Let $(E,\mathcal{A},\mu)$ be probability space and $X$ be a separable Banach space. the associated dual space is denoted by $X^*$ and the usual duality between $X$ and $X^*$ by $\langle.,.\rangle$. Let $H$ be a countable, dense subset in $X^*$ for weak$-*$ topology
we suppose, we have a problem, as :
Show that there exists a nul set $M\in\mathcal{A}$ such that $$ A(t)\subseteq B(t)\qquad \forall t\in E\smallsetminus M.\qquad(∗) $$
with, for $t\in E$ : we pose $A(t)$ and $B(t)$ two subset of $X$ such that there exists two function $f_A,f_B:E\times X^*\to \mathbb{R}$ such that: $$ A(t)=\bigcap_{x^*\in H}\{x\in X:\langle x^*,x\rangle\leq f_A(t,x^*) \} $$ $$ B(t)=\bigcap_{x^*\in H}\{x\in X:\langle x^*,x\rangle\leq f_B(t,x^*) \} $$ Problem:
The negation of $(*)$ is : there exists a non nul set $N\in\mathcal{A}$ such that $$ A(t)\nsubseteq B(t)\qquad \forall t\in N. $$ then $$ \forall t\in N,\exists x^*_t\in H\text{ such that : }f_A(t,x^*)>f_B(t,x^*) $$ Can we say that : Since $H$ is countable, then there exist $x^*\in H$ and a nonnull set $N^{'}$ such that $$ f_A(t,x^*)>f_B(t,x^*) \qquad\forall t\in N^{'} $$