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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?

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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.

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    $\begingroup$ 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. $\endgroup$ – Julian Newman Aug 3 '16 at 16:13

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