Completion of Measure - Clarifying the Wikipedia Example to Motivate Completion of Lebesgue Measure Preface:
I refer to the following link:
https://en.wikipedia.org/wiki/Complete_measure
Here, the author states a motivation for the notion of "completion of measure", in that if one defines a 1-D Lebesgue measure space: $(\mathbb{R}, \mathbf{B}, \lambda)$, then the naive product sigma algebra $\mathbf{B} \otimes\mathbf{B}$ would be incomplete.
They state an example, where if one had some $A\subset\mathbb{R}$ that were immeasurable (in the Lebesgue sense), then the evaluation of say, $\lambda^2(\{0\}\times A) = \lambda(\{0\})\cdot \lambda(A)$ would be undefined (since $A$ is immeasurable! It doesn't make sense to apply Lebesgue measure to this term).
As far as I can understand they appear to resolve this by paradox by stating that $\{0\}\times A \subset \{0\}\times \mathbb{R}$, and that it is possible to evaluate $\lambda^2(\{0\}\times \mathbb{R})$ instead (I assume using the convention that $0\cdot \infty = 0$). Dealing with $\{0\}\times \mathbb{R}$ instead seems to act like a kind of "extension" or "completion" to me intuitively to circumvent the evaluation of immeasurable-$A$.
Question:

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*Does the example even technically make sense? If we consider the Lebesgue measure space then $A$-immeasurable shouldn't be part of it no? Since we are only concerned with Lebesgue measurable $\sigma$-algebra? (even though I am aware that the subsets of Lebesgue sets can indeed be immeasurable e.g. Vitali sets). As far as I understand every set in the 1-D Lebesgue $\sigma$-algebra should be Lebesgue measurable, so how could we accidentally end up with immeasurable-$A$ in the product measure space?


*Perhaps built on Q1. How does their proposed method for performing the completion "remove" the effect of these immeasurable subsets? It seems to be a method for making sure subsets of null sets should evaluate to zero (nothing to do with considering the immeasurable sets as far as I can tell?)
 A: The motivation given in the Wikipedia article obfuscates the real reason for why we introduce complete measure spaces.
Suppose one constructs an outer measure $\mu^*$ on the measurable space $(X, \mathcal{P}(X))$, where $\mathcal{P}(X)$ is the power set of $X$. We say that a set $A \subset X$ is $\mu^*$-measurable if the Carathéodory condition holds:
$$\mu^* (E) = \mu^*(E \cap A) + \mu^* (E \cap A^c)$$
for all $E \subset X$. If, $\mu^*(A) = 0$, then monotonicity of $\mu^*$ implies that $A$ and all its subsets are $\mu^*$-measurable. If $\mathcal{M}$ is the $\sigma$-algebra of $\mu^*$-measurable subsets of $X$, then $(X, \mathcal{M}, \mu^*)$ forms a complete measure space.
Oftentimes, however, one may wish to further restrict the $\sigma$-algebra $\mathcal{M}$ to a smaller, more manageable, one. For instance, one often defines Lebesgue measure on the Borel sets, as this measure space is compatible with the topology of the real line. However, one readily sees that the Borel $\sigma$-algebra is not complete (e.g. by observing that many subsets of the Cantor set are not Borel-measurable).
Behind those technical remarks are the real reason why we introduce completeness:   sometimes it helps us avoid some pesky technical measurability details involving null sets! However, it does come at a cost: sometimes the completion of a $\sigma$-algebra is too unwieldy, as is the case with the Lebesgue $\sigma$-algebra.
With that motivation in mind let me answer your questions:

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*The example makes sense from the perspective that $\{0\} \times A$ being a "thin" set should have measure zero (indeed, it has outer measure zero being a subset of $\{0\} \times \mathbb{R}$). This motivation is something that can also be done in 1-dimension (cf. Borel vs Lebesgue $\sigma$-algebras). You also correctly point out that a Vitali set would not even be in the Lebesgue $\sigma$-algebra of $\mathbb{R}$. So really, what the example is actually showing is that the product of two complete $\sigma$-algebras need not be complete (i.e. $\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R}) \neq \mathcal{L}(\mathbb{R}^2)$), rather than motivating completeness.

*Their method is correct: you simply declare subsets of null sets to be null as well. This is the standard construction of the completion of a $\sigma$-algebra. As you correctly identify, this doesn't really say anything about general non-measurable sets.

