If we first start with the set:
\begin{align}
\left\{f: \left[-T, T\right] \to \mathbb{R}^{n} \ | \ \text{f is continuous on } [-T,T]\right\}
\end{align}
we end up with a gigantic set that is not complete (https://en.wikipedia.org/wiki/Complete_metric_space).
If we restrict this space too,
\begin{align}
\left\{f: \left[-T, T\right] \to \mathbb{R}^{n} \ | \ ||f||_{\mathcal{L}_{2}} < \infty, \text{f is continuous on } [-T,T]\right\}
\end{align}
then the space is complete.
We (usually) only deal with causal signals (https://blog.oureducation.in/classification-of-signals/), i.e., a function $f(t)$ where $f(t) = 0$ for all $t<0$. Here is a post talking about LTI systems that highlights that if the signal isn't causal then the output is dependent on future values of the input (https://dsp.stackexchange.com/a/35265). This is a reason for looking at causal signals.
We are also interested in functions that are defined from $[0,\infty)$ and not just the interval $[0,T]$. Therefore, if we consider the space
\begin{align}
\mathcal{L}_{2}^{n}[0,\infty ) := \left\{ f: \mathbb{R} \to \mathbb{R}^{n} \ | \ ||f||_{\mathcal{L}_{2}} < \infty, (\forall t <0) f(t) = 0 \right\}
\end{align}
where $\left< f, g \right>_{\mathcal{L}_2} = \int^{\infty}_{-\infty} \left< f(t), g(t) \right>_{2}dt$ and $||f||_{\mathcal{L}_{2}} = \sqrt{\left< f,f\right>}$. Then $\mathcal{L}_{2}^{n}[0,\infty )$ is a Hilbert space (https://en.wikipedia.org/wiki/Hilbert_space). Being a Hilbert space comes with many nice properties that allow us to take the Fourier transform of $f$ and move between time and frequency domain (also https://en.wikipedia.org/wiki/Plancherel_theorem).
Unfortunately, this space is a little too restrictive because it doesn't include commonly used signals such as the unit step, i.e., $f(t) = 1(t)$. Therefore, to include these other signals we extend the $\mathcal{L}_{2}$ space to,
\begin{align}
\mathcal{L}_{2e}^{n}[0,\infty ) := \left\{ f: \mathbb{R} \to \mathbb{R}^{n} \ | \ (\forall T>0)\ ||f_{T}||_{\mathcal{L}_{2}} < \infty, (\forall t <0) f(t) = 0 \right\}
\end{align}
where $f_T(t) = f(t)$ for $t\in[0,T]$ and $0$ otherwise.
By extending the space we pay a price. $\mathcal{L}_{2e}^{n}[0,\infty )$ is not a normed vector space. Fortunately, it does maintain some useful properties so all is not lost. When looking at systems we are interested in their gain. There is a result that states that:
An operator $M$ is $\mathcal{L}_{2}$ stable with finite gain iff $M$ is $\mathcal{L}_{2e}$ stable with finite gain.
This provides a bridge for looking at $\mathcal{L}_2$ and $\mathcal{L}_{2e}$ signals.
