Measure space

A measure space $(𝑋, Σ, 𝜇)$ consists of a measurable space $(𝑋, 𝜎)$ and a measure $𝜇$ on that space. A measurable space $(𝑋, Σ)$ consists of a set $𝑋$ and a σ-algebra on that set $𝜎$ . A measure $𝜎$ on a measurable space $(𝑋, Σ)$ is a function $𝜇 : 𝜎 \to [- \infty, \infty]$ satisfying #m/def/measure

non-negativity (unless Signed measure): $𝜇 (𝐴) \geq 0$ for all $𝐴 \in Σ$ .
empty set has zero measure¹: $𝜇 (\emptyset) = 0$
σ-additivity: $𝜇 (𝐴 \cup 𝐵) = 𝜇 (𝐴) + 𝜇 (𝐵)$ for any $𝐴, 𝐵 \in Σ$ with $𝐴 \cap 𝐵 = \emptyset$ . By induction the same holds for any countable collection of pairwise disjoint sets.

Thus a measure space generalises volume in the same way that a metric space generalises length.

Properties

From the above axioms it follows

monotonicity: $𝐴, 𝐵 \in Σ, 𝐴 \subseteq 𝐵 ⟹ 𝜇 (𝐴) \leq 𝜇 (𝐵)$
countable subadditivity: Let ${𝐸 𝑖} \infty 𝑖 = 1 \subseteq Σ$ be a countable (or finite²) sequence of measurable sets, then

𝜇 (\infty ⋃ 𝑖 = 1 𝐸 𝑖) \leq \infty \sum 𝑖 = 1 𝜇 (𝐸 𝑖)

Proof of 1–2

Let $𝐴, 𝐵 \in Σ$ be measurable sets such that $𝐴 \subseteq 𝐵$ . Then by ^M5 $𝐴 ∖ 𝐵 \in Σ$ and $𝐴 = 𝐵 \cup (𝐴 ∖ 𝐵)$ , so by σ-additivity $𝜇 (𝐴) = 𝜇 (𝐵) + 𝜇 (𝐴 ∖ 𝐵)$ , wherefore $𝜇 (𝐴) \leq 𝜇 (𝐵)$ , proving monotonicity.

Now let Let ${𝐸 𝑖} \infty 𝑖 = 1 \subseteq Σ$ be a countable sequence of measurable sets. Then

𝜇 (\infty ⋃ 𝑖 = 1 𝐸 𝑖) = 𝜇 (𝐸 1 ∖ \infty ⋃ 𝑖 = 2 𝐸 𝑖) + 𝜇 (\infty ⋃ 𝑖 = 2 𝐸 𝑖)

and by monotonicity

𝜇 (𝐸 1 ∖ \infty ⋃ 𝑖 = 2 𝐸 𝑖) \leq 𝜇 (𝐸 1)

hence

𝜇 (\infty ⋃ 𝑖 = 1 𝐸 𝑖) = 𝜇 (𝐸 1) + 𝜇 (\infty ⋃ 𝑖 = 2 𝐸 𝑖)

applying this argument inductively proves countable subadditivity.

#state/tidy | #lang/en | #SemBr

If at least one $𝐸 \in Σ$ has finite measure, then this follows from σ-additivity since $𝜇 (𝐸) = 𝜇 (𝐸 \cup \emptyset) = 𝜇 (𝐸) + 𝜇 (\emptyset)$ . ↩
Just give the sequence trailing $\emptyset$ . ↩

Measure space

Properties

Footnotes