29.6. MAPPINGS THAT ARE NOT ONE TO ONE 1021

29.6 Mappings that are not One to OneLet h : Rn → Rm be Lipschitz. We drop the requirement that h be one to one. Again, letA be the set on which Dh(x) exists. Let k be as used earlier in Theorem 29.4.3. ThusJk(x) ̸= 0 for all x ∈ A the set where Dh(x) exists. Thus there is a sequence of disjointBorel sets {Ek} whose union is A such that k is Lipschitz on Ek. Let S be given by

S≡{

x ∈ A, such that U (x)−1 does not exist}

Then S is a Borel set and so letting Sk j ≡ S∩Ek ∩B(0, j) , the change of variables formulaabove implies

H n (h(Sk j))≤H n (k(Sk j

))=∫

k(Sk j)dH n =

∫AXSk j (x)J∗k(x)dmn ≤ δmn

(Sk j)

where k is chosen with ε small enough that J∗k(x)< δ . δ is arbitrary, so H n(h(Sk j))

= 0and so H n (h(S∩Ek)) = 0. Consequently H n (h(S)) = 0. This is stated as the followinglemma. Note how this includes the earlier Sard’s theorem.

Lemma 29.6.1 For S defined above, H n (h(S)) = 0.

Thus mn (N) = 0 where N is the set where Dh(x) does not exist. Then by Lemma29.1.2

H n (h(S∪N))≤H n (h(S))+H n (h(N)) = 0. (29.6.25)

Let B≡ Rn \ (S∪N).Recall Lemma 29.4.1 above which said that for each x ∈ A+ the set where U (x) is

invertible there is a Borel set F containing x on which h is one to one. In fact it was oneof countably many sets of the form E (T,c, i) . By enumerating these sets as done earlier,referring to them as Ek, one can let F1 ≡ E1, and if F1, · · · ,Fn have been chosen, Fn+1 ≡En+1 \∪n

i=1Fi to obtain the result of the following lemma.

Lemma 29.6.2 There exists a sequence of disjoint measurable sets, {Fi}, such that

∪∞i=1Fi = B⊆ A+

and h is one to one on Fi.

The following corollary will not be needed right away but it is of interest. Recall thatA is the set where h is differentiable and A+ is the set where det

(Dh(x)∗Dh(x)

)> 0. Part

of Lemma 29.4.1 is reviewed in the following corollary.

Corollary 29.6.3 For each Fi in Lemma 29.6.2, h−1 is Lipschitz on h(Fi).

Now let g : h(Rn)→ [0,∞] be H n measurable. By Theorem 29.5.3,∫h(A)

Xh(Fi) (y)g(y)dH n =∫

Fi

g(h(x))J∗ (x)dm. (29.6.26)

29.6. MAPPINGS THAT ARE NOT ONE TO ONE 102129.6 Mappings that are not One to OneLet h: R” > R” be Lipschitz. We drop the requirement that h be one to one. Again, letA be the set on which Dh(x) exists. Let k be as used earlier in Theorem 29.4.3. ThusJk (x) # 0 for all x € A the set where Dh(x) exists. Thus there is a sequence of disjointBorel sets {E,} whose union is A such that k is Lipschitz on E;. Let S be given byS= {x € A, such that U(x)! does not exist}Then S is a Borel set and so letting Sx; = SNE; 1B (0, j) , the change of variables formulaabove impliesKH" (h(Sxj)) < 4” (K (Skj)) = his. dH” = | Koj (x) Sek (x) dit < Srp (Sx;)where k is chosen with € small enough that J..k (x) < 6. 6 is arbitrary, so #” (hh (S;;)) =0and so .#” (h(SME;)) =0. Consequently #7” (h(S)) =0. This is stated as the followinglemma. Note how this includes the earlier Sard’s theorem.Lemma 29.6.1 For S defined above, 7" (h(S)) = 0.Thus m,(N) = 0 where N is the set where Dh(x) does not exist. Then by Lemma29.1.2KH" (h(SUN)) < 4” (h(S)) +4" (h(N)) =0. (29.6.25)Let B=R"\ (SUN).Recall Lemma 29.4.1 above which said that for each x € A* the set where U (x) isinvertible there is a Borel set F containing x on which h is one to one. In fact it was oneof countably many sets of the form E (T,c,i). By enumerating these sets as done earlier,referring to them as E;, one can let F, = FE), and if F,,--- ,F, have been chosen, F,41 =En+1 \U#_,F; to obtain the result of the following lemma.Lemma 29.6.2 There exists a sequence of disjoint measurable sets, {F;}, such thatUe) =BCAtTand h is one to one on F,.The following corollary will not be needed right away but it is of interest. Recall thatA is the set where h is differentiable and A* is the set where det (Dh (x)* Dh (x)) > 0. Partof Lemma 29.4.1 is reviewed in the following corollary.Corollary 29.6.3 For each F; in Lemma 29.6.2, h~! is Lipschitz on h(F;).Now let g : h(R”) — [0,->] be #” measurable. By Theorem 29.5.3,[ uw (VB (yd Hl" = / a (h(x) Js (x) dm. (29.6.26)Jh(A) SF;