Lebesgue Measure

A measure assigns a size to sets. Lebesgue measure is the measure on the real line — the one that gives an interval its ordinary length, and then extends that notion of length to a vast collection of far stranger sets. It is the foundation on which the Lebesgue integral, and hence modern probability, is built.

The starting point is the only sensible one: the measure of an interval is its length,

\lambda\big([a, b]\big) = b - a.

From that single rule, plus the demand that measure behave sensibly under countable unions, everything else is forced.

The defining properties

Lebesgue measure \lambda on the Borel sets is pinned down by:

Length: \lambda([a,b]) = b - a.
Countable additivity: for disjoint sets A_1, A_2, \dots, \lambda\big(\bigcup_i A_i\big) = \sum_i \lambda(A_i).
Translation invariance: \lambda(A + x) = \lambda(A) — sliding a set along the line doesn't change its size.

These force some memorable facts. A single point \{a\} = [a,a] has measure a - a = 0. And by countable additivity, any countable set is a countable union of points, so it too has measure zero.

The rationals have length zero

Here is the result that first makes Lebesgue measure feel powerful. The rational numbers \mathbb{Q} are dense — between any two reals sits a rational — yet they are countable. So as a countable union of single points, their total measure is

\lambda(\mathbb{Q}) = \sum_{q \in \mathbb{Q}} \lambda(\{q\}) = \sum 0 = 0.

The rationals are everywhere and yet occupy no length at all. All the "length" of the real line lives in the uncountably many irrationals. This is exactly why the Lebesgue integral succeeds where the older Riemann integral chokes: a function that misbehaves only on a measure-zero set (like the rationals) can be integrated cleanly, because that set contributes nothing.

Worked numbers. The interval [-2, 5] has length 5-(-2)=7. A pair of points \{-1\}\cup\{4\} still has measure 0+0=0. And \lambda(\mathbb{Q}\cap[0,1])=0 even though those rationals sit densely inside a set of length one — density is not length.

Here's a hands-on proof that \mathbb{Q} has measure zero. List the rationals q_1, q_2, q_3, \dots and, for any tiny \varepsilon > 0, cover q_n with an interval of length \varepsilon / 2^n. Every rational is now covered, and the total length is \sum_n \varepsilon/2^n = \varepsilon — as small as you please. A set you can trap inside intervals of arbitrarily small total length has measure zero. The trick is that the lengths form a convergent geometric series.

Not every set is measurable. Using the axiom of choice one can build a non-measurable set (a Vitali set) to which no consistent length can be assigned. That is precisely why measure is defined only on a σ-algebra, not on all subsets.
Measure zero does not mean "small" or "few". The rationals are infinite and dense yet have measure zero; conversely the Cantor set is uncountable yet also has measure zero. Cardinality and measure are different rulers.

Start with [0,1] and delete the open middle third (1/3, 2/3); from each remaining closed interval delete its open middle third; repeat forever. What is left is the middle-thirds Cantor set C. At stage n you have removed 2^{n-1} intervals of length 3^{-n}, so the total length removed is \sum_{n=1}^{\infty} 2^{n-1}/3^n = 1. Hence \lambda(C)=0. But C is uncountable (it is in bijection with \{0,1\}^{\mathbb{N}} via ternary expansions that use only the digits 0 and 2). An uncountable set can still have no length — cardinality and measure really are different rulers.

Drag ε, watch the cover shrink

Below, a handful of rationals in [0,1] are each given a small interval of length \varepsilon/2^{n}. The coloured bars are those intervals, stacked slightly so they stay visible when they overlap. Slide \varepsilon down: every rational is still covered, but the total length of the cover — the sum of the bar lengths — equals \varepsilon and can be made as small as you like. That is measure zero, drawn.