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Let E and F be measurable sets, \(f \in L(E, \mu)\) and \(\mu(E \,\Delta\, F) = 0\), then :

  • A.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu = \int_{F} f\, d\mu\)
  • B.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu \lt \int_{F} f\, d\mu\)
  • C.\(f \in L(F, \mu)\) and \(\int_{E} f\, d\mu \gt \int_{F} f\, d\mu\)
  • D.\(f \notin L(F, \mu)\)

Answer: A

\(f \in L(F, \mu)\) and \(\int_{E} f\,d\mu = \int_{F} f\,d\mu\).

\(\mu(E \,\Delta\, F) = 0\) means E and F differ by a null set: \(E \setminus F\) and \(F \setminus E\) both have measure zero. Integrals ignore null sets, so

\[ \begin{aligned} \int_{F} f &= \int_{E \cap F} f + \int_{F \setminus E} f \\ &= \int_{E \cap F} f + 0 \end{aligned} \]

and similarly \(\int_{E} f = \int_{E \cap F} f\). Since \(|f|\) is integrable over E, it is integrable over \(E \cap F\) and hence over F. A strict inequality between the two integrals, or non-integrability over F, is therefore impossible.

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