Le $f,g : \mathbb{R} → \mathbb{R}$ be extended integrable functions. Let $λ = f dµ$ and $ρ = g dµ$ for Lebesgue measure $µ$. Give necessary and sufficient conditions on $f,g$ for $λ ⊥ ρ$ and necessary and sufficient conditions on $f,g$ for $λ << ρ$.
Where $λ ⊥ ρ$ means $λ$ and $ρ$ are $\textbf{mutually singular}$ as stated in Foland's Real Analysis, that is, We say that two signed measures $λ$ and $ρ$ on $(X, M)$ are mutually singular if there exist $E, F \in M $such that $E \cap F = \emptyset$, $E \cup F = X$, $E$ is null for $λ$, and F is null for $ρ$. Informally speaking, mutual singularity means that $λ$ and $ρ$ ''live on disjoint sets."
And, $λ << ρ$ means $\textbf{absolutely continuous}$. That is, if $λ$ is a signed measure and $ρ$ is a positive measure on $(X, M)$, We say that $λ$ is absolutely continuous with respect to $ρ$ and write $λ<<ρ$ if $λ(E) = 0$ for every $E \in M$ for which $ρ(E) = 0$.
I am not sure how to imply the definition to the question, could anyone help please. Thanks a lot!
Let $\lambda \perp \rho$, and let $E \subset \mathbb{R}$ be such that $\lambda(E)=0$ and $\rho(E^c)=0$. Writing $\lambda(E)=\int_E f \mathop{d\mu}$ and $\rho(E^c)=\int_{E^c} g \mathop{d\mu}$, what does this imply about the value of $f$ on $E$ ($\mu$-almost everywhere)? and of $g$ on $E^c$ ($\mu$-almost everywhere)? Can you show the converse?
Let $\lambda \ll \rho$. Then $\rho(E)=0$ implies $\lambda(E)=0$. Again writing this statement in terms of $\mu$-integrals, can you see what this implies about $f$ and $g$?