Compactness criteria for integral operators in $L\sp \infty$ and $L\sp 1$ spaces

Let $(S,\Sigma ,\mu )$ be a positive measure space, $k:S \times S \to \mathbb{R}$ be a measurable function such that the kernel $\vert k\vert$ induces a bounded integral operator on ${L^\infty }(S,\Sigma ,\mu )$ (equivalently, that ${\text{ess}}.{\sup _{s \in S}}\vert k(s,t)\vert d\mu (t) < \infty$), and for $s \in S$ let ${k_s}(t) = k(s,t)$. We show that it is sufficient for the integral operator T induced by k on ${L^\infty }(S,\Sigma ,\mu )$ to be compact, that there exists a locally $\mu$-null set $N \in \Sigma$ such that the set $\{ {k_s}:s \in S\}$ is relatively compact in ${L^1}(S,\Sigma ,\mu )$, and that this condition is also necessary if $(S,\Sigma ,\mu )$ is separable. In the case of Lebesgue measure on a subset of ${\mathbb{R}^n}$, we use Riesz's characterisation of compact sets in ${L^1}({\mathbb{R}^n})$ to provide a more tractable form of this criterion.