A measure on ℝ is like length. I will present something more general than length because it is
no trouble to do so and the generalization is useful in many areas of mathematics such as
probability.
This outer measure will end up determining a measure known as the Lebesgue Stieltjes
measure.
Definition 19.8.1The following definition is important.
F (x+) ≡ yli→mx+F (y), F (x− ) = ly→ixm− F (y)
Thus one of these is the limit from the left and the other is the limit from the right.
In probability, one often has F
(x)
≥ 0, F is increasing, and F
(x+)
= F
(x)
. This is the case where F is
a probability distribution function. In this case, F
(x)
≡ P
(X ≤ x)
where X is a random variable. In this
case, lim_{x→∞}F
(x)
= 1 but we are considering more general functions than this including the
simple example where F
(x)
= x. This last example will end up giving Lebesgue measure on
ℝ.
Definition 19.8.2P
(S)
denotes the set of all subsets of S.
Theorem 19.8.3Let F be an increasing function defined on ℝ. This will be called an integratorfunction. There exists a function μ : P
(ℝ )
→
[0,∞ ]
which satisfies the following properties.
If A ⊆ B, then 0 ≤ μ
(A)
≤ μ
(B)
,μ
(∅)
= 0.
μ
(∪∞ A )
k=1 i
≤∑_{i=1}^{∞}μ
(A )
i
μ
([a,b])
= F
(b+)
− F
(a− )
,
μ
((a,b))
= F
(b− )
− F
(a+ )
μ
((a,b])
= F
(b+)
− F
(a+)
μ
([a,b))
= F
(b− )
− F
(a− )
.
Proof: First it is necessary to define the function μ. This is contained in the following
definition.
Definition 19.8.4For A ⊆ ℝ,
( ∞ )
μ (A ) = inf{∑ (F (b− )− F (a +)) : A ⊆ ∪ ∞ (a ,b)}
(j=1 i i i=1 i i)
In words, you look at all coverings of A with open intervals. For each of these open coverings, you add
the “lengths” of the individual open intervals and you take the infimum of all such numbers
obtained.
Then 1.) is obvious because if a countable collection of open intervals covers B, then it also covers A.
Thus the set of numbers obtained for B is smaller than the set of numbers for A. Why is μ
(∅)
= 0? Pick a
point of continuity of F. Such points exist because F is increasing and so it has only countably many points
of discontinuity. Let a be this point. Then ∅⊆
(a − δ,a + δ)
and so μ
(∅)
≤ F
(a+ δ)
−F
(a− δ)
for every
δ > 0. Letting δ → 0, it follows that μ
(∅)
= 0.
Consider 2.). If any μ
(Ai)
= ∞, there is nothing to prove. The assertion simply is ∞≤∞. Assume then
that μ
(Ai)
< ∞ for all i. Then for each m ∈ ℕ there exists a countable set of open intervals,