Why is $\phi_\mu(x)= \int_{\widehat{G}} \xi(x)d\mu(\xi)$ a continuous map? Let $G$ be a locally compact Hausdorff abelian group. Let $\widehat{G}$ be its dual group, consisting of the unitary characters $G \to \mathbb{T}$. If $\mu \in M(\widehat{G})$ (= complex Radon measures on $\widehat{G}$), define
$$\phi_\mu: G \to \mathbb{C}: x\mapsto\int_{\widehat{G}} \xi(x) d\mu(\xi).$$
Note that this integral exists because
$$\int_{\widehat{G}}|\xi(x)|d|\mu|(\xi)= |\mu|(\widehat{G}) < \infty.$$
Why is the map $\phi_\mu$ continuous? I.e. assume that we have the net convergence $x_\alpha \to x$ in $G$. Then we should be able to show that
$$\int_\widehat{G} \xi(x_\alpha) d \mu(\xi)\to \int_\widehat{G} \xi(x)d\mu(\xi).$$
I tried to estimate
$$\left|\int_\widehat{G} \xi(x_\alpha)d\mu(\xi)-\int_\widehat{G} \xi(x)d\mu(\xi)\right|\le \int_\widehat{G}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi).$$
Now, I am not sure how to proceed. Any hints/help are highly appreciated!
Reference: Folland's book "A course in abstract harmonic analysis", second edition p103.
 A: Let $z_0 \in G$ and $\varepsilon > 0$ be given. By inner regularity, one can find  a compact set $K \subset \hat G$ such that $|\mu|(K^c) \le \varepsilon$. The mapping $(z,\xi) \longmapsto \xi(z)$ is continuous because $G$ is LCH and $\hat G$ is equipped with the compact-open topology. Then, for every $\xi_0 \in K$ there are open sets $U \ni z_0$ and $V\ni \xi_0$ such that $|\xi(z)-\xi_0(z_0)| \le \varepsilon$ whenever $z \in U$ and $\xi \in V$.
Such open sets $V$ cover $K$, so by compactness there are $\xi_1,\ldots,\xi_n \in K$ and open sets $U_i \ni z_0$, $V_i \ni \xi_i$ such that
$$ K \subseteq \bigcup_{i=1}^n V_i ~~\text{ and }~~ \forall z\in U_i,\forall \xi \in V_i ~,~ |\xi(z)-\xi_i(z_0)| \le \varepsilon $$ Now let $U = \bigcap_{i=1}^n 
 U_i$ and observe that for any $z\in U$ and $\xi \in K$, we have
$$ |\xi(z)-\xi(z_0)| \le |\xi(z)-\xi_i(z_0)| + |\xi_i(z_0)-\xi(z_0)|  \le 2\varepsilon $$where $i$ is chosen so that $\xi \in V_i$. Finally, whenever  $z \in U$, we have
$$ \begin{split} |\phi_\mu(z)-\phi_\mu(z_0)| &\le \int_{K} |\xi(z)-\xi(z_0)| \,|\mu|(\mathrm d\xi) + \int_{K^c} |\xi(z)-\xi(z_0)| \,|\mu|(\mathrm d\xi)\\
&\le 2\varepsilon |\mu|(K) + 2|\mu|(K^c)\\
&\le 2\varepsilon \|\mu\|_\text{TV} + 2\varepsilon
\end{split} $$
The last quantity can be as small as we want, hence $\phi_\mu$ is continuous at $z_0$.
A: The first relevant point to observe is that the topology of the dual of a locally compact abelian group $G$
is defined as the topology of uniform convergence on compact sets.  In other words, a net $\{\xi _\alpha \}_\alpha $  in $\widehat G$ converges to
some $\xi $ in $\widehat G$ iff
$$
  \sup_{x\in  K}|\xi _\alpha (x)-\xi (x)| \to 0
  \qquad\qquad{(\dagger)}
  $$
for every compact subset $K\subseteq G$.
Next we need to focus on the precise meaning of Pontryagin duality, namely the result according to which
$\, \widehat {\!\widehat G}$
coincides with $G$.  To begin with, this says that every character on $\widehat G$ is of the
form
$$
  \Phi_x: \xi \in \widehat G \mapsto  \xi (x)\in \mathbb T,
  $$
for some $x\in G$,
so that the map
$$
  \Phi: x\in  G \mapsto  \Phi_x\in \, \widehat {\!\widehat G}
  $$
turns out to be a group isomorphism.  Pontryagin duality does not stop there as it also states that $\Phi$  is a topological
homeomorphism!  Interpreting the latter fact from the point of view of the definition of the topology on the dual group
$\, \widehat {\!\widehat G}$,
we see that a net $\{x_\alpha \}_\alpha $  in $G$ converges to
some $x$ in $G$ iff
$$
  \sup_{\xi\in  K}|\xi (x_\alpha )-\xi (x)| \to 0
  $$
for every compact subset $K\subseteq \widehat G$.  It is perhaps a  good idea to contrast this with $(\dagger)$,
paying  special attention to the precise position of the subscript $\alpha $!
This said, given $\varepsilon >0$, and using the regularity of the measure $\mu $, and hence also of $|\mu |$, choose some compact
$K\subseteq \widehat G$ such that $|\mu |(\widehat G\setminus K)<\varepsilon $.  Next, choose some $\alpha _0$ such that for all $\alpha \geq \alpha _0$, one has
that
$$
  \sup_{\xi\in  K}|\xi (x_\alpha )-\xi (x)| <\varepsilon .
  $$
For all such $\alpha $ one then has that
$$
  \left|\int_{\widehat{G}} \xi(x_\alpha)d\mu(\xi)-\int_{\widehat{G}} \xi(x)d\mu(\xi)\right|\le
  \int_{\widehat{G}}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi) = $$$$ =
  \int_{K}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi)  +   \int_{\widehat{G}\setminus K}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi)
\leq $$$$\leq
  \varepsilon |\mu |(K)+ 2   |\mu |(\widehat G \setminus K) \leq    \varepsilon |\mu |(K)+ 2  \varepsilon .
  $$
