352 CHAPTER 13. LEBESGUE MEASURE

≤∞

∑i=1

mn

(h(

B̂i

))≤

∞

∑i=1

mn (B(h(xi) ,2krxi))

=∞

∑i=1

mn (B(xi,2krxi)) = (2k)n∞

∑i=1

mn (B(xi,rxi))

≤ (2k)n mn (V )≤ (2k)nε.

Since ε > 0 is arbitrary, this shows mn (h(Tk)) = 0. Now

mn (h(T )) = limk→∞

mn (h(Tk)) = 0.

This proves the lemma.

Lemma 13.5.2 Let h satisfy 13.5.11. If S is a Lebesgue measurable subset of U, thenh(S) is Lebesgue measurable.

Proof: Let Sk = S∩B(0,k) ,k ∈ N. By inner regularity of Lebesgue measure, thereexists a set, F , which is the countable union of compact sets and a set T with mn (T ) = 0such that

F ∪T = Sk.

Then h(F) ⊆ h(Sk) ⊆ h(F)∪ h(T ). By continuity of h, h(F) is a countable union ofcompact sets and so it is Borel. By Lemma 13.5.1, mn (h(T )) = 0 and so h(Sk) is Lebesguemeasurable because of completeness of Lebesgue measure. Now h(S) = ∪∞

k=1h(Sk) andso it is also true that h(S) is Lebesgue measurable. This proves the lemma.

In particular, this proves the following corollary.

Corollary 13.5.3 Suppose A is an n×n matrix. Then if S is a Lebesgue measurable set, itfollows AS is also a Lebesgue measurable set.

Lemma 13.5.4 Let R be unitary (R∗R = RR∗ = I) and let V be a an open or closed set.Then mn (RV ) = mn (V ) .

Proof: First assume V is a bounded open set. By Corollary 13.4.6 there is a disjointsequence of closed balls, {Bi} such that V =∪∞

i=1Bi∪N where mn (N) = 0. Denote by xi thecenter of Bi and let ri be the radius of Bi. Then by Lemma 13.5.1 mn (RV ) = ∑

∞i=1 mn (RBi) .

Now by invariance of translation of Lebesgue measure, this equals ∑∞i=1 mn (RBi−Rxi) =

∑∞i=1 mn (RB(0,ri)) . Since R is unitary, it preserves all distances and so RB(0,ri) = B(0,ri)

and therefore,

mn (RV ) =∞

∑i=1

mn (B(0,ri)) =∞

∑i=1

mn (Bi) = mn (V ) .

This proves the lemma in the case that V is bounded. Suppose now that V is just an openset. Let Vk =V ∩B(0,k) . Then mn (RVk) = mn (Vk) . Letting k→ ∞, this yields the desiredconclusion. This proves the lemma in the case that V is open.