“…Indeed, for any 𝑔 ∈ 𝐿 𝑞 0 (𝐵), ∫ 𝐵 [ F(𝑥) − 𝐹(𝑥) ] 𝑔(𝑥) 𝑑𝜇(𝑥) = 0, which, together with the fact that ℎ − 𝑚 𝐵 (ℎ) ∈ 𝐿 𝑞 0 (𝐵) for any ℎ ∈ 𝐿 𝑞 (𝐵), further implies that, for any ℎ ∈ 𝐿 𝑞 (𝐵),∫ 𝐵 [ F(𝑥) − 𝐹(𝑥) ] [ℎ(𝑥) − 𝑚 𝐵 (ℎ)] 𝑑𝜇(𝑥) = 0. (6.15)Moreover, for any 𝐻 ∈ 𝐿 𝑞 () and 𝐺 ∈ 𝐿 𝑞 ′ (), by the definitions of both 𝑚 𝐵 (𝐻) and 𝑚 𝐵 (𝐺), we obtain∫ 𝐵 [𝑚 𝐵 (𝐻)𝐺(𝑥) − 𝑚 𝐵 (𝐺)𝐻(𝑥)] 𝑑𝜇(𝑥) = ∫ 𝐵 {𝑚 𝐵 (𝐻)[𝐺(𝑥) − 𝑚 𝐵 (𝐺)] + 𝑚 𝐵 (𝐺)[𝑚 𝐵 (𝐻) − 𝐻(𝑥)]} 𝑑𝜇(𝑥) = 0,(6.16)which, together with ℎ ∈ 𝐿 𝑞 (𝐵) and 𝐹, F ∈ 𝐿 𝑞 ′ (𝐵), further implies that∫ 𝐵 𝑚 𝐵 (ℎ) [ F(𝑥) − 𝐹(𝑥) ] 𝑑𝜇(𝑥) = ∫ 𝐵 ℎ(𝑥)𝑚 𝐵(15) and (6.17), we conclude that, for any ℎ ∈ 𝐿 𝑞 (𝐵),∫ 𝐵 ℎ(𝑥) [ F(𝑥) − 𝐹(𝑥) − 𝑚 𝐵 ( F − 𝐹 )] 𝑑𝜇(𝑥) = 0,which further implies that, for almost every 𝑥 ∈ 𝐵,F(𝑥) − 𝐹(𝑥) = 𝑚 𝐵 ( F − 𝐹 ) (𝑥).Now, we take a sequence {𝐵 𝑗 } 𝑗∈ℕ of balls such that 𝐵 1 ⊂ 𝐵 2 ⊂ ⋯ ⊂ 𝐵 𝑗 ⊂ ⋯ and ⋃ 𝑗∈ℕ 𝐵 𝑗 = . Then, by (6.14), we know that there exists a sequence {𝐹 𝑗 } 𝑗∈ℕ of measurable functions such that, for any 𝑗 ∈ ℕ, 𝐹 𝑗 ∈ 𝐿 𝑞 ′ (𝐵 𝑗 ) and, for any 𝑔 ∈ 𝐿𝑞 0 (𝐵 𝑗 ), 𝑇(𝑔) = ∫ 𝐹 𝑗 (𝑥)𝑔(𝑥) 𝑑𝜇(𝑥).…”