26.5. RADEMACHER’S THEOREM 955

It follows that at Lebesgue points of ∇u, the above expression is o(|x−y|) and so at allsuch points u is differentiable. As to 26.5.15, this follows from an application of Corollary26.2.6 to f (t) = u(x+tv).

Note that for a.e. x,Dvu(x) = ∇u(x) ·v. If you have a line with direction vector v, doesit follow that Du(x+ tv) exists for a.e. t? We know the directional derivative exists a.e. tbut it might not be clear that it is ∇u(x) ·v.

For |w| = 1, denote the measure of Section 13.9 defined on the unit sphere Sp−1 as σ .Let Nw be defined as those t ∈ [0,∞) for which Dwu(x+ tw) ̸= ∇u(x+ tw) ·w.

B≡{

w ∈ Sp−1 : Nw has positive measure}

This is contained in the set of points of Rp where the derivative of v(·) ≡ u(x+ ·) fails toexist.Thus from Section 13.9 the measure of this set is∫

B

∫Nw

ρn−1dρdσ (w)

This must equal zero from what was just shown about the derivative of the Lipschitz func-tion v existing a.e. and so σ (B) = 0. The claimed formula follows from this. Thus weobtain the following corollary.

Corollary 26.5.6 Let u be Lipschitz. Then for any x and v ∈ Sp−1 \Bx where σ (Bx) = 0,it follows that for all t,

u(x+tv)−u(x) =∫ t

0Dvu(x+ sv)ds =

∫ t

0∇u(x+ sv) ·vds

In the all of the above, the function u is defined on all of Rp. However, it is always thecase that Lipschitz functions can be extended off a given set. Thus if a Lipschitz function isdefined on some set Ω, then it can always be considered the restriction to Ω of a Lipschitzmap defined on all of Rp.

Theorem 26.5.7 If h : Ω→ Rm is Lipschitz, then there exists h : Rp→ Rm which extendsh and is also Lipschitz.

Proof: It suffices to assume m = 1 because if this is shown, it may be applied to thecomponents of h to get the desired result. Suppose

|h(x)−h(y)| ≤ K |x−y|. (26.5.16)

Defineh(x)≡ inf{h(w)+K |x−w| : w ∈Ω}. (26.5.17)

If x ∈Ω, then for all w ∈Ω,

h(w)+K |x−w| ≥ h(x)

by 26.5.16. This shows h(x) ≤ h(x). But also you could take w = x in 26.5.17 whichyields h(x)≤ h(x). Therefore h(x) = h(x) if x ∈Ω.