Corollary 1

— If $\lambda$ and $\mu$ are measures, then $\nu$ is a measure.

For, let $x = (s, t)$ be a point of $X$, and let $U$ and $V$ be neighborhoods of $s, t$ respectively, such that $\lambda^\bullet(U) < +\infty$, $\mu^\bullet(V) < +\infty$; the set $U \times V$ is a neighborhood of $x$, and $\nu^\bullet(U \times V) = \lambda^\bullet(U)\mu^\bullet(V) < +\infty$ by (4); the encumbrance $\nu^\bullet$ is thus locally bounded, and the premeasure $\nu$ is a measure.

This result extends at once to complex measures.