## Idea

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.

## Definitions

Let $X$ be a measurable space (so $X$ consists of a set $\mid X\mid$ and a $\sigma$-algebra ${ℳ}_{X}$), 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 $X$.

###### Definition

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 $\nu$:

$\mu \left(A\right)={\int }_{A}f\nu ={\int }_{x\in A}f\left(x\right)\nu \left(\mathrm{d}x\right).$\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.)

## Properties

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

###### Theorem

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 $\nu$.

###### Theorem

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

###### Theorem

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.

###### Proofs

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 $\mu$.

## Notation

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

$\mu \left(A\right)={\int }_{A}f\nu ,$\mu(A) = \int_A f \nu ,

or equivalently

${\int }_{A}\mu ={\int }_{A}f\nu .$\int_A \mu = \int_A f \nu .

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

$\mu =f\nu ;$\mu = f \nu ;

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

$f=\mu /\nu .$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

${\int }_{A}\mathrm{d}\mu ={\int }_{A}f\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\nu ,$\int_A \mathrm{d}\mu = \int_A f \,\mathrm{d}\nu ,

$f=\mathrm{d}\mu /\mathrm{d}\nu$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 ‘$\mathrm{d}$’:

${\int }_{x\in A}\mu \left(\mathrm{d}x\right)={\int }_{x\in A}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\nu \left(\mathrm{d}x\right);$\int_{x \in A} \mu(\mathrm{d}x) = \int_{x \in A} f(x) \,\nu(\mathrm{d}x) ;

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

which does not give a symbol for $f$ directly. If instead of $\mu \left(\mathrm{d}x\right)$ one unwisely writes $\mathrm{d}\mu \left(x\right)$, 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

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

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

$\mu /\nu =F\prime =\mathrm{d}F/\mathrm{d}t.$\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

$\mu =\mathrm{d}F$\mu = \mathrm{d}F

and

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

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

## References

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

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)