Let $\mu$ and $\lambda$ be two positive $\sigma$-finite measures such that $\mu \ll \lambda$ and $\lambda \ll \mu$, then $L^p(\mu)$ and $L^p(\lambda)$ are isometric.
My attempt: Since $\mu \ll \lambda$, by Radon-Nikodym theorem, exists $g \in L^1(\lambda)$ such that $d\mu = g d\lambda$, i.e. $\int \chi_E d\mu = \int g \chi_E d\lambda$ for every measurable $E$. Then, $$ \int f d\mu = \int gf d\lambda \quad (I) $$ holds for every simple measurable function. Then, $(I)$ holds for functions in $L^\infty$ (since simple functions are dense in $L^\infty$) and then $(I)$ holds for every $f \in L^p(\mu)$, since $L^\infty$ is dense in $L^p$.
Using the same argument, since $\lambda \ll \mu$, exists $h \in L^1(\mu)$ such that $$ \int f d\lambda = \int hf d\mu \quad(II) $$ holds for every $f \in L^p(\lambda)$.
I don't know how to proceed. Help?
If $f \in L^{p}(\mu)$ then $fg^{1/p} \in L^{p}(\lambda)$ and is it straightforward to verify that the map $f \to fg^{1/p}$ is an isometric isomorphism for any $p \in [1,\infty)$. The identity map is an isometric isomorphism from $L^{\infty}(\mu)$ on to $L^{\infty}(\lambda)$.