26.4. LIPSCHITZ FUNCTIONS 947

Let V ⊆ (a,b) be an open set containing Npq such that m(V )<m(Npq)+ε . By assumption,if x ∈ Npq, there exist arbitrarily small h such that

f (x+h)− f (x)h

< p,

These intervals [x,x+h] are then a Vitali covering of Npq. It follows from Corollary 13.4.6that there is a disjoint union of countably many, {[xi,xi +hi]}∞

i=1 which cover all of Npqexcept for a set of measure zero. Thus also the open intervals {(xi,xi +hi)}∞

i=1 also coverall of Npq except for a set of measure zero. Now for points x′ of Npq so covered, there arearbitrarily small h such that

f (x′+h′)− f (x′)h′

> q

and [x′,x′+h′] is contained in one of these original open intervals (xi,xi +hi). By the Vitalicovering theorem again, Corollary 13.4.6, it follows that there exists a countable disjointsequence

{[x′j,x

′j +h′j

]}∞

j=1which covers all of Npq except for a set of measure zero, each

of these[x′j,x

′j +h′j

]being contained in some (xi,xi +hi) . Then it follows that

qm(Npq) ≤ q∑j

h′j ≤∑j

f(x′j +h′j

)− f

(x′j)≤∑

if (xi +hi)− f (xi)

≤ p∑i

hi ≤ pm(V )≤ p(m(Npq)+ ε)

Since ε > 0 is arbitrary, this shows that qm(Npq) ≤ pm(Npq) and so m(Npq) = 0. Nowtaking the union of all Npq for p,q ∈Q, it follows that for a.e. x,D+ f (x) = D+ f (x) and sothe derivative from the right exists. Similar reasoning shows that off a set of measure zerothe derivative from the left also exists. You just do the same argument using D− f (x) andD− f (x) to obtain the existence of a derivative from the left. Next you can use the sameargument to verify that D− f (x) = D+ f (x) off a set of measure zero. This is outlined next.Define a new Npq,

Npq ≡{

x ∈ (a,b) : D+ f (x)> q > p > D− f (x)}

Let V be an open set containing Npq such that m(V )< m(Npq)+ ε. For each x ∈ Npq thereare arbitrarily small h such that

f (x)− f (x−h)h

< p

Then as before, there is a countable disjoint sequence of closed intervals contained inV,{[xi−hi,xi]}∞

i=1 such that their union includes all of Npq except a set of measure zero.Thus this is also true of the open intervals {(xi−hi,xi)}∞

i=1. Then for the points of Npqcovered by these open intervals x′, there are arbitrarily small h′ such that

f (x′+h′)− f (x′)h′

> q.