PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Measure of a set difference when the smaller set has finite measure

Statement

Let $μ$ be a measure and let $A \subseteq B$ be measurable with $μ (A) < + \infty$ . Then

$μ (B) = μ (A) + μ (B ∖ A) .$

If $μ (B) < + \infty$ , all terms are real and hence $μ (B ∖ A) = μ (B) - μ (A)$ . If $μ (B) = + \infty$ , then $μ (B ∖ A) = + \infty$ .

Facts & Assumptions

Given: Measurable sets $A \subseteq B$ for a measure $μ$ , with $μ (A) < + \infty$ .

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

If $C \subseteq D$ are measurable, then $μ (C) \leq μ (D)$ (Measures are monotone).

Proof

technique · direct

1.1givenL1

The disjoint measurable sets $A$ and $B ∖ A$ have union $B$ , so $μ (B) = μ (A) + μ (B ∖ A)$ .

2.1givenstep 1.1L2algebra

If $μ (B) < + \infty$ , monotonicity makes both summands in step 1.1 finite, and cancellation in $R$ gives $μ (B ∖ A) = μ (B) - μ (A)$ .

3.1givenstep 1.1step 2.1algebra∎

If $μ (B) = + \infty$ , then $μ (B ∖ A)$ cannot be finite, because its sum with the finite number $μ (A)$ would be finite; hence it is $+ \infty$ . Together with steps 1.1 and 2.1 this proves every asserted case, including $A = \emptyset$ and $A = B$ .

Depends on

Dependency tree · 2 levels

Sources

S. Axler, Measure, Integration & Real Analysis, Theorem 2.57