This is motivated by the set-up in exercise 47 in Folland's book: let $X$ be any uncountable linearly-ordered set such that for each $x \in X$, the set $\{y \in X : y < x\}$ is countable (for example, the set of all binary sequences under dictionary ordering). Let $\mathcal{M}$ be the $\sigma$-algebra of countable/co-countable subsets of $X$, and let the measures $\mu$ be defined as $$ \mu(A) = \begin{cases} 0 & \text{$A$ is countable} \\ 1 & \text{$A$ is co-countable} \end{cases}$$
Let $E = \{ (x, y) \in X \times X : y < x\}$. I want to show that $E$ is not measurable. In the exercise, one shows that the iterated integrals of $\chi_E$ over $X\times X$ are not equal, so by the Tonelli theorem it must be that $E$ is non-measurable. Is there a more direct way to show this? For example if we denote the outer-measure induced by $(\mu \times\mu)$ by $(\mu \times \mu)^\ast$, is it true and possible to show that that $(\mu \times \mu)^\ast(E) = (\mu \times \mu)^\ast(E^c) = 1$? This would contradict that $(\mu \times \mu)(X\times X) = 1$.