17.3. THE AREA FORMULA 457

Note that it is not clear whether c ∈ E(T,c, i) because of the above two requirements.What is going on here is that we are looking for b such that Dh(b) is sufficiently close toone of those T which also are in a piece of B. Thus we start with one of those T and oneof those points c and look for all b, if any, which do the right things. In one dimension, theT and Dh(b) would just be slopes. There are countably many of these pieces of B beingdenoted as E (T,c, i).

The union of these E (T,c, i) is all of B because if b ∈ B,

|h(a)−h(b)−Dh(b)(a−b)|< ε |U (b)(a−b)| (17.5)

whenever a ∈ B(b, 2

i

)provided i is sufficiently large. Thus also, by Lemma 17.2.1, there

is T ∈S such that the above holds for U (b) replaced with T and a ∈ B(b, 2

i

)and also

17.3, 17.4. Thus b ∈ E (T,c, i), so indeed the union of these sets is B.Now let a,b ∈ E (T,c, i) . Since a,b ∈ E (T,c, i) , a,b are within 1/i of c and so a is

within 2/i of b and so 17.2 holds because of the definition of E (T,c, i). Therefore, from17.2 and the inequalities which follow, 17.3 and 17.4,

(1−3ε) |T (a−b)| ≤ |h(a)−h(b)| ≤ (1+3ε) |T (a−b)| (17.6)

Indeed, from 17.5, 17.4,

|h(a)−h(b)| < |U (b)(a−b)|+ ε |U (b)(a−b)|= (1+ ε) |U (b)(a−b)|< (1+ ε)2 |T (a−b)|< (1+3ε) |T (a−b)|

The other side of 17.6 is similar.There are countably many of these E(T,c, i) each being a Borel set. Therefore, B is a

disjoint union of these sets called {Ek} where I will denote the special T as Tk correspond-ing to Ek. Thus from 17.6 and the definition of Hausdorff measure, it follows that

H n (h(Ek)) ∈ [(1−3ε)H n (TkEk) ,(1+3ε)H n (TkEk)]

= [(1−3ε)mn (TkEk) ,(1+3ε)mn (TkEk)] (17.7)

From 17.3 and 17.4 and b ∈ Ek,

U (b)(B(0,1))⊆ (1+ ε)Tk (B(0,1)) , U (b)(B(0,1))⊇ (1− ε)Tk (B(0,1))

which implies on taking the Lebesgue measure that

(1− ε)n |det(Tk)| ≤ det(U (b))≤ (1+ ε)n |det(Tk)|

Therefore, from 17.7,

H n (h(Ek)) ∈ [(1−3ε) |det(Tk)|mn (Ek) ,(1+3ε) |det(Tk)|mn (Ek)]

=

[∫Ek

(1−3ε) |det(Tk)|dmn,∫

Ek

(1+3ε) |det(Tk)|dmn

]⊆[∫

Ek

(1−3ε)

∣∣∣∣det(U (x))

(1+ ε)n

∣∣∣∣dmn (x) ,∫

Ek

(1+3ε)

∣∣∣∣det(U (x))

(1− ε)n

∣∣∣∣dmn (x)]

(17.8)

Note that this does not assume h is one to one on B.