$\lim_{n \to \infty}n \left[ \int_0^n \frac{g(t)f(\frac{t}{n})}{n}dt\right]$ 
$*$ Let $f(x)$ be a continuous function in $[0,1]$ and let $g(x)$ be
  continuous in $[0,\infty]$ such that
  $\int_0^\infty\left|g(x)\right|<\infty$, calculate the following
  integral.
$$\lim_{n\to \infty} \left[ n \int_0^1 g(nx)f(x)dx \right]$$
$*$ Now let $f_0(x)$ be an arbitrary integrable function in $[0,a]$
define $f_n(x)=\int_0^x f_{n-1}(t)dt$
Prove ${f_n(x)}$ converges uniformly to $0$ in $[0,a]$

The first is at follows
By change of variables: $nx=t, ndx = dt$
The integral becomes $$\lim_{n \to \infty}n \left[ \int_0^n \frac{g(t)f(\frac{t}{n})}{n}dt\right]=\lim_{n \to \infty}\left[ \int_0^n g(t)f(\frac{t}{n})dt\right]$$
This is where I get fuzzy, suppose I want to inter-exchange the limit and integral, I learnt that I can do it as long as the integrand converges uniformly and the limit of the integral are unrelated to $n$, how do I do it in this case?
The 2nd part, I have no idea how to even start. 
Edit: I have not studied Lebesgue integral yet, this question is suppose to be solvable using uniform convergence, interchanging limit and integral, Dini's and iterated integrals. The full material can be found (in English) here. 
I proved the 2nd part, but I am not sure if its correct, can anyone check that?
added a bounty.
 A: For the first: Use the Dominated Convergence Theorem. Specifically, define $f_n\colon [0,\infty)\to\mathbb{R}$
$$
f_n(t) \stackrel{\rm def}{=} g(t)f\left(\frac{t}{n}\right)\mathbf{1}_{t\leq n}
$$
for $n\geq 1$. It is continuous, integrable with
$$
\int_0^n g(t)f\left(\frac{t}{n}\right) dt = \int_0^\infty f_n(t)dt
$$
and 


*

*Convergence pointwise: for all $t\in[0,\infty)$ $\lim_{n\to\infty}f_n(t) = F(t)$ with $F(t) = g(t)f(0)$

*Domination: since $f$ is continuous on $[0,1]$, it is bounded, and thus for all $n$, $\lvert f_n(t)\rvert \leq \lVert f\rVert_\infty\cdot\vert g(t)\rvert$; and the RHS is integrable by assumption on $g$.
Therefore, by the DCT we have
$$
\int_0^\infty f_n(t)dt \xrightarrow[n\to\infty]{} \int_0^\infty F(t)dt = f(0)\int_0^\infty g(t)dt
$$
A: Your 2nd part looks fine. For the first, without Lebesgue, it's a two step $\epsilon$-argument: 
Let  $I = \int_0^\infty |g(t)|dt <+\infty$ and $M=\sup_{[0,1]}|f|$. 
Given  $\epsilon>0$ there is $R<+\infty$ so that $\int_R^\infty |g(t)| dt <\epsilon/(4M)$.
Then by continuity of $f$ there is $N<\infty$ so that $\displaystyle |f(x)-f(0)| < \frac{\epsilon}{2I}$ for $0 \le x \le \frac{R}{N}$. Then for all $n>N$:
$$ |\int_0^n g(t) (f(t/n)-f(0)) dt |\le \int_R^\infty |g(t)| 2M dt +
 \int_0^R |g(t)| \frac{\epsilon}{2I} dt \le \epsilon$$
so the limit is zero. You readily deduce the convergence of your integral towards $f(0)\int_0^\infty g(t) dt$.
A: For the 2nd part, I was able to prove that $|f_n(x)| \le \frac{Mx^n}{n!}$ by induction where $M=sup_{x \in [0,a]} |f_0(x)|$
$$f_1(x)=\int_0^xf_0(t)dt \le \int_0^xMdt=Mx$$
Assume $|f_{n}(x)| \le \frac{Mx^n}{n!}$, to prove for $n+1$ simply by monotonically of integral apply $\int_0^x$ to both sides
$$f_{n+1}:=\int_0^xf_n(t)dt\underbrace{\le}_{*}\int_0^x|f_n(t)|dt\le \int_0^x\frac{Mt^n}{n!}dt\le \frac{Mx^{x+1}}{(n+1)!}$$


*

*since the integral of integrable function on finite interval is continuous and therefore always integrable, we can assume the absolute value of the function is also integrable.


We can also calculate that for any $a\in \mathbb{R}$ that $\frac{Mx^n}{n!} \to 0$ and therefore by sequence converges uniformly to $0$.
I still can't prove the first part without dominated convergence theorem, but I would like if someone could verify this part.
