In this post I prove a result on absolutely continuous ergodic measures.

Set Up

Let \(X\) be a topological space. We will restrict ourselves to studying the Borel \(\sigma\)-algebra on \(X\) for now (although the results should be true for other \(\sigma\)-algebras). Recall that this is the \(\sigma\)-algebra generated by the open sets of \(X\). We will fix a transformation \(T : X \rightarrow X\) for now. Recall that a measure is said to be ergodic with respect to \(T\) (or vice versa, \(T\) is ergodic with respect to \(\mu\)) if the following hold:

• \(\mu(T^{-1}(A)) = \mu(A)\) for all \(A\) measurable.

• If \(T^{-1}(A) \subseteq A\) then we either have \(\mu(A) = 0\) or \(1\).

The first condition above is generally referred to as measure preserving (so \(T\) is measure preserving with respect to \(\mu\)). There are many equivalent definitions for a measure preserving transformation to be ergodic. The one of interest to us is the following.

Observation: Let \(T : X \rightarrow X\) be a measure preserving transformation with respect to the measure \(\mu\). We have that \(T\) is ergodic with respect to \(\mu\) if and only if every \(f \in L^1(\mu)\) which is \(T\)-invariant is constant almost everywhere.

Proof: We break this up into steps.

\((\implies):\) Assume that \(T\) is ergodic. Let \(f \in L^1(\mu)\) be \(T\)-invariant, so \(f \circ T = f\) almost everywhere. Assume that \(f\) is not constant almost everywhere for contradiction. There is a constant \(c \in \mathbb{R}\) so that if we let \(A := \{f \leq c\}\) then \(0 < \mu(A) < 1\). Since \(f \circ T = f\) almost everywhere, we get that

\[T^{-1}(A) = \{ x \in X \ | \ f(T(x)) \geq c\} = A,\]

at least up to a set of measure \(0\). This gives us the contradiction, since ergodicity forces either \(\mu(A) = 0\) or \(1\).

\((\impliedby):\) Assume that the condition holds for every \(f \in L^1(\mu)\). In particular, it holds for characteristic functions. Recall \(\chi_A \circ T(x) = \chi_{T^{-1}(A)}(x),\) so \(\chi_A\) being \(T\)-invariant implies that \(\chi_{T^{-1}(A)} = \chi_A\) which implies that \(\chi_A\) is constant almost everywhere, hence it is almost everywhere \(0\) or \(1\). Now we integrate this to get that \(\mu(A) = 0\) or \(\mu(A) = 1\). \(\blacksquare\)

Let \(\lambda\) be a measure on \(X\). We say that a measure \(\mu\) is absolutely continuous with respect to \(\lambda\), which we denote by \(\mu \ll \lambda\), if for every measurable \(A\) with \(\lambda(A) = 0\) we have \(\mu(A) = 0\).

Our goal is to prove the following result.

Proposition: If \(\mu\) is an ergodic \(T\)-invariant probability measure and \(\lambda\) is a \(T\)-invariant probability measure with \(\lambda \ll \mu\), then \(\lambda = \mu\).

In other words, ergodic measures are “minimal” with respect to this domination property.

To prove this, we will need the following theorem from measure theory.

Theorem: If \(\lambda \ll \mu\) then there is a measurable density function \(\rho : X \rightarrow [0,\infty)\) so that

\[\lambda(A) = \int_A \rho d\mu.\]

This \(\rho\) is called the Radon-Nikodym derivative.

We will also need the following exercise from measure theory.

Observation: Let \(\mu\) be a probability measure on \(X\). Suppose that \(f, g\) are integrable. If \(\int_A fd\mu = \int_A g d\mu\) for every measurable set \(A\), then \(f = g\) almost everywhere.

Proof: It suffices to show that if \(\int_A f d\mu = 0\) for every measurable set, then \(f =0\) almost everywhere. Observe first that this implies that

\[\int_A | f | d\mu = 0\]

for every measurable set \(A\). To see why, let \(A_0 = \{f > 0\}\) and \(A_1 = \{f \leq 0\}\). Then we have

\[0 = \int_{A \cap A_0} f d\mu = \int_{A \cap A_0} |f| d\mu,\] \[0 = \int_{A \cap A_1} f d\mu = - \int_{A \cap A_1} |f| d\mu,\]

hence

\[\int_A |f| d\mu = \int_{A \cap A_0} |f| d\mu + \int_{A \cap A_1} |f| d\mu = 0.\]

Now suppose that \(f\) is not zero almost everywhere. There is a constant \(c > 0\) so that

\[A := \{|f| > c\}\]

has strictly positive measure. Now we integrate

\[0 = \int_A |f| > \int_A c = \mu(A) c > 0.\]

This is a contradiction, so we must have that \(f\) is zero almost everywhere. \(\blacksquare\)

Proof of Result

We now give a short proof.

Observe by the usual density argument for characteristic functions we have that for any \(f\) which is integrable the following holds:

\[\int \rho f d\mu = \int f d\lambda.\]

Now assume that \(f\) is integrable. We see that

\[\int_X \rho f d\mu = \int_X f d\lambda = \int_X f d(\lambda \circ T^{-1}) = \int_X f \circ T d\lambda = \int_X \rho (f \circ T) d\mu = \int_X \rho \circ T^{-1} f d\mu.\]

Now if we take the characteristic functions of measurable sets, we see that for every measurable set \(A\) the following holds:

\[\int_A \rho d\mu = \int_A \rho \circ T^{-1} d\mu.\]

Thus by the observation in the last section we have that \(\rho = \rho \circ T^{-1}\) almost everywhere. Ergodicity of \(\mu\) implies that \(\rho\) is a constant almost everywhere, say \(\rho \equiv c\) almost everywhere. Finally, these are probability measures, so

\[1 = \lambda(X) = \int_{X} \rho d\mu = c \mu(X) = c.\]

Hence \(\rho \equiv 1\) almost everywhere and we have that \(\mu = \lambda\). \(\blacksquare\)