On weak star convergent sequences of functions that are not strongly convergent anywhere Can we find example of a sequence of functions $g_n:(0,1)\rightarrow\mathbf{R}$ such that $g_n$ converges weak-star in ${\rm L}^{\infty}(0,1)$ as $n\rightarrow+\infty$, but such that for every measurable set $A\subseteq (0,1)$ such that $\lambda(A)>0$ the sequence $g_n\vert_{A}$ has no strongly convergent subsequence in ${\rm L}^{\infty}(A)$? In fact, I am dealing with a sequence of $\theta$-Holder continuous functions, for some fixed $0<\theta<1$ which is independent on $n$. Any chance of finding an example which satisfies all assumptions? Or maybe there is no such example? Thanks in advance.  

Remark on previous comment by pozz: I point out that constant functions actualy have bounded oscillation property. In this previous comment we need to take into account that ${\rm sup}_{J^n_i}g_n\geq b$ is never satisfied, provided $g_n=c$, and provided we choose $b>c$. So there exists no sub-interval at all which satisfies both inf and sup estimate, as required in the definition of "bounded oscillation" property. Therefore, $g_n:=c$ is both strongly convergent and satisfies bounded oscillation property. The same argument applies to any sequence of functions which take values within some fixed discrete set (which is independent of $n$). Notice, however, that such sequences of "simple" functions are no longer continuous. Sorry for not answering the question, but maybe this observation helps. 
