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Suppose $f:[0,T]\to X$ is a measurable map where $X$ is Hilbert space. Suppose also that $R(t):X \to X^*$ is an isometric isomorphism with $$\lVert R(t)f(t)\rVert_{X^*} = \lVert f(t) \rVert_X$$ also measurable.

Can we deduce $$t \mapsto R(t)f(t)$$ is measurable?

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As long as $R(t)$ is otherwise arbitrary, the answer is no. Take a non-measurable set $E$ and put $$ R(t) = \begin{cases}R&t \in E\\-R & t \not\in E\end{cases} $$ for some isometric isomorphism $R$. Together with a constant $f$, this yields a non-measurable $R(t)\,f(t)$.

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