1576 CHAPTER 48. MULTIFUNCTIONS AND THEIR MEASURABILITY

Lemma 48.6.1 Let F be those points where u = 0. Then for a.e. x ∈ F,∇u(x) = 0. Hereu ∈W 1,p (Ω) and we assume ∂Ω has measure zero.

Proof: It suffices to consider u = 0 on F contained in the interior of Ω. First I showthat u,xn = 0 a.e.

u(t1, · · · , tn−1, tn)−u(t1, · · · , tn−1, tn−h)h

=1h

∫ tn

tn−hu,xn (t1, · · · , tn−1,s)ds

Then a.e. tn is a Lebesgue point of u,xn for F(t1,··· ,tn−1)where F(t1,··· ,tn−1) consists of thosepoints of F where (t1, · · · , tn−1) is fixed. Also let tn be a point of density of F(t1,··· ,tn−1). Ofcourse m1 a.e. points of F(t1,··· ,tn−1) are points of density. Therefore, there exists a sequencehk→ 0+ such that tn−hk→ tn as k→∞ and (tn−hk)∈ F(t1,··· ,tn−1). Otherwise there wouldbe some open set about tn which excludes points of F(t1,··· ,tn−1) which would imply that tnis not actually a point of density. Then using the fundamental theorem of calculus, we getfor such points which are points of F(t1,··· ,tn−1) the fact that u,xn (t1, · · · , tn−1, tn) = 0. Thusfor a.e. tn ∈ F(t1,··· ,tn−1),u,xn (t1, · · · , tn−1, tn) = 0. Thus u,xn (t1, · · · , tn−1, tn) = 0 for a.e. tn inF(t1,··· ,tn−1). Similar reasoning holds for differentiation with respect to the other variables.Thus ∇u = 0 a.e. on F .

Lemma 48.6.2 Let V be a closed subset of W 1,p (Ω) , p> 1 and let k∈V. Then max(k,u)∈V and if un→ u in V, then max(un,k)→max(u,k) in V.

Proof: We consider ψ (r) = |r| ,ψε (r) =√

ε + r2. Then for φ ∈C∞c (Ω) ,∫

Ω

ψ (u(x))φ ,xk (x) = limε→0

∫Ω

ψε (u(x))φ ,xk (x)

= − limε→0

∫Ω

u(x)√ε +u2 (x)

u,xk (x)φ (x)

= −∫

Ω

ξ (u(x))u,xk (x)φ (x)

where ξ (r) = 1 if r > 0,−1 if r < 0 and 0 if r = 0. Thus ψ (u) ,xk = ξ (u(x))u,xk (x) a.e.and so ψ ◦ u is clearly in W 1,p (Ω). Of course max(u,k) = |k−u|+(k+u)

2 so this shows thatmax(u,k) is in W 1,p (Ω) .

Next suppose un→ u in W 1,p (Ω) . Does

ξ (un)un,xk→ ξ (u)u,xk ?

Let G = {x : u(x) ̸= 0} . A subsequence, still denoted by un converges pointwise a.e . to uand un,xk→ u,xk pointwise a.e. Therefore, off a set of measure zero, ξ (un (x)) = ξ (u(x))for all n large enough on G. Also,(∫

Ω

∣∣ξ (un)un,xk−ξ (u)u,xk

∣∣p)1/p

≤(∫

Ω

∣∣ξ (un)un,xk−ξ (un)u,xk

∣∣p)1/p

+

(∫Ω

|ξ (un)−ξ (u)|p∣∣u,xk

∣∣p)1/p

(*)