1092 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

and so µ (A) = 0 and A is µ measurable. It follows µ (N (a,b)) = 0 because

µ (N (a,b))≤∞

∑m=1

µ (N (a,b)∩B(0,m)) = 0.

DefineN0 ≡

{x /∈ Z : Dµ λ (x)> Dµ λ (x)

}.

Thus µ (N0) = 0 because

N0 ⊆ ∪{N (a,b) : 0≤ a < b, and a,b ∈Q}

Therefore, N0 is also µ measurable and has µ measure zero. Letting N ≡ N0 ∪Z, it fol-lows Dµ λ (x) exists on NC. We can assume also that N is a Gδ set. It remains to verifyXNC (·)Dµ λ (·) is finite a.e. and is µ measurable.

LetI =

{x : Dµ λ (x) = ∞

}.

Then by Lemma 31.3.2

λ (I∩B(0,m))≥ aµ (I∩B(0,m))

for all a and since λ is finite on bounded sets, the above implies µ (I∩B(0,m)) = 0 foreach m which implies that I is µ measurable and has µ measure zero since

I = ∪∞m=1I∩B(0,m) .

Now the issue is measurability. Let λ be an arbitrary Radon measure. I need show thatx→ λ (B(x,r)) is measurable. Here is where it is convenient to have the balls be closedballs. If V is an open set containing B(x,r) , then for y close enough to x,B(y,r)⊆V alsoand so,

lim supy→x

λ (B(y,r))≤ λ (V )

However, since V is arbitrary and λ is outer regular or observing that B(x,r) the closedball is the intersection of nested open sets, it follows that

lim supy→x

λ (B(y,r))≤ λ (B(x,r))

Thus x→ λ (B(x,r)) is upper semicontinuous and so,

x→λ (B(x,r))µ (B(x,r))

is measurable. Hence XNC (x)Dµ (λ )(x) = limri→0 XNC (x) λ (B(x,r))µ(B(x,r)) is also measurable.

Typically I will write Dµ λ (x) rather than the more precise XNC (x)Dµ λ (x) since thevalues on the set of measure zero N are not important due to the completeness of themeasure µ .