nLab Radon–Nikodym derivative

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.

The Radon–Nikodym theorem

Definition

Suppose $X$ is a set, $\Sigma$ is a σ-algebra of subsets of $X$, $\mu\colon\Sigma\to[0,\infty]$ is a countably additive measure, and $\nu\colon\Sigma\to\mathbf{R}$ is a finitely additive map. We say that $\nu$ is absolutely continuous with respect to $\mu$ if for any $\epsilon\gt0$ there is $\delta\gt0$ such that $|\nu E|\lt\epsilon$ for any $E\in\Sigma$ such that $\mu(E)\lt\delta$. We say that $\nu$ is truly continuous with respect to $\mu$ if for any $\epsilon\gt0$ we can find $E\in\Sigma$ and $\delta\gt0$ such that $\mu(E)$ is finite and $|\nu(F)|\lt\epsilon$ whenever $F\in\Sigma$ and $\mu(E\cap F)\lt\delta$.

Remark

If we assume $\nu$ to be countably additive, then the definition of absolutely continuity simplifies as follows: $\nu$ is absolutely continuous with respect to $\mu$ if and only if for any $E\in\Sigma$ we have $\nu(E)=0$ whenever $\mu(E)=0$.

Remark

$\nu$ is truly continuous with respect to $\mu$ if and only if $\nu$ it is countable additive, $\nu$ absolutely continuous with respect to $\mu$, and for any $E\in\Sigma$ such that $\nu(E)\gt0$ we can find $F\in\Sigma$ such that $F\subset E$, $\mu(F)$ is finite, and $\nu(F)\gt0$.

Theorem

Suppose $X$ is a set, $\Sigma$ is a σ-algebra of subsets of $X$, $\mu\colon\Sigma\to[0,\infty]$ is a countably additive measure, and $\nu\colon\Sigma\to\mathbf{R}$ is a function. Then there is a $\mu$-integrable function $f$ such that $\nu(E)=\int_E f$ for all $E\in\Sigma$ if and only if $\nu$ is finitely additive and truly continuous with respect to $\mu$.

Corollary

In the context of the above theorem, if $(X,\Sigma,\mu)$ is σ-finite?, then there is a $\mu$-integrable function $f$ such that $\nu(E)=\int_E f$ for all $E\in\Sigma$ if and only if $\nu$ is countably additive and absolutely continuous with respect to $\mu$. If $\mu(X)$ is finite, it suffices to require that $\nu$ is finitely additive.

Example

The following example shows that the condition of being truly continuous is necessary in the non-σ-finite case. Consider an uncountable set $X$, a σ-algebra $\Sigma$ of subsets of $X$ that are countable or have countable complement, the counting measure $\mu\colon\Sigma\to[0,\infty]$, and the measure $\nu\colon\Sigma\to[0,1]$ that vanishes on countable sets and takes value 1 on all uncountable subsets. Then $\nu$ is absolutely continuous with respect to $\mu$ because $\mu$ only vanishes on the empty set. However, $\nu$ is not truly continuous with respect to $\mu$ because $\mu$ takes finite values only on finite sets, but for any such a set $E$ we have $\nu(X\setminus E)=1$, which yields a contradiction if $\epsilon\lt1$.

Example

The measure space in the last example is quite pathological: it is not localizable?. A localizable? example can be constructed if and only if real-valued-measurable cardinals? exist. In this case, take $X$ to be a real-valued-measurable cardinal, $\mu\colon 2^X\to[0,\infty]$ to be the counting measure, and $\nu\colon 2^X\to[0,1]$ a probability measure that vanishes on all countable subsets of $X$. Then $\nu$ is absolutely continuous with respect to $\mu$ because only the empty subset of $X$ has $\mu$-measure 0. However, $\nu$ is not truly continuous with respect to $\mu$ because $\mu$ takes finite values only on finite sets, but for any such a set $E$ we have $\nu(X\setminus E)=1$, which yields a contradiction if $\epsilon\lt1$.

Remark

Truly continuous countably (or finitely) additive measures on a measure space $(X,\Sigma,\mu)$ form a module over the complex *-algebra of measurable maps $X\to\mathbf{R}$ modulo equality almost everywhere. The Radon–Nikodym theorem then says that this module is a free module of rank 1. Furthermore, generators of this module can be identified with truly continuous measures $\nu$ such that $\mu$ is absolutely continuous with respect to $\nu$.

The Lebesgue decomposition

Theorem

Suppose $X$ is a set, $\Sigma$ is a σ-algebra of subsets of $X$, $\mu\colon\Sigma\to[0,\infty]$ is a countably additive measure, and $\nu\colon\Sigma\to\mathbf{R}$ is a countably additive function. Then there is a unique decomposition

$$\nu=\nu_s+\nu_t+\nu_e,$$

where $\nu_t$ is truly continuous with respect to $\mu$, $\nu_e$ is absolutely continuous with respect to $\mu$ and vanishes on every measurable set of finite $\mu$-measure, and $\nu_s$ is singular with respect to $\mu$, meaning there is a set $F\in\Sigma$ such that $\mu(F)=0$ and $\nu(E)=0$ for all $E\in\Sigma$ such that $E\subset X\setminus F$. In particular, setting $\nu_a=\nu_t+\nu_e$ yields a unique decomposition $\nu=\nu_s+\nu_a$ of $\nu$ as a sum of a singular and absolutely continuous measure.

Definitions

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 $X$.

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 $\nu$:

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

(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

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.

References

A comprehensive treatment can be found in Chapter 23 of

• David H. Fremlin, Measure Theory.

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.