A σ-algebra answers only half the question. It tells us which subsets of a space X we are allowed to measure — a family \mathcal{F} closed under complements and countable unions, rich enough for analysis, small enough to dodge the non-measurable monsters. But a list of "measurable sets" says nothing about how much each one contains. Naming the sets you may weigh is not the same as owning a scale.

This page installs the scale. A measure \mu is the map that assigns to every set in \mathcal{F} a size in [0, \infty] — a length, an area, a probability, a count — subject to one decisive rule: the size of a disjoint countable pile is the sum of the sizes of its pieces. A σ-algebra and a measure on it, riding on a set X, form a measure space (X, \mathcal{F}, \mu) — the single object on which the whole of integration, and (when \mu(X) = 1) all of probability, is built.

The definition

Fix a measurable space (X, \mathcal{F}). A measure on it is a function

(the value +\infty is genuinely allowed — the whole line has infinite length) satisfying two axioms:

• (M1) — the empty set weighs nothing: \mu(\varnothing) = 0.
• (M2) — countable additivity (σ-additivity): for any sequence A_1, A_2, A_3, \dots \in \mathcal{F} that is pairwise disjoint (A_i \cap A_j = \varnothing for i \ne j), \mu\!\left(\bigcup_{n=1}^{\infty} A_n\right) = \sum_{n=1}^{\infty} \mu(A_n).

The triple (X, \mathcal{F}, \mu) is a measure space. Everything else on this page — every property you would want a "size" to have — is squeezed out of these two lines.

The word that carries all the weight is countable. Axiom (M2) is not merely finite additivity (split a set in two, the sizes add). It promises the sum survives an infinite sequence of disjoint pieces — and that is exactly the promise that lets sizes commute with limits, the property Riemann's theory so badly lacked. A single disjoint union A \sqcup B is the easy special case; the theory lives in the tail.

Everything follows from two axioms

The axioms look thin. Watch how much they carry — each property below is forced by (M1) and (M2) alone, with a one-line proof.

Finite additivity

Take disjoint A_1, \dots, A_k \in \mathcal{F} and pad the sequence with empty sets: A_{k+1} = A_{k+2} = \dots = \varnothing. These are still pairwise disjoint, so (M2) applies, and by (M1) every padded term contributes 0:

So finite additivity is a corollary — which is why we bothered to demand the countable version: the strong axiom hands us the weak one for free, but not conversely.

Monotonicity

Suppose A \subseteq B, both in \mathcal{F}. Split B into the disjoint pieces A and B \setminus A (both measurable — a σ-algebra is closed under differences). By finite additivity,

because \mu(B \setminus A) \ge 0. Bigger set, bigger (or equal) measure — no set can shrink by gaining points. As a bonus, when \mu(A) < \infty we may subtract to get \mu(B \setminus A) = \mu(B) - \mu(A).

Countable subadditivity

Now drop the disjointness. For any A_1, A_2, \dots \in \mathcal{F} (overlapping freely),

The trick is disjointification: set B_1 = A_1 and B_n = A_n \setminus (A_1 \cup \dots \cup A_{n-1}). The B_n are disjoint, each B_n \subseteq A_n, and \bigcup B_n = \bigcup A_n. So by (M2) then monotonicity,

Equality when the sets are disjoint (that is (M2)); a strict deficit whenever they overlap — you paid for the overlap twice on the right.

Continuity from below

A measure respects increasing limits. If A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots with union A = \bigcup_n A_n (written A_n \uparrow A), then

Peel the tower into disjoint rings C_1 = A_1, C_n = A_n \setminus A_{n-1}. Then A_k = \bigcup_{n \le k} C_n and A = \bigcup_n C_n, so by (M2) \mu(A) = \sum_n \mu(C_n) = \lim_k \sum_{n\le k}\mu(C_n) = \lim_k \mu(A_k). Countable additivity is precisely this continuity — the two are equivalent given finite additivity.

Continuity from above (with a catch)

The decreasing version holds too — but only if you start from finite measure. If A_1 \supseteq A_2 \supseteq \dots with A_n \downarrow A = \bigcap_n A_n, and \mu(A_1) < \infty, then

Apply continuity from below to the growing complements A_1 \setminus A_n \uparrow A_1 \setminus A, and use \mu(A_1 \setminus A_n) = \mu(A_1) - \mu(A_n) (legal because \mu(A_1) < \infty). The finiteness is not a technicality — the next box shows the theorem is false without it.

The measures you already know (and some you don't)

Abstract axioms deserve concrete inhabitants. Every one of these is a bona fide measure on some (X, \mathcal{F}).

• Lebesgue measure \lambda on \mathbb{R}. The one we were chasing: it extends length, \lambda([a,b]) = b - a, to the whole Borel (and Lebesgue) σ-algebra, is translation-invariant, and assigns the rationals measure 0. Building it is the work of the next few pages; that it is a countably additive measure is its defining achievement.
• Counting measure. On any X with \mathcal{F} = 2^X, put \mu(A) = |A| — the number of elements of A (with \mu(A) = \infty for infinite A). The empty set has 0 elements, and cardinalities of disjoint sets add, so both axioms hold. On \mathbb{N} this is the measure that turns integration into ordinary summation.
• The Dirac measure \delta_x (a point mass). Fix a point x \in X and ask only "is x in the set?": \delta_x(A) = \begin{cases} 1, & x \in A, \\ 0, & x \notin A. \end{cases} All of the mass sits on a single point. \delta_x(\varnothing) = 0 since x \notin \varnothing, and for disjoint sets x lands in at most one of them, so the sum on the right has at most one non-zero term — additivity holds.
• A discrete probability measure. Choose points x_1, x_2, \dots and weights p_n \ge 0 with \sum_n p_n = 1, and stack point masses: \mu = \sum_{n} p_n\, \delta_{x_n}, \qquad \mu(A) = \sum_{n \,:\, x_n \in A} p_n. This is a weighted mixture of Dirac measures — the measure behind a fair die (x_n \in \{1,\dots,6\}, each p_n = \tfrac16) or any discrete random variable.

A measure with \mu(X) = 1 is a probability measure: the total mass is exactly one unit, so \mu(A) reads as "the probability of A". Every axiom of probability — that the whole sample space has probability 1, that disjoint events add — is just a measure with total mass 1. Probability theory is measure theory with a normalization.

Null sets, "almost everywhere", and completeness

Sets of measure zero are the dust the theory learns to ignore. A set N \in \mathcal{F} is a null set (or μ-null) if \mu(N) = 0. Under Lebesgue measure the rationals, every finite or countable set, and even the uncountable Cantor set are null — negligible for the purposes of integration.

A property is said to hold almost everywhere (abbreviated a.e., or almost surely in probability) if the set of points where it fails is null. Dirichlet's function is 0 a.e. because it is non-zero only on \mathbb{Q} — a null set — which is precisely why its Lebesgue integral is 0. "Almost everywhere" is the phrase that lets us change a function on a null set without changing a single integral.

One subtlety earns a name. A measure is complete if every subset of a null set is itself measurable (and hence null): whenever \mu(N) = 0 and M \subseteq N, we insist M \in \mathcal{F}. This seems obvious — a subset of something negligible should be negligible — but a raw σ-algebra need not contain all such M. The Borel sets under Lebesgue measure are incomplete; one completes them by adjoining every subset of every Borel null set, giving the (complete) Lebesgue σ-algebra. Completeness removes annoying "is this scrap even measurable?" caveats from later theorems.

See it: the parts add up to the whole

Countable additivity is a statement about an infinite disjoint pile, so let us watch one converge. Split [0, 1) into disjoint intervals

so I_n has Lebesgue measure \lambda(I_n) = 2^{-n}. The pieces are pairwise disjoint and their union is all of [0,1), so (M2) demands \sum_n 2^{-n} = 1. Drag the slider to lay down more pieces and watch the running total \sum_{k=1}^{n} \lambda(I_k) = 1 - 2^{-n} creep up to the measure of the whole interval — never overshooting, always converging.

This is also continuity from below in miniature: the partial unions A_n = I_1 \cup \dots \cup I_n = [0, 1 - 2^{-n}) increase up to [0,1), and their measures 1 - 2^{-n} increase up to 1. Additivity and continuity are the same fact wearing two hats.