Radon–Nikodym derivative

Given two measures $\mu, \nu$ on the same measurable space, their Radon–Nikodym derivative is essentially their ratio $\mu/\nu$, although this is traditionally written $\mathrm{d}\mu/\mathrm{d}\nu$ because of analogies with differentiation. This ratio or derivative is a measurable function which is defined up to equality almost everywhere with respect to the divisor $\nu$. It only exists iff $\mu$ is absolutely continuous with respect to $\nu$.

Integration on a general measure space can be seen as the process of multiplying a measure by a function to get a measure. Then the Radon–Nikodym derivative is the reverse of this: dividing two measures to get a function.

Let $X$ be a measurable space (so $X$ consists of a set ${|X|}$ and a $\sigma$-algebra $\mathcal{M}_X$), and let $\mu$ and $\nu$ be measures on $X$, valued in the real numbers (and possibly taking infinite values) or in the complex numbers (and taking only finite values). Let $f$ be a measurable function $f$ (with real or complex values) on $X$.

The function $f$ is a **Radon–Nikodym derivative** of $\mu$ with respect to $\nu$ if, given any measurable subset $A$ of $X$, the $\mu$-measure of $A$ equals the integral of $f$ on $A$ with respect to $\nu$:

$\mu(A) = \int_A f \nu = \int_{x \in A} f(x) \nu(\mathrm{d}x) .$

(The latter two expressions in this equation are different notations for the same thing.)

These properties are basic to the concept; the notation is as in the definition above.

Let $f$ be a Radon–Nikodym derivative of $\mu$ with respect to $\nu$, and let $g$ be a measurable function on $X$. Then $g$ is a Radon–Nikodym derivative of $\mu$ with respect to $\nu$ if and only if $f$ and $g$ are equal almost everywhere with respect to $\nu$.

If a Radon–Nikodym derivative of $\mu$ with respect to $\nu$ exists, then $\mu$ is absolutely continuous with respect to $\nu$.

If $\mu$ is absolutely continuous with respect to $\nu$ and both $\mu$ and $\nu$ are $\sigma$-finite, then a Radon–Nikodym derivative of $\mu$ with respect to $\nu$ exists.

For fairly elementary proofs, see Bartels (2003).

(This last theorem is not as general as it could be.)

Note the repetition of ‘with respect to $\nu$’ in various guises; let us fix $\nu$ (assumed to be $\sigma$-finite) and take everything with respect to it. Then it is convenient to treat all measurable functions up to equality almost everywhere; and given any absolutely continuous $\mu$ (also assumed to be $\sigma$-finite), we speak of the Radon–Nikodym derivative of $\mu$.

See also the discussion of notation at measure space.

Using the simplest notation for integrals, the definition of Radon–Nikodym derivative reads

$\mu(A) = \int_A f \nu ,$

or equivalently

$\int_A \mu = \int_A f \nu .$

In other words, the measure $\mu$ is the product of the function $f$ and the measure $\nu$:

$\mu = f \nu ;$

and so $f$ is the ratio of $\mu$ to $\nu$:

$f = \mu/\nu .$

So this is the simplest notation for the Radon–Nikodym derivative.

However, this notation for integrals is uncommon; one is more likely to see

$\int_A \mathrm{d}\mu = \int_A f \,\mathrm{d}\nu ,$

which leads to

$f = \mathrm{d}\mu/\mathrm{d}\nu$

for the Radon–Nikodym derivative. But none of these ‘$\mathrm{d}$’s are really necessary.

We can also use a fuller notation with a dummy variable as the object of the symbol ‘$\mathrm{d}$’:

$\int_{x \in A} \mu(\mathrm{d}x) = \int_{x \in A} f(x) \,\nu(\mathrm{d}x) ;$

this leads to

$f(x) = \mu(\mathrm{d}x)/\nu(\mathrm{d}x) ,$

which does not give a symbol for $f$ directly. If instead of $\mu(\mathrm{d}x)$ one unwisely writes $\mathrm{d}\mu(x)$, then this gives the previous notation for the Radon–Nikodym derivative.

Now let $\nu$ be Lebesgue measure on the real line and let $F$ be an upper semicontinuous function on the real line, so that $F$ defines a Borel measure? $\mu$ generated by

$\mu({]-\infty,a]}) \coloneqq F(a) .$

Then $F$ is absolutely continuous? if and only if $\mu$ is absolutely continuous, in which case the derivative $F'$ exists almost everywhere and is a Radon–Nikodym derivative of $\mu$. That is,

$\mu/\nu = F' = \mathrm{d}F/\mathrm{d}t .$

The presence of ‘$\mathrm{d}$’ on the right-hand side inspires people to put it on the left-hand side as well; but this is spurious, since we really want to write

$\mu = \mathrm{d}F$

and

$\nu = \mathrm{d}t ,$

where $t$ is the identity function on the real line.

Some fairly elementary proofs prepared for a substitute lecture in John Baez's introductory measure theory course are here:

- Toby Bartels (2003):
*The Radon Nikodym Theorem*; web.

The strategy there is based on:

- Richard Bradley (1989): An Elementary Treatment of the Radon-Nikodym Derivative, American Mathematical Monthly 96(5), 437–440.

Created on July 14, 2012 12:52:34
by Toby Bartels
(98.23.132.98)