measure μ inTheorem 19.8.3, on which μ is a measure. Then every open interval is in ℱ. So are all open and closedsets. Furthermore, if E is anyset in ℱ
μ(E) = sup {μ(K ) : K compact, K ⊆ E} (19.18)
(19.18)
μ (E ) = inf{μ(V) : V is an open set V ⊇ E } (19.19)
(19.19)
Proof: The first task is to show
(a,b)
∈ℱ. I need to show that for every S ⊆ ℝ,
( C)
μ(S) ≥ μ(S ∩(a,b)) +μ S ∩(a,b) (19.20)
(19.20)
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 19.20 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 19.20 holds whenever S is an open interval. Now it is clear 19.20 also
holds if μ
Since ε is arbitrary, this shows 19.20 holds for any S and so any open interval is in ℱ.
It follows any open set is in ℱ. This follows from the fact that any open set in ℝ is the countable union
of open intervals. See Theorem 11.2.8 for example. There can be no more than countably many disjoint
open intervals because the rational numbers are dense and countable. 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. Then the regularity of this measure
follows right away from Corollary 19.4.8 because the measure is finite on any open interval.
■
Definition 19.10.2When the integrator function is F
(x)
= x, the Lebesgue Stieltjes measure justdiscussed is known as one dimensionalLebesgue measure and is denoted as m.
Proposition 19.10.3For m Lebesgue measure, m
([a,b])
= m
((a,b))
= 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 19.8.3. 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 by Theorem 11.2.8, 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