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Outer measure splits exactly over finite Carathéodory-measurable partitions
Statement
Let be pairwise disjoint Carathéodory measurable subsets of , where . For every ,
If are pairwise disjoint Carathéodory measurable sets, then outer measure splits every test set over those pieces and the remaining complement.
Facts & Assumptions
Given: An outer measure , a natural number , pairwise disjoint Carathéodory measurable sets , and a test set .
A set is Carathéodory measurable for when for every . (Carathéodory measurable sets)
The Carathéodory measurable subsets of form an algebra of subsets. (Carathéodory measurable sets form an algebra)
Proof
At the finite union is empty and the finite sum is the empty sum , so the formula is ; moreover every partial union is measurable by [L1].
Assume the displayed formula at and put .
Since is disjoint from , [F1] gives . Substituting this equality into the induction hypothesis in step 1.2 gives the formula for , including empty pieces and infinite values without cancellation.
Step 1.1 is the base case and step 2.1 proves the successor case, so the formula holds for every .
Depends on
Used by
Dependency tree · two levels
9 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
- G. Folland, Real Analysis, 2nd ed., proof of Theorem 1.11 (standard reference, not scraped)
- T. Tao, An Introduction to Measure Theory, proof of Theorem 1.7.3 (standard reference, not scraped)