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Let $M$ be a smooth compact $(n-1)$ dimensional submanifold in $\mathbb{R}^n$. Let $H$ be the $(n-1)$ dimensional Hausdorff measure. Let $f(x,y,t)$ is a function for $x\in\mathbb{R}^n$, $y\in\mathbb{R}^n$ and $t\in\mathbb{R}$. For any $x,z\in\mathbb{R}^n$ we assume that $f(x,x+tz,t)\rightarrow g(x,z)$ for a function $g$ as $t\rightarrow 0$, where $|g|$ is bounded by some integratable function on $\mathbb{R}^n\times \mathbb{R}^n$.

Let $\Gamma_x$ be the tangent plane of $M$ at $x\in M$, and $\Gamma_{x,t} = \{z: x+tz\in M\}$ for any scalar $t$. Here $\Gamma_{x,t}$ is growing from $M$ as $t$ is getting smaller. It seems $\Gamma_{x,t}$ will eventually become $\Gamma_x$ as $t\rightarrow0$. As a simple example, a circle with infinite radius becomes a line, as discussed in this link: https://math.stackexchange.com/questions/82220/a-circle-with-infinite-radius-is-a-line.

If this is correct, then is there any problem in the following derivation?

\begin{align*} &\int_M \int_M f(x,y,t) dH(y)dH(x)\\ =&t^{n-1}\int_M \int_{\Gamma_{x,t}} f(x,x+tz,t) dH(z)dH(x)\\ =&t^{n-1}\int_M \int_{\Gamma_x} g(x,z) dH(z)dH(x) + o(t^{n-1}), \end{align*} as $t\rightarrow 0$. We have used the transformation $y=x+tz$ in the second step.

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    $\begingroup$ Yes zooming at a point on a Riemannian manifold always Gromov-Hausdorff converges to the tangent space. More generally given a metric space $X$ Gromov called "tangent cone" the limit space obtained by zooming, providing it exists. $\endgroup$
    – YCor
    Commented Feb 3, 2017 at 23:36
  • $\begingroup$ Thank you YCor. Can Gromov-Hausdorff convergence lead to the convergence of the above integration? $\endgroup$
    – Bill J
    Commented Feb 4, 2017 at 0:24

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