Make the dominated convergence theorem rigourous I have the following when I apply the dominated convergence theorem :


  Let's say I want to find : $\lim_{n \to +\infty}\int_0^1 t^n \mathrm{d}t$ using the DCT (I know we can just integrate but it's an example to show where I have a problem).


So we defined the set of function : $f_n : [0,1] \to \mathbb{R} : t \mapsto t^n$. We have : $f_n \leq 1$ and $f_n$ converges to the piewise continuous function : $g = 0$ on $[0,1[$ and $g(1) = 1$.

Now the problem is that when I interchange limit and integration : 

$$\lim_{n \to \infty} \int_0^1 f_n = \int_0^1 \lim_{n \to \infty} f_n = \int_0^1 g$$

Then clearly $\int_0^1 g(x) \mathrm{d}x = 0$ but how can I show this rigourously ? Since $g(1) = 1$ I don't know how to prove it's equal to $0$.
Maybe I cant take an $\epsilon$ and say : $\int_0^1 g \leq  \int_0^{1-\epsilon} 0 + \int_{1-\epsilon}^1 1 $

But it seems a lot of effort to prove an obvious fact.

Thank you ! 

N.B : All these integrals are Riemann not Lesbegue
First, you have to understand that the Lebesgue integral is a generalization of the Riemann integral. What we mean by that is that if a function is Riemann integrable, it is Lebesgue integrable and the two notions of integration are he same, ie : $\int_a^b f(x)dx=\int_{[a,b]}f dm$ where $m$ is the Lebesgue measure. A powerful theorem for this equivalence is the following :
$f$ is Riemann integrable iff its set of discontinuities has Lebesgue measure 0. Now in your example, it is easy to see that you only have 1 point where the function is discontinuous and so the two notions agree. 
Suppose now you dont know that this theorem exists and that you only care about Riemann integrals. The important thing you have to remember is that integration never depends on the value of a function at a point ! Namely, you can change countably many points of your function (for example, let the function be equal to $\pi$ at all rational points !) and the resulting value of the integral will be the same. Your idea is good ! Simply take an $\epsilon$ neighborhood around 1, integrate as you wanted to do and let $\epsilon\to 0$. The intuition behind that is that the area under a single point is always as small as you want it to be !
I hope I clarified your uncertainties, if you have any questions leave them in the comment. 

PS : The power of the Lebesgue Integration theory lives in its powerful convergence theorems (monotone convergence and dominated convergence). They not only yield results that were previously unobtainable with the Riemann theory of integration but also reduce the labor in proving classical results (such as a function being Riemann integrable).
To answer your question: It's trivial to verify from the definition of the integral that $\int_0^1 f(t)\,dt=0$.

But it's very curious that you talk about DCT in the context of Riemann integrals! Because the statement usually known as "DCT" is false for the Riemann integral - hence one wonders exactly what theorem  you're referring to.

A weak version of DCT for integrals on $[0,1]$:


  
    DCT. Suppose $f_n$ and $g$ are integrable on $[0,1]$, $|f_n|\le g$, and $f(t)=\lim_nf_n(t)$ exists for every $t\in[0,1]$. Then $f$ is integrable on $[0,1]$ and $\int_0^1 f=\lim_n\int_0^1f_n$.
  


Counterexample for the Riemann integral: In general if $A$ is a set define the function $\chi_A$ by $$\chi_A(t)=\begin{cases}1,&amp;(t\in A),
\\0,&amp;(t\notin A).\end{cases}$$

Say $E=\{r_1,r_2,\dots\}$ is a countable dense subset of $[0,1]$. Let $E_n=\{r_1,\dots,r_n\}$. Let $f_n=\chi_{E_n}$, $g=1$.  Then $$\lim_nf_n(t)=f(t)=\chi_E(t)$$for every $t$, but $f$ is not Riemann integrable (because for example it is discontinuous everywhere).