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The Radon-Nikodym Theorem

May 27, 2019

Let $\mu$ and $\nu$ be two measures on the same measurable space $(\Omega, \Sigma)$. We say that $\mu$ is absolutely continuous with respect to $\nu$ (written $\mu \ll \nu$) if $\mu(E) = 0$ whenever $\nu(E) = 0$ for $E \in \Sigma$. Given $\nu$, there is a simple way to generate a new measure which is absolutely continuous with respect to $\nu$: take any measurable function $g \colon \Omega \to \R$ and define

$$\mu_g(E) = \int_E g\, d\nu$$

It is a standard exercise in measure theory that $\mu_g$ is a measure and that $\mu_g(E) = 0$ whenever $\nu(E) = 0$.

The Radon-Nikodym theorem asserts that under mild hypotheses every measure $\mu$ which is absolutely continuous with respect to $\nu$ has this form for a unique measurable function $g$, called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$, often written $\frac{d\mu}{d\nu}$. The theorem is an important part of the structure theory of measures, and the Radon-Nikodym derivative comes up frequently in applications. Indeed, I was reminded of this theorem because of its role in another blog post that I was writing about KL-divergence.

The plan for this post is to give a proof of the Radon-Nikodym theorem and then discuss some applications and examples. The proof, which I learned in graduate school, uses Hilbert spaces to do most of the work; I think it is originally due to von Neumann.

Warm-up

The proof of the theorem in its full generality inevitably uses some reasonably serious measure theory and/or functional analysis. But I think it is instructive to first work it out in a simple special case: measures on finite sets.

So let $\Omega = \br{x_1, \ldots x_n}$ be a finite set and let $\Sigma$ be the power set of $\Omega$. A measure $\mu$ on $\Omega$ is determined by its values on each one point set $\br{x_i}$, so it is nothing more than a vector in $\R^n$ whose entries are all nonnegative. Let us use the notation $\mu_i$ for the components of $\mu$, meaning $\mu_i = \mu(\br{x_i})$.

Given two measures $\mu$ and $\nu$, we have that $\mu$ is absolutely continuous with respect to $\nu$ if and only if $\mu_i = 0$ whenever $\nu_i = 0$. We would like to find a measurable function $g \colon \Omega \to \R$ - which is also just a vector in $\R^n$ - with the property that:

$$\mu(E) = \int_E g\, d\nu$$

for every $E \in \Sigma$. Applying this to the one point sets $E_i = \br{x_i}$, we get:

$$\mu_i = \mu(E_i) = \int_{E_i} g\, d\nu = g_i \nu_i$$

This yields a simple formula for the Radon-Nikodym derivative:

$$\frac{d\mu}{d\nu}(x_i) = \begin{cases} \frac{\mu_i}{\nu_i} & \nu_i \neq 0 \\ 0 & \text{otherwise} \end{cases}$$

In fact this formula works for any $\sigma$-algebra on a finite set - not just the power set - so the theorem on finite sets is quite simple and explicit. Moreover if we look at this calculation in the right way it gives us a hint for how to handle the general case. The equation $\mu_i = g_i \nu_i$ implies that for any function $f$ on $\Omega$ we have

$$\int_\Omega f\, d\mu = \int_\Omega f g\, d\nu = \inner{f, g}_\nu$$

where $\inner{\cdot, \cdot}_\nu$ is the $L^2$-inner product determined by $\nu$. So another way to construct $g$ is to apply the Riesz representation theorem to the linear functional $f \mapsto \int_\Omega f\, d\mu$ on the finite dimensional Hilbert space $L^2(\Omega, \nu)$. This idea doesn’t quite work as stated in the general case because there is no guarantee that the linear functional is bounded, but it can be fixed by working instead over the Hilbert space $L^2(\Omega, \mu + \nu)$. The details appear in the next section.

Proof of the Radon-Nikodym theorem

Without further ado, let us jump into the proof.

Let $\mu$ and $\nu$ be finite measures on a measurable space $(\Omega, \Sigma)$. If $\mu$ is absolutely continuous with respect to $\nu$ then there is a unique $g \in L^1(\Omega, \nu)$ with the property that:

$$\mu(E) = \int_E g\, d\nu$$

for every $E \in \Sigma$.

Consider the Hilbert space $H = L^2(\mu + \nu)$. Define a linear functional $\phi \colon H \to \R$ by

$$\phi(f) = \int_\Omega f\, d\mu$$

Using the Schwarz inequality and the fact that $\mu \leq \mu + \nu$, we get:

$$\abs{\phi(f)}^2 \leq \left( \int_\Omega 1 \cdot \abs{f}\, d\mu \right)^2 \leq \mu(\Omega) \norm{f}_{L^2(\mu)}^2 \leq \mu(\Omega) \norm{f}_H^2$$

This shows that $\phi$ is a bounded linear functional, so by the Riesz representation theorem there is a function $h \in H$ with the property that:

$$\begin{equation} \label{riesz1} \int_\Omega f\, d\mu = \phi(f) = \inner{f, h} = \int_\Omega fh\, d(\mu + \nu) \end{equation}$$

It follows that:

$$\int_\Omega f\, d(\mu + \nu) = \int_\Omega f\, d\nu + \int_\Omega fh\, d(\mu + \nu)$$

Rearranging, we get:

$$\begin{equation} \label{riesz2} \int_\Omega f\, d\nu = \int_\Omega f(1 - h)\, d(\mu + \nu) \end{equation}$$

Now, by $\eqref{riesz1}$ we have for any measurable set $E$:

$$\nu(E) = \int_\Omega 1_E\, d\nu = \int_\Omega 1_E h\, d(\mu + \nu) = \int_E h\, d(\mu + \nu)$$

Consider the case $E = h^{-1}(-\infty, 0]$. Then $\int_E h\, d(\mu + \nu) \leq 0$ since $h \leq 0$ on $E$, so we must have $\nu(E) = 0$. A similar argument using the $h^{-1}[1, \infty)$ and $\eqref{riesz2}$ implies that $0 < h < 1$ almost everywhere with respect to $\nu$. Consequently the function $g = \frac{h}{1 - h}$ is $\nu$-measurable; I claim that $g$ satisfies the conditions in the statement of the theorem.

Indeed, for any measurable set $E$ we have:

$$\begin{align*} \int_E g\, d\nu &= \int_\Omega g 1_E\, d\nu \\ &= \int_\Omega g 1_E (1 - h)\, d(\mu + \nu) \quad \text{by \eqref{riesz2}} \\ &= \int_\Omega h 1_E\, d(\mu + \nu) \\ &= \int_\Omega 1_E\, d\mu \quad \text{by \eqref{riesz1}} \\ &= \mu(E) \end{align*}$$

(This also proves that $g$ really is $\nu$-integrable: $g$ is the increasing limit of the $\nu$-integrable functions $g 1_{E_n}$ where $E_n = h^{-1}(0, 1 - \frac{1}{n})$, and the computation above shows that $\int_\Omega g 1_{E_n}\, d\nu = \mu(E_n) \to \mu(\Omega)$. So $g$ is integrable by the monotone convergence theorem.)

This completes the proof that the desired function $g$ exists, so it remains only to prove that $g$ is unique. Suppose $g’$ is another $\nu$-integrable function with the property that $\mu(E) = \int_E g’\, d\nu$ for every measurable set $E$. Let $E$ denote the subset of $\Omega$ where $g > g’$; we have:

$$0 = \mu(E) - \mu(E) = \int_E (g - g')\, d\nu$$

This forces $\nu(E) = 0$. Repeating this argument on the set where $g’ > g$ implies that $g = g’$ almost everywhere with respect to $\nu$.

Concluding remarks

The Radon-Nikodym theorem admits some modest generalizations. It can be proved for $\sigma$-finite measures by applying the theorem to an increasing union of sets on which both $\mu$ and $\nu$ are finite. It also extends to signed measures by decomposing them into their positive and negative parts, and it extends to complex measures by considering their real and imaginary parts.

But for the moment I am mostly interested in applications to probability theory, for which the formulation of the theorem presented here is sufficient. In follow-up posts I plan to explore two such applications in particular: the conditional expectation of a random variable and the KL-divergence.