Let $(\Omega, \mathcal F)$ be a measurable space. Let $\mu$ and $\nu$ be two $\sigma$-finite measures on $(\Omega,\mathcal F).$ Define two norms $\|\cdot\|_{\mu}$ and $\|\cdot\|_{\nu}$ on $L^1(\mu)$ and $L^1(\nu)$ respectively defined by $\|f\|_{\mu} : = \int_{\Omega} |f|\ d\mu,\ f \in L^1(\mu)$ and $\|f\|_{\nu} : = \int_{\Omega} |f|\ d\nu,\ f \in L^1 (\nu).$ Is there any necessary and sufficient condition for the normed linear spaces $(L^1(\mu),\|\cdot\|_{\mu})$ and $(L^1(\nu),\|\cdot\|_{\nu})$ to be isometrically isomorphic?
What I have found is as follows $:$
If $\nu \lt \lt \mu$ (or $\mu \lt \lt \nu$) and if the corresponding Radon-Nikodym derivative $\frac {d\nu} {d\mu} = \alpha$ (say) be such that $\alpha \gt 0$ and $\alpha \in [c,C]$ for some $0 \lt c \lt C$ a.e. $\mu$ (and hence a.e. $\nu$) then $L^1(\mu) = L^1 (\nu)$ and the map $f \mapsto \frac {f} {\alpha}$ gives the required isometric isomorphism from $(L^1(\mu),\|\cdot\|_{\mu})$ to $(L^1(\nu),\|\cdot\|_{\nu}).$
So the above gives a sufficient condition for the two normed linear spaces to be isometrically isomorphic. I have also found a weaker converse of the above result which is as follows $:$
If $L^1(\mu) = L^1(\nu)$ and the normed linear spaces $(L^1(\mu),\|\cdot\|_{\mu})$ and $(L^1(\nu),\|\cdot\|_{\nu})$ are isometrically isomorphic then $\nu$ is absolutely continuous with respect to $\mu$ and vice-versa and the corresponding Radon-Nikodym derivative is bounded by $[c,C]$ a.e. $\mu$ (resp. $\nu$) for some $0 \lt c \lt C.$
How do I proceed if $L^1(\mu) \neq L^1(\nu)\ $? Any help in this regard will be appreciated.
Thanks for your time.