A σ-algebra
tells us which sets are measurable. A measure tells us how
big each one is. Given a sample space
\Omega and a σ-algebra
\mathcal{F}, a measure is a function
\mu : \mathcal{F} \to [0, \infty]
assigning a non-negative size to each measurable set, subject to two requirements: the
empty set has zero size, \mu(\emptyset) = 0, and
countable additivity (σ-additivity) — for any
pairwise disjoint sets
A_1, A_2, \dots \in \mathcal{F},
\mu\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \mu(A_i).
The triple (\Omega, \mathcal{F}, \mu) is called a
measure space.
Properties that follow, step by step
Two requirements were assumed — that \mu(\emptyset) = 0 and that
\mu is countably additive on disjoint sets. Watch how much
structure they force out. We derive each property using nothing but those two facts (and
that \mu never goes below zero).
1. Finite additivity. Countable additivity is stated for an infinite list
of disjoint sets, but a finite disjoint list
A_1, \dots, A_n is a special case: pad it with infinitely many
copies of the empty set, which are disjoint from everything and from each other and change
no union. Then
\mu(A_1 \sqcup \dots \sqcup A_n) = \mu\!\left(A_1 \sqcup \dots \sqcup A_n \sqcup \emptyset \sqcup \emptyset \sqcup \cdots\right),
and countable additivity applied to the padded list gives
= \mu(A_1) + \dots + \mu(A_n) + \mu(\emptyset) + \mu(\emptyset) + \cdots = \sum_{i=1}^{n} \mu(A_i),
once we know the tail of \mu(\emptyset) terms contributes
nothing — which is the very next point.
2. The empty set has measure zero. (We listed
\mu(\emptyset) = 0 as a requirement; here is why no other value
is even consistent with additivity.) The empty set is disjoint from itself, so it is a
disjoint union of countably many copies of itself:
\emptyset = \emptyset \sqcup \emptyset \sqcup \emptyset \sqcup \cdots.
Applying countable additivity to that union, and writing
m = \mu(\emptyset) for short,
m = \mu(\emptyset) = \sum_{i=1}^{\infty} \mu(\emptyset) = \sum_{i=1}^{\infty} m.
The only number m \ge 0 for which adding it to itself
infinitely often returns the same finite number is m = 0:
any m > 0 would make the sum
+\infty \ne m. (This is why a measure must take at least one
finite value somewhere — for a probability measure
\mu(\Omega) = 1 < \infty does the job.) Hence
\mu(\emptyset) = 0.
3. Monotonicity. Suppose A \subseteq B. We can
carve B into the part inside A and the
part of B left over; these two pieces are disjoint and together
make all of B:
B = A \sqcup (B \setminus A).
Finite additivity (the two-set case from point 1) turns this disjoint split into a sum:
\mu(B) = \mu(A) + \mu(B \setminus A).
The leftover piece has non-negative measure,
\mu(B \setminus A) \ge 0, so dropping it can only decrease the
right-hand side:
\mu(B) = \mu(A) + \mu(B \setminus A) \ge \mu(A).
That is monotonicity:
A \subseteq B \;\Longrightarrow\; \mu(A) \le \mu(B).
4. Countable subadditivity. For an arbitrary family
A_1, A_2, \dots (now allowed to overlap), additivity does not
apply directly. The trick is to disjointify: peel each set down to the part
that is genuinely new, having removed everything already covered by earlier sets,
B_1 = A_1, \qquad B_n = A_n \setminus (A_1 \cup \dots \cup A_{n-1}).
By construction the B_n are pairwise disjoint, each
B_n \subseteq A_n, and they cover exactly the same union, since
every point of \bigcup_i A_i is "claimed" by the first
A_n that contains it:
\bigcup_{i=1}^{\infty} A_i = \bigsqcup_{i=1}^{\infty} B_i.
Now the right side is disjoint, so countable additivity applies, and then
monotonicity (point 3) replaces each \mu(B_i) by the no-smaller
\mu(A_i):
\mu\!\left(\bigcup_{i} A_i\right) = \mu\!\left(\bigsqcup_{i} B_i\right) = \sum_{i} \mu(B_i) \le \sum_{i} \mu(A_i).
That is countable subadditivity — overlap can only cause double-counting,
never undercounting.
The axiom could have asked only for finite additivity. Why insist on
countably many disjoint pieces? Because that single strengthening is what makes
limits behave — it is the load-bearing axiom of the whole theory.
Concretely, lengths of intervals must add up the way our intuition demands. The unit
interval (0, 1] is a countable disjoint union of shrinking
pieces,
(0, 1] = \bigsqcup_{n=1}^{\infty} \left(\tfrac{1}{n+1},\, \tfrac{1}{n}\right],
and we want 1 = \sum_{n} \left(\tfrac1n - \tfrac1{n+1}\right) —
a genuinely infinite sum that only countable additivity is entitled to assert. Finite
additivity would let length be consistent on any finite chopping but say nothing
about this telescoping limit. Without it, the
integral would lose its
convergence theorems, and the whole edifice — including probability's continuity along
limits — would collapse.
A measure may assign \mu(\Omega) = \infty — Lebesgue measure on
the whole real line is infinite, since \mathbb{R} is unbounded.
That is harmless provided the space can be exhausted by countably many pieces of
finite measure. A measure is σ-finite when
\Omega = \bigcup_{n=1}^{\infty} E_n, \qquad \mu(E_n) < \infty \text{ for every } n.
Lebesgue measure is σ-finite via E_n = [-n, n], each of finite
length 2n. σ-finiteness is the technical hypothesis behind the
big theorems (Radon–Nikodym, Fubini): it lets us reason about an infinite space one finite
slab at a time. Every probability measure is trivially σ-finite, since
\mu(\Omega) = 1 < \infty already.
One might hope to assign a length to every subset of
\mathbb{R}. You cannot — at least not while keeping length
translation-invariant and countably additive. Giuseppe Vitali (1905) used the
axiom of choice to pick one
representative from each coset of \mathbb{Q} in
[0, 1], building a set V whose
rational translates tile the line in countably many disjoint congruent copies. If
V had a length \mu(V), countable
additivity would force the total to be both 0 (if
\mu(V) = 0) and \infty (if
\mu(V) > 0) — never the finite length of the interval it sits
in. The escape is exactly why we built a
σ-algebra
first: we only ever demand \mu on the measurable sets, quietly
leaving the pathological ones out.