Weak continuity of a vector valued function [closed]

Question

Let $f:[0,1]\to \ell_\infty[0,1]$ be defined by $f(t)=\chi_{[0,t]}$. Is it true that $f$ is weakly continuous 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 $[0,1]$.

Answer 1

"Weakly continuous almost everywhere" means there is a negligible set of $t$'s outside which $t\mapsto x^*(f(t))$ is continuous for all $x^*\in\ell_\infty^*$, not the other way round. But $\delta_s(f(t))$ depends continuously on $t$ everywhere except when $t$ crosses $s$ : no common negligible set is suitable for all $x^*\in\ell_\infty^*$.

What Tanmoy Paul proves is: for all $x^*\in\ell_\infty^*$, $t\mapsto x^*(f(t))$ is continuous almost everywhere. That's true, but doesn't answer the question as stated.

Answer 2

This function is weakly continuous almost everywhere. One can show that $f$ is Riemann Int. over $[0,1]$. Now if the value of this integral is $x$, for some $x\in \ell_\infty[0,1]$, then for any $x^*\in \ell_\infty[0,1]^*$ $x^*(x)=\int_{[0,1]}x^*f(t)d\lambda(t)$. Since $f$ is Riemann Int. the integral in RHS is scalar valued Riemann Int. Thus $x^*f$ is Riemann Int which leads to that $x^*f$ is continuous almost everywhere.