Let $(M,g)$ be a Riemannian manifold, $r>0$ be its injecitivity radius, $p$ be a point on $M$. Let $\phi:T_p(M)\to \mathbf R$ be a function supported in $$B_{T_p M}(0_p,r):=\{X\in T_p M:\|X\|_g<r \}.$$ It is easy to see that the exponential map $\exp_p:B_{T_p M}(0_p,r)\to M$ is a diffeomorphism to its image. Then the following family of scaling functions of $\phi$ is well-defined on $M$, \begin{align} \phi_p^\lambda(q):= \begin{cases} \lambda^{-d}\phi(\lambda^{-1}\exp_p^{-1}(q)), & q\in \exp(B_{T_p M}(0_p,\lambda r)), \\ 0, & \text{otherwise}. \end{cases} \quad \lambda\in (0,1] \end{align}
My question is: does the following equality holds $$\int_M \phi_p^\lambda\ dV_g = \int_{T_pM}\phi\ dx?$$ Here $dV_g$ is the volume form on $M$, $dx$ is the lebesgue measure on $T_pM$.
Yes. It is simply derived as follows: \begin{equation} \begin{split} \int_M \phi_p^\lambda\ dV_g &= \int_{\exp(B_{T_p M}(0_p,\lambda r))} \lambda^{-d}\phi(\lambda^{-1}\exp_p^{-1}(q))\ dV_g(q) \\ (x=\exp_p^{-1} q) &= \int_{B_{T_p M}(0_p,\lambda r)}\lambda^{-d}\phi(\lambda^{-1}x)\ dx^1\wedge\cdots\wedge dx^d \\ (v=\lambda^{-1}x) &= \int_{B_{T_p M}(0_p,r)}\phi(v)\ dv \\ &= \int_{T_pM} \phi(v) dv. \end{split} \end{equation} Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = \delta_{ij}$.