I am a bit dumbfounded by this problem as it has always been given in the build up to an application of the Radon-Nikodym theorem.
Let $(\Omega, \mathcal{A}, P)$ is a probability space. Suppose that $Y:\Omega \rightarrow \mathbb{R}$, $Y(\Omega) = \mathbb{R}$, is a continuous random variable with density $f(y)$. Define $$\mathcal{C} = \{Y^{-1}(B): B \in \mathcal{B}\}$$ where $\mathcal{B}$ is the Borel $\sigma$-field defined on $\mathbb{R}$. Next, define $\mu(Y^{-1}(B)) = \lambda(B)$ for any $Y^{-1}(B) \in \mathcal{C}$ where $\lambda$ is the Lebesgue measure in $\mathbb{R}$. Show that $P \ll \mu$.
We know that $P \ll \mu$ if $\mu(A) = 0$ implies $P(A) = 0$ but I am not certain how to prove this relationship without more context about $P$.
Thanks in advanced for the help!
As pointed out by humanStampedist we cannot, in general, define a measure $\mu$ on $\mathcal C$ by $\mu(Y^{-1} (B))=\lambda (B)$. I would interpret the problem as follows: suppose there exists a measure $\mu$ on $\mathcal C$ such that $\mu(Y^{-1} (B))=\lambda (B)$. Then $P<<\mu$ on $\mathcal C$. To prove this let $\mu(Y^{-1}(B))=0$. Then $\lambda (B)=0$. Hence $P(Y^{-1}(B))=\int_B f(x) \, dx=0$.