We are given a real separable non-reflexive Banach space $(\mathfrak X, ||\cdot||_{\mathfrak X})$, and we consider its topological dual $\mathfrak X'$ endowed with $\sigma(\mathfrak X',\mathfrak X)$, i.e. the weak-star topology. Consider then $\mathcal B(\sigma(\mathfrak X',\mathfrak X))$, i.e. the Borel sigma-algebra generated by the weak-star topology.
For $x\in \mathfrak X$, let $\ell_x:\mathfrak X'\to \mathbb R$, $\phi \stackrel{\ell}{\mapsto} \phi(x)$ be the functionals which generate the weak-star topology. Consider then the sigma-algebra generated by all these functionals, i.e. $\sigma(\ell_x; x\in \mathfrak X)$.
Question: $\sigma(\ell_x; x\in \mathfrak X)= \mathcal B(\sigma(\mathfrak X',\mathfrak X))$ ?
I know that this is true on the pre-dual if it is separable and we use the weak-topology. But on the dual with the weak-star topology things quickly get murkier.