= Solution
The integrands are absolutely integrable because $f$ has compact support, $g\in L^1(\mathbb R)$, and $|\varphi_Z|\leq1$. <Fubini's theorem> therefore permits us to insert $\varphi_Z(s)=\mathbb E[e^{isZ}]$ and interchange the expectation and integrals:
$$
\begin{aligned}
&\iint g(\varepsilon s)f(x)e^{-isx}\varphi_Z(s)\,ds\,dx\\
&=\mathbb E\left[
\iint g(\varepsilon s)f(x)e^{is(Z-x)}\,ds\,dx
\right].
\end{aligned}
$$
With $r=\varepsilon s$, the inner integral in $s$ is
$$
\int g(\varepsilon s)e^{is(Z-x)}\,ds
=\frac1\varepsilon
\widehat g\left(\frac{Z-x}{\varepsilon}\right).
$$
Now set $t=(Z-x)/\varepsilon$. The change of variables $x=Z-\varepsilon t$ gives
$$
\iint g(\varepsilon s)f(x)e^{-isx}\varphi_Z(s)\,ds\,dx
=\int\widehat g(t)\mathbb E[f(Z-\varepsilon t)]\,dt,
$$
which is the required identity.
Choose any $g$ with $g,\widehat g\in L^1$, $g(0)\ne0$, and for which <Fourier inversion theorem> applies, for example a <Gaussian function>. The left side factors through the <Fourier transform> of $f$ as
$$
\int g(\varepsilon s)\widehat f(-s)\varphi_Z(s)\,ds.
$$
On the right side, $f(Z-\varepsilon t)\to f(Z)$ and
$$
|\widehat g(t)\mathbb E[f(Z-\varepsilon t)]|
\leq\lVert f\rVert_\infty|\widehat g(t)|.
$$
The <dominated convergence theorem> and Fourier inversion at zero give
$$
\int\widehat g(t)\mathbb E[f(Z-\varepsilon t)]\,dt
\longrightarrow
2\pi g(0)\mathbb E[f(Z)].
$$
Consequently the <Fourier-cutoff recovery of expectations from a characteristic function> is
$$
\boxed{
\mathbb E[f(Z)]
=\lim_{\varepsilon\downarrow0}
\frac{1}{2\pi g(0)}
\int g(\varepsilon s)\widehat f(-s)\varphi_Z(s)\,ds.}
$$
The right side depends only on $\varphi_Z$. Equal <characteristic functions> therefore give equal expectations for every compactly supported continuous $f$, and part a yields the <uniqueness theorem for characteristic functions>.
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