$u\in W^{1,p}(B)$ implies $u\in W^{1,p}(\partial B_t)$ for almost $t\in (0,1]$? Suppose that $u\in W^{1,p}(B)$ where $B=B(0,1)\subset\mathbb{R}^N$ and $p\geq 1$. It was showed on this post (as an application of Fubini's theorem) that for almost $t\in (0,1]$  $$u_{|\partial B_t},(\nabla u)_{|\partial B_t}\in L^p(\partial B_t)$$
where $B_t=B(0,t)$. My question is: Is there any simple way to see that $u\in W^{1,p}(\partial B_t)$ for those balls where $u_{|\partial B_t},(\nabla u)_{\partial B_t}$ are defined and if $v\in L^p(\partial B_t)^N$ is the gradient of $u$ in the distributional sense, then $v=(\nabla u)_{|\partial B_t}$?
Update: I have an idea. Let $u_n\in C^1(\overline{B})$ such that $u_n\to u$ in $W^{1,p}(B)$. Note that byt Fatou's Lemma: $$\int_0^1\liminf\int_{\partial B}(|u_n(r\omega)-u(r\omega)|^p+|\nabla u_n(r\omega)-\nabla u(r\omega)|^p)r^{N-1}drd\Gamma\leq \\ \lim\int_B(|u_n(x)-u(x)|^p+|\nabla u_n(x)-\nabla u(x)|^p)dx \to 0$$
Hence, for almost $t\in (0,1]$, there exist a subsequence of $u_n$ not relabeled such that $$\int_{\partial B_t}(|u_n(y)-u(y)|^p+|\nabla u_n(y)-\nabla u(y)|^p)d\Gamma_t\to 0$$
Can I conclude what I have estated from the last convergence?
Thank you 
 A: To avoid unnecessary complication, let me straigthen up the spheres. Then the problem becomes the following: 
$\newcommand{\R}{\mathbb R}$

If $u(x,y) \in W^{1,p}(\R^n \times \R^k)$, then $v_x := u(x,\cdot) \in W^{1,p}(\R^k)$ for a.e. $x \in \R^n$, with weak gradient $\nabla v_x (y) = \nabla_y u(x,y)$. 

To see this, we first test the definition of weak gradient with functions of the form $\varphi(x)\eta(y)$, where $\varphi \in C_c^\infty(\R^n)$, $\eta \in C_c^\infty(\R^k)$, and consider only derivatives in the last $k$ directions: 
$$
\int_{\R^{n+k}} \nabla_y u(x,y) \varphi(x) \eta(y) \, dx dy 
= - \int_{\R^{n+k}} u(x,y) \varphi(x) \nabla \eta(y) \, dx dy 
$$
Let me rephrase it as follows: the function 
$$ I_\eta(x) := \int_{\R^k} \left( \nabla_y u(x,y) \eta(y) + u(x,y) \nabla \eta(y) \right) \, dy $$
satisfies 
$$ \int_{\R^n} I_\eta(x) \varphi(x) \, dx = 0 \quad \text{for every } \varphi \in C_c^\infty(\R^n). $$
By the fundamental lemma of Calc. Var., this means that for each $\eta \in C_c^\infty(\R^k)$ we have $I_\eta(x) = 0$ for a.e. $x \in \R^n$. 
We would like to reverse the order of quantifiers. To this end we choose a countable subset $\eta_j \in C_c^\infty(\R^k)$ dense in the $C^1$ norm. Then for a.e. $x \in \R^n$ we have 
$$ \int_{\R^k} \nabla_y u(x,y) \eta_j(y) \, dy = - \int_{\R^k} u(x,y) \nabla \eta_j(y) \, dy \quad \text{for } j=1,2,\ldots $$
By density, we obtain this identity for the same set of $x$'s and all possible $\eta$'s. 
