Let λ ∈ M ( X ; C ) , μ ∈ M ( Y ; C ) \lambda \in \mathscr{M}(X; \mathbf{C}),\ \mu \in \mathscr{M}(Y; \mathbf{C}) λ ∈ M ( X ; C ) , μ ∈ M ( Y ; C )
∣ λ ⊗ μ ∣ = ∣ λ ∣ ⊗ ∣ μ ∣ .
|\lambda \otimes \mu| = |\lambda| \otimes |\mu|.
∣ λ ⊗ μ ∣ = ∣ λ ∣ ⊗ ∣ μ ∣. Let f ∈ K + ( X × Y ) , g ∈ K ( X × Y ; C ) f \in \mathscr{K}_+(X \times Y),\ g \in \mathscr{K}(X \times Y; \mathbf{C}) f ∈ K + ( X × Y ) , g ∈ K ( X × Y ; C ) ∣ g ∣ ≤ f |g| \leq f ∣ g ∣ ≤ f
∣ ⟨ g , λ ⊗ μ ⟩ ∣ = ∣ ∫ d λ ( x ) ∫ g ( x , y ) d μ ( y ) ∣ ≤ ∫ d ∣ λ ∣ ( x ) ∫ ∣ g ( x , y ) ∣ d ∣ μ ∣ ( y ) = ⟨ ∣ g ∣ , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ ≤ ⟨ f , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ .
\begin{align*}
|\langle g, \lambda \otimes \mu \rangle| &= \left| \int d\lambda(x) \int g(x, y) d\mu(y) \right| \\
&\leq \int d|\lambda|(x) \int |g(x, y)|\ d|\mu|(y) \\
&= \langle |g|, |\lambda| \otimes |\mu| \rangle \leq \langle f, |\lambda| \otimes |\mu| \rangle.
\end{align*}
∣ ⟨ g , λ ⊗ μ ⟩ ∣ = ∫ d λ ( x ) ∫ g ( x , y ) d μ ( y ) ≤ ∫ d ∣ λ ∣ ( x ) ∫ ∣ g ( x , y ) ∣ d ∣ μ ∣ ( y ) = ⟨ ∣ g ∣ , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ ≤ ⟨ f , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ . It follows that ⟨ f , ∣ λ ⊗ μ ∣ ⟩ ≤ ⟨ f , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ \langle f, |\lambda \otimes \mu| \rangle \leq \langle f, |\lambda| \otimes |\mu| \rangle ⟨ f , ∣ λ ⊗ μ ∣ ⟩ ≤ ⟨ f , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩
∣ λ ⊗ μ ∣ ≤ ∣ λ ∣ ⊗ ∣ μ ∣ .
|\lambda \otimes \mu| \leq |\lambda| \otimes |\mu|.
∣ λ ⊗ μ ∣ ≤ ∣ λ ∣ ⊗ ∣ μ ∣. On the other hand, let u ∈ K + ( X ) , v ∈ K + ( Y ) u \in \mathscr{K}_+(X),\ v \in \mathscr{K}_+(Y) u ∈ K + ( X ) , v ∈ K + ( Y ) ε > 0 \varepsilon > 0 ε > 0 u 1 ∈ K ( X ; C ) , v 1 ∈ K ( Y ; C ) u_1 \in \mathscr{K}(X; \mathbf{C}),\ v_1 \in \mathscr{K}(Y; \mathbf{C}) u 1 ∈ K ( X ; C ) , v 1 ∈ K ( Y ; C ) ∣ u 1 ∣ ≤ u , ∣ v 1 ∣ ≤ v |u_1| \leq u,\ |v_1| \leq v ∣ u 1 ∣ ≤ u , ∣ v 1 ∣ ≤ v
∣ ⟨ u 1 , λ ⟩ ∣ ≥ ⟨ u , ∣ λ ∣ ⟩ − ε , ∣ ⟨ v 1 , μ ⟩ ∣ ≥ ⟨ v , ∣ μ ∣ ⟩ − ε
|\langle u_1, \lambda \rangle| \geq \langle u, |\lambda| \rangle - \varepsilon,\quad |\langle v_1, \mu \rangle| \geq \langle v, |\mu| \rangle - \varepsilon
∣ ⟨ u 1 , λ ⟩ ∣ ≥ ⟨ u , ∣ λ ∣ ⟩ − ε , ∣ ⟨ v 1 , μ ⟩ ∣ ≥ ⟨ v , ∣ μ ∣ ⟩ − ε (§ 1, No. 6). It follows that
∣ u 1 ⊗ v 1 ∣ ≤ u ⊗ v |u_1 \otimes v_1| \leq u \otimes v ∣ u 1 ⊗ v 1 ∣ ≤ u ⊗ v and
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ ≥ ∣ ⟨ u 1 ⊗ v 1 , λ ⊗ μ ⟩ ∣ = ∣ ⟨ u 1 , λ ⟩ ⟨ v 1 , μ ⟩ ∣ ≥ ( ( ⟨ u , ∣ λ ∣ ⟩ − ε ) ( ⟨ v , ∣ μ ∣ ⟩ − ε ) ) .
\begin{align*}
\langle u \otimes v, |\lambda \otimes \mu| \rangle &\geq |\langle u_1 \otimes v_1, \lambda \otimes \mu \rangle| = |\langle u_1, \lambda \rangle \langle v_1, \mu \rangle| \\
&\geq ((\langle u, |\lambda| \rangle - \varepsilon)(\langle v, |\mu| \rangle - \varepsilon)).
\end{align*}
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ ≥ ∣ ⟨ u 1 ⊗ v 1 , λ ⊗ μ ⟩ ∣ = ∣ ⟨ u 1 , λ ⟩ ⟨ v 1 , μ ⟩ ∣ ≥ ((⟨ u , ∣ λ ∣ ⟩ − ε ) (⟨ v , ∣ μ ∣ ⟩ − ε )) . Since ε \varepsilon ε
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ ≥ ⟨ u , ∣ λ ∣ ⟩ ⟨ v , ∣ μ ∣ ⟩ = ⟨ u ⊗ v , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ .
\langle u \otimes v, |\lambda \otimes \mu| \rangle \geq \langle u, |\lambda| \rangle \langle v, |\mu| \rangle = \langle u \otimes v, |\lambda| \otimes |\mu| \rangle.
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ ≥ ⟨ u , ∣ λ ∣ ⟩ ⟨ v , ∣ μ ∣ ⟩ = ⟨ u ⊗ v , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ . Taking (8) into account, we see that
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ = ⟨ u ⊗ v , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ .
\langle u \otimes v, |\lambda \otimes \mu| \rangle = \langle u \otimes v, |\lambda| \otimes |\mu| \rangle.
⟨ u ⊗ v , ∣ λ ⊗ μ ∣ ⟩ = ⟨ u ⊗ v , ∣ λ ∣ ⊗ ∣ μ ∣ ⟩ . Since every function in K ( X ; C ) \mathscr{K}(X; \mathbf{C}) K ( X ; C ) K ( Y ; C ) \mathscr{K}(Y; \mathbf{C}) K ( Y ; C ) K + ( X ) \mathscr{K}_+(X) K + ( X ) K + ( Y ) \mathscr{K}_+(Y) K + ( Y ) u ∈ K ( X ; C ) u \in \mathscr{K}(X; \mathbf{C}) u ∈ K ( X ; C ) v ∈ K ( Y ; C ) v \in \mathscr{K}(Y; \mathbf{C}) v ∈ K ( Y ; C ) K ( X ; C ) ⊗ C K ( Y ; C ) \mathscr{K}(X; \mathbf{C}) \otimes_{\mathbf{C}} \mathscr{K}(Y; \mathbf{C}) K ( X ; C ) ⊗ C K ( Y ; C ) K ( X × Y ; C ) \mathscr{K}(X \times Y; \mathbf{C}) K ( X × Y ; C )
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.