Let $f:[0,1]\to \ell_\infty[0,1]$ be defined by $f(t)=\chi_{[0,t]}$. Is it true that $f$ is weakly measurable almost everywhere w.r.t. Lebesgue measure ? Here $\ell_\infty[0,1]$ represents the function space $L_\infty(\Omega,\Sigma,\mu)$ where $\Omega=[0,1], \Sigma=\mathcal{P}([0,1])$ and $\mu$ be the counting measure on $[0,1]$.