35.3. WEAK DERIVATIVES IN Lploc 1237

Therefore, m(Em) = 0, a contradiction. Thus

m(E)≤∞

∑m=1

m(Em) = 0

and so, since ε > 0 is arbitrary,

m({x : f (x)> 0}) = 0.

Similarly m({x : f (x)< 0}) = 0. This proves the lemma.This lemma allows the following definition.

Definition 35.3.4 Let U be an open subset of Rn and let u∈ L1loc (U) . Then Dα u∈ L1

loc (U)if there exists a function g ∈ L1

loc (U), necessarily unique by Lemma 35.3.3, such that forall φ ∈C∞

c (U), ∫U

gφdx = Dα u(φ)≡∫

U(−1)|α| u(Dα

φ)dx.

Then Dα u is defined to equal g when this occurs.

Lemma 35.3.5 Let u ∈ L1loc (Rn) and suppose u,i ∈ L1

loc (Rn), where the subscript on the ufollowing the comma denotes the ith weak partial derivative. Then if φ ε is a mollifier anduε ≡ u∗φ ε , it follows uε,i ≡ u,i ∗φ ε .

Proof: If ψ ∈C∞c (Rn), then∫

u(x−y)ψ ,i (x)dx =∫

u(z)ψ ,i (z+y)dz

= −∫

u,i (z)ψ (z+y)dz

= −∫

u,i (x−y)ψ (x)dx.

Therefore,

uε,i (ψ) = −∫

uε ψ ,i =−∫ ∫

u(x−y)φ ε (y)ψ ,i (x)d ydx

= −∫ ∫

u(x−y)ψ ,i (x)φ ε (y)dxdy

=∫ ∫

u,i (x−y)ψ (x)φ ε (y)dxdy

=∫

u,i ∗φ ε (x)ψ (x)dx.

The technical questions about product measurability in the use of Fubini’s theorem may beresolved by picking a Borel measurable representative for u. This proves the lemma.

What about the product rule? Does it have some form in the context of weak deriva-tives?