I have a question related to Lebesgue Integration Theory.
The Radon-Nikodym derivative is defined as follows: If $\mu$ and $ν$ are two $σ-$finite measures on $(Ω, F)$, such that $ν ≪ \mu$, then there is a non-negative $F$-measurable function $ρ$ such that $$ν(E) = \int_{E}ρd\mu $$for every $E \in F$
($ν ≪ \mu$ means absolute continuity of $ν$ with respect to $\mu$). Moreover $ρ$ is unique up to $\mu-$almost everywhere denoted by $\frac{dν}{d\mu}$.
The lecture notes say that the following is a routine computation: Suppose $\mu$ and $ν$ are two $σ-$finite measures on $(Ω, F)$, such that $ν ≪ \mu$. Let $f$ be an $F-$measurable function. Then $f$ is integrable with respect to $ν$ if and only if $f\frac{dν}{d\mu}$ is integrable with respect to $\mu$ but I can't really how this follows. Also, I'd be glad if you could explain the difference between almost averywhere and $\mu$-almost everywhere.
Any help is appreciated.
It can be shown that for a nonnegative measurable $f$ we have in this context:$$\int fd\nu=\int f\rho d\mu\tag1$$
If $f\rho$ is integrable wrt $\mu$ then:$$\int f_+d\nu=\int f_+\rho d\mu=\int(f\rho)_+d\mu<\infty$$ Likewise we find that $\int f_-d\nu<\infty$ and conclude that $f$ is integrable wrt $\nu$.
If conversely $f$ is measurable and integrable wrt $\nu$ then: $$\int(f\rho)_+d\mu=\int f_+\rho d\mu=\int f_+d\nu<\infty$$ Likewise we find that $\int(f\rho)_-d\mu<\infty$ and conclude that $f\rho$ is integrable wrt $\mu$.
The term "almost everywhere" is used if it is clear from context which measure this concerns. In this situation that is not quite obvious because two measures ($\mu$ and $\nu$) are involved.