Proposition 4

— Let $\lambda \in \mathscr{M}(X; \mathbf{C}), \ \mu \in \mathscr{M}(Y; \mathbf{C})$. Then

with the convention that the second member is to be replaced by 0 whenever one of the factors is 0 and the other is $+\infty$. In particular, if $\lambda$ and $\mu$ are bounded then $\lambda \otimes \mu$ is bounded.

By the above Proposition 3, and §1, No. 8, Cor. 1 of Prop. 10, we may limit ourselves to the case that $\lambda$ and $\mu$ are positive measures. If $\lambda = 0$ or $\mu = 0$, the result is trivial; let us therefore assume that $\lambda \neq 0$ and $\mu \neq 0$. Let us first prove that $\|\lambda \otimes \mu\| \leq \|\lambda\| \cdot \|\mu\|$. We may suppose $\lambda$ and $\mu$ to be bounded. For every $f \in \mathscr{K}_+(X \times Y)$,

and

for every $x \in X$, therefore

which proves our assertion. On the other hand, let

be two real numbers $\geqslant 0$. There exist $g \in \mathscr{K}_+(X) , \ h \in \mathscr{K}_+(Y)$ such that

Then $g \otimes h \in \mathscr{K}_+(X \times Y) , \ \| g \otimes h \| \leqslant 1$ and $\langle g \otimes h , \lambda \otimes \mu \rangle \geqslant \alpha \beta ;$ therefore $\| \lambda \otimes \mu \| \geqslant \alpha \beta$ and finally $\| \lambda \otimes \mu \| \geqslant \| \lambda \| \cdot \| \mu \|$, which completes the proof.