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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!

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Converrgencce in measure can not be even descibed in terms of topology. In other words there is no topology which gives rise convergencce in measure. –  Norbert Dec 29 '12 at 18:19
    
@Norbert: Thanks! Isn't it that convergence in measure can be induced from the quasinorm $\int \frac{|f|}{1+|f|}$? Or is it just for convergence in probability? C.f. Pavel M's comments following this post. –  Tim Dec 29 '12 at 18:31
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@Norbert You may be thinking of convergence a.e. –  user53153 Dec 29 '12 at 20:34
    
@PavelM M, you are right. This is convergence aproximately everywhere –  Norbert Jan 4 '13 at 17:31
    
You could be interested in weak $L^p$ spaces. –  Davide Giraudo Jan 18 '13 at 20:57
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