Discussion
Let R be the set of real numbers, \(\mathbf{B}\) the Borel \(\sigma\)-field. Let '\(m\)' denotes the Lebesgue measure and \(\mu\) the measure defined by the function \(F(x) = (1 - e^{-x})\) if \(x \gt 0\) and \(F(x) = 0\) elsewhere.
Consider the statements :
(I) \(\mu \lt\lt m\), (i.e. \(\mu\) is absolutely continuous w.r.t. \(m\))
(II) \(m \lt\lt \mu\)
(III) \(\dfrac{d\mu}{dm}(x) = e^{-x}\ x \gt 0\), \(\dfrac{d\mu}{dm}(x) = 0\), \(x \le 0\)
Which of the above statements are true ?
Answer: B
The true statements are (I) and (III) only.
\(\mu\) is the exponential distribution, with density \(e^{-x}\) on \(x \gt 0\) and 0 elsewhere. Whenever \(m(A) = 0\), \(\mu(A) = \int_{A} e^{-x}1_{x \gt 0}\,dx = 0\), so \(\mu \ll m\) (I). By the Radon-Nikodym theorem the derivative \(d\mu/dm\) is that density, \(e^{-x}\) for \(x \gt 0\) and 0 for \(x \le 0\) (III).
(II) fails: \(\mu((-\infty, 0]) = 0\) but \(m((-\infty, 0]) = \infty\), so a \(\mu\)-null set need not be \(m\)-null.
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