[proofplan]
We construct $X_t$ as the $L^2$ limit of finite linear combinations of the white-noise variables. The key estimate is that the $L^2$ distance between two finite partial sums is exactly $\sigma^2$ times the squared $\ell^2$ tail of the coefficient sequence, because different white-noise coordinates are uncorrelated. Absolute summability of $(\psi_j)$ implies square summability, so the finite sums are Cauchy in $L^2$. We then compute expectations and covariances at the finite-sum level and pass to the $L^2$ limit using Cauchy-Schwarz continuity of expectation and covariance.
[/proofplan]
[step:Convert absolute summability into square summability of the coefficients]
Define the coefficient mass $A \in [0,\infty)$ by
\begin{align*}
A = \sum_{j \in \mathbb Z} |\psi_j|.
\end{align*}
Since $|\psi_j| \leq A$ for every $j \in \mathbb Z$, we have
\begin{align*}
\sum_{j \in \mathbb Z} \psi_j^2 \leq A \sum_{j \in \mathbb Z} |\psi_j| = A^2 < \infty.
\end{align*}
Thus $(\psi_j)_{j \in \mathbb Z} \in \ell^2(\mathbb Z)$.
[guided]
We first need square summability because $L^2$ convergence is controlled by second moments. The hypothesis gives absolute summability, not square summability directly, so we prove the implication explicitly.
Define
\begin{align*}
A = \sum_{j \in \mathbb Z} |\psi_j|.
\end{align*}
The number $A$ is finite by hypothesis. For each fixed $j \in \mathbb Z$, the single term $|\psi_j|$ is bounded by the whole non-negative sum, hence $|\psi_j| \leq A$. Multiplying this inequality by $|\psi_j|$ gives
\begin{align*}
\psi_j^2 = |\psi_j|^2 \leq A|\psi_j|.
\end{align*}
Summing over $j \in \mathbb Z$ yields
\begin{align*}
\sum_{j \in \mathbb Z} \psi_j^2 \leq A \sum_{j \in \mathbb Z} |\psi_j| = A^2 < \infty.
\end{align*}
Therefore the coefficient sequence belongs to $\ell^2(\mathbb Z)$. This is the estimate that will make the finite partial sums Cauchy in $L^2$.
[/guided]
[/step]
[step:Build the finite partial sums and prove they are Cauchy in $L^2$]
Fix $t \in \mathbb Z$. For each finite set $F \subset \mathbb Z$, define the finite partial-sum [random variable](/page/Random%20Variable) $S_{F,t}:\Omega \to \mathbb R$ by
\begin{align*}
S_{F,t}(\omega) = \sum_{j \in F} \psi_j Z_{t-j}(\omega).
\end{align*}
Let $F,G \subset \mathbb Z$ be finite. Define the coefficient function $c_{F,G}:\mathbb Z \to \mathbb R$ by $c_{F,G}(j)=1$ for $j\in F\setminus G$, $c_{F,G}(j)=-1$ for $j\in G\setminus F$, and $c_{F,G}(j)=0$ otherwise. Then
\begin{align*}
S_{F,t}-S_{G,t}=\sum_{j\in F\triangle G} c_{F,G}(j)\psi_j Z_{t-j}.
\end{align*}
Since $\mathbb E[Z_{t-j}]=0$ and $\mathbb E[Z_{t-j}Z_{t-k}]=\sigma^2\mathbb{1}_{\{j=k\}}$, bilinearity of expectation gives
\begin{align*}
\mathbb E[(S_{F,t}-S_{G,t})^2] = \sigma^2 \sum_{j \in F \triangle G} c_{F,G}(j)^2\psi_j^2 = \sigma^2 \sum_{j \in F \triangle G} \psi_j^2.
\end{align*}
Here $F \triangle G = (F\setminus G)\cup(G\setminus F)$ denotes the symmetric difference, and $c_{F,G}(j)^2=1$ on $F\triangle G$.
Let $\varepsilon>0$. Since $\sum_{j \in \mathbb Z}\psi_j^2<\infty$, there exists a finite set $K\subset\mathbb Z$ such that
\begin{align*}
\sigma^2 \sum_{j \in \mathbb Z\setminus K}\psi_j^2 < \varepsilon^2.
\end{align*}
Whenever $F$ and $G$ are finite subsets of $\mathbb Z$ containing $K$, we have $F\triangle G\subset \mathbb Z\setminus K$, hence
\begin{align*}
\|S_{F,t}-S_{G,t}\|_{L^2(\Omega,\mathcal F,\mathbb P)}^2 = \mathbb E[(S_{F,t}-S_{G,t})^2] \leq \sigma^2 \sum_{j \in \mathbb Z\setminus K}\psi_j^2 < \varepsilon^2.
\end{align*}
Thus $(S_{F,t})_F$ is a Cauchy net in $L^2(\Omega,\mathcal F,\mathbb P)$. Since $L^2(\Omega,\mathcal F,\mathbb P)$ is complete, there exists $X_t \in L^2(\Omega,\mathcal F,\mathbb P)$ such that $S_{F,t} \to X_t$ in $L^2$ as $F$ increases through finite subsets of $\mathbb Z$. This is precisely the $L^2(\Omega,\mathcal F,\mathbb P)$ convergence of
\begin{align*}
\sum_{j \in \mathbb Z}\psi_j Z_{t-j}.
\end{align*}
[/step]
[step:Pass the zero mean from finite sums to the limit]
For every finite $F\subset\mathbb Z$,
\begin{align*}
\mathbb E[S_{F,t}] = \sum_{j \in F}\psi_j\mathbb E[Z_{t-j}] = 0.
\end{align*}
Since $S_{F,t}\to X_t$ in $L^2$, Cauchy-Schwarz gives
\begin{align*}
|\mathbb E[X_t]-\mathbb E[S_{F,t}]| = |\mathbb E[X_t-S_{F,t}]| \leq \|X_t-S_{F,t}\|_{L^2(\Omega)}.
\end{align*}
The right-hand side tends to $0$, so $\mathbb E[X_t]=0$.
[/step]
[step:Compute finite-sum covariances and identify the limiting series]
Fix $t,h\in\mathbb Z$. For finite sets $F,G\subset\mathbb Z$, use the already defined random variables $S_{F,t+h}$ and $S_{G,t}$. Since both finite sums have mean $0$,
\begin{align*}
\operatorname{Cov}(S_{F,t+h},S_{G,t}) = \mathbb E[S_{F,t+h}S_{G,t}].
\end{align*}
Expanding the product and using the white-noise covariance relation gives
\begin{align*}
\mathbb E[S_{F,t+h}S_{G,t}] = \sigma^2 \sum_{\substack{k\in F, \, j\in G, k=j+h}} \psi_k\psi_j.
\end{align*}
Equivalently,
\begin{align*}
\operatorname{Cov}(S_{F,t+h},S_{G,t}) = \sigma^2 \sum_{j\in G:\,j+h\in F}\psi_{j+h}\psi_j.
\end{align*}
The infinite series on the right is absolutely convergent, because $|\psi_{j+h}| \leq A$ for every $j\in\mathbb Z$ and hence
\begin{align*}
\sum_{j\in\mathbb Z}|\psi_{j+h}\psi_j| \leq A\sum_{j\in\mathbb Z}|\psi_j| = A^2 < \infty.
\end{align*}
We now identify the limit of the finite covariance sums. Let $\varepsilon>0$. By absolute convergence, there exists a finite set $H\subset\mathbb Z$ such that
\begin{align*}
\sigma^2\sum_{j\in\mathbb Z\setminus H}|\psi_{j+h}\psi_j| < \varepsilon.
\end{align*}
Define the shifted finite set $H+h := \{j+h : j\in H\}\subset\mathbb Z$. If $G$ contains $H$ and $F$ contains $H+h$, then the index set $\{j\in G : j+h\in F\}$ contains $H$, and therefore
\begin{align*}
\left|\sigma^2\sum_{j\in\mathbb Z}\psi_{j+h}\psi_j - \sigma^2\sum_{j\in G:\,j+h\in F}\psi_{j+h}\psi_j\right| \leq \sigma^2\sum_{j\in\mathbb Z\setminus H}|\psi_{j+h}\psi_j| < \varepsilon.
\end{align*}
Thus the finite sums converge to
\begin{align*}
\sigma^2\sum_{j\in\mathbb Z}\psi_{j+h}\psi_j
\end{align*}
as $F$ and $G$ increase through finite subsets of $\mathbb Z$.
[/step]
[step:Use $L^2$ continuity of covariance to pass to the process]
We show that covariance passes to the $L^2$ limit. Since $S_{F,t+h}\to X_{t+h}$ and $S_{G,t}\to X_t$ in $L^2(\Omega,\mathcal F,\mathbb P)$, and all four random variables have mean $0$, we estimate
\begin{align*}
|\operatorname{Cov}(X_{t+h},X_t)-\operatorname{Cov}(S_{F,t+h},S_{G,t})| = |\mathbb E[X_{t+h}X_t-S_{F,t+h}S_{G,t}]|.
\end{align*}
Add and subtract $S_{F,t+h}X_t$ inside the expectation and apply the triangle inequality and Cauchy-Schwarz:
\begin{align*}
|\mathbb E[X_{t+h}X_t-S_{F,t+h}S_{G,t}]| \leq \|X_{t+h}-S_{F,t+h}\|_{L^2(\Omega)}\|X_t\|_{L^2(\Omega)} + \|S_{F,t+h}\|_{L^2(\Omega)}\|X_t-S_{G,t}\|_{L^2(\Omega)}.
\end{align*}
The first factor in the first product tends to $0$. The sequence $\|S_{F,t+h}\|_{L^2(\Omega)}$ remains bounded because it converges to $X_{t+h}$ in $L^2$, and the second factor in the second product tends to $0$. Hence
\begin{align*}
\operatorname{Cov}(X_{t+h},X_t) = \lim_{F,G}\operatorname{Cov}(S_{F,t+h},S_{G,t}) = \sigma^2\sum_{j\in\mathbb Z}\psi_{j+h}\psi_j.
\end{align*}
[/step]
[step:Conclude weak stationarity from the mean and covariance formula]
For every $t\in\mathbb Z$, we have proved $\mathbb E[X_t]=0$. For every $t,h\in\mathbb Z$, we have also proved
\begin{align*}
\operatorname{Cov}(X_{t+h},X_t) = \sigma^2\sum_{j\in\mathbb Z}\psi_{j+h}\psi_j.
\end{align*}
The right-hand side depends on $h$ and not on $t$. Therefore $(X_t)_{t\in\mathbb Z}$ has constant mean and autocovariance depending only on the lag. This is exactly weak stationarity, with autocovariance function
\begin{align*}
\gamma:\mathbb Z &\to \mathbb R
\end{align*}
given by
\begin{align*}
\gamma(h) = \sigma^2\sum_{j\in\mathbb Z}\psi_{j+h}\psi_j.
\end{align*}
[/step]
