Using Lebesgue's dominated convergence theorem to show a function is continuous. I have a function 
$U(t)=\int_\mathbb{R} u(x) \cos(xt)dx$ 
and I am trying to use Lebesgue's dominated convergence theorem to show $U(t)$ is continuous for all $t \in \mathbb{R}$
This is the proof. 

Notes that $x \to \cos(xt)u(x)$ is measurable for every $t \in \mathbb{R}$

How do we know $x \to \cos(xt)u(x)$? And how do we know it is measurable?

Also note that $|\cos(xt)u(x)| \leq |u(x)|$ and that $|u|$ is by
  assumptions integrable

I assume this is from the Lebesgue dominated convergence theorem?

Thus $U(t)$ is well defined for every $t \in \mathbb{R}$ Now choose a
  sequence $t_n \to t$.
Since $\cos(t)$ is continuous we get $\cos(xt_n)u(x) \to \cos(xt)U(x)$
  and as $|\cos(xt)u(x)| \leq |u(x)|$

Is this also from Lebesgue dominated convergence theorem? 

Hence $U(t_n) \to U(t)$ proving continuity at every $t \in \mathbb{R}$

 A: Because $u$ is integrable (and thus measurable), the product $x \mapsto \cos(xt) u(x)$ of two measurable functions is measurable.
Let $(t_n)$ be a sequence such that $t_n \to t_0$. Now we can consider the sequence of measurable functions $f_n(x) = \cos(x t_n) u(x)$. Now
$$
\lim_{n \to \infty} U(t_n) = \lim_{n \to \infty} \int u(x) \cos (x t_n) \, dx = \lim_{n \to \infty} \int f_n(x) \, dx\,.
$$
Because $|f_n(x)| \leq |u(x)|$ for all $x$, $|u|$ is integrable and $cos(t)$ is continuous, the dominated convergence theorem says that
$$
\lim_{n \to \infty} \int f_n(x) \, dx\ = \int \lim_{n \to \infty} f_n(x) \, dx = \int u(x) \cos (x t_0) \, dx = U(t_0)\,.
$$
Thus
$$
\lim_{n \to \infty} U(t_n) = U(t_0)\,,
$$
which means that $U$ is continuous.
A: Just a few additions to the other answer:

Also note that $|\cos(xt)u(x)| \leq |u(x)|$

This follows because $|\cos(xt)| \leq 1$ for any $xt \in \mathbb{R}$
$u$ must have been assumed to be integrable in your question (you didnt put it above in the question

Hence $U(t_n) \to U(t)$ proving continuity at every $t \in \mathbb{R}$

Lebesgue dominated convergence theorem proves that $U(t_n)$ tends to $U(t)$ in the $L^1$ norm. Hence is continuous. 
