Show $$\int_{-\infty}^{\infty}\,f(u,t)dG(u)$$ is a ch.f. where $G$ is a d.f. ; and $f(u,\cdot)$ is a ch.f. for each $u$ and $f(\cdot,t)$is continuous for each $t$.
Note that ch.f. means "characteristic function" and d.f. means "distribution function". It's a exercise from Kai Lai Chung's probability theory book.
My idea is that: in the special case $G$ is discrete, the problem is reduced to the following easy problem:
If $f_n$ are ch.f.'s and $\alpha_n \ge 0$, $\sum_{n=1}^{\infty}\alpha_n=1$, then
$$\sum_{n=1}^{\infty}\alpha_nf_n$$ is a ch.f.
I think the crucial point is how to use"$f(\cdot,t)$ is continuous for each $t$"
For each $u$ consider the random variable $X_u: \Omega_u\to \Bbb{R} $ here $\Omega_u= \Bbb{R}$ but we consider it endowed with the probability $\mu_u$ associated with the characteristic function $f(u,\cdot)$. The probability space is $(\Bbb{R}, \mathbb{B}, \mu_u)$
Extend this to $(\mathbb{R}^{(-\infty,\infty)}, \mathcal{B}(\mathbb{R})^{\otimes (-\infty,\infty)}, \mu^{\otimes (-\infty,\infty)})$ (use kolmogorov extension for intervals )
Now we have $\tilde{X}_u: \mathbb{R}^{(-\infty,\infty)} \to \Bbb{R}$ $\tilde{X}_u(\tilde{\omega}) = X_u(\tilde{\omega}(u))$with law given by $\mu_u$. (notation $\tilde{\omega} \in \mathbb{R}^{(-\infty,\infty)}$, $\tilde{\omega}: \Bbb{R} \to \Bbb{R}$) Let $Y$ be a random variable with density function given by $G$ and with law $\nu$
Extend one more time your space $(\mathbb{R}^{(-\infty,\infty)}\times \Bbb{R}, \mathcal{B}(\mathbb{R})^{\otimes (-\infty,\infty)}\otimes \Bbb{R}, \mu^{\otimes (-\infty,\infty)}\otimes \nu)$ (notation $\hat{\omega} \in \mathbb{R}^{(-\infty,\infty)}\otimes \Bbb{R}$, $\tilde{\omega}: \Bbb{R} \cup \{\Delta\} \to \Bbb{R}$)
$\hat{X}_u: \mathbb{R}^{(-\infty,\infty)}\times \Bbb{R} \to \Bbb{R}$ $\hat{X}_u(\hat{\omega}) = X_u(\tilde{\omega}(u))$ with law given by $\mu_u$.
$\hat{Y}: \mathbb{R}^{(-\infty,\infty)}\times \Bbb{R} \to \Bbb{R}$ $\hat{Y}(\hat{\omega}) = Y(\tilde{\omega}(\Delta))$ with law given by $\nu_u$.
Define $Z:\mathbb{R}^{(-\infty,\infty)}\times \Bbb{R} \to \Bbb{R}$ by $Z(\hat{\omega}) = X_{Y(\hat{\omega}(\Delta))}(\hat{\omega})$
Now we calculate
\begin{align*} \Bbb{E}[e^{itZ}] &= \int_{\Bbb{R}^{(-\infty,\infty)}\times \Bbb{R}} e^{it Z(\hat{\omega})}\, d\mu^{\otimes (-\infty,\infty)}\otimes \nu \\ &= \int_{\Bbb{R}}\int_{\Bbb{R}^{(-\infty,\infty)}} e^{it X_{Y(\hat{\omega}(\Delta))}(\hat{\omega})}\, d\mu^{\otimes (-\infty,\infty)}(\tilde{\omega}) \,d \nu(\hat{\omega}(\Delta))\\ &= \int_{\Bbb{R}} \Bbb{E}\bigg[ e^{it X_{Y(\hat{\omega}(\Delta))}(\hat{\omega})}\bigg]\, d\mu^{\otimes (-\infty,\infty)}(\tilde{\omega}) \,d \nu(\hat{\omega}(\Delta))\\ &= \int_{\Bbb{R}} f(\hat{\omega}(\Delta),t) \,d \nu(\hat{\omega}(\Delta))\\ &= \int_{\Bbb{R}} f(u,t) \,d G(u)\\ \end{align*}
which is a characteristic function.
remark: I didn't use continuity of $f(u,t)$. I'm not sure this is needed.