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.
Definition 7.8.1The following definition is important.
F (x+) ≡ lim F (y), F (x− ) = lim F (y)
y→x+ y→x −
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 7.8.2P
(S)
denotes the set of all subsets of S.
Theorem 7.8.3Let F be an increasing function defined on ℝ. This will becalled an integrator function. There exists a function μ : P
(ℝ)
→
[0,∞ ]
which satisfiesthe following properties.
If A ⊆ B, then 0 ≤ μ
(A)
≤ μ
(B)
,μ
(∅)
= 0.
μ
(∪∞ Ai)
k=1
≤∑_{i=1}^{∞}μ
(Ai)
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 7.8.4For A ⊆ ℝ,
{∑∞ }
μ (A ) = inf (F (bi− )− F (ai+)) : A ⊆ ∪ ∞i=1(ai,bi)
i=1
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,