Radon-Nikodym

When one measure is another one reweighted, and the density that does the reweighting.

The idea

Theorem (Radon–Nikodym theorem).

Let $\mu$ and $\nu$ be $\sigma$-finite measures on the same measurable space, and suppose $\nu \ll \mu$, meaning every set with $\mu(A) = 0$ also has $\nu(A) = 0$. Then there is a non-negative measurable function $f$ with $\nu(A) = \int_{A} f \, d\mu \qquad \text{for every measurable } A,$ and any two such functions agree outside a set of $\mu$-measure $0$.

We write $f = \dfrac{d\nu}{d\mu}$ and call it the Radon–Nikodym derivative of $\nu$ with respect to $\mu$.

The theorem says when one measure is another reweighted: $\nu$ is $\mu$ weighted by a density $f$ exactly when $\nu \ll \mu$. One direction is immediate — if such an $f$ exists, then integrating it over a set of $\mu$-measure $0$ gives $0$, so $\nu \ll \mu$ must hold. The content is the converse: absolute continuity only compares null sets, and out of that comparison the theorem produces an actual function.

The name derivative records two facts: $f$ is a rate, the amount of $\nu$ per unit of $\mu$, and it satisfies a chain rule, since $\nu \ll \mu \ll \lambda$ gives $\dfrac{d\nu}{d\lambda} = \dfrac{d\nu}{d\mu}\cdot\dfrac{d\mu}{d\lambda}$. But $f$ is not a limit of ratios at a point: only the integral identity defines it, which is why uniqueness holds only up to a $\mu$-null set.

Ways to work on it

• Walkthrough. When one measure has a density with respect to another, and why a probability density function is a Radon–Nikodym derivative.
• Practice. Decide when one measure is absolutely continuous with respect to another, and compute the resulting densities.
• Hardest. Change a Gaussian's mean and variance by a change of measure, then decide what a change of measure cannot do.

Not sure where to start? Take the ten-question placement test.