The question I want to do says:
Let $f(u,t) : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function, such that for each $u$, $f(u, \cdot)$ is a characteristic function, and such that for each $t$, $f(\cdot, t)$ is continuous. Show that
$$F(t):= \int_{\mathbb{R}} f(u,t) dG(u) $$
is again a characteristic function for any distribution function $G$.
My attempt is as follows. I'm not sure if it is correct, as I don't use the continuity of $f$; my argument appears to only require measurability.
Let $(X_u)_{u \in \mathbb{R}}$ be a family of random variables such that the characteristic function of $X_u$ is $f(u, \cdot)$ for each $u \in \mathbb{R}$. Define a new random variable $Y$ by its law:
$$P(Y \in A) := \int_{\mathbb{R}} P(X_u \in A) G(du) $$
for Borel $A$. Putting this differently, for Borel $A$ we have
$$E(\textbf{1}_A(Y)) = \int_{\mathbb{R}} E(\textbf{1}_A(X_u)) G(du) $$
and by a monotone class argument this equation readily extends for bounded measurable functions.
In particular, writing for fixed real $t$, $g_t(x):= \exp(itx)$ we obtain
$$E(\exp(itY)) =E(g_t(Y)) = \int_{\mathbb{R}} E(g_t(X_u)) G(du) = \int_{\mathbb{R}} f(u,t) G(du). $$
This shows that $F$ is the c.f. of $Y$.
My questions are: 1. Is this argument correct? 2. If so, why do I not require continuity, but just measurability?
Many thanks for your help.
PS I am not keen on applying Bochner's theorem, as this, though completely correct and technically easy in this case, does not motivate the origin of $F$.
The $X_u$ should not be independent: if they were, I think that $X_Z$ would not be measurable.