I do think that $L^0$ is nice usage. As is well-known $\lim_{p \to \infty} \|\cdot \|_{L^p} = \|\cdot\|_\infty$ for certain spaces or functions. The case for $L^0$ is not that pretty, but at least still nice.
Recall the distribution function $\mu$ of $f$ given by, $\mu(\alpha) := \mu_f(\alpha) := \mu\{|f|>\alpha\}.$
Fubini gives that, $\|f\|_{L^p}^p = p \int_0^\infty \mu_f(\alpha) \alpha^p \frac{\mathrm{d}\alpha}{\alpha}.$
We can define the Lorentz spaces in a similar way. And indeed, for a finite measure space, we have if $p < q$ that $L^q \subseteq L^p.$ Hence, it is natural to define $L^0$ as, $L^0 = \bigcup_{p > 0} L^p.$ We would like to have that $L^0$ is also complete as a metric space, otherwise the notation would be quite deceiving indeed. For this we need a notation of convergence. On $L^p$ for $0 < p < 1$ it is not the norm that induces the metric, but it is $\|\cdot\|_p^p$.
So, for $0 < p < 1$ we have, $d_p(f, g) = p \int_0^\infty \mu\{|f - g|>\alpha\} \alpha^p \frac{\mathrm{d}\alpha}{\alpha}.$
$\varepsilon$-neighborhoods $N^p_\varepsilon$ of $f$ in $L^p$ are then given by $N^p_\varepsilon(f) = \Biggl\{g : p \int_0^\infty \mu\{|f - g|>\alpha\} \alpha^p \frac{\mathrm{d}\alpha}{\alpha} < \varepsilon \Biggr\}.$
Too be continued, I wanted to give a brief remark, but I have decided otherwise in due progress.