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

Question

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. $f(x_0)=0$.

Answer 1

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.