RNP is invariant under Banach-space isomorphism

Transport a $Y$-valued vector measure to $X$. Assume $X$ has RNP, let $(\mathrm{\Omega },\mathcal{A},\mu )$ be a finite measure space, and let $\nu$ be a bounded-variation $Y$-valued measure with $\nu \ll \mu$. Put $\tilde{\nu }=T^{-1}\circ \nu$. By [L1], it is norm-countably additive, $\mathrm{\mid }\tilde{\nu }\mathrm{\mid }(E)\le \mathrm{\parallel }{T}^{-1}\mathrm{\parallel }\mathrm{\mid }\nu \mathrm{\mid }(E)$, and $\tilde{\nu }\ll \mu$.

Transport the density back to $Y$. By [L2], choose a Bochner density $f$ of $\tilde{\nu }$. Then [L3] makes $Tf$ Bochner integrable and gives ${\int }_{E}Tf=T({\int }_{E}f)=T\tilde{\nu }(E)=\nu (E)$ for every measurable $E$. Thus $Y$ has RNP.

Apply the same implication to the inverse. [L1, step 2.1] If $Y$ has RNP, apply steps 1.1--2.1 with the bounded isomorphism $T^{-1}:Y\to X$ to obtain RNP for $X$. Hence the two properties are equivalent. For zero spaces, zero measures, and empty control spaces, every transported object and density is zero, so both directions remain valid.