PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21

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The two-set measure identity $μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B)$

Statement

For measurable sets $A$ and $B$ in a measure space,

$μ (A \cup B) + μ (A \cap B) = μ (A) + μ (B) .$

The equality is valid in $[0, + \infty]$ , including when one or both sides equal $+ \infty$ .

Facts & Assumptions

Given: A measure $μ$ and measurable sets $A, B$ .

[L1]

A measure is additive on every finite pairwise disjoint measurable family (Measures on sigma-algebras).

Proof

technique · direct

1.1given

Put $C = A ∖ B$ , $D = A \cap B$ , and $F = B ∖ A$ . These sets are measurable and pairwise disjoint, with $A = C \cup D$ , $B = D \cup F$ , and $A \cup B = C \cup D \cup F$ .

2.1step 1.1L1

Finite additivity gives $μ (A) = μ (C) + μ (D)$ , $μ (B) = μ (D) + μ (F)$ , and $μ (A \cup B) = μ (C) + μ (D) + μ (F)$ .

3.1step 2.1algebra∎

Adding $μ (D) = μ (A \cap B)$ to the last equality and regrouping nonnegative extended sums gives the displayed identity; no subtraction occurs, so infinite values and the cases $A = \emptyset$ , $B = \emptyset$ , or $A = B$ are included.

Depends on

Measures on sigma-algebras

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Measure, Integration & Real Analysis, Theorem 2.61 (standard reference, not scraped)

The two-set measure identity μ(A∪B)+μ(A∩B)=μ(A)+μ(B)μ(A∪B)+μ(A∩B)=μ(A)+μ(B)