Let $B$ be the unit ball in $\mathbb{R}^N$ with center in origin and consider the space $L^p(B)$ with Lebesgue measure ($1<p<\infty$). Let $B_t\subset B$ be a concentric ball of radius $t\in (0,1]$. For fixed $t$ we can define $L^p(\partial B_t$) in the same sense as $L^p(B)$, by using the surface measure of $\partial B_t$.
My question is: Let $u\in L^p(B)$. Is true that for almost every $t\in (0,1]$ (Lebesgue measure), it does make sense to talk about the (classical) restrictuon of $u$ to $\partial B_t$, to wit, $u_{|\partial B_t}$ and $u_{|\partial B_t}\in L^p(\partial B_t)$?
Thank you
By the change of variables theorem for Lebesgue integration on $\mathbb{R}^n$ we have $$ \int_{B}|f(x)|^pdx=\int_0^1t^{n-1}\Bigl(\int_{\partial B}|f(t\,x)|^p\,d\sigma\Bigr)\,dt=\int_0^1\Bigl(\int_{\partial B_t}|f(x)|^p\,d\sigma_t\Bigr)\,dt<\infty, $$ where $d\sigma$ is surface measure on $\partial B$ and $d\sigma_t$ is surface measure on $\partial B_t$. Thus $$ \int_{\partial B_t}|f(x)|^p\,d\sigma_t<\infty\quad $$ for almost every $t\in(0,1\,]$.