Let $ h: U \rightarrow R^{n-m} $ be a $ C^1 $-function on $ \overline{U} $, where $ 0 \in U \subset R^m $ is an open set. Suppose $ h(0)=0 $ and $ \nabla h (0)=0 $. Then if $ M \subset R^n $ is the graph of $ h $ we have that the (standard) tangent plane $ T_0(M)= R^m =\{x: x_{m+1}, \ldots x_{n}=0 \} $. I want to prove that
$$ \lim_{\lambda \rightarrow 0}\int_{M}f(y/\lambda)\lambda^{-m}dH^m(y)=\int_{T_0(M)}f(y)dH^m(y) $$
for every $ f $ continous function with compact support in $ R^n $, where $ H^m $ is the m-dimensional Haussdorf measure.
This question is related to the equality of the approximate tangent spaces and standard tangent spaces for smooth submanifolds.
Thanks
I prefer to change the variable $u=y/\lambda$ on the left, thus rescaling $M$ instead of $f$: $$ \int_{M}f(y/\lambda)\lambda^{-m}dH^m(y)=\int_{\lambda^{-1}M} f(u)\,dH^m(u)\tag1$$ The manifold $\lambda^{-1}M$ is the graph of function $h_\lambda(y)=\lambda^{-1}h(\lambda y)$. As $\lambda\to 0$, we have $h_\lambda \to 0$ uniformly on compact sets. Also, $\nabla h_\lambda(y)=\nabla h(\lambda y)\to 0$ uniformly on compact sets.
Writing the integral of $f$ over $\lambda^{-1}M$ using the parametrization $y\mapsto (y,h_\lambda(y))$ we find that $$\int_{\lambda^{-1}M} f(u)\,dH^m(u)=\int_{T_0(M)} f(y,h_\lambda(y))\sqrt{1+\|\nabla h_\lambda\|^2} dH^m(y) \tag2$$ The integrand on the right converges to $f$ uniformly.