# Need confirmation of the definition of $\mathcal{L}^2$ convergence

So we say $f_n(t)$ converges in $\mathcal{L}^2$ to $f(t)$, more specifically we mean: $\lim\limits_{n\rightarrow \infty}\int_0^t|f_n(s) - f(s)|^2 ds = 0$, hence if in terms of the $\lim\limits_{n\rightarrow \infty}$ notation instead of the usual notation of $\rightarrow^{\mathcal{L}^2}$, does this also mean: $f_n(t)$ converges to $f(t)$ in $\mathcal{L}^2$ $\Leftrightarrow$ (iff) $\lim\limits_{n\rightarrow 0} \int_0^t f_n(s)^2ds = \int_0^t f(s)^2 ds$?

in other words, in Euclidean norm, we say $f(t)$ is the limit of $f_n(t)$ when $\lim_{n\to\infty}f_n(t)=f(t)$, but how about when we say $f(t)$ is the $\mathcal{L}^2$ limit of $f_n(t)$? we can only express it as $f_n(t) \to^{\mathcal{L}^2} f(t)$? can we express $f(t) = \lim\limits_{n\to\infty}...$?

Let me use $X$ to denote the measure space and $\mu$ to denote the measure.
$L^2$ convergence is just like any other form of convergence in a metric space. So let's write this out formally: $$\lim_{n\to\infty} f_n = f ~\text{in}~ L^2$$
means $$\lim_{n\to\infty}\Vert f_n-f\Vert_2^2 = 0.$$ Here $\Vert\cdot\Vert_2$ is the $L^2$ norm. I haven't used anything specific to $L^2$ yet, this works for all metric spaces (using a different form of course). Let's expand what this means: using the definition of $\Vert\cdot\Vert_2$, $$\lim_{n\to\infty}\int_X |f_n-f|^2 d\mu = 0.$$ So this is what it means for $f_n\to f$ in $L^2$.