What are some commonly used strategies to prove that $(f_n)_{n\in\mathbb{N}}\to f$ in $\Vert\cdot\Vert_2$ norm? That is, for $f, f_n:[a,b]\to\mathbb{R}$,
$$\lim_{n\to\infty}\left(\int_{[a,b]}|f_n-f|^2\,\text{d}\lambda\right)^{1/2}=0$$
I'm specifically having trouble figuring out if what I did in a specific exercise is correct or not. Consider $(f_n)_{n\in\mathbb{N}}:[0,2]\to\mathbb{R}$ given by
$$f_n(x)=\frac{1}{1+x^{2n}}$$
It's not difficult to find that $f_n\to f$ pointwise for $f$ defined as
$$f(x)=\begin{cases}1,&\text{if }x\in[0,1),\\ \frac12,&\text{if }x=1,\\0,&\text{if }x\in(1,2].\end{cases}$$
I tried to find whether or not $f_n\to f$ in $\Vert\cdot\Vert_2$ norm, here's what I did:
$$\lim_{n\to\infty}\int_{[0,2]}\left|f_n-f\right|^2\,\text{d}\lambda=\lim_{n\to\infty}\left(\int_{[0,1)}\left|\frac1{1+x^{2n}}-1\right|^2\,\text{d}\lambda + \int_{(1,2]}\left|\frac1{1+x^{2n}}\right|^2\,\text{d}\lambda\right)$$
because we can discard the integral over $\{1\}$ for being a null-measure set. Then, both integrands are bounded functions over a finite measure interval, so by the bounded convergence theorem, we should be able to commute the limit and integral, so the expression above equals zero. Is there anything right about this argument?
Note that $|f_n(x)| \le 1$ for all $n,x$ and since $f_n(x) \to f(x)$, using the dominated convergence theorem we see that $\|f-f_n\|_2 \to 0$.