Let $\rho(f) := \mu(|f| \geq \epsilon)$ where $\epsilon >0$ is a constant and $\mu$ is a measure and $f$ is a measurable function. Then
- $f\equiv 0 \rightarrow \rho(f) = 0$
- when $|a| \neq 0$, $\rho(af) = \mu(|f| \geq \epsilon/|a|)$ not necessarily equals $|a| \rho(f)$. But it holds when $|a|=1$.
- it is not true that $\rho(f+g) = \mu(|f+g| \geq \epsilon) \leq \mu(|f| \geq \epsilon) + \mu(|g| \geq \epsilon) = \rho(f) + \rho(g)$ because $\{|f+g| \geq \epsilon\}$ might be nonempty, while $\{|f| \geq \epsilon\}$ and $\{|g| \geq \epsilon\}$ can be empty.
What kinds of generalized norm is it? BTW: the generalized norms that I have heard of are quasinorms (which generalize the triangle inequality in some way), F-norms (which generalize the positive homogenity in some way) and seminorms (which generalize the positive definiteness in some way).
Motivation: I am trying to see if convergence in measure can be understood wrt the family of "generalized norms" $\{\rho_\epsilon, \epsilon >0\}$.
Thanks and regards!