3
$\begingroup$

Let $(X,\Sigma)$ be a measurable space [which we can assume to be a standard Borel space if we wish].

Let $\mathcal{S}$ be a set of probability measures on $(X,\Sigma)$. [If we wish, we can assume that $\mathcal{S}$ is an element of the $\sigma$-algebra on the space of probability measures on $(X,\Sigma)$ generated by the evaluation mappings $\{\mu \mapsto \mu(A) : A \in \Sigma\}$.]

The set of measures $\mathcal{S}$ is said to be mutually singular if for any distinct $\mu_1,\mu_2 \in \mathcal{S}$, the measures $\mu_1$ and $\mu_2$ are mutually singular. So, for example, the set of Borel measures on $[0,1]$ given by $$ \mathcal{S} \ := \ \{\delta_x : x \in [0,1]\} \cup \{\mathrm{Lebesgue}\} $$ is mutually singular.

Now I can think of some (seemingly) highly natural ways of strengthening the definition of a "mutually singular" collection of measures, such that the above example no longer qualifies:

Definition. We say that $\mathcal{S}$ is A-strongly mutually singular if there exists a function $L \colon \mathcal{S} \to \Sigma$ such that for any distinct $\mu_1,\mu_2 \in \mathcal{S}$, $\mu_1(L(\mu_1))=1$ and $\mu_1(L(\mu_2))=0$.

Definition. We say that $\mathcal{S}$ is measurably A-strongly mutually singular if there exists a function $L \colon \mathcal{S} \to \Sigma$ as in the previous definition, with the additional property that the set $\{(\mu,x):x \in L(\mu)\}$ is a measurable subset of $\mathcal{S} \times X$ (where $\mathcal{S}$ is equipped with the $\sigma$-algebra generated by the evaluation maps $\{\mu \mapsto \mu(A) : A \in \Sigma\}$).

Definition. We say that $\mathcal{S}$ is B-strongly mutually singular if for any two mutually singular probability measures $Q_1$ and $Q_2$ on $\mathcal{S}$ (equipped with the $\sigma$-algebra generated by $\{\mu \mapsto \mu(A) : A \in \Sigma\}$), the probability measures $$ A \mapsto \int_\mathcal{S} \mu(A) \, Q_1(d\mu) \hspace{5mm} \textrm{and} \hspace{5mm} A \mapsto \int_\mathcal{S} \mu(A) \, Q_2(d\mu) $$ on $(X,\Sigma)$ are mutually singular.

[I suspect that under the standardness assumptions on $X$ and $\mathcal{S}$, being measurably A-strongly mutually singular implies being B-strongly mutually singular; but I haven't yet managed to prove it.]

My questions:

Have any of the above (or similar) stronger forms of "mutually singular" been studied before? Do they have standard names?

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a measurable map admitting uncountably many ergodic probability measures, and let $\mathcal{S}$ be the set of ergodic probability measures of $f$. Is $\mathcal{S}$ necessarily "strongly" mutually singular under any of the above (or similar) definitions?

$\endgroup$

1 Answer 1

4
$\begingroup$

The set of all $f$-invariant Borel probability measures on $\mathbb{R}$ is $A$-strongly mutually singular. To see this let $L(\mu)$ be the set of all points $x$ such that $\frac{1}{n}\sum_{k=0}^{n-1}\phi(f^k(x)) \to \int \phi\,d\mu$ for all compactly supported continuous $\phi \colon \mathbb{R}\to \mathbb{R}$. Note that by the Weierstrass approximation theorem $\frac{1}{n}\sum_{k=0}^{n-1}\phi(f^k(x)) \to \int \phi\,d\mu$ for all $\phi \in C([-N,N])$ iff this holds for all polynomial functions $\phi \in C([-N,N])$ with rational coefficients. Thus there exists a sequence $(\phi_m)$ of compactly supported continuous functions such that $\frac{1}{n}\sum_{k=0}^{n-1}\phi(f^k(x)) \to \int \phi\,d\mu$ for all compactly supported continuous $\phi \colon \mathbb{R}\to \mathbb{R}$ if and only if $\frac{1}{n}\sum_{k=0}^{n-1}\phi_m(f^k(x)) \to \int \phi_m\,d\mu$ for all $m \geq 1$. Using the Birkhoff ergodic theorem it follows that $\mu(L(\mu))=1$ for all ergodic invariant $\mu$. On the other hand if $\nu_1,\nu_2$ are distinct ergodic invariant measures then there exists continuous compactly supported $\phi \colon \mathbb{R} \to\mathbb{R}$ such that $\int \phi\,d\nu_1 \neq \int \phi\,d\nu_2$, and therefore $L(\nu_1)$ belongs to the exceptional set for the Birkhoff ergodic theorem applied to $\phi$, $\nu_2$ and in particular has $\nu_2$-measure zero.

I believe that with a little more effort this could be extended to measurable $A$-strong mutual singularity just by using the definition of convergence to express the set $\{(\mu,x):x\in L(\mu)\}$ using countably many set operations.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you! Alternatively: let $\mathcal{C} \subset \mathcal{B}(\mathbb{R})$ be a countable $\pi$-system generating $\mathcal{B}(\mathbb{R})$, and let $L(\mu)$ be the set of points $x$ such that for all $C \in \mathcal{C}$, $\frac{1}{n}\sum_{k=0}^{n-1}\mathbf{1}_C(f^k(x)) \to \mu(C)$. Measurable $A$-strong mutual singularity seems clear (using this same $L$); and moreover, since the sets $\{L(\mu):\mu \in \mathcal{S}\}$ are mutually disjoint, it should be easy to derive $B$-strong mutual singularity as a consequence. $\endgroup$ Aug 3, 2016 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.