Consider the following function:
$$g\left(x\right)=\begin{cases}
\frac{1}{x}, & \text{if} \ x\in\left(1,\infty\right)\\
0, & \text{if} \ x \in \left(-\infty, 1\right]
\end{cases}$$
Determine the set $\{p\in\left[0, \infty\right] : g\in\mathcal L^p\left(\lambda\right)\}$.
I know that the set is equal to $\left(1, \infty\right]$. But I am not sure how to calculate the cases $p=0$ and $p=\infty$.
In my textbook we have the following definitions for $p=0$ and $p=\infty$:
$$\mathcal L^\infty\left(\mu\right)=\{f\in\mathcal M\left(\mathcal E\right) : \exists R>0 : |f| \leq R \ \ \mu\text{-almost everywhere}\}$$
and
$$\mathcal L^0\left(\mu\right)=\{f\in\mathcal M\left(\mathcal E\right) : \lim_{t\rightarrow\infty}\mu\left(\{|f|\geq t\}\right)=0\}$$
where $\mathcal M\left(\mathcal E\right)=\{f : X \rightarrow\mathbb R : f \ \ \text{is} \ \mathcal E \text{-}\mathcal B\left(\mathbb R\right)\text{- measurable}\}$.
I will appreciate any help.
Hints:
$|g(x)|≤ 1$
$|1/x| ≥ t \iff |x| < 1/t$ ... what is the size of the set $\{x\in(1,\infty) : |x|<1/t\}$ ?