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Let \(\mu^{*}\) be an outer measure on \(H(\mathbf{R})\). Then \(E \in H(\mathbf{R})\) is \(\mu^{*}\)-measurable if :

where \(H(\mathbf{R})\)-hereditary \(\sigma\)-ring.

  • A.for each \(A \in H(\mathbf{R})\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
  • B.for some \(A \in H(\mathbf{R})\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
  • C.for each \(A \in H(\mathbf{R})\), \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)
  • D.for some \(A \in H(R)\), \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\)

Answer: A

E is \(\mu^{*}\)-measurable if for each \(A\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\).

This is Caratheodory's criterion: E must split every test set A additively. The family of such sets is a \(\sigma\)-ring on which \(\mu^{*}\) is a measure.

The inequality \(\mu^{*}(A) \le \mu^{*}(A \cap E) + \mu^{*}(A \cap E^{C})\) always holds, by subadditivity of an outer measure, so options C and D say nothing about E. Requiring the equation for only some A (option B) is far too weak: \(A = \emptyset\) satisfies it for every E.

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