Every ergodic invariant measure of one dimensional dynamical system is a Dirac measure

Recently, I have learned some theorems about attractors and invariant measures. In the book I am reading, there is a theorem presented without its proof. I am interested in how to prove it.

Recall some definitions we will use below. Let we have dynamical system $$\{\varphi_{t},t\in\mathbb{R}\}$$. Then,

1. Invariant measure $$\mu$$: for any $$A\in\mathcal{B}(\mathbb{R})$$, $$\mu(A)=\mu(\varphi_{t}^{-1}(A))$$, $$t\in\mathbb{R}.$$
2. Ergodic meausre $$\mu$$: for any $$A\in\mathcal{B}(\mathbb{R})$$ with $$\varphi_{t}^{-1}(A)=A, t\in\mathbb{R}$$, one has $$\mu(A)=0$$ or $$1$$.

The theorem whose proof I am interested in is as follows:

Theorem Consider the ODE $$\dot{x}(t)=f(x(t))$$, wit some $$x(0)=x_0\in\mathbb{R},f\in C^{1}(\mathbb{R})$$ and let $$\varphi_t$$ be the associated dynamical system. If $$\mu$$ is an ergodic invariant measure of this dynamical system, then $$\mu=\delta_{x_0}$$, where $$x_0$$ is an equilibrium, i.e. $$f(x_0)=0$$.

• Note that the flow of the specific example can be found explicitly by integration. But the only invariant sets are the zeros of $f$
– lcv
Commented Mar 7 at 11:10

Let $$U$$ be a connected component of the complementary of the set of equilibrium points. Then $$U$$ is an open interval, on which $$f$$ does not vanish.

The general theory of ordinary differential equations tells you that the restriction of $$(\phi_t)_t$$ to $$U$$ is conjugated to the translation flow on $$\mathbb{R}$$, which notoriously has no finite invariant measures, so neither does the flow on $$U$$.

Therefore, any finite invariant measure is supported on the set of equilibrium points, and the claims follows easily.

• What do you mean with "the flow is conjugated with the translation flow on $\mathbb R$"?
– lcv
Commented Mar 10 at 8:42
• That there exists a diffeomorphism $h$ from $\mathbb{R}$ to $U$ such that for all $t,s$, $h(t+s)= \phi_s(h(t))$. In loose words, acting by the flow on $U$ is the same thing as acting on $\mathbb{R}$ by translations, up to a diffeomorphism.
– Plop
Commented Mar 10 at 10:57
• But that's general to all (regular) flows no?
– lcv
Commented Mar 10 at 11:09
• No, it’s not true for the translation flow on $\mathbb{R}/\mathbb{Z}$.
– Plop
Commented Mar 10 at 19:22