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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.2012-12-29
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    @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](http://math.stackexchange.com/q/266216/1281).2012-12-29
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    @Norbert You may be thinking of convergence a.e.2012-12-29
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    @PavelM M, you are right. This is convergence aproximately everywhere2013-01-04
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    You could be interested in weak $L^p$ spaces.2013-01-18

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