If $f$ is Lebesgue integrable on an open set $U$ is it integrable over the surface of a submanifold contained in $U$?

Question

Let $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open and $M\subseteq U$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$

Let $f\in\mathcal L^1(U)$ and $\sigma_M$ denote the surface measure on $\mathcal B(M)$. Are we able to show that $\left.f\right|_M\in\mathcal L^1(\sigma_M)$?

Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. Maybe we can show $$\sigma_M(B)\le\lambda^{\otimes d}(B)\;\;\;\text{for all }B\in\mathcal B(M)\tag1$$ and use this to conclude the desired claim.

In this regard, we may note that, trivially, $U$ is a $d$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ and $$\sigma_U=\left.\lambda^{\otimes d}\right|_U\tag2.$$

Remark: It might be useful to note that there is the following characterization of the surface measure: $\sigma_M$ is the unique measure on $\mathcal B(M)$ with $$\left.\sigma_M\right|_\Omega=\sigma_\Omega\tag3$$ for every open subset (in the subspace topology) $\Omega$ of $M$.

Answer 1

sketch of a counterexample:

Let $d=2$, $U=(-9,9)\times (-9,9)$ and let $M$ be the $1$-dimensional submanifold which is described by the circle with radius $1$ and center $(1,0)$. For the function $f$ we choose $$ f(x,y)= \begin{cases} x^{-2/3} &: x>0 \\ 0 &: x\leq 0 \end{cases}, $$ where $(x,y)\in\mathbb R^2$.

Then one can show that $f\in L^1(U)$, but not $f\in L^1(\sigma_M)$.

(hint: for showing that $f\not\in L^1(\sigma_M)$ estimates of the form $c_1s^2\leq 1-\cos s \leq c_2 s^2$ can be useful for $s\in\mathbb R$ where $|s|$ is small.)

Answer 2

If $k<d$ the answer is no. $M$ has full surface measure but which has zero Lebesgue measure, while for $U\setminus M$ it is the converse. So they are mutually singular. In particular, your inequality (1) does not hold. When $M$ is compact and $C^1$, it has finite surface measure since this will be the case locally around each point in $M$ (and then use compactness).