A hypersurface jump is not $W^{1,p}$

The set $Q_{+}=\{x_n>0\}\cap Q=Y\times (0,1)$ is a box, so [F3] makes it measurable with finite measure $2^{n-1}$, and [F2] makes $u={\mathbf{1}}_{Q_{+}}$ measurable. For finite $p$, ${\int }_Q\mathrm{\mid }u{\mathrm{\mid }}^p={\lambda }_n(Q_{+})=2^{n-1}<\mathrm{\infty }$; for $p=\mathrm{\infty }$, $\mathrm{\mid }u\mathrm{\mid }\le 1$, so $u\in L^p(Q)$ for every $1\le p\le \mathrm{\infty }$, and in particular $u\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(Q)$ by [F4].

Let $\phi \in C_c^{\mathrm{\infty }}(Q)$. Since $\mathrm{supp}\phi$ is compact in $Q$, [F5] and [F6] apply to the integrable function $u{\mathrm{\partial }}_n\phi$ and reduce the integral to the iterated integral over $Y\times I$. For fixed $y\in Y$ the function $t\mapsto \phi (y,t)$ is $C^1$ on the interval $[0,1]$ and vanishes at $t=1$, so [F7] gives ${\int }_0^1{\mathrm{\partial }}_t\phi (y,t)dt=\phi (y,1)-\phi (y,0)=-\phi (y,0)$. Hence ${\int }_{Q}u{\mathrm{\partial }}_n\phi dx={\int }_Y({\int }_0^1{\mathrm{\partial }}_t\phi (y,t)dt)dy=-{\int }_Y\phi (y,0)dy,$ and by the distributional sign convention the normal derivative of $T_u$ is the surface functional $⟨{\mathrm{\partial }}_n{T}_u,\phi ⟩=-{\int }_{Q}u{\mathrm{\partial }}_n\phi dx={\int }_Y\phi (y,0)dy\mathrm{.}$

Suppose $v\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(Q)$ represented that normal derivative, that is ${\int }_{Q}u{\mathrm{\partial }}_n\phi dx=-{\int }_{Q}v\phi dx$ for every test $\phi$ by [F8]. Combined with step 1.2 this means ${\int }_{Q}v\phi dx={\int }_Y\phi (y,0)dy\text{for every }\phi \in C_c^{\mathrm{\infty }}(Q)\mathrm{.}$

Fix $\psi$ as in [F9] and let $c={\int }_Y\psi >0$. Let $\eta :\mathbb{R}\to [0,1]$ be smooth, equal to $1$ on $[-1\mathrm{/}2,1\mathrm{/}2]$ and supported in $I$; for $0<\delta <1\mathrm{/}2$ put ${\eta }_{\delta }(t)=\eta (t\mathrm{/}\delta )$ and ${\phi }_{\delta }=\psi \otimes {\eta }_{\delta }$, which is a test by [F10]. Step 2.1 evaluated at ${\phi }_{\delta }$ gives ${\int }_{Q}v{\phi }_{\delta }dx={\int }_Y\psi (y){\eta }_{\delta }(0)dy=c$ for every $\delta$. On the other hand $v\psi \in L^1(K)$ for the compact product $K=\mathrm{supp}\psi \times [-1\mathrm{/}2,1\mathrm{/}2]$ by [F4], and the supports of the ${\phi }_{\delta }$ are contained in the slabs $A_{\delta }=\{-\delta <t<\delta \}\cap K$ with ${\lambda }_n(A_{\delta })\le 2\delta \cdot {\lambda }_{n-1}(\mathrm{supp}\psi )\to 0$ by [F3]. Since $\mathrm{\mid }{\phi }_{\delta }\mathrm{\mid }\le 1$, [F11] applied to $v\psi$ on $K$ yields $\mid {\int }_{Q}v{\phi }_{\delta }dx\mid \le {\int }_{A_{\delta }}\mathrm{\mid }v\psi \mathrm{\mid }dx⟶0,$ contradicting the constant value $c>0$. Hence no locally integrable $v$ represents ${\mathrm{\partial }}_n{T}_u$.

It remains to note the tangential directions. For $j<n$ and any test $\phi$, the same Fubini reduction gives ${\int }_{Q}u{\mathrm{\partial }}_j\phi ={\int }_{I}H(t)({\int }_Y{\mathrm{\partial }}_j\phi (y,t)dy)dt=0$, because [F7] applied along the $j$-th coordinate of the compactly supported function $y\mapsto \phi (y,t)$ makes the inner integral vanish; so the tangential distributional derivatives are represented by the zero function. That does not remove the obstruction of step 3.1: by [F8] membership of $u$ in $W^{1,p}(Q)$ for any $1\le p\le \mathrm{\infty }$ would require an $L^p$ class with a locally integrable representative for the multi-index $\alpha =e_n$, which step 3.1 rules out. Therefore $u\mathrm{\notin }{W}^{1,p}(Q)$ for every $1\le p\le \mathrm{\infty }$, including both endpoints, although $u\in L^p(Q)$ for every $p$ by step 1.1. The assumption used is Countable Choice [F1], spent through the box-measure, product-completion, Fubini and one-dimensional fundamental-theorem interfaces; no full Axiom of Choice occurs. $\mathrm{\square }$