266 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

9. Let { fn} be a sequence of real or complex valued measurable functions. Let

S = {ω : { fn(ω)} converges}.

Show S is measurable. Hint: You might try to exhibit the set where fn convergesin terms of countable unions and intersections using the definition of a Cauchy se-quence.

10. Let (Ω,S ,µ) be a measure space and let f be a nonnegative measurable functiondefined on Ω. Also let φ : [0,∞)→ [0,∞) be strictly increasing and have a continuousderivative and φ (0) = 0. Suppose f is bounded and that 0≤ φ ( f (ω))≤M for somenumber, M. Show that ∫

Ω

φ ( f )dµ =∫

∞

0φ′ (s)µ ([s < f ])ds,

where the integral on the right is the ordinary improper Riemann integral. Hint:First note that s→ φ

′ (s)µ ([s < f ]) is Riemann integrable because φ′ is continuous

and s→ µ ([s < f ]) is a nonincreasing function, hence Riemann integrable. From thesecond description of the Lebesgue integral and the assumption that φ ( f (ω))≤M,argue that for [M/h] the greatest integer less than M/h,

∫Ω

φ ( f )dµ = suph>0

[M/h]

∑i=1

hµ ([ih < φ ( f )])

= suph>0

[M/h]

∑i=1

hµ([

φ−1 (ih)< f

])= sup

h>0

[M/h]

∑i=1

h∆i

∆iµ([

φ−1 (ih)< f

])where ∆i =

(φ−1 (ih)−φ

−1 ((i−1)h)). Now use the mean value theorem to write

∆i =(φ−1)′ (ti)h

=1

φ′ (

φ−1 (ti)

)h

for some ti between (i−1)h and ih. Therefore, the right side is of the form

suph

[M/h]

∑i=1

φ′ (

φ−1 (ti)

)∆iµ

([φ−1 (ih)< f

])where φ

−1 (ti) ∈(φ−1 ((i−1)h) ,φ−1 (ih)

). Argue that if ti were replaced with ih,

this would be a Riemann sum for the Riemann integral∫φ−1(M)

0φ′ (t)µ ([t < f ])dt =

∫∞

0φ′ (t)µ ([t < f ])dt.