I was reading some notes on Sobolev spaces and came across this statement.
"The Meyers-Serrin theorem 1.18 is false for $p = \infty$. Indeed, if $u_i \in C^\infty(\Omega)\cap W^{1,∞}(\Omega)$ such that $u_i \rightarrow u$ in $W^{1,∞}(\Omega)$, then $u ∈ C^1(\Omega)$ (exercise)."
I thought about this for a bit and I feel like I think I have a counter-example. Let $u$ be zero on all of $\mathbb{R}$ except $u(0) =1$. Then $u \in W^{1,\infty}(\Omega)$ and the weak derivative of $u$ is the zero function. Let he sequence $u_i$ just be the constant sequence that is just the 0 function. Then it we have that $\text{ess sup}|u_i - u| = 0$ and $\text{ess sup}|Du_i - Du| = 0$ for all $i$ and so $u_i \rightarrow u$ in $W^{1,\infty}(\Omega)$. But $u$ is clearly not continuous so $u \notin C^1(\mathbb{R})$.
Is my reasoning correct, or is there something I'm missing?
Here u is continuous means, up to zero measure, u is continuous. In other words, we can alter values of u on zero measure set, since they represent the same thing in Sobolev space.