An outer measure can underestimate the sum of sizes of disjoint pieces because each piece may need a different efficient cover. Carathéodory’s criterion selects a set E when it splits every test set T exactly: μ*(T)=μ*(T intersect E)+μ*(T outside E). The test set T is arbitrary, not assumed measurable. Subadditivity always gives one inequality; measurability requires the reverse inequality as well.
Predict
Predict why complements should pass the same test. Replacing E by its complement merely exchanges the two terms. Predict also why null sets should pass: their intersection with any T has outer measure zero, while removing them cannot increase outer measure. These observations suggest that the selected class will be closed under logical operations and will include all subsets of null sets.
Experiment
Keep the two inequality sources distinct.
Statement
Reason
μ*(T)≤μ*(T∩E)+μ*(T outside E)
Outer subadditivity
Reverse inequality for every T
Measurability requirement
μ*(T∩N)=0 when N is outer-null
Monotonicity
Complement passes whenever E passes
Same two pieces
For a test set of finite outer measure eight, a proposed split of sizes three and six would violate measurability. Infinite values require extended nonnegative sums, never subtraction of infinities.
Understand
Let M be the sets satisfying the splitting criterion. The whole space and complements belong to M. If E, F belong, split T first by E and then split T outside E by F. This gives μ*(T)≥μ*(T∩E)+μ*((T outside E)∩F)+μ*(T outside (E∪F)). The first two terms dominate μ*(T∩(E∪F)) by subadditivity, proving E∪F measurable. Thus finite unions and differences are allowed. For disjoint E_j in M, repeated splitting gives μ*(T)≥Σ through n of μ*(T∩E_j)+μ*(T outside union E_j), because the remaining finite-stage part contains the full complement. Taking the supremum of partial sums and using subadditivity on the union proves the countable-union criterion.
Apply
Arbitrary countable unions can be disjointified using the already proved finite differences, so M is a sigma-algebra. Set T equal to a disjoint union of measurable E_j in the preceding inequality to obtain μ*(T)≥Σμ*(E_j); outer subadditivity supplies equality. Therefore μ=μ* restricted to M is a measure. If μ*(N)=0, every subset A of N has outer measure zero. For arbitrary T, its piece inside A has measure zero and its outside piece has outer measure at most μ*(T), proving the required reverse inequality. Hence A is measurable. This proves completeness of the constructed measure.
Continue
The theorem is abstract: no intervals or topology were used after assuming the outer-measure axioms. It converts a covering-based size into a complete measure on a precisely defined sigma-algebra. It does not prove that all subsets are measurable, nor that this sigma-algebra equals the Borel sets of a particular topology. For Lebesgue measure, Borel measurability and completion will be established separately. A careful construction keeps those assertions distinct rather than using the word measurable for several unproved properties at once.
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If μ*(N)=0 and A⊆N, find μ*(A).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A measurable E splits a test set T with μ*(T∩E)=3 and μ*(T minus E)=5. Find μ*(T).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove the complement of a Carathéodory measurable set is measurable.
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Prove every subset of an outer-null set is measurable.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Explain how finite splitting yields countable disjoint additivity.
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Why must the defining equality hold for every subset T, not only measurable T?
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Reflect and continue
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