298 CHAPTER 12. THE CONSTRUCTION OF MEASURES

With this lemma the next theorem which is the main result follows easily.

Theorem 12.6.4 Let f ≥ 0 be measurable and let φ be a C1 function defined on [0,∞)which satisfies φ

′ (t)> 0 for all t > 0 and φ (0) = 0. Then∫φ ( f )dµ =

∫∞

0φ′ (t)µ ([ f > t])dm.

Proof: By Theorem 11.3.9 on Page 241 there exists an increasing sequence of nonneg-ative simple functions, {sn}which converges pointwise to f . By the monotone convergencetheorem and Lemma 12.6.2,∫

φ ( f )dµ = limn→∞

∫φ (sn)dµ = lim

n→∞

∫∞

0φ′ (t)µ ([sn > t])dm

=∫

∞

0φ′ (t)µ ([ f > t])dm

This theorem can be generalized to a situation in which φ is only increasing and con-tinuous. In the generalization I will replace the symbol φ with F to coincide with earliernotation.

Lemma 12.6.5 Suppose s≥ 0 is a measurable simple function,

s(ω)≡n

∑k=1

akXEk (ω)

where the ak are the distinct nonzero values of s,a1 < a2 < · · · < an. Suppose F is anincreasing function defined on [0,∞),F (0) = 0,F being continuous at 0 from the right andcontinuous at every ak. Then letting µ be a measure and (Ω,F ,µ) a measure space,∫

(0,∞]µ ([s > t])dν =

∫Ω

F (s)dµ.

where the integral on the left is the Lebesgue integral for the measure ν given as the Radonmeasure representing the functional ∫

∞

0gdF

for g ∈Cc ([0,∞)) .

Proof: This follows from the following computation and Proposition 12.5.1. Since Fis continuous at 0 and the values ak,∫

∞

0µ ([s > t])dν (t) =

n

∑k=1

∫(ak−1,ak]

µ ([s > t])dν (t)

=n

∑k=1

∫(ak−1,ak]

n

∑j=k

µ (E j)dF (t) =n

∑j=1

µ (E j)j

∑k=1

ν ((ak−1,ak])

=n

∑j=1

µ (E j)j

∑k=1

(F (ak)−F (ak−1)) =n

∑j=1

µ (E j)F (a j)≡∫

Ω

F (s)dµ

Now here is the generalization to nonnegative measurable f .