Lebesgue Measure and Integration Questions and Answers

Answer: A

All three statements are true.

(I) \(\mu(A) = \int_{A} e^{-x^{2}/2}\, dm\) is the integral of a non-negative measurable function over A, which is a measure on the Borel sets (countable additivity by monotone convergence).

(II) By definition \(e^{-x^{2}/2}\) is the Radon-Nikodym derivative \(d\mu/dm\), unique up to sets of Lebesgue measure zero.

(III) The density is strictly positive everywhere, so \(\mu(A) = 0\) forces \(m(A) = 0\); hence \(m \ll \mu\) as well, and the two measures are equivalent.

Options leaving out (II) or (III) are therefore incomplete.

Answer: C

The statement that is not true is \([0, \infty) \in \mathbf{F}\).

F is the countable-cocountable class. The interval \([0, \infty)\) is uncountable, and its complement \((-\infty, 0)\) is uncountable too, so neither condition holds and \([0, \infty) \notin \mathbf{F}\).

The other statements are true. R belongs to F because its complement is empty, hence countable. F is closed under countable unions: if all \(B_{n}\) are countable the union is countable, and if some \(B_{m}\) has countable complement then the complement of the union lies inside \(B_{m}^{c}\) and is countable. Finally \(Q^{c}\), the irrationals, has complement Q, which is countable, so \(Q^{c} \in \mathbf{F}\). In fact F is a sigma-algebra.