# Given a Borel measure $\mu$ on a compact metric space, is there a continuous map $T$ such that $T$ is $\mu$ ergodic?

Given a continuous (surjective) map of a compact metric space, the Kryloff-Bogolioubuff argument shows that there is at least $T$ invariant measure and hence one or uncountably many measures for which $T$ is ergodic. Is every Borel measure ergodic (and invariant) for some continuous map?

An ergodic invariant measure associated to a periodic orbit is uniform over that orbit. Therefore, a finitely supported measure that is not uniform over its support is not an ergodic invariant measure for any measurable transformation.

Update: I suspect that the answer remains negative even if you assume the measure to be non-atomic, but I don't know a counter-example. However, if you do not require the map to be continuous (but merely measurable), then the answer becomes positive: every non-atomic probability measure on a compact metric space is ergodic for some measurable transformation.

Namely, if $X$ is a complete separable metric space and $\mu$ is a non-atomic Borel probability measure on $X$, then the measure space $(X,\mu)$ is isomorphic mod $0$ to the unit interval with the Lebesgue measure. Take any map on the unit interval for which the Lebesgue measure is ergodic (say an irrational rotation) and lift it to $X$.

• I found essentially the same thing with a measure that's uniform over some portion of the unit interval and putting the rest of the mass on a single point. What if we require no point masses or even non-atomic? – 3-in-441 Dec 26 '16 at 22:02

Any constant map is continuous and the only invariant sets are $\emptyset$ and the whole space. Hence, any measure is ergodic with respect to a constant map.

But I suspect that you meant invariant ergodic measures. Nice question but that is a whole different matter and I am strictly answering your question. However, note that ergodic theory doesn't start with invariant measures. For example, a measure is ergodic if and only if any function that is invariant almost everywhere is in fact constant almost everywhere. This does not require the measure to be invariant!

Just a little correction: Krylov-Bogolubov's theorem requires a compact metric space.

• Thanks for the response. I did include compact metric space in the title and will edit for clarity. I did mean invariant measure! Do you have any idea if it is true? Thanks for pointing this out to me. – 3-in-441 Dec 24 '16 at 0:07
• I don't know. Some years ago a co-author suggested your question (or perhaps without ergodicity but with invariance, it was long ago and I don't remember well). After a few discussions we went on to other things. I regret that as I recall we weren't able to add anything that I could transmit. – John B Dec 24 '16 at 0:25