224 CHAPTER 9. THE LEBESGUE INTEGRAL

Corollary 9.12.11 Suppose f is measurable with respect to ∏pi=1 Ei and suppose for

some permutation, (i1, · · · , ip) ,∫· · ·∫| f |dµ i1 · · ·dµ ip

< ∞. Then f ∈ L1(∏

pi=1 Xi

).

Proof: By Theorem 9.12.9,∫Rp | f |dλ =

∫· · ·∫| f (x)|dµ i1 · · ·dµ ip

< ∞ and so f is inL1 (Rp). ■

You can of course consider the completion of a product measure by using the outermeasure approach described earlier. This could be used to get p dimensional Lebesguemeasure.

9.13 The Brouwer Fixed Point TheoremI found this proof of the Brouwer fixed point theorem in Evans [16] and Dunford andSchwartz [14]. The main idea which makes proofs like this work is Lemma 6.11.2 whichis stated next for convenience.

Lemma 9.13.1 Let g : U →Rp be C2 where U is an open subset of Rp. Then it followsthat ∑

pj=1 cof(Dg)i j, j = 0,where here (Dg)i j ≡ gi, j ≡ ∂gi

∂x j. Also, cof(Dg)i j =

∂ det(Dg)∂gi, j

.

Definition 9.13.2 Let h be a function defined on an open set, U ⊆ Rp. Then h ∈Ck(U)

if there exists a function g defined on an open set, W containng U such that g = h

on U and g is Ck (W ) .

Lemma 9.13.3 There does not exist h ∈ C2(

B(0,R))

with h : B(0,R)→ ∂B(0,R)

which has the property that h(x) =x for all x∈ ∂B(0,R)≡{x : |x|= R} Such a functionis called a retract.

Proof: First note that if h is such a retract, then for all x ∈ B(0,R), det(Dh(x)) = 0.This is because if det(Dh(x)) ̸= 0 for some such x, then by the inverse function theorem,h(B(x,δ )) is an open set for small enough δ but this would require that this open set isa subset of ∂B(0,R) which is impossible because no open ball is contained in ∂B(0,R).Here and below, let BR denote B(0,R).

Now suppose such an h exists. Let λ ∈ [0,1] and let pλ (x)≡ x+λ (h(x)−x) . Thisfunction, pλ is a homotopy of the identity map and the retract h. Define the function I (λ )by I (λ )≡

∫B(0,R) det(Dpλ (x))dx. Then using the dominated convergence theorem,

I′ (λ ) =∫

B(0,R)∑i. j

∂ det(Dpλ (x))

∂ pλ i, j

∂ pλ i j (x)

∂λdx

=∫

B(0,R)∑

i∑

j

∂ det(Dpλ (x))

∂ pλ i, j(hi (x)− xi), j dx

=∫

B(0,R)∑

i∑

jcof(Dpλ (x))i j (hi (x)− xi), j dx

Now by assumption, hi (x) = xi on ∂B(0,R) and so one can integrate by parts, in theiterated integrals used to compute

∫B(0,R) and write

I′ (λ ) =−∑i

∫B(0,R)

∑j

cof(Dpλ (x))i j, j (hi (x)− xi)dx = 0.