I am given a seperable Banach space and an interval $J$. Pettis' theorem tells that for such a space $Y$, we have: $f : J \rightarrow Y$ is measurable iff for any bounded linear functional $g \in Y^*$, the real function $J \ni t \mapsto g(f(t))$ is measurable.
Now I am supposed to use this theorem to show that any function $f \in C([0,1],Y)$ with a separable Banach space $Y$ is measurable.
My confusion: Given such a continuous $f$, and noting that we have the Borel sigma algebras on the interval $[0,1]$ and $Y$. Is such a continuous map not automatically continuous, since the preimage of open maps is open ($f$ is continuous) and the open maps generate the Borel sigma algebra?
Do I miss something or does this indeed follow automatically by continuity?
Thanks in advance!