36.3. COMPARISON WITH H n 1263

Then from the definition of the determinant and matrix multiplication,

det(A∗A) = ∑i1,··· ,in

sgn(i1, · · · , in)p

∑k1=1

ak1i1ak11

p

∑k2=1

ak2i2ak22

· · ·p

∑kn=1

akninaknn

= ∑J∈Λ(p,n)

∑{k1,··· ,kn}=J

∑i1,··· ,in

sgn(i1, · · · , in)ak1i1ak11ak2i2ak22 · · ·akninaknn

= ∑J∈Λ(p,n)

∑{k1,··· ,kn}=J

∑i1,··· ,in

sgn(i1, · · · , in)ak1i1ak2i2 · · ·aknin ·ak11ak22 · · ·aknn

= ∑J∈Λ(p,n)

∑{k1,··· ,kn}=J

sgn(k1, · · · ,kn)det(AJ)ak11ak22 · · ·aknn

= ∑J∈Λ(p,n)

det(AJ)det(AJ)

and this proves the lemma.It follows from this lemma that

Ji (u) = det(Dh(u)∗Dh(u)

)1/2.

From 36.3.10 and the area formula, the functional equals

Λ f ≡∞

∑i=1

∫giΓi

f ψ i (hi (u))Ji (u)du

=∞

∑i=1

∫Γi

f ψ i (y)dH n =∫

Γ

f (y)dH n.

Now H n is a Borel measure defined on Γ which is finite on all compact subsets of Γ. Thisfiniteness follows from the above formula. If K is a compact subset of Γ, then there existsan open set, W whose closure is compact and a continuous function with compact support,f such that K ≺ f ≺W . Then H n (K)≤

∫Γ

f (y)dH n < ∞ because of the above formula.

Lemma 36.3.2 µ = H n on every µ measurable set.

Proof: The Riesz representation theorem shows that∫Γ

f dµ =∫

Γ

f dH n

for every continuous function having compact support. Therefore, since every open set isthe countable union of compact sets, it follows µ = H n on all open sets. Since compactsets can be obtained as the countable intersection of open sets, these two measures are alsoequal on all compact sets. It follows they are also equal on all countable unions of compactsets. Suppose now that E is a µ measurable set of finite measure. Then there exist sets,