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Lebesgue Outer Measure and the Carathéodory Criterion

Now we actually build a measure on the real line. We define Lebesgue outer measure by covering a set with intervals as efficiently as possible, see why it is only subadditive, and use Carathéodory's splitting test to carve out exactly the measurable sets — on which it becomes a genuine measure.

Covering a set with intervals

The size of an interval (a,b) is obviously b − a. To measure an arbitrary set E ⊆ ℝ, cover it by a countable collection of open intervals and add up their lengths; then take the most efficient such cover. Formally, the Lebesgue outer measure is m*(E) = inf { Σ (b_k − a_k) : E ⊆ ⋃ (a_k, b_k) }, the infimum over all countable interval covers of E.

This m* is defined for every subset of ℝ, which is a real advantage. It is monotone (a bigger set needs bigger covers) and countably subadditive (concatenate the covers). It also gives the right answer on intervals: m*([a,b]) = b − a. The proof that m*([a,b]) ≥ b − a uses compactness — the Heine–Borel reduction of a cover to a finite subcover — and is the one genuinely nontrivial computation here.

Carathéodory's splitting test

Carathéodory's brilliant move is to define measurability by how a set splits everyone else. We say E is [[measurable-set|measurable]] (in the Carathéodory sense) if it cuts every test set A cleanly: m*(A) = m*(A ∩ E) + m*(A ∩ E^c) for all A ⊆ ℝ. The Carathéodory criterion demands that E partition the outer measure of every set additively — the part inside E plus the part outside E must add back to the whole.

One inequality is automatic: subadditivity already gives m*(A) ≤ m*(A ∩ E) + m*(A ∩ E^c) for every E and A. So measurability is really the single reverse inequality m*(A) ≥ m*(A ∩ E) + m*(A ∩ E^c). This asymmetry makes verifications easier than they look — you only ever need to check the ≥ direction.

Claim: every set N with m*(N) = 0 is Caratheodory-measurable.

Let A be any test set. We must show
      m*(A) >= m*(A and N) + m*(A and N^c).     (the only direction needed)

Step 1 (monotonicity).  A and N is a subset of N, so
      m*(A and N) <= m*(N) = 0,   hence  m*(A and N) = 0.

Step 2 (monotonicity again).  A and N^c is a subset of A, so
      m*(A and N^c) <= m*(A).

Step 3 (add).
      m*(A and N) + m*(A and N^c) = 0 + m*(A and N^c) <= m*(A).

That is exactly the >= inequality. Combined with automatic subadditivity:
      m*(A) = m*(A and N) + m*(A and N^c).
So N is measurable, and m(N) = m*(N) = 0.                 QED
Every outer-measure-zero set passes the Carathéodory test — the first family we know is measurable.

What the criterion buys us

Carathéodory's theorem then states the payoff: the collection of measurable sets is a sigma-algebra, and m* restricted to it is a genuine, countably additive measure — namely Lebesgue measure m. One verifies closure under complement (the definition is visibly symmetric in E and E^c) and under countable unions (a careful induction plus a limiting argument), and additivity drops out of the very splitting condition.