That seems an unnecessary approach here. First prove the theorem for non-negative functions, then extend to all functions as you've indicated.
For nonnegative $f$, note that if $h$ is bounded and has finite support on $A$, then you can define $$h_E(x) =\begin{cases} h(x) & x\in A\\0 & x \in E \setminus A\end{cases}$$
Then $h_E$ is bounded and has finite support on $E$. Further, $h_E\chi_A = h_E$ and $h_E|_A = h$. Conversely if $h$ is bounded and has finite support on $E$, then $h\chi_A|_A$ is bounded and has finite support on $A$.
Edit: as requested, where I've tried to dot all the i's and cross all the t's.
For notational convenience, for measurable sets $B$ and measurable non-negative functions $g$, define ${\scr M}(g, B)$ to be the set of all bounded measurable functions $h$ of finite support on $B$ such that $0 \le h \le g$.
Let $f$ be a non-negative measurable function on $E$ and $A$ a measurable subset of $E$. Then $\chi_Af$ is also a measurable function on $E$.
Now if $h \in {\scr M}(\chi_Af,E)$ and $x \in E \setminus A$, then $0 \le h(x) \le \chi_A(x)f(x) = 0\cdot f(x) = 0$. Hence $\operatorname{supt} h \subseteq A$. And for $x \in A$, $0 \le h(x) \le \chi_A(x)f(x) = 1\cdot f(x) = f(x)$, so $0 \le h \le f$ on $A$. Therefore $h|_A \in {\scr M}(f,A)$. And, $$\int_E h = \int_A h + \int_{E\setminus A} h = \int_A h$$ since $h = 0$ on $E\setminus A$.
Conversely, if $h' \in {\scr M}(f,A)$, then define $h : E \to \bar{\Bbb R}$ by $$h(x) = \begin{cases} h'(x) & x \in A\\0 & x \notin A\end{cases}$$
Then $h$ is measurable, $\operatorname{supt} h = \operatorname{supt} h'$ and therefore is finite, and if $h' < M$, then $M > 0$, so $h < M$ as well. And lastly $$h(x) = \begin{cases} h'(x) \le f(x) = \chi_A(x)f(x) & x \in A\\0 = \chi_A(x)f(x) & x \notin A\end{cases}$$
Hence $h \in {\scr M}(\chi_Af,E), h' = h|_A$ and therefore $$\int_E h = \int_A h'.$$ Therefore the restriction map $T\ :\ {\scr M}(\chi_Af,E) \to {\scr M}(f,A)\ :\ h \to h_A$ is a bijection, and $\int_E h = \int_A T(h)$. So $$\begin{align}\int \chi_Af &= \sup\left\{\int_E h \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\
&=\sup\left\{\int_A T(h) \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\
&=\sup\left\{\int_A T(h) \ :\ T(h) \in {\scr M}(f,A)\right\}\\
&=\sup\left\{\int_A h' \ :\ h' \in {\scr M}(f,A)\right\}\\
&=\int_A f\end{align}$$
When $f$ may be negative, $$\int_E\chi_Af = \int_E\chi_Af^+ -\int_E\chi_Af^- = \int_A f^+ - \int_Af^- = \int_Af$$
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.
Best Answer
If $U$ is open in $\mathbb R$ then $f^{-1}(U)$ is open $M$ which means there is an open set $V$ in $\mathbb R^{n}$ such that $f^{-1}(U)=M\cap V$. This makes $f^{-1}(U)$ a measurable set. Hence $f$ is measurable. Integrability is not true in general. If $M=\mathbb R$ and $f(x)=\frac 1 {1+|x|}$ then $f$ is not integrable.
If $M$ has finite Lebesgue measure and $f$ is bounded by $C$ then $\int|f| \leq C\lambda (M) <\infty$ where $\lambda$ is the Lebesgue measure, so $f$ is Lebesgue integrable.