Kenneth Kuttler

532 CHAPTER 20. PRODUCT MEASURES

12. Give a different proof of the above as follows. First suppose f is a simple function,f (ω) = ∑

nk=1 akXEk (ω) where the ak are strictly increasing, φ (a0) = a0 ≡ 0. Then

explain carefully the steps to the following argument.∫Ω

φ ◦ f dµ =n

∑i=1

∫φ(ai)

φ(ai−1)µ ([φ ◦ f > t])dt =

∑i=1

∫φ(ai)

φ(ai−1)

∑k=i

µ (Ek)dt

∑i=1

∑k=i

µ (Ek)∫ ai

ai−1

φ′ (t)dt =

∑i=1

∫ ai

ai−1

φ′ (t)

∑k=i

µ (Ek)dt

∑i=1

∫ ai

ai−1

φ′ (t)µ ([ f > t])dt =

∫∞

0φ′ (t)µ ([ f > t])dt

Note that this did not require the measure space to be σ finite and comes directlyfrom the definition of the integral.

13. Give another argument for the above result as follows.∫φ ◦ f dµ =

∫∞

0µ ([φ ◦ f > t])dt =

∫∞

0µ([

f > φ−1 (t)

])dt

and now change the variable in the last integral, letting φ (s) = t. Justify the easymanipulations.

532 CHAPTER 20. PRODUCT MEASURES12. Give a different proof of the above as follows. First suppose f is a simple function,f (@) =Lp_) ax Xe, (@) where the a; are strictly increasing, @ (ao) = a9 = 0. Thenexplain carefully the steps to the following argument.[scree = Ef" wivor>aar=¥ [0 Sacto= yyu Sh eind=¥ [" OY nevai=lk=i aj-| k=i= rife L([f >t))dt= [eon ([f >t])dtNote that this did not require the measure space to be o finite and comes directlyfrom the definition of the integral.13. Give another argument for the above result as follows.Jee fdp = [we of >t))dt= [Hl o oarand now change the variable in the last integral, letting @(s) = +. Justify the easymanipulations.