Suppose $E$ a vector space of continuous function from $[-1,1]$ to $\mathbb{C}$, we define the norm:
$$||f||_1= \displaystyle\int^1_{-1}|f(t)|dt$$
and we define a sequence such as:
$$ f_n(t)= \begin{cases} -1&\text{if } -1\le t\le \frac{-1}{n}\\ nt& \text{if }\frac{-1}{n}\le t\le \frac{1}{n} \\ 1&\text{if } \frac{1}{n}\le t\le 1 \end{cases}$$
How can I prove that:
$\displaystyle \lim_{n\to+\infty}\int^{-\alpha}_{-1}|f_n(t)+1|dt=0$ and $\displaystyle \lim_{n\to+\infty}\int^{1}_{\alpha}|f_n(t)-1|dt=0$
I would recommend that you draw the function for a general $ n $ or at least think about why the value of these two integrals may be $ 0 $.
It would be interesting to know how you came around these integrals. Was it measure theory?
If so, you may take a look at the Dominated convergence theorem and how to apply it to the two functions given by (or even easier: use the Beppo-Levi-Theorem/Monotone convergence theorem):
$$ g_n(t) = | f_n(t) + 1 | \\ h_n(t) = | f_n(t) - 1 | $$
First: all functions here are measurable over the Borel-$\sigma $-Algebra of $ \mathbb{R} $ as they are continuous (check that!). The next step would be to think about the dominants you could find. A step towards this is to evaluate $ f_n(t) $ for $ \frac{1}{n}, -\frac{1}{n} $ and $ \frac{1}{2n} $. Also, check if your dominant is independent from $ n $ and whether or not it is integrable from $ -1 $ to $ \alpha $ or $ \alpha $ to $ 1 $.
The last step is to find functions $ g(t) $ and $ h(t) $, so that $ \lim\limits_{n \rightarrow \infty} g_n(t) = g(t) \;\; \forall t \in [-1, 1] $ (pointwise convergence) (the same for $ h_n(t) $) and to integrate over $ g(t), h(t) $ instead of $ g_n(t), h_n(t) $.
That should do the job.