Intuition on the Lebesgue-Radon-Nikodym Theorem

In my first blog post, I discuss the proof of the Lebesgue-Radon-Nikodym Theorem as given in Folland’s Real Analysis.

To begin with, this is really the combination of the Radon-Nikodym theorem and Lebesgue’s decomposition theorem. Loosely speaking, Radon-Nikodym allows us to relate a signed measure which is absolutely continuous with respect to a positive measure through an integral. Meanwhile, Lebesgue decomposition, given two $\sigma$-finite signed measures, allows us to decompose either one into two parts such that one part is absolutely continuous with respect to the other measure, at which point we can apply Radon-Nikodym.

More rigorously, Radon-Nikodym states: let $\sigma$-finite $\nu$ and $\mu$ be signed and positive measures respectively on $(X,M)$ where $\nu$ is absolutely continuous with respect to $\mu$. Then, there exists a finite valued measurable function $f$ on $X$ such that $\mu(E) = \int_E f d\mu$ for every measurable $E$, and $f$ is unique $\mu$-a.e. On the other hand, Lebesgue decomposition says that for any two $\sigma$-finite signed measures $\mu$ and $\nu$, we can decompose $\nu$ into the sum of two $\sigma$-finite signed measures $\nu_0$ and $\nu_1$ which are respectively absolutely continuous and singular with respect to $\mu$.

The proof can be reduced to the case where $\nu$ and $\mu$ are finite positive measures; this is because if they are $\sigma$-finite measures, then $X$ is a countable union of intersections of $\mu$- and $\nu$-finite sets, say ${A_j}$. Then, we can take $\mu_j(E) = \mu(E \cap A_j)$ and $\nu_j(E)$ similarly, which reduces to the case of finite positive measures, giving us $d\nu_j = d\lambda_j + f_j d\mu_j$ for each $j$, and then we simply take $\lambda = \sum \lambda_j$ and $f = \sum f_j$. Also, if $\nu$ is a signed measure, we simply apply the previous case to $\nu^+$ and $\mu^-$ and subtract the results.

Having said that, we begin the proof for the case of finite positive measures by choosing $f$ as the ‘supremum’ of functions whose integrals with respect to $\mu$ are bounded by $\nu$. Then, the measure $d\lambda = d\nu - f d\mu$ is singular with respect to $\mu$, or else the maximality of $\int f d\mu$ is contradicted. Also, if ${\nu_j}$ is a sequence of positive measures, then ‘additivity’ holds for singularity and absolute continuity with respect to some $\mu$ (this was left as an exercise in Folland) which allows us to prove uniqueness $\mu$-a.e.