quantum mechanics - In a Bell scenario, why can correlations be nonlocal only if there are at least two measurement settings to choose from?

In (Brunner et al. 2013), the authors mention (end of pag. 6) that a set of correlations $p(ab|xy)$ can be nonlocal only if $\Delta\ge2$ and $m\ge2$, where $\Delta$ is the number of measurement outcomes for the two parties (that is, the number of different values that $a$ and $b$ can assume), and $m$ the number of measurement settings that one can choose from (that is, the number of values of $x$ and $y$).

A probability distribution $p(ab|xy)$ is here said to be nonlocal if it can be written as

$$p(ab|xy)=\sum_\lambda q(\lambda) p(a|x,\lambda)p(b|y,\lambda).\tag1$$ This means that if either there is only one possible measurement outcome, or only one possible measurement setting, then all probability distributions can be written as in (1).

The $\Delta=1$ (only one measurement outcome) case is trivial: if this is the case, denoting with $1$ the only possible measurement outcome, we have $p(11|xy)=1$ for all $x,y$. Without even needing the hidden variable, we can thus just take $p(1|x)=p(1|y)=1$ and we get the desired decomposition (1).

The $m=1$ case, however, is less trivial. In this case the question seems equivalent to asking whether an arbitrary probability distribution $p(a,b)$ can be written as

$$p(a,b)=\sum_\lambda q(\lambda)p(a|\lambda)p(b|\lambda).$$ The paper does not mention any reference to support this fact. How can this be proven?

Answer

Make a $\lambda_{a,b}$ for every pair $(a,b)$.

Then make $q(\lambda_{a,b}) = p(a,b)\,$, and

$p(a|\lambda_{a,b}) = p(b|\lambda_{a,b}) = 1.$