${\mathbf{M}}$ as an array, there must be at least one zero in each block.

Case \[nowiring\], however, is impossible. Consider the state ${\mathbf{P}}$ defined by $P(a|x) = \delta_{a, a_x}$. ${\mathbf{P}}$ is clearly an allowed state, as it is positive, normalised and non-signalling. However, if case 2 holds then $${\mathbf{M}}.{\mathbf{P}} = \sum_{ax} M(a|x) P(a|x) = 0.$$ This implies that ${\mathbf{M}}.{\mathbf{P}} = 0$ for all states ${\mathbf{P}}$. But the only way to achieve this is to take ${\mathbf{M}}= 0$, which contradicts the initial assumption that ${\mathbf{M}}$ is non-zero. Therefore case \[yeswiring\] is the only possibility. Subnormalised basic measurements can be subtracted from ${\mathbf{M}}$ until the zero vector remains. Hence any single-system measurement can be expressed as a finite sum of subnormalised basic measurements. Since $\overline{{\mathbf{M}}}.{\mathbf{P}} = 1$ for any state ${\mathbf{P}}$ and normalised $\overline{{\mathbf{M}}}$, this procedure yields a decomposition of $\overline{{\mathbf{M}}}$ as a convex combination of basic measurements.

### Bi-partite system measurements

Consider a (non-zero) subnormalised ${\mathbf{M}}$, associated with a bipartite system, with components $M(ab|xy)$. (In this subsection, fiducial measurements are labelled $x, y$, and outcomes $a, b$, rather than $x_1, x_2$ and $a_1,a_2$ respectively.) One of the following must hold:

1.  It is possible to subtract a (non-zero) subnormalised basic measurement 