The (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set

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I could probably write the statement a bit better

Statement

Suppose that [ilmath]\mu[/ilmath] is either a measure (or a pre-measure) on the [ilmath]\sigma[/ilmath]-ring (or ring), [ilmath]\mathcal{R} [/ilmath] then:

• for all [ilmath]A\in\mathcal{R} [/ilmath] and for all countably infinite or finite sequences [ilmath](A_i)\subseteq\mathcal{R} [/ilmath] we have:
  • [ilmath]A\subseteq\bigcup_i A_i\implies\mu(A)\le\sum_{i}\mu(A_i)[/ilmath]

Note: this is slightly different to sigma-subadditivity (or subadditivity) which states that [ilmath]\mu\left(\bigcup_i A_i\right)\le\sum_i\mu(A_i)[/ilmath] (for a pre-measure, we would require [ilmath]\bigcup_i A_i\in\mathcal{R} [/ilmath] which isn't guaranteed for countably infinite sequences)

Proof

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See outline

Outline:

1. Construct a new sequence, [ilmath](B_i)[/ilmath] where the elements are pairwise disjoint
2. Define [ilmath]B[/ilmath] as the union of all [ilmath](B_i)[/ilmath]
3. As [ilmath]A\subseteq B[/ilmath] we can use the monotonicity of a measure/pre-measure.