Radon–Nikodym theorem

Remarks

A standard proof first treats finite measures. Consider the measurable $g \ge 0$ with $\int_A g \, d\mu \le \nu(A)$ for all $A$, and take one that maximises $\int_X g \, d\mu$. If the leftover measure $\nu - g \mu$ were not zero, a Hahn decomposition of $\nu - g\mu - \varepsilon \mu$ would allow $g$ to be increased, a contradiction. Von Neumann gave another proof through the Riesz representation of a bounded linear functional on $L^2(\mu + \nu)$. The σ-finite case follows by cutting $X$ into countably many pieces of finite measure.

Example: if $\nu$ is the standard normal distribution on $\R$ and $\mu$ is Lebesgue measure, then $f(x) = e^{-x^2/2} / \sqrt{2\pi}$.

The hypothesis of σ-finiteness of $\mu$ cannot be dropped. Let $\mu$ be counting measure on the Borel sets of $[0, 1]$ and $\nu$ Lebesgue measure. Then $\nu \ll \mu$, because only the empty set has $\mu$-measure $0$. A density $f$ would satisfy $f(x) = \nu(\{x\}) = 0$ for every $x$, which gives $\nu = 0$, a contradiction.

The function $f$ is the measure with density description of $\nu$: $\nu = f \mu$.

Going deeper

For a student

The theorem is the source of the Radon–Nikodym derivative $d\nu / d\mu$. In probability it gives conditional expectation: the conditional expectation of an integrable variable given a sub-σ-algebra is a Radon–Nikodym derivative. It also identifies the dual of $L^p(\mu)$ with $L^q(\mu)$ for $1 \le p < \infty$ and σ-finite $\mu$. Together with the Lebesgue decomposition it describes every σ-finite measure relative to a fixed one.