Suppose $\phi = \{\phi_{t}\}_{t\in \mathbb{R}}$ is a flow on a probability space $(\Omega, \mathcal{F},\mu)$. Suppose $\mu$ is stationary with respect to $\phi$, that is, $\mu(A) = \mu(\phi_{t}^{-1}(A))$ holds for every $A \in \mathcal{F}$ and $t \in \mathbb{R}$. We say that $\phi$ is ergodic if every measurable set $F \in \mathcal{F}$ satisfying $\phi_{t}(F) = F$ has either probability $\mu(F) = 0$ or $\mu(F) =1$.
I know that this definition implies the following result: $\phi$ is ergodic iff the only functions $f \in L^{2}(\Omega,d\mu)$ satisfying $f\circ \phi_{t} = f$ are the constant functions.
I want to prove the following: $\phi$ is ergodic if, and only if for every $f \in L^{1}(\Omega,d\mu)$ the time average: $$\bar{f}(x) := \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^{T}(f\circ\phi_{t})(x)dt \tag{1}\label{1}$$ exists and is a constant $\bar{f}$, given by the expectation: $$\bar{f} = \int f(x)d\mu(x). \tag{2}\label{2}$$
I already proved one direction. If $f \in L^{1}(\Omega,d\mu)$, by Birkhoff Ergodic Theorem the time average limit (\ref{1}) exists and equals its expectation. In particular, this condition implies $\bar{f}\circ \phi_{t} = \bar{f}$. Hence, if $\phi$ is ergodic, because $L^{1}(\Omega,d\mu) \subset L^{2}(\Omega,d\mu)$ we get $\bar{f}(x) = \bar{f}$ is a constant and equals its expectation.
However, I am stuck at proving the converse, that is, if the time average $\bar{f}(x)$ in (\ref{1}) exists, and is a constant given by (\ref{2}), then $\phi$ is ergodic. Any help is welcome!