I am trying to understand the concept of uniform continuity as it pertains to characteristic functions.
First my understanding of uniform continuity:
Def:
$$\forall x_0, \forall \epsilon>0, \exists \delta>0,\hspace{4mm} \text{if}\hspace{4mm} |x-x_0|<\delta \hspace{4mm}\text{then} \hspace{4mm} |f(x)-f(x_0)|<\epsilon$$
I see this as fixing an $\epsilon$ and then find an interval of size $2\delta$ such that as I slide this interval across the x axis then all points $f(x), f(y)$ of x and y in this interval will be within $2\epsilon$ distance of each other. In layman terms I see this as the rate at which any two points $f(x)$ and $f(y)$ approach each other is very similar irrespective of the points selected.
My understanding may be wrong please correct me if I am wrong.
Now for the characteristic function:
$$|\varphi(t)-\varphi(s)|=\bigg|\int e^{itX}-e^{isX}\mu(dx)\bigg|\leq\int |e^{itX}-e^{isX}|\mu(dx) \leq\int|e^{iX(t-s)}-1|\mu(dx) \leq\int 2\mu(dx)=2$$
then by the dominated convergence theorem
$$\lim_{t\to s}|\varphi(t)-\varphi(s)|=0$$
But why does this imply uniform continuity not just continuity?
Another way of stating your correct intuition: uniform continuity results when you can prove that you can take an $\epsilon$ that does not depend on $s,t$, except through the difference $|t-s|<\delta$. A careful analysis of your proof shows that in fact
$$ \lim_{t\to s} \int | e^{iX(t-s)}-1| \mu(dx) \to 0$$ and this upper bound $ \int | e^{iX(t-s)}-1| \mu(dx)$ only depends on $|t-s|$, as needed.