Can every set of real numbers be assigned a sensible length? It feels like it should — but in 1905 Giuseppe Vitali used the axiom of choice to build a subset of [0,1) that has no consistent length at all. This Vitali set is the canonical non-measurable set, and it is the reason a measure cannot be defined on every subset of the line — only on a well-behaved σ-algebra.

The construction

Work inside [0,1). Call two points equivalent if their difference is rational: x \sim y \iff x - y \in \mathbb{Q}. This chops [0,1) into equivalence classes (cosets of \mathbb{Q}). Each class is countable (it's a single point plus rational shifts), but there are uncountably many classes.

Now invoke the axiom of choice: from each class, pick exactly one representative. Collect those representatives into a set V — the Vitali set. (You cannot write V down; choice only promises it exists.)

Here is why V cannot have a length. Translate it by every rational q \in [0,1), wrapping around mod 1, to get copies V_q. Two facts collide:

• The copies are disjoint, and together they tile all of [0,1) — so their measures must add up to 1.
• Translation doesn't change length, so every V_q has the same measure m as V, and there are countably many of them.

By countable additivity, the total is \textstyle\sum_{q} m. But a sum of countably many equal numbers is only ever 0 (if m = 0) or \infty (if m > 0) — never 1:

Both cases contradict "= 1". So V can have no length. It is non-measurable.

Why measure lives on a σ-algebra

The Vitali set is the precise reason Lebesgue measure is defined on the Borel (or Lebesgue) sets rather than on the whole power set \mathcal{P}(\mathbb{R}). If we demanded a length for every subset, V would break the rules. Restricting to a σ-algebra — a collection closed under complements and countable unions, but not containing pathological sets like V — is what lets measure theory (and hence all of rigorous probability) work at all.