Let $\mathcal{M} \subseteq (0,1)^n \subseteq \mathbb{R}^n$ be a d dimensional differentiable manifold and $f: [0,1]^n \rightarrow \mathbb{R}^{n-d}$ be a smooth function such that $\mathcal{M} = f^{-1}(\{0\}) \cap (0,1)^n$, where $0$ is a regular value of $f$. Further, assume that $\mu_{\mathcal{M}}(\mathcal{M}) < \infty$, where $\mu_{\mathcal{M}}$ denotes the usual measure on a manifold $\mathcal{M}$ defined using chart maps, i.e. $ \mu_{\mathcal{M}}(A \cap \Phi(U)) = \int_{\Phi^{-1}(A)} {\det( J_{\Phi}^{\top} J_{\Phi})}^{\frac{1}{2}} d\lambda_d$ for a chart $\Phi: U \rightarrow \Phi(U)$, where $\lambda_d$ refers to the $d$-dimensional Lebesgue integral.
Now my question is:
Does it hold for some continuous function $g:[0,1]^d \rightarrow \mathbb{R}$ that \begin{equation} \frac{ \int_{(0, 1)^k}k_m(\lVert f(x)\rVert) \cdot g(x) d\lambda_d(x) }{\int_{(0, 1)^k} k_m(\lVert f(x)\rVert) d\lambda_d(x)} \xrightarrow[]{m \rightarrow \infty} \frac{ \int_{\mathcal{M}} g(x) d\mu_{\mathcal{M}}(x)}{\mu_{\mathcal{M}}(\mathcal{M})}, \end{equation} where $k_m(t) = k(m \cdot t)$ is some kernel function. I am only interested in the case $k(t) = \exp(-t^2)$.
To me this seems quite intuitive, since the function $k_m(\lVert f(x)\rVert)$ is very close to zero away from the manifold $\mathcal{M}$. If anyone knows some sort of statements that give insight into the limit relationships between the Lebesgue measure and such a manifold measure or the Hausdorff measure would also help a lot.
Some ideas I have for proving this:
1.) First, prove it locally using some diffeomorphism that maps $\mathcal{M}$ to a flat space.
2.) Use the implicit function theorem to transform $\lVert f(x)\rVert$ to something like $\lVert a - h(b) \rVert$ for $x = (a, b)$ and $h$ the function from the implicit function theorem.
3.) Use the uniform continuity of $f$ and $g$.