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Let (X, M, \(\mu\)) be a measurable space and \(E \in M\) be a measurable set. Let \(f\) be a non-negative measurable function with support contained in E. Define \(\lambda(A) = \int_{A \cap E} f\, d\mu\).

Which of the following statements is false ?

  • A.\(\lambda\) is a measure on M
  • B.\(\lambda\) is a measure on \(M(E) = \{A \cap E / A \in M\}\)
  • C.\(\lambda\) is absolutely continuous with respect to \(\mu\) on M(E)
  • D.\(\lambda\) is singular with respect to \(\mu\) on M(E)

Answer: D

The false statement is that \(\lambda\) is singular with respect to \(\mu\) on M(E).

\(\lambda(A) = \int_{A \cap E} f\, d\mu\) is a measure on M (countable additivity follows from the monotone convergence theorem), and its restriction to \(M(E)\) is a measure too, so A and B hold. If \(\mu(A) = 0\) then \(\lambda(A) = 0\), so \(\lambda \ll \mu\) (C); \(f\) is its Radon-Nikodym derivative.

A measure that is absolutely continuous with respect to \(\mu\) can be singular with respect to it only if it is the zero measure. For a non-negative \(f\) that is not zero almost everywhere on E, \(\lambda\) is not singular, so D is the false statement.

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