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From Wikipedia:

The vector space of (equivalence classes of) measurable functions on $(S, Σ, μ)$ is denoted $L^0(S, Σ, μ)$.

This doesn't seem connected to the definition of $L^p(S, Σ, μ), \forall p \in (0, \infty)$ as being the set of measurable functions $f$ such that $\int_S |f|^p d\mu <\infty$. So I wonder if I miss any connection, and why use the notation $L^0$ if there is no connection?

Thanks and regards!

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    @Tim No, there is no particular reason to mention things like that in a book. I came up with these properties thinking of what is needed to show that (a) $\int \psi(|f-g|)$ is a metric; (b) convergence in this metric is equivalent to convergence is measure.2012-12-29

4 Answers 4

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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.

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    Correction: I wrote "largest topology smaller", but it should be "smallest topology larger"2012-12-28
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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]

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    @P$a$vel Thank you, I see your point.2013-01-03
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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.

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    @JonasTeuwen: I know this post was made some time ago, but do you have any thoughts on how to arrive at the $L^{0}$ topology of convergence in measure as a "limit" of the $L^{p}$ topologies?2014-06-30