Radon-Nikodym Theorem

• categories: Measure Theory, Theorem

Statement:
Let $(X,\mathcal{F},\mu )$ be a Sigma-Finite Measure space and $\nu$ be another $\sigma$-finite measure on the same measurable space such that $\nu \ll \mu$ (i.e., $\nu$ is absolutely continuous with respect to $\mu$). Then there exists a measurable function $f:X\to [0,\infty )$, called the Radon-Nikodym derivative, such that:

$$\nu (A)={\int }_{A}fd\mu \text{for all }A\in \mathcal{F}.$$

The function $f$ is denoted by $\frac{d\nu }{d\mu }$ and satisfies:

$$\frac{d\nu }{d\mu }\in L^1(\mu )\text{and}\nu (A)={\int }_A\frac{d\nu }{d\mu }d\mu .$$

Intuition:
The Radon-Nikodym theorem allows us to express the measure $\nu$ as being “scaled” by a density function $f$ with respect to the reference measure $\mu$. This is analogous to expressing one probability distribution in terms of another via a probability density function.

Key Properties:

• The Radon-Nikodym derivative $\frac{d\nu }{d\mu }$ is unique $\mu$-almost everywhere.
• If $\mu$ is a probability measure, $f$ can be interpreted as the conditional density of $\nu$ relative to $\mu$.

Proof Outline:

1. Construct a sequence of sets $\{A_n\}$ with $\mu (A_n)<\infty$ that cover $X$.
2. Apply the Hahn Decomposition Theorem to decompose $\nu -\lambda f\mu$ into positive and negative parts, where $f$ is chosen iteratively.
3. Prove that this construction results in a valid Radon-Nikodym derivative using monotone convergence arguments.