256 CHAPTER 10. CHANGE OF VARIABLES

10.7 SeparabilityWhen dealing with a Radon measure, (complete, Borel, regular, and finite on compact sets)one can assert that the Lp spaces are separable. Recall this means that they have a countabledense subset.

Theorem 10.7.1 For p≥ 1 and µ a Radon measure, Lp(Rn,µ) is separable. Recallthis means there exists a countable set, D , such that if f ∈ Lp(Rn,µ) and ε > 0, there existsg ∈D such that ∥ f −g∥p < ε .

Proof: Let Q be all functions of the form cX[a,b) where [a,b) ≡ [a1,b1)× [a2,b2)×·· ·× [an,bn), and both ai, bi are rational, while c has rational real and imaginary parts. LetD be the set of all finite sums of functions in Q. Thus, D is countable. In fact D is dense inLp(Rn,µ). To prove this it is necessary to show that for every f ∈ Lp(Rn,µ), there existsan element of D , s such that ∥s− f∥p < ε. If it can be shown that for every g ∈ Cc (Rn)there exists h ∈ D such that ∥g−h∥p < ε , then this will suffice because if f ∈ Lp (Rn) isarbitrary, Theorem 9.4.2 implies there exists g ∈Cc (Rn) such that ∥ f −g∥p ≤ ε

2 and thenthere would exist h ∈Cc (Rn) such that ∥h−g∥p <

ε

2 . By the triangle inequality,

∥ f −h∥p ≤ ∥h−g∥p +∥g− f∥p < ε.

Therefore, assume at the outset that f ∈Cc (Rn).Let Pm consist of all sets of the form [a,b)≡∏

ni=1[ai,bi)where ai = j2−mand bi =( j+

1)2−m for j an integer. Thus Pm consists of a tiling of Rn into half open rectangles havingdiameters 2−mn

12 . There are countably many of these rectangles; so, letPm = {[ai,bi)} for

i≥ 1, and Rn = ∪∞i=1[ai,bi). Let cm

i be complex numbers with rational real and imaginaryparts satisfying

| f (ai)− cmi |< 2−m, |cm

i | ≤ | f (ai)|. (10.12)

Let sm(x) = ∑∞i=1 cm

i X[ai,bi) (x) . Since f (ai) = 0 except for finitely many values of i, theabove is a finite sum. Then 10.12 implies sm ∈ D . If sm converges uniformly to f then itfollows ∥sm− f∥p→ 0 because |sm| ≤ | f | and so

∥sm− f∥p =

(∫|sm− f |p dµ

)1/p

=

(∫spt( f )

|sm− f |p dµ

)1/p

≤ [εmn (spt( f ))]1/p

whenever m is large enough.Since f ∈ Cc (Rn) it follows that f is uniformly continuous and so given ε > 0 there

exists δ > 0 such that if |x−y|< δ , | f (x)− f (y)|< ε/2. Now let m be large enough thatevery box in Pm has diameter less than δ and also that 2−m < ε/2. Then if [ai,bi) is oneof these boxes of Pm, and x ∈ [ai,bi),

| f (x)− f (ai)|< ε/2

and| f (ai)− cm

i |< 2−m < ε/2.

Therefore, using the triangle inequality, it follows that

| f (x)− cmi |= |sm (x)− f (x)|< ε