Basicaly the title, but to be more precise: Let $(X, \mathcal{X} , \mu)$ be a measure space. $L^2(\mu)$ the space of $\mu$-square integrable functions modulo the nullfunctions. Let $\vert \vert \cdot \vert \vert$ be a fixed Norm on $L^2(\mu)$ such that, there exists a constant $c>0$.
$c^{-1} \vert \vert f \vert \vert \leq \vert \vert f \vert \vert_{L^2(\mu)} \leq c \vert \vert f \vert \vert $
for all $f \in L^2(\mu)$.
Now does a measure $\nu$ on $X$ exist such that $\vert \vert f \vert \vert = \vert \vert f \vert \vert_{L^2(\nu)}$ for all $f \in L^2(\mu)$ ?
The answer is NO. Hint: consider $(0,1)$ with Lebesgue measure and define $\|f\|_1=\int |f|+\|f\|_{L^{2}(\mu)}$. If such a measure $\nu$ exists in this case then we get $\nu (E)=\mu (E)+\mu(E)^{2}+2\mu(E)\sqrt {\mu (E)}$ (by taking $f= I_E$). I will leave it to you to show that $\nu$ is not countably additive.