I was having a closer look at the Sobolev space $W^{1, \infty}([0, T])$. I was wondering if $u'$ has a representative that is continuous a.e.. I know that $W^{1, \infty}([0, T]) \hookrightarrow C([0, T])$, but does that tell me anything about the derivatives...
2026-05-05 20:43:29.1778013809
Discontinuities of $u'$ where $u \in W^{1, \infty}([0, T])$
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Since $v = u'$ can be any bounded measurable function, your question is equivalent to asking if every bounded measurable function $v$ has a representative that is continuous a.e. The answer is certainly no; one counterexample is $v = 1_C$ where $C$ is a "fat Cantor set" which is nowhere dense but has positive measure. If $w = v$ a.e. then the set where $w=0$ is still dense, and the set where $w=1$ still has positive measure, so $w$ is discontinuous at every point of that set.