= Solution
On the same <measurable space>, \b[<absolute continuity of measures>] means
$$
\boxed{\widetilde{\mathbb P}\ll\mathbb P\quad\Longleftrightarrow\quad\mathbb P(A)=0\implies\widetilde{\mathbb P}(A)=0\text{ for every }A\in\mathcal F.}
$$
The <Radon-Nikodym theorem> then supplies a nonnegative <measurable> density $Z$ with $\mathbb E_{\mathbb P}Z=1$ and $\widetilde{\mathbb P}(A)=\mathbb E_{\mathbb P}[Z\mathbf1_A]$. The converse follows directly from this formula. Strict positivity of the <Radon-Nikodym derivative> gives equivalence of the two <probability measures>; <absolute continuity of measures> alone need not give equivalence.
Back to article page
