Understanding the $\mathcal{L}^0(\mu)$ space

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After having introduced the $\mathcal{L}^p(\mu)$ spaces for $p\in(0,\infty)$, my book has a little remark that we can expand the definition to $p=0$.

Let $(X,\mathcal{E},\mu)$ be a measure space.

Let $\mathcal{M}(\mathcal{E})=\{f:X\to \mathbb{R}\ |\ f \text{ is } \mathcal{E}-\mathcal{B}(\mathbb{R})\text{-measurable}\}$ i.e. the set of measurable functions.

For $p=0$ we define the $\mathcal{L}^p(\mu)$ space as:

$$ \mathcal{L}^0(\mu)=\{f\in \mathcal{M}(\mathcal{E})|\ \lim_{t\to\infty}\mu(\{ |f|\geq t\})=0 \}$$

The book then simply state that $\mathcal{L}^0(\mu)$ is a vector-space (left as an exercise to the reader), and by markov's inequality $\mathcal{L}^r(\mu)\subseteq \mathcal{L}^0(\mu)$ for all $r\in (0,\infty)$. That is it.

I am now trying to understand what this space is.

My idea is that this atleast includes excludes functions where $f(x) \to \pm \infty$ for $x\to \pm \infty$, since for all $t$ there will be an infinite interval were the function is larger.

We can have a function with vertical asymptotes, but i cant think of a way to make them satisfy $\lim_{t\to\infty}\mu(\{ |f|\geq t\})>0$ without having uncountably many vertical asymptotes, and i do not think that is possible.

Am i heading in the right direction when thinking about $\mathcal{L}^0(\mu)$, or how do we intuitively think about this space?

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"My idea is that this at least $\color{red}{includes}$ functions where $f(x)\to\pm\infty$ for $x\to\infty$, ..."

I guess that you really meant to exclude them. Mind you that the space $\mathcal{L}^{0}(\mu)$ defined in OP does depend on $\mu$ as well. For example,

  1. If $\mu$ is the Lebesgue measure on $\mathbb{R}^d$, then $\mathcal{L}^{0}(\mu)$ exclues any measurable functions $f: \mathbb{R}^d \to \mathbb{R}$ such that $|f(x)| \to \infty$ as $|x| \to \infty$. Of course there are other examples, such as $\tan(\cdot)$ on $\mathbb{R}$.

  2. On the other hand, if $\mu$ is a finite measure, then $\mathcal{L}^{0}(\mu) = \mathcal{M}(\mathcal{E})$ by the continuity of $\mu$ from above.

In fact, it follows that

$$ f \notin \mathcal{L}^{0}(\mu) \qquad \Leftrightarrow \qquad \mu(\{|f| \geq t\}) = \infty \quad \text{for all} \quad t. $$