Kenneth Kuttler

376 CHAPTER 13. LEBESGUE MEASURE

which contradicts the above inequality.Therefore, ĝ(y) ̸= a for any y ∈ Σ. So consider ĝ(Φ(y)) for y ∈ f(B).

≤ r10

+ |g(Φ(y))−g(f(a))− (g(y)−g(f(a)))| (13.12.30)

This last equals 0 if |y− c| ≥ 2ε . On the other hand, if |y− c|< 2ε,

y ∈ B(f(a) ,2ε) ,Φ(y) ∈ ∂B(c,2ε)

so both y,Φ(y) are in B(f(a) ,4ε) and so this last term in 13.12.30 is no larger than|g(Φ(y)−g(f(a)))|+ |g(y)−g(f(a))|< r

10 +r

10 and so for all y ∈ f(B) ,

|ĝ(Φ(y))−g(y)| ≤ 3r10

Now note that for x ∈ B, from what was just shown,

|ĝ(Φ(f(x)))−x|= |ĝ(Φ(f(x)))−g(f(x))| ≤ 3r10

It follows that for every x ∈ B,a+x− ĝ(Φ(f(x))) ∈ B and so by the Brouwer fixed pointtheorem, there is a fixed point x and hence

a+x− ĝ(Φ(f(x))) = x

so ĝ(Φ(f(x))) = a contrary to what was just shown that there is no solution to ĝ(y) = afor y ∈ Σ.

With the lemma, it is easy to prove the invariance of domain theorem which is as fol-lows.

Theorem 13.12.3 Let U be an open set in Rn and let f : U → f(U) ⊆ Rn. Then f(U) isalso an open set in Rn.

Proof: For a ∈U, let a ∈ Ba ⊆U, where Ba is a closed ball centered at a. Then fromLemma 13.12.2, f(a) ∈ Vf(a) an open subset of f(Ba) . Hence f(U) = ∪a∈UVf(a) which isopen.

13.13 Besicovitch Covering TheoremThe Besicovitch covering theorem is one of the most amazing and insightful ideas that Ihave ever encountered. It is simultaneously elegant, elementary and profound. The nextsection is an attempt to present this wonderful result.

When dealing with probability distribution functions or some other Radon measure,it is necessary to have a better covering theorem than the Vitali covering theorem whichworks well for Lebesgue measure. However, for a Radon measure, if you enlarge the ballby making the radius larger, you don’t know what happens to the measure of the enlarged