1036 CHAPTER 29. THE AREA FORMULA

so (I A0 I

)−1( I +AB 0B I

)(I A0 I

)=

(I 0B I +BA

)which shows that the two matrices(

I +AB 0B I

),

(I 0B I +BA

)are similar and so they have the same determinant. Thus

det(I +AB) = det(I +BA)

Note that the two matrices are different sizes.

Corollary 29.9.2 Let A be an m×n real matrix. Then

det(I +AA∗) = det(I +A∗A) .

It is convenient to define the following [47] for a measure space (Ω,S ,µ) and f : Ω→[0,∞], an arbitrary function, maybe not measurable.∫ ∗

f dµ ≡∫ ∗

Ω

f dµ ≡ inf{∫

Ω

gdµ : g≥ f , and g measurable}

This is just like an outer measure. It resembles an old idea found in Hobson [66] called ageneralized Stieltjes integral.

Lemma 29.9.3 Suppose fn ≥ 0 and limsupn→∞

∫ ∗ fndµ = 0. Then there is a subsequencefnk such that fnk (ω)→ 0 a.e. ω.

Proof: For n large enough,∫ ∗ fndµ < ∞. Let n be this large and pick gn ≥ fn, gn

measurable, such that ∫ ∗fndµ +n−1 >

∫gndµ.

Thuslimsup

∫gndµ = liminf

∫gndµ = lim

∫gndµ = 0.

If n = 1,2, · · · , let kn > max(kn−1,n) be such that∫

gkndµ < 2−n. Thus

µ([gkn ≥ n−1]

)≤ 2−nn and

∞

∑n=1

µ([gkn ≥ n−1]

)< ∞

so for all N,

µ(∩∞

n=1∪m≥n [gkm ≥ m−1])≤

∞

∑n=N

µ([gkm ≥ m−1]

)≤

∞

∑n=N

n2−n.