10.3. CHANGE OF VARIABLES, LINEAR MAPS 241

= limr→0

1mp (B(x,r))

∫B(x,r)

∣∣ f XB(0,k) (y)− f XB(0,k) (x)∣∣dmp (y) = 0.

The last claim holds because∣∣∣∣ f (x)− 1mp (B(x,r))

∫B(x,r)

f (y)dmp (y)∣∣∣∣

≤ 1mp (B(x,r))

∫B(x,r)

| f (y)− f (x)|dmp (y) ■

Definition 10.2.4 Let E be a measurable set. Then x ∈ E is called a point ofdensity if

limr→0

mp (B(x,r)∩E)mp (B(x,r))

= 1

Proposition 10.2.5 Let E be a measurable set. Then mp a.e. x∈E is a point of density.

Proof: This follows from letting f (x) = XE (x) in Corollary 10.2.3. ■

10.3 Change of Variables, Linear MapsThis is about changing the variables for linear maps where Fp denotes the Lebesgue mea-surable sets.

Theorem 10.3.1 In case h : Rp → Rp is Lipschitz, satisfying the Lipschitz condi-tion ∥h(x)−h(y)∥ ≤ K ∥x−y∥ , then if T is a set for which mp(T ) = 0, it follows thatmp (h(T )) = 0. Also if E ∈Fp, then h(E) ∈Fp.

Proof: By the Lipschitz condition, ∥h(x+v)−h(x)∥ ≤ K ∥v∥ and you can simplylet T ⊆V where mp (V )< ε/(K p5p) . Then there is a countable disjoint sequence of balls{Bi} such that

{B̂i}

covers T and each ball Bi is contained in V each having radius no morethan 1. Then the Lipschitz condition implies h

(B̂i)⊆ B(h(xi) ,5K) and so

m̄p (h(T ))≤∞

∑i=1

mp(h(B̂i))≤ 5pK p

∑i=1

mp (Bi)≤ K p5pmp (V )< ε

Since ε is arbitrary, this shows that h(T ) is measurable and mp (h(T )) = 0.Now let E ∈Fp, mp (E)< ∞. Then by of the measure and Theorem 8.7.4, there exists

F which is the countable union of compact sets such that E = F ∪N where N is a set ofmeasure zero. Then from the first part, h(E \F) ⊆ h(N) and this set on the right hasmeasure zero and so by completeness of the measure, h(E \F) ∈ Fp and so h(E) =h(E \F)∪h(F) ∈ Fp because F = ∪kKk, each Kk compact. Hence h(F) = ∪kh(Kk)which is the countable union of compact sets, a Borel set, due to the continuity of h. Forarbitrary E, h(E) = ∪∞

k=1h(E ∩B(0,k)) ∈Fp. ■Of course an example of a Lipschitz map is a linear map. ∥Ax−Ay∥= ∥A(x−y)∥ ≤

∥A∥∥x−y∥ . Therefore, if A is linear and E is Lebesgue measurable, then A(E) is alsoLebesgue measurable. This is convenient.

Lemma 10.3.2 Every open set U in Rp is a countable disjoint union of half open boxesof the form Q≡∏

pi=1[ai,ai +2−k) where ai = l2−k for l some integer.

10.3. CHANGE OF VARIABLES, LINEAR MAPS 2411= lim m, (B(a,r)) bee |f 25(0,0) (y) —fZ2B00,%) (x)| dm, (y) = 0.The last claim holds because0) ay |, f wamolo)Mp (B(ax,r)).Be i r) f(y) —f(#)|dm, (y)<Definition 10.2.4 Let E be a measurable set. Then x € E is called a point ofdensity ifB Eham 2? (x,r)M ) =|r+0 Mp (B(a,r))Proposition 10.2.5 Let E be a measurable set. Then mp a.e. x € E is a point of density.Proof: This follows from letting f (a) = 2x (x) in Corollary 10.2.3. HH10.3 Change of Variables, Linear MapsThis is about changing the variables for linear maps where .¥,, denotes the Lebesgue mea-surable sets.Theorem 10.3.1 in case h: R? - R? is Lipschitz, satisfying the Lipschitz condi-tion ||h(a) —h(y)|| < K || —y||, then if T is a set for which m,(T) = 0, it follows thatmp (h(T)) =0. Also if E € Fp, then h(E) € Fy.Proof: By the Lipschitz condition, ||h (a+ v)—h(a)|| < K||v|| and you can simplylet T C V where my (V) < €/(K?5?). Then there is a countable disjoint sequence of balls{B;} such that {B;} covers T and each ball B; is contained in V each having radius no morethan 1. Then the Lipschitz condition implies h (B;) C B(h (aj) ,5K) and so8iip (A(T) < Ymp (Ie (Bi) <5?K? Y- mp (Bi) < KPS?my (V) <ei=l i=1Since € is arbitrary, this shows that h (7) is measurable and m, (h(T)) = 0.Now let E € F,, mp (E) < o. Then by of the measure and Theorem 8.7.4, there existsF which is the countable union of compact sets such that FE = F UN where N is a set ofmeasure zero. Then from the first part, h(E \F) C h(N) and this set on the right hasmeasure zero and so by completeness of the measure, h(E\F) € Fp, and so h(£) =h(E\F)UR(F) € Fy because F = UgKx, each Ky compact. Hence h(F) = Ugh (Kx)which is the countable union of compact sets, a Borel set, due to the continuity of h. Forarbitrary E, h(E) = Up_,h(ENB(0,k)) € Fy.Of course an example of a Lipschitz map is a linear map. ||Ax — Ay]|| = ||A (a — y)|| <\|A|| |la — y||. Therefore, if A is linear and E is Lebesgue measurable, then A (E) is alsoLebesgue measurable. This is convenient.Lemma 10.3.2 Every open set U in R? is a countable disjoint union of half open boxesof the form Q = T]_, [a;,ai + 2-*) where a; = 12~* for | some integer.