326 CHAPTER 14. THE LEBESGUE INTEGRAL

Definition 14.3.4 Let µ = µF be defined on B (R) as follows.

µ (E)≡ supn∈N

∫ n

−nXE∩[−n,n] (x)dF = lim

n→∞

∫ n

−nXE∩[−n,n] (x)dF

Does this definition make sense? I need to verify the limit exists because{∫ n

−nXE∩[−n,n] (x)dF

}∞

n=1

is an increasing sequence. If unbounded, then the limit is ∞ and if bounded, it is the reallimit of the sequence. Thus the definition does make sense.

Lemma 14.3.5{∫ n−n XE∩[−n,n] (x)dF

}∞

n=1 is increasing in n. Also, if E is a bounded set

contained in [−(n−1) ,n−1] , then∫ n+1−(n+1)XE∩[−(n+1),n+1] (x)dF =

∫ n−n XE∩[−n,n] (x)dF.

Proof:∫ n+1−(n+1)XE∩[−(n+1),n+1] (x)dF =

∫ −n

−(n+1)XE∩[−(n+1),n+1] (x)dF +

∫ n

−nXE∩[−(n+1),n+1] (x)dF

+∫ n+1

nXE∩[−(n+1),n+1] (x)dF ≥

∫ n+1

−(n+1)XE∩[−n,n] (x)dF

If E ⊆ [−(n−1) ,n−1] , then∫ n+1

−(n+1)XE∩[−(n+1),n+1] (x)dF =

∫ n+1

−(n+1)XE∩[−(n−1),n−1] (x)dF

=∫ −n

−(n+1)XE∩[−(n−1),n−1] (x)dF +

∫ n

−nXE∩[−(n−1),n−1] (x)dF

+∫ n+1

nXE∩[−(n−1),n−1] (x)dF

=∫ n

−nXE∩[−(n−1),n−1] (x)dF =

∫ n

−nXE∩[−n,n] (x)dF

because on [−(n+1) ,−n] and [n,n+1] all the sums in defining the integrals are 0. Thus,if E is bounded, the integrals giving the measure as a limit are eventually constant andµ (E)< ∞.

Next I need to verify that this is a measure. Let {Ei}∞

i=1 be disjoint sets in B (R). Then∪∞

i=1Ei ∈B (R) and so, from the monotone convergence theorem for generalized Riemannintegrals,

µ (∪∞i=1Ei) ≡ sup

n

∫ n

−nX∪∞

i=1Ei∩[−n,n] (x)dF = supn

limm→∞

∫ n

−nX∪m

i=1Ei∩[−n,n] (x)dF

= supn

limm→∞

∫ n

−n

m

∑i=1

XEi∩[−n,n]dF = supn

supm

∫ n

−n

m

∑i=1

XEi∩[−n,n]dF