934 CHAPTER 26. INTEGRALS AND DERIVATIVES

The rx are all bounded because

m(B(x,rx))<1α

∫B(x,rx)

| f | dm <1α|| f ||1.

By the Vitali covering theorem, there are disjoint balls B(xi,ri) such that

S⊆ ∪∞i=1B(xi,5ri)

and1

m(B(xi,ri))

∫B(xi,ri)

| f | dm > α.

Therefore

m(S) ≤∞

∑i=1

m(B(xi,5ri)) = 5n∞

∑i=1

m(B(xi,ri))

≤ 5n

α

∞

∑i=1

∫B(xi,ri)

| f | dm

≤ 5n

α

∫Rn| f | dm,

the last inequality being valid because the balls B(xi,ri) are disjoint. This proves the theo-rem.

Note that at this point it is unknown whether S is measurable. This is why m(S) and notm(S) is written.

The following is the fundamental theorem of calculus from elementary calculus.

Lemma 26.1.3 Suppose g is a continuous function. Then for all x,

limr→0

1m(B(x,r))

∫B(x,r)

g(y)dy = g(x).

Proof: Note thatg(x) =

1m(B(x,r))

∫B(x,r)

g(x)dy

and so ∣∣∣∣g(x)− 1m(B(x,r))

∫B(x,r)

g(y)dy∣∣∣∣

=

∣∣∣∣ 1m(B(x,r))

∫B(x,r)

(g(y)−g(x))dy∣∣∣∣

≤ 1m(B(x,r))

∫B(x,r)

|g(y)−g(x)|dy.

Now by continuity of g at x, there exists r > 0 such that if |x−y| < r, |g(y)−g(x)| < ε .For such r, the last expression is less than

1m(B(x,r))

∫B(x,r)

εdy < ε.

This proves the lemma.