Ergodicity for the mean of a linear process without finite second moment Suppose that $\{X_k:k\in\mathbb Z\}$ is a linear process, i.e. a sequence of random variables such that
$$
X_k=\sum_{j=0}^\infty\psi_j\varepsilon_{k-j}
$$
for each $k\in\mathbb Z$, where $\{\psi_j:j\ge0\}$ is a real sequence such that $\sum_{j=0}^\infty|\psi_j|<\infty$ and $\{\varepsilon_k:k\in\mathbb Z\}$ are independent and identically distributed random variables with $\operatorname E|\varepsilon_0|<\infty$.

Is the linear process $\{X_k:k\in\mathbb Z\}$ ergodic for the mean, i.e. does the average $n^{-1}\sum_{k=1}^nX_k$ converge almost surely (or in probability) to $\operatorname EX_0$ as $n\to\infty$?

If we assume that $\operatorname E\varepsilon_0^2<\infty$, then $\sum_{j=0}^\infty|\psi_j|<\infty$ implies the ergodicity for the mean (see p. 52 of Time Series Analysis by James D. Hamilton), but I'm interested in the case when we only assume that $\operatorname E|\varepsilon_0|<\infty$.
Any help is much appreciated!
 A: Denote by $\mu$ the expected value of $\varepsilon_0$ and let 
\begin{equation}
S_{n,j} := \frac{1}{n} \sum_{k=1}^n \varepsilon_{k-j} \quad  , \quad \bar{X}_n = \frac{1}{n} \sum_{k=1}^n X_k
\end{equation}
We have, by definition of $X_k$ : 
\begin{equation}
\bar{X}_n = \sum_{j=0}^{\infty} \psi_j S_{n,j} ,
\end{equation}
and thus : 
\begin{equation}
|\bar{X}_n - \mathbb{E}X_0| \leq \sum_{j=0}^{\infty} |\psi_j| |S_{n,j} - \mu|.
\end{equation}
Because $S_{n,j}$ satisfies the assumptions of the strong law of large numbers, we have
\begin{equation}
\mathbb{E}|S_{n,j} - \mu| \to 0 \text{ as } n \to \infty
\end{equation}
and this is true for any fixed $j$.
On the other hand, we have 
$$\mathbb{E}|S_{n,j} - \mu| \leq 2 \mathbb{E}|\varepsilon_0| $$
By the discrete dominated convergence theorem we have :
\begin{equation}
\mathbb{E}|\bar{X}_n - \mathbb{E}X_0| \to 0
\end{equation}
thus giving the convergence in probability of the whole sequence and a.s. convergence up to a subsequence.
On a side note, this post seems interesting, but the missing factor of $n$ in the sum may be a trouble. 
