How come the product $V\times\{0\}$ is measurable with $V$ non measurable?

Question

Take a non Lebesgue measurable set $V\subset\mathbb{R}$ (say the Vitali set). The Lebesgue outer measure $\mu\big[V\times\{0\}\subset\mathbb{R}^2\big]=0$ and that directly implies that $V\times\{0\}$ is measurable. How is this possible?

Answer 1

Any set of the form $V\times\{0\}$ has zero outer measure (it is contained in the line, which famously has zero (outer) measure). Any set of zero outer measure is measurable. Perhaps one could say that the one-dimensional weirdness of $V$ is irrelevant, because it is so small in the product set $V\times\{0\}$.

It is important to maintain the distinction between measures and outer measures in order to make all this sensible. Every set has an outer measure. The concept of measure is restricted to measurable sets. The outer measure is more flexible, so it is best to argue with it.

Answer 2

Lebesgue measure is a complete measure. In particular since $$ V\times{0}\subset\mathbb{R^2}\times\{0\} $$ where $\mathbb{R}^2\times \{0\}$ is measurable (it is closed) and moreover $m(\mathbb{R}^2\times\{0\})=0$, it follows that $V\times \{0\}$ is measurable.

In fact when constructing Lebesgue measure using the Caratheodory- Hahn Extension theorem, we extend a premeasure $\mu$ defined on the algebra of finite disjoint unions of rectangles, $\mathcal{A}$, uniquely to a measure $m$ defined on the borel sigma algebra $\mathcal {B}=\sigma(\mathcal{A})$. But in fact the theorem extends this measure to an algebra bigger than $\mathcal{B}$, call it $\mathcal{L}$, those sets that are measurable with respect to Caratheodory's criterion. So we have the inclusions $$ \mathcal{A}\subset\mathcal{B}\subset\mathcal{L} $$ It can be shown that $\mathcal{L}=\{B\cup N\mid B\in\mathcal{B}, N\subset E, m(E)=0\quad\text{for some E$\in\mathcal{B}$}\}$.