Definition of $L^0$ space

Note that when we restrict ourselves to the probability measures, then this terminology makes sense: $L^p$ is the space of those (equivalence classes of) measurable functions $f$ satisfying $$\int |f|^p<\infty.$$ Therefore $L^0$ should be the space of those (equivalence classes of) measurable functions $f$ satisfying $$\int |f|^0=\int 1=1<\infty,$$ that is the space of all (equivalence classes of) measurable functions $f$. And it is indeed the case.

If the measure of $S$ is finite, the $L^p$ spaces are nested: $L^{p}\subset L^q$ whenever $p\ge q$. The smaller the exponent, the larger the space. Since the space of measurable functions contains all of the $L^p$ for $p>0$, one may be tempted to denote it by $L^0$.

This temptation should be resisted and the notation $L^0$ banished from usage. [/rant]

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.