14.1. MEASURES 323

Next suppose µ (F1)< ∞ and {Fn} is a decreasing sequence. Then F1 \Fn is increasingto F1 \F and so by the first part,

µ (F1)−µ (F) = µ (F1 \F) = limn→∞

µ (F1 \Fn) = limn→∞

(µ (F1)−µ (Fn))

This is justified because µ (F1 \Fn)+µ (Fn) = µ (F1) and all numbers are finite by assump-tion. Hence µ (F) = limn→∞ µ (Fn).

Now I will specify a collection of subsets of R which I will refer to as measurable.

Definition 14.1.5 Given an integrator function F let

S ≡ {E ⊆ R : XE ∈ R∗ [p,q] for all closed intervals [p,q]}

I will call these sets “measurable”.

Then these so called measurable sets satisfy the following:

Lemma 14.1.6 The following hold.

1. All open intervals are in S .

2. If {Ei}∞

i=1 are disjoint sets in S then ∪∞i=1Ei ∈S also.

3. If E ∈S , then EC ≡ R\E is also in S .

Proof: Consider 1. Let (a,b) be an open interval. There are several possibilities for(a,b)∩ [p,q]

1. (a,b)∩ [p,q] = (α,β )⊆ [p,q],

2. (a,b)∩ [p,q] = (α,q]⊆ [p,q],

3. (a,b)∩ [p,q] = [p,β )⊆ [p,q] , or

4. (a,b)∩ [p,q] = [p,q] .

In case 1., let ψn vanish off (a,b), be continuous, and increase to X(α,β ). Then ψn ∈R([p,q]) so it is in R∗ ([a,b]) and by the monotone convergence X(α,β ) ∈ R∗ ([a,b]).

Next consider 2. Is X(α,q] ∈ R∗ [p,q]? Yes, and a similar argument to 1.) will hold. justget a sequence of functions continuous on [p,q] and increasing to X(α,q] and use a similarargument.

Case 3.) is entirely similar.The last case is obvious from Lemma 9.7.2. In fact,

∫ qp X[p,q]dF = F (q+)−F (p−),

this being the ordinary Riemann Stieltjes integral.Letting E ≡ ∪∞

i=1Ei, the Ei being disjoint, why is XE ∈ R∗ [p,q]? It follows from whatwas done above that X∪m

i=1Ei∩[a,b] ∈ R∗ [p,q] because, since the sets are disjoint, X∪mi=1Ei

equals ∑mi=1 XEi and it was shown that the sum of functions in R∗ [p,q] is in R∗ [p,q]. Now

apply the monotone convergence theorem as m→ ∞ to conclude that X∪∞i=1Ei ∈ R∗ [p,q] .

Next consider 3. If E ∈S , then XEC = 1−XE and the two summands on the rightare in R∗ [p,q] so the function on the left is also.