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

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?

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.

• So for any non measurable set $V\subset\mathbb{R}^n$ $V\times\{0\}$ has "volume"/measure zero right? – John Cataldo Mar 26 '18 at 15:11
• @StanislasHildebrandt Yes! It is very useful to make a distinction between a measure and an outer measure in cases like this. Are you familiar with both? – Joonas Ilmavirta Mar 26 '18 at 15:12
• I mean we only defined outer measure in class and now we call it just "measure" if what is being measured is measurable – John Cataldo Mar 26 '18 at 15:13
• That is the basic idea; if you restrict an outer measure to the collection of sets that are measurable (as in they satisfy the Caratheodory criterion), you obtain a measure. – Xander Henderson Mar 26 '18 at 15:17
• @StanislasHildebrandt Ok, then the terminology shouldn't feel too alien. Now you don't know at first whether what you are dealing with is measurable, so it's best to start with the outer measure. – Joonas Ilmavirta Mar 26 '18 at 15:23

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}$}\}$.