Folland analysis 2.46, why is the diagonal an element of the product sigma algebra?

This is from Folland 2.46

Let $$X=Y=[0,1]$$, $$\mathcal{M}=\mathcal{N}=\mathcal{B}_{[0,1]}$$, $$\mu=$$Lebesgue measure, and $$\nu=$$counting measure. If $$D=\{(x,x):x\in[0,1]\}$$ is the diagonal in $$X\times Y$$, then what is $$(\mu\times\nu)(D)$$?

So this question has been asked on ME and the answer is that the product measure of the diagonal is $$\infty$$ which can be argued using outer measure with rectangles as our basic covering sets. My question is that we know outer measures always exist for any sets, but how do we actually know the diagonal is an element of the product sigma-algebra in the first place and thus is actually a measurable set (product measure)? Does that not require us to show the following: $$\mu(A) = \mu(A \cap D) + \mu(A \cap D^{c}).$$

for all $$A$$ in the product space.

I wanted to ask because many of the solutions presented seems to skip this step. Is this not necessary?

Edit: So just to clarify how Folland defines product measure, we first define a premeasure on the disjoint union of rectangles which is an algebra , then this algebra induces an outer measure and by Caretheodory's extension theorem, we can extend this to a unique product measure.

$$D$$ is a closed set, hence an element of the Borel algebra (because the Borel algebra of a product is the product of the Borel algebras if the underlying spaces are nice enough).
Let $$X^\prime = X \cap \mathbb Q$$, $$Y^\prime = Y \cap \mathbb Q$$. $$X^\prime \times Y^\prime$$ is countable. For any $$(x,y) \in X^\prime \times Y^\prime \setminus D$$, we can pick up $$r(x,y) \in \mathbb Q$$ such that for $$B((x,y), r(x,y)) = (x- r(x,y), x+r(x,y)) \times (y- r(x,y), y+r(x,y))$$ we have
$$B((x,y), r(x,y)) \subseteq X \times Y \setminus D.$$
The union of the $$B((x,y), r(x,y))$$ is a countable union of elements of $$\mathcal M \times \mathcal N$$ and is equal to $$X \times Y \setminus D$$, proving that $$D$$ is an element of the product $$\sigma$$-algebra.