Lp and Sobolev classes do not determine point values

Put $E=\{x_0\}$. By [F1], $E$ is measurable and $\mathrm{\mid }E\mathrm{\mid }=0$, so [F2] makes $u_1={\mathbf{1}}_E$ measurable. Both functions are locally integrable; for every compact $K\subseteq \mathrm{\Omega }$, ${\int }_K\mathrm{\mid }{u}_1\mathrm{\mid }dx={\int }_{K\cap E}1dx=0$ by [F6]. They agree at every $x\mathrm{\ne }{x}_0$, hence almost everywhere. For $1\le p<\mathrm{\infty }$, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{u}_1{\mathrm{\mid }}^{p}dx={\int }_{E}1dx=0$ again by [F6]; for $p=\mathrm{\infty }$, every positive superlevel set $\{\mathrm{\mid }{u}_1\mathrm{\mid }>\epsilon \}$ is empty or $E$, so its measure is zero and $\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}=0$. Thus $[u_0]=[u_1]$ in every $L^p$ by [F3], including both endpoints.

For every multi-index $\alpha$, zero has weak derivative zero because both sides of its test identity vanish. The functions $u_0,u_1,0,0$ are locally integrable and $u_0=u_1$ almost everywhere, so [F5] transfers the identity to $u_1$: zero is a weak derivative of both representatives at every order. At $\alpha =0$ the derivative class is their common zero $L^p$ class by [F7]; for $\mathrm{\mid }\alpha \mathrm{\mid }>0$ it is the zero $L^p$ class. By [F4], both belong to $W^{k,p}$ with identical derivative classes through order $k$, so every term in the Sobolev norm is zero. This includes $k=0$, $p=1$, and $p=\mathrm{\infty }$.

The representatives have different point values, $u_0(x_0)=0$ and $u_1(x_0)=1$, although [F3] and [F4] identify them as the same $L^p$ and $W^{k,p}$ elements. A value at $x_0$ therefore cannot be assigned from either class alone. After fixing $x_0$, the construction makes no choices; the stated Countable Choice hypothesis is exactly ${\mathrm{A}\mathrm{C}}_{\omega }$ by [F8] and is carried only through the null-set, representative-independence, and Sobolev-class interfaces. No full Axiom of Choice is used.