When working with Dominik Kwietniak and Jakub Konieczny, the question appeared:

Let $X$ and $Y$ be two subshifts on the same alphabet, $M(X)$, $M(Y)$ the sets of shift-invariant measures on $X$ and $Y$, respectively, $M_e(X)$ and $M_e(Y)$ be the set of the ergodic ones.

The Besicovitch measure $\bar{d}$ considered on the set of all shift-invariant measures can be lifted in a way the Hausdorff distance is established and we can define the distance between the sets of ergodic measures, or all invariant measures, as follows:

$$\bar{d}(M_e(X),M_e(Y))=\max\left(\sup_{\mu\in M_e(X)}\inf_{\nu\in M_e(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M_e(Y)}\inf_{\mu\in M_e(X)}\bar{d}(\mu,\nu)\right),$$

$$\bar{d}(M(X),M(Y))=\max\left(\sup_{\mu\in M(X)}\inf_{\nu\in M(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M(Y)}\inf_{\mu\in M(X)}\bar{d}(\mu,\nu)\right).$$

If we are not wrong (if someone disagree I can add the details), it is true that $$\bar{d}(M_e(X),M_e(Y))\le \bar{d}(M(X),M(Y)).$$

**The question is whether**
$$\bar{d}(M(X),M(Y))\le \bar{d}(M_e(X),M_e(Y)).$$

The reason why we "claim" it is the observation that it is true when on the left side are the convex hulls of the ergodic measures instead of all invariant measures.

The idea of the proof is to take for given $\mu\in M(X)$ its ergodic decomposition $\mu=\int_{M_e(X)}\mu'd\lambda(\mu')$. Every ergodic $\mu'$ assign with an ergodic measure $\nu'$ on $Y$ that satisfies $$\bar{d}(\mu',\nu')<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$$ It defines a map $f$ from $M_e(X)$ to $M_e(Y)$, $f(\mu')=\nu'$. Now integrate $$\nu=\int_{M_e(X)}f(\mu') d\lambda(\mu')$$ and prove that $\bar{d}(\mu,\nu)<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$

But I do not know how to show measurability of the map $f$. The definition of the mapping is not very constructive and is not unique at all.

Any comment, suggestion would be appreciated!