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Suppose $X$ is a measure space with measure $\mu$. Given a strictly increasing continuous (or sufficiently nice) function $\phi:[0, \infty)\to [0, \infty)$ with $\phi(0)=0$. Is it true that we can find a norm $\|\cdot\|$ on the space of measurable functions (or at least a subspace of "sufficiently nice" functions) satisfying the following two conditions?

  1. $\|f\|\le \|g\|$ if $|f|\le |g|$ (pointwise).

  2. $\|\chi_A\|=\phi(\mu(A))$, where $\chi_A$ is the characteristic function.

Such a norm seems to be constructible by integrals if $\phi$ is simple. E.g. $\phi(u)=\sqrt{u+u^2}$, we can take $\|f\|=\sqrt{(\int_X |f|d\mu)^2+\int_X |f|^2 d\mu}$. But otherwise I am stuck. If this is not true in general, what conditions on $\phi$ can we ensure such an existence?

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  • $\begingroup$ I think at a minimum you need $\phi$ to be concave. For instance consider $\phi(t) = t^2$. If we take $X =\mathbb{R}$ with Lebesgue measure, we would have to have $\|\chi_{[0,1)} + \chi_{[1,2]}\| = \|\chi_{[0,2]}\| = 4 > 2 = \|\chi_{[0,1)}\| + \|\chi_{[1,2]}\|$ contradicting the triangle inequality for the norm. $\endgroup$
    – Nate Eldredge
    Commented Nov 18, 2017 at 22:46

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What you are asking about is known as a symmetric (or rearrangement invariant) Banach space of measurable functions (not to be confused with Riemannian symmetric spaces) and the associated fundamental function, see the recent book Foundations of symmetric spaces of measurable functions by Rubshtein et al. For instance, the Lorentz Banach spaces are the ones that correspond to concave functions $\phi$ with $\phi(0)=0$.

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