This is false. The easiest way to prove this is to put $\xi=N$. Uniform convergence would imply that $|\int [e^{iy}-1] \phi (y) dy |<\epsilon$ for every $\epsilon >0$ which need not be true.
Local uniform convergence: There exists $T$ such that $\int_{|x| >T} |\phi (x)| d x<\epsilon$. Now $\int_{-T}^{T} |e^{i\xi y/N}-1|dy \leq \frac {|\xi T|} N \int_{-T}^{T} |\phi (y)|dy $. Can you finish?
I have used the inequality $|e^{ia} -1| \leq |a|$ for all $a \in \mathbb R$.
How does one use DCT in this argument?: If $\int_{-T}^{T} [e^{i\xi y/N}-1] \phi (y) dy$ does not tend to $0$ uniformly on some bounded set then there exist $\epsilon >0$ and sequence $\xi_n, N_n$ such that $(\xi_n)$ is bounded and $|\int_{-T}^{T} [e^{e\xi y/N}-1] \phi )(y) dy |\geq \epsilon$ for all $n$. Now use the fact that the integrand is bounded by a constant times $|\phi|$ to get a contradiction to DCT.
If you are already aware of Lebesgue's criteria for Riemann integrability ($f$ is Riemann integrable iff $f$ is continuous everywhere outside a set of measure zero) then of course, you can deduce the integrability of $f$ from this. Suppose $D_n$ is the set of discontinuities of $f_n$ and let $D=\bigcup_nD_n$. Since $f_n$ converges uniformly to $f$, then $f$ is continuous on $[a,b]\setminus D$ (continuity is presented by uniform convergence) The countable union of sets of measure $0$ is also of measure $0$. Then $f$ is continuous outside $D$ (a set of measure $0$) and hence, $f$ is Riemann integrable.
If you are not aware of Lebesgue's result, you can prove integrability of $f$ through Darboux criteria: Here is a sketch of how one may approach this:
Given $\varepsilon>0$, there is $N_\varepsilon\in\mathbb{N}$ such that
$\|f-f_n\|_u<\varepsilon/3(b-a)$ for all $n\geq N_\varepsilon$.
Since $f_N$ is Riemann integrable, there is a partition $\mathcal{P}_\varepsilon=\{a=x_0<\ldots<b_n=x_n\}$ of $[a,b]$ such that
$U(f_N,P)-L(f_N,P)<\varepsilon/3$
It follows that
$\sup_{x\in[x_j,x_{j+1}]}f(x)=\sup_{x\in[x_j,x_{j+1}]}\big(f(x)-f_N(x)+f_N(x)\big)<\frac{\varepsilon}{3}(b-a) + \sup_{x\in[x_j,x_{j+1}]}f_N(x)$
$\inf_{x\in[x_j,x_{j+1}]}f(t)=\inf_{x\in[x_j,x_{j+1}]}\big(f(x)-f_N(x)+f_N(x)\big)> -\frac{\varepsilon}{3}(b-a) + \inf_{x\in[x_j,x_{j+1}]}f_N(x)$
Then
$$\begin{align}
U(f,P_\varepsilon)&=\sum^n_{j=1}\big(\sup_{x\in[x_j,x_{j+1}]}f(x)\big)\,(x_j-x_{j-1})\\
&<\frac{\varepsilon}{3}+\sum^n_{j=1}\big(\sup_{x\in[x_j,x_{j+1}]}f_N(x)\big)\,(x_j-x_{j-1})=\frac{\varepsilon}{3}+U(f_N,P_\varepsilon)
\end{align}$$
and
$$\begin{align}
L(f,P_\varepsilon)&=\sum^n_{j=1}\big(\inf_{x\in[x_j,x_{j+1}]}f(x)\big)\,(x_j-x_{j-1})\\
&>\frac{\varepsilon}{3} +\sum^n_{j=1}\big(\inf_{x\in[x_j,x_{j+1}]}f_N(x)\big)\,(x_j-x_{j-1})=-\frac{\varepsilon}{3}+L(f_N,P_\varepsilon)
\end{align}$$
Hence
$$ U(f,P_\varepsilon) - L(f,P_\varepsilon)<\frac23\varepsilon +U(f_N,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon
$$
This shows that $f$ is Riemann integrable.
To conclude, notice that since $|f(x)-f_n(x)|<\frac{\varepsilon}{3}(b-a)$ for all $x\in[a,b]$ and all $n\geq N_\varepsilon$
$$
\Big|\int^b_a(f-f_n)\Big|\leq\int^b_a|f-f_n|\leq \varepsilon/3,\qquad n\geq N_\varepsilon
$$
This shows that $\lim_n\int^b_af_n=\int^b_af$.
Best Answer
Convergence
By IBP, for all $a_n \in (0,1)$ $$\begin{align}f_n(x)&=\frac{n-1}{x} \int_{0}^1 t^x(1-t)^{n-2}dt \le \frac{n-1}{x} \left[ \int_{0}^{a_n} t^xdt +\int_{a_n}^1 (1-t)^{n-2}dt\right]\\ &=\frac{(n-1)a_n^{x+1}}{x(x+1)}+\frac{1}{x}(1-a_n)^{n-1}=:A_n(x) \end{align}$$ Choosing $a_n=(n-1)^{-1/(x/2+1)}$, we can see that $$\lim_n A_n(x)=0$$ Hence the pointwise convergence.
Not uniform
We have $$f_n(x) \ge \int_0^{1/2} t^{x-1} (1/2)^{n-1}dt=\frac{2^{-x-n+1}}{ x}$$ Thus $$f_n(2^{-n}) \ge 2^{-2^{-n}+1} \ge 1$$ Hence the conclusion.