= Solution
For a discrete-time <weakly stationary process>, the <spectral density of a stationary process>, when it exists, is a nonnegative function satisfying
$$
\gamma(h)=\int_{-\pi}^{\pi}e^{ih\omega}f_X(\omega)\,d\omega.
$$
If $\sum_h|\gamma(h)|<\infty$, then
$$
f_X(\omega)=\frac1{2\pi}\sum_{h\in\mathbb Z}\gamma(h)e^{-ih\omega}.
$$
A general second-order stationary process has a <spectral measure of a stationary time series>, which need not have a <spectral density>: a centered random constant has an atom at zero. Thus the <spectral density> calculation below assumes that $f_X$ exists; the corresponding measure identity holds without that extra assumption.
Absolute summability of the filter coefficients gives <mean-square convergence> because
$$
\left\|\sum_{s\in E}a_sX_{t-s}\right\|_2
\leq\|X_0\|_2\sum_{s\in E}|a_s|.
$$
Consequently the bilateral filtered process is well defined, with constant <mean> $m\sum_sa_s$. Its <covariance> is
$$
\gamma_Y(h)=\sum_{r,s\in\mathbb Z}a_ra_s\gamma_X(h-r+s).
$$
The sum is absolutely convergent, since $|\gamma_X(j)|\leq\gamma_X(0)$ and $\sum|a_s|<\infty$. It depends only on lag, proving <weak stationarity> of $Y$.
Insert the spectral representation and interchange summation and integration using absolute summability and the finite total spectral mass. For real coefficients,
$$
\begin{aligned}
\gamma_Y(h)
&=\int_{-\pi}^{\pi}e^{ih\omega}
\left(\sum_ra_re^{-ir\omega}\right)
\left(\sum_sa_se^{is\omega}\right)f_X(\omega)\,d\omega\\
&=\int_{-\pi}^{\pi}e^{ih\omega}|A(e^{i\omega})|^2f_X(\omega)\,d\omega.
\end{aligned}
$$
Therefore the <spectral density transformation under a linear filter> is
$$
\boxed{f_Y(\omega)=|A(e^{i\omega})|^2f_X(\omega).}
$$
For a bilateral sequence, $A$ is guaranteed to converge on the unit circle. Its negative powers need not converge inside the disk: for example, $a_{-j}=2^{-j}$ makes the series diverge at $z=1/4$. Only the unit-circle values are used here, so the unnecessarily broad domain printed for $A$ does not affect the proof.
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