Kenneth Kuttler

12.11. COMPLETION OF MEASURES 325

is measurable with respect to S , it follows that x→∫

Y f (x,y)dλ is also measurable withrespect to S . A similar conclusion can be drawn about y→

∫X f (x,y)dµ . Thus the two

iterated integrals make sense. Since 12.10.50 holds for fn, another application of the Mono-tone Convergence theorem shows 12.10.50 holds for f . This proves the theorem.

Corollary 12.10.15 Let f : X×Y → C be S ×F measurable. Suppose either∫X

∫Y| f |dλdµ or

∫Y

∫X| f |dµdλ < ∞

Then f ∈ L1(X×Y,µ×λ ) and∫X×Y

f d(µ×λ ) =∫

∫Y

f dλdµ =∫

∫X

f dµdλ (12.10.51)

with all integrals making sense.

Proof: Suppose first that f is real valued. Apply Theorem 12.10.14 to f+and f−. Then12.10.51 follows from observing that f = f+− f−; and that all integrals are finite. If f iscomplex valued, consider real and imaginary parts. This proves the corollary.

Suppose f is product measurable. From the above discussion, and breaking f down intoa sum of positive and negative parts of real and imaginary parts and then using Theorem11.3.9 on Page 241 on approximation by simple functions, it follows that whenever f isS ×F measurable, x→ f (x,y) is µ measurable, y→ f (x,y) is λ measurable.

12.11 Completion Of MeasuresSuppose (Ω,F ,µ) is a measure space. Then it is always possible to enlarge the σ algebraand define a new measure µ on this larger σ algebra such that

(Ω,F ,µ

)is a complete

measure space. Recall this means that if N ⊆ N′ ∈F and µ (N′) = 0, then N ∈F . Thefollowing theorem is the main result. The new measure space is called the completion ofthe measure space.

Theorem 12.11.1 Let (Ω,F ,µ) be a σ finite measure space. Then there exists a uniquemeasure space,

(Ω,F ,µ

)satisfying

1.(Ω,F ,µ

)is a complete measure space.

2. µ = µ on F

3. F ⊇F

4. For every E ∈F there exists G ∈F such that G⊇ E and µ (G) = µ (E) .

5. For every E ∈F there exists F ∈F such that F ⊆ E and µ (F) = µ (E) .

Also for every E ∈F there exist sets G,F ∈F such that G⊇ E ⊇ F and

µ (G\F) = µ (G\F) = 0 (12.11.52)

12.11. COMPLETION OF MEASURES 325is measurable with respect to .Y, it follows that x + fy f(x,y)d/ is also measurable withrespect to .. A similar conclusion can be drawn about y > fy f(x,y)du. Thus the twoiterated integrals make sense. Since 12.10.50 holds for f,,, another application of the Mono-tone Convergence theorem shows 12.10.50 holds for f. This proves the theorem.Corollary 12.10.15 Let f :X x Y > C be Y x F measurable. Suppose eitherJ [ \plaaan or [| \plduaa <0Then f € L'(X x Y,u x A) andfd(u x2) =| [ farau = | [fanaa (12.10.51)XxYwith all integrals making sense.Proof: Suppose first that f is real valued. Apply Theorem 12.10.14 to ft and f~. Then12.10.51 follows from observing that f = f* — f~; and that all integrals are finite. If f iscomplex valued, consider real and imaginary parts. This proves the corollary.Suppose f is product measurable. From the above discussion, and breaking f down intoa sum of positive and negative parts of real and imaginary parts and then using Theorem11.3.9 on Page 241 on approximation by simple functions, it follows that whenever f isSY x F measurable, x > f (x,y) is measurable, y > f (x,y) is A measurable.12.11 Completion Of MeasuresSuppose (Q,.¥, 1) is a measure space. Then it is always possible to enlarge the o algebraand define a new measure Hf on this larger o algebra such that (Q, Ff) is a completemeasure space. Recall this means that if N C N’ € ¥ and f{(N’) =0, then N € ¥. Thefollowing theorem is the main result. The new measure space is called the completion ofthe measure space.Theorem 12.11.1 Let (Q,4,) be a o finite measure space. Then there exists a uniquemeasure space, (Q oF ,L) satisfying1, (Q Q,F .H) is a complete measure space.L234. For every E € F there exists G € F such that GD E and u(G) =I (E).5. For every E € F there exists F € F such that F CE and u(F) = (E).Also for every E € F there exist sets G,F € ¥ such that GD E D F andu(G\F)=1(G\F)=0 (12.11.52)