1246 CHAPTER 35. WEAK DERIVATIVES

≤∞

∑i=1

mn

(h(

B̂i

))

≤∞

∑i=1

α (n)(Lip(h))n 5nrni = 5n (Lip(h))n

∞

∑i=1

mn (Bi)

≤ (Lip(h))n 5nmn (V )≤ ε (Lip(h))n 5n.

Since ε is arbitrary, this proves the lemma.With the conclusion of this lemma, the next lemma is fairly easy to obtain.

Lemma 35.6.8 If A is Lebesgue measurable, then h(A) is mn measurable. Furthermore,

mn (h(A))≤ (Lip(h))n mn (A). (35.6.9)

Proof: Let Ak = A∩B(0,k) ,k ∈N. Let V ⊇ Ak and let mn (V )< ∞. By Lemma 35.6.6,there is a sequence of disjoint balls {Bi} and a set of measure 0, T , such that

V = ∪∞i=1Bi∪T, Bi = B(xi,ri).

By Lemma 35.6.7,mn (h(Ak))≤ mn (h(V ))

≤ mn (h(∪∞i=1Bi))+mn (h(T )) = mn (h(∪∞

i=1Bi))

≤∞

∑i=1

mn (h(Bi))≤∞

∑i=1

mn (B(h(xi) ,Lip(h)ri))

≤∞

∑i=1

α (n)(Lip(h)ri)n = Lip(h)n

∞

∑i=1

mn (Bi) = Lip(h)n mn (V ).

Therefore,mn (h(Ak))≤ Lip(h)n mn (V ).

Since V is an arbitrary open set containing Ak, it follows from regularity of Lebesguemeasure that

mn (h(Ak))≤ Lip(h)n mn (Ak). (35.6.10)

Now let k → ∞ to obtain 35.6.9. This proves the formula. It remains to show h(A) ismeasurable.

By inner regularity of Lebesgue measure, there exists a set, F , which is the countableunion of compact sets and a set T with mn (T ) = 0 such that

F ∪T = Ak.

Then h(F) ⊆ h(Ak) ⊆ h(F)∪ h(T ). By continuity of h, h(F) is a countable union ofcompact sets and so it is Borel. By 35.6.10 with T in place of Ak,

mn (h(T )) = 0